User talk:Quondum/Archive 2

One of Jacobson's problems on quaternions
I have no idea if you're interested in trying this out, but this problem's solution has evaded me now for a while. In BA1, in his section on quaternions, one of Jacobson's exercises reads: if S is a sub-division ring of ℍ, and $$xSx^{-1}\subseteq S$$ for all nonzero x in ℍ, show that either S=ℍ or else S is contained in the center of ℍ (which as we know is ℝ). (Said another way, if it might be insightful, S* is a normal subgroup of ℍ*, where the * denotes the nonzero elements. This also says that elements of ℍ "almost" commute with elements of s, because for every s in S, xs=s'x  for some element s'  of S.)

My experience tells me that a proof would begin "Suppose S≠ℍ", and would then proceed to use that to show that $$ysy^{-1}s^{-1}-1=0$$ for all y. If this is shown, then by rearranging it, you can see that it says ys=sy.

The only way that S≠ℍ" condition has seemed useful in this direction is this: if x is not in S, then xs in S implies s=0. Now $$ysy^{-1}s^{-1}-1=0$$ is in S by the hypotheses, but I could see no reason why $$x(ysy^{-1}s^{-1}-1)$$ would be in S, so I seem to have run into a dead end.

At this point, I reminded myself that this is probably not true for all division rings, and that I'm probably not using one of the special properties of ℍ, so I thought involving the multiplicative norm and the fact that $$b^{-1}=\overline{b}/|b|^2 $$ would probably be a good idea. My second approach was to show that S≠ℍ would help me compute that $$|ysy^{-1}s^{-1}-1|=0$$, which would establish that ys=sy as well.

Computing that results in $$2-(ysy^{-1}s^{-1}+\overline{ysy^{-1}s^{-1}})$$, but now I'm not sure why the S≠ℍ hypothesis can help us show that the real part of $$ysy^{-1}s^{-1}$$ is 1! So another dead end...

I was hoping a fresh set of eyes might seem something simple I overlooked. I really hope (and think) there is a way to do it without working out long computations.

A third strategy I forgot to mention was to check and see if you could prove $$s=\overline{s}$$ for every s in S, showing that S is a subset of the reals. I have not made progress on this one, either. Rschwieb (talk) 13:34, 30 October 2012 (UTC)


 * I presume that by "sub-division ring" you mean a division ring that is also a subring. (Ah, yes: p.100 exercise 9 – "a division subring".)
 * Using the special properties of ℍ might be significant, or at least useful. Any single element of S that is not in ℝ can have its vector part of rotated into every direction in three dimensions using the sandwiching operation xsx−1, if x can be chosen freely from ℍ \ {0}, generating a continuous sphere. This set then generates the entire ℍ through the ring operations. Only elements of S in ℝ cannot generate elements outside ℝ via either the sandwiching or group operations. It remains to be shown in detail why my two assertions (generating a sphere, and generating from this the whole ℍ) are true. — Quondum 18:59, 30 October 2012 (UTC)
 * Yeah, I found myself thinking geometrically about the problem too (which I find to be an exciting change for me!) At the moment I'm unable to synthesize all of the inputs, though. We're essentially given a fact about the multiplicative groups of these rings, but somehow that will react with the additive ring structure, plus some other specific property about ℍ to give us the result. This is not the first Jacobson exercise that's held me up for a long time... It seems like he tends to pick ones that only have a very narrow path to proof. Rschwieb (talk) 19:29, 30 October 2012 (UTC)
 * I think that this one can be can be shown easily algebraically too, using x= αi, αj, or αk . One does not have to think geometrically. I imagine this is likely what he'd intended. — Quondum 19:43, 30 October 2012 (UTC)
 * I take that back. A successful arithmetic approach seems to approximate the geometric approach, with more general rotors required, e.g. x=1+αi+βj+γj, α,β,γ∈ℝ. There are presumably two real degrees of freedom required. I wonder whether there is some simpler approach? — Quondum 06:30, 31 October 2012 (UTC)
 * Another thought: the knowledge that S is a normal subgroup probably does not universally imply that it is either the whole ring or in the center. Given this, it would be necessary to use further properties of ℍ. — Quondum 08:48, 31 October 2012 (UTC)
 * Yeah, I think you're up to speed. That's the weird part: The whole of ℍ is trivially stabilized, but if S is just a part of ℍ, the same stability collapses to something trivial (central elements). Using central elements to sandwich only ever results in the identity map! I still haven't seen the connection I need... Rschwieb (talk) 13:42, 31 October 2012 (UTC)
 * What is it you're still looking for? It seems we have a solution to the problem, albeit somewhat clumsy. Are you looking for the minimum structure of ℍ that is needed and an elegant way to derive the result? — Quondum 14:10, 31 October 2012 (UTC)

Solution?! I didn't see it yet :) The connection I'm looking for is why the existence of an x outside of S forces elements of S to commute with everything. Algebraically I just can't see what that would happen. From a different direction, if you want to suppose there exists s and x such that sx≠xs, and then somehow show you can generate anything you want in ℍ, that would also be a solution... did you say that this approach with the ijk basis works somehow? Rschwieb (talk) 15:04, 31 October 2012 (UTC)
 * Yes. The logic is this. Given any single s=a⋅1+b⋅i+c⋅j+d⋅k, you can find an x such that s′=xsx−1=a⋅1+e⋅i+f⋅j+g⋅k for every possible e,f,g that adheres to the constraint |s|=|s′|. Now provided that s∉ℝ, this gives a a set of elements continuous (in a sphere) in three dimensions centered on a⋅1, and the ring operations with this set of elements generate the entire ℍ (S must be closed under the ring operations). Therefore, if there exists s∈S with s∉ℝ then S=ℍ.  Otherwise all s∈ℝ i.e. S⊆ℝ. Which is the conclusion we want.
 * We did not even use the fact that S was specified to be a division ring. This might of course be a clue to another approach, for example, one may be able to prove that any normal division subring of a division ring is always the whole ring or in its center, but that is just a speculation. It may be a much neater argument if you can find it, though.
 * An interesting variation: if we use the quaternion ring over rationals, the argument remains identical. — Quondum 16:26, 31 October 2012 (UTC)
 * I haven't seen the details yet, but I think I will adopt the "Suppose S is not contained in ℝ" approach, because it's a better way to think about it than the version I was using (there is y, s, ys≠sy). Not being in ℝ means that it would have nonzero imaginary part, which I could think of as a 3-d vector. My geometric thinking asks: "using sandwiching, isn't it possible to rotate and rescale that vector into whatever 3-d vector I want?" However I'm noticing that sandwiching always preserves the real part of the sandwiched vector, so you can only get the things in ℍ with the same real part that way... Rschwieb (talk) 19:33, 31 October 2012 (UTC)
 * ...until you use the ring operations. Subtract suitably chosen points on the sphere (a ring operation), and you get i,j,k with multipliers of zero to up to 2r, where r2=b2+c2+d2. Square one of these (a ring operation), and you get ... wait for it ... any element in ℝ from zero to a given limit (4r2).  Now we can add any finite number of each, and thus generate the entire ℍ. So the set of points on the the sphere with fixed real part and a given non-zero radius generates the entire ℍ. — Quondum 19:51, 31 October 2012 (UTC)
 * Sorry, but this is all too schematic for me... it glosses over everything I'm trying to pin down... Moments later while trying to write the next few sentences, some of your solution said started my cogs turning the right way, (but some parts still don't make sense to me). Anyhow, here's what I came up with. We will assume that I believe the rotation argument, so that for any x in S, S contains all other quaternions with the same real part. Now let s be nonreal in S. Then all other quaternions with the same real part as s are in s, including the real number Re(s). Because S is a division ring, it contains the inverse of Re(s). Multiply s by this so that we have a new element with real part 1, and nonzero imaginary part. Since S contains 1, S contains a pure imaginary quaternion. By the 'rotation argument', S contains all imaginary quaternions. But since S contains 1, this means S contains ℍ!


 * That convinces me, as long as I can convince myself about the 'rotation argument' :) Rschwieb (talk) 20:21, 31 October 2012 (UTC)
 * No! Damn, I slipped into thinking S was an ℝ algebra again (something I have been trying to avoid all along). My argument above only establishes that S contains all the imaginary quaternions, plus all the ones with rational real parts, and all the rational multiples of Re(s), and its powers... etc. I'm still missing the thing I've been missing all along. I don't see how to produce the full continuum of real parts for things in S. I think you were saying that it is possible, and I think the reason that I don't see it is because my intution about rotation via quaternions is not fully developed. So, my question to you is: can you show me concretely how to get whatever real part I want? Rschwieb (talk) 20:29, 31 October 2012 (UTC)
 * Rotation does not generate all elements with a given vector part, only all s with a given real part and a vector part with magnitide r. This is a very much smaller part of ℍ than you are saying (the surface of a sphere). It is the ring operations that take it from there. — Quondum 20:34, 31 October 2012 (UTC)
 * Your last edit slipped in unnoticed by me. Concretely:
 * Given a specific s=a+bi+cj+dj, rotation produces all quaternions for which a remains unchanged, but b,c,d vary constrained by b2+c2+d2=r2, r fixed. By freely choosing pairs of points on this sphere and subtracting, I can get any vectors in a ball: ei+fj+gk where e2+f2+g2≤4r2. The key here is that we have a continuum of vectors throughout the ball. By squaring each of these, we also get a continuum of real numbers in an interval. — Quondum 20:51, 31 October 2012 (UTC)

Ohh, ok, now the ≤2r comment makes sense to me. I fully agree that we can produce whatever (Euclidean) length vector we want with length ≤2|r|. Then, you can rotate it to be fully real to bring all reals ≤2|r| into S. Then, by adding s to itself to increase the magnitude of the real part, you can bring in all positive real numbers. Since S contains -1, you have all the real numbers. I'm fully convinced by the battleplan (but I still don't understand your 'squaring' comment! :) ) At any rate, the geometric viewpoint you helped me with was fantastically useful. I'm going to look to see if there's some algebraic translation that Jacobson might have intended beginners to follow, though. It's highly unlikely he would have expected a solution like this from the information he gave. Rschwieb (talk) 21:09, 31 October 2012 (UTC)
 * I hope you aren't thinking that you can use sandwiching to rotate a vector to be fully real: you can't. That is what I used squaring for, a ring operation. And I did not use that s contains −1: the ring operation of repeated addition and subtraction S→ℤS being closed ensures that negative reals are included in S (i.e. one does not have to refer to −1, nor even prove that +1 is in S; it already comes for free with I=$[0,4r^{2}]$, I→ℤI).
 * On what Jacobson may have expected a beginner to use, I'm not sure. It might be something we've missed entirely, but from this discussion it is clear that it relies crucially on the continuity that the stabilizing sandwiching operation inherits from ℍ: no other property anywhere gives this. Thus, the beginners will have been expected to use this property in some guise. She may have been expected to reason along the following lines: using the sandwiching operation s↦xsx−1 for x∈{i,j,k} plus the multiplier $1⁄2$∈S which is contained in any division ring, she can build a projection operation on the ring that extracts each of the four components independently (use of $1⁄2$ is not needed, but makes thinking easier). Thus, if she assumes s∈S and s∉ℝ, she can extract at least one nonzero vector component. Let us take this as βi, β≠0, though it does not matter whether it is i,j or k: the argument is identical for any one of them being nonzero. Next she needs x=cosθ+jsinθ, x−1=cosθ−jsinθ, so sandwiching gives βi↦(cosθ+jsinθ)βi(cosθ−jsinθ)=(cos2θ⋅β)i−(sin2θ⋅β)k. Using one of the projections, she gets (2cos2θ⋅β)i. She squares this and varies θ to get a real number in the range $[−4β^{2},0]$, which generates ℝ via addition and subtraction. Given that she can use the sandwiching to produce βi,βj,βk with suitable choice of θ, and that she has constructed ℝ⊆S, she has a basis and scalar multiplication from which she can now trivially construct ℍ. The rotation argument has been simplified to the pattern familiar to anyone who's familiar with complex algebra, and as I've shown that it is necessary in some form, I don't see any less abstract way of dealing with this. Each step is within the scope of a "beginner"; the key point is discovering the (necessary) use of rotation to generate a continuum, and the rest is pretty obvious (I hope). Any more abstract argument that avoids rotation will really uses more advanced results of which I am not aware. — Quondum 05:51, 1 November 2012 (UTC)

As I tried to rerun the argument through my head last night, I found that I didn't understand the picture of the sphere at all. And with your comment that you can't rotate to the real part, it's certain that I don't understand rotations properly as well. But reading what you just wrote, I'd like to try again:

Given any nonreal quaternion q, the imaginary part of it can represent a 3-d vector in the space with i,j,k as axes. Via sandwiching, this can be rotated in any manner of direction, including one that is a scalar multiple of i (or of j, or of k). (Before, I had been thinking of a "real axis" too, but of course that has no place in the 3-d picture.) I gather now that I don't understand the limits of which quaternions q can be rotated to. I have seen certainly that all of them share the same real part. Certainly the rotated version of q must also have the same norm as q, and since the real part is fixed, it looks like this means that the norm of the complex part of q also is invariant under sandwiching. (The tips of these vectors trace out the sphere?) So, is it true to say that the set of sandwiched images of q consists exactly of the quaternions of the same real part, whose imaginary parts have the same length as q's imaginary part?Rschwieb (talk) 17:53, 1 November 2012 (UTC)
 * Yes. To everything in this last paragraph. — Quondum 18:28, 1 November 2012 (UTC)
 * OK, then I think I have the full chain of reasoning. If there is a quaternion with nonzero imaginary part in S, then we may infer the existence of a nonzero pure imaginary quaternion z in S, and by sandwiching that, we deduce that all imaginary quaternions which have the same length as z are in S. By what you pointed out, using subtraction we may produce every pure imaginary quaternion of length shorter than z (actually, all of them less than twice the length of z). By vector addition, we can infer that there are arbitrarily long pure imaginary quaternions in S, and hence all imaginary quaternions are in S. Finally (an obvious step that I overlooked before) since iℝ is in S, we may multiply everything in iℝ by i, showing that all of ℝ is in S. Thus all of ℍ is in S.
 * Sounds like a complete argument, yes? (For myself, being unconfident with the rotation portion, I would have to write a little more on that.) Rschwieb (talk) 19:35, 2 November 2012 (UTC)
 * Yup, seems to be. On the rotation portion, as I said, you could use three specific rotations to produce given values plus one rotatioon in a single plane to produce the continuum over a real interval. Though I keep getting this ghastly feeling that there is a simple argument along the lines that you originally proposed that proves the result for arbitrary division rings, possibly using that there are no zero divisors, and some relationship between stabilizing and centering. — Quondum 21:27, 2 November 2012 (UTC)
 * That is a good follow-up question: checking to see how often this happens. I wouldn't be surprised if it held for all quaternion division rings over other fields (did you see the Generalized Hurwitz Theorem in BA1 page 447?) Rschwieb (talk) 00:58, 3 November 2012 (UTC)
 * On reading that, I'd say perhaps for all composition algebras that have an anisotropic quadratic form (i.e that are division rings, and hence have no zero divisors). In particular, I have a feeling that the same argument of "rotation" (sandwiching) allows one to generate the entire algebra from a single element not in the center of the algebra. In this one would have to show that every element of the center can be generated via the construction that I gave (squaring and multiplication by ℤ). This approach seems to be algebra-centric rather than ring-centric. (It relies heavily on the field properties and may extend to non-rings, such as the octonians. The concept of "stabilizer" would have to be defined for nonassociative algebras as x stabilizes a set S if xS = Sx.) — Quondum 06:04, 3 November 2012 (UTC)

Infuriatingly, the very next exercise in Jacobson is: show that it is true for all division rings. So, I'm back in my original state of frustration with not being able to see the division ring-theoretic reason. Rschwieb (talk) 15:04, 3 November 2012 (UTC)
 * Eish, you're right. I've got a feeling something along the same lines must be used: that any element of the ring can be generated from a non-central element using the given operations. Any element can be reached as a product in a division ring. I guess we're missing something obvious ... — Quondum 16:08, 3 November 2012 (UTC)
 * Looking at it more, the argument shows promise of generalizing to arbitrary division rings. I haven't gone through it rigorously, but the arguiment seems to go along the following lines.
 * We are given that D is a division ring with center C, and S is a division subring of D, plus that ∀d ∈ D×, ∀s ∈ S: dsd−1 ∈ S.
 * If D is commutative, the desired result follows trivially.
 * If D is not commutative, ∃a,b ∈ D: x = ab − ba ≠ 0. Suppose a = c + y, where xc = cx and xy = −yx (i.e. c is the part of a that commutes with x, and y is the part that anticommutes with x. Thus ax − xa = (c + y)x − x(c + y) = 2yx, so y = (a − xax−1)/2. Put z = xy = (xy − yx)/2. Thus xz = xxy = −xyx = −zx i.e. z anticommutes with x. Similarly, yz = yxy = −xyy = −zy, i.e. y and z anticommute. Thus {x, y, z} is an anticommuting set.  Now we can find the subset of D that commutes with all of x, y and z, and this should be a division subring. If this subring is not commutative, we can repeat the process until we find a field, which will be the center of D, i.e. C, though the generalized Hurwitz theorem suggests that we would have reached a field on the first pass (though it would be nice to prove this here).
 * Given any s ∈ S \ C, we can split it into four parts according to which of x, y and z the part commutes/anticommutes with: the classic quaternion decomposition. The idea is then to show that one can choose elements of D to make "rotors" from, expressed as an element in C1 + Cx + Cy + Cz, and sandwich a nonzero "imaginary" part of s, and that one can build up the whole of D in this fashion.  I have not gone through this exercise, but as you can see, it does mirror my process with quaternions very closely, but now in a more rigorous way as applied to an arbitrary division ring.
 * Does this make sense to you? I think your comment still applies: a beginner could not have been expected to have followed this process. — Quondum 19:56, 7 November 2012 (UTC)


 * I've kept working on this in bits of spare time. I also found a solution on math.stackexchange, which I have avoided reading. (Thus I have no idea if the solution is complete.) I'm still convinced my original approach makes sense (that is, suppose x isn't in S, and for arbitrary y in D and s in S, use that to show ysy-1-s=0). If $$ysy^{-1}-s\neq 0$$, then it has an inverse t in S, which could convert this nonequality to an equality: $$ysy^{-1}t-st=tysy^{-1}-ts=1$$. I keep feeling the solution is right around the corner...
 * Here's one little lemma I came up with: Under the conditions of the problem we are working on, if S is proper, and everything outside of S commutes with everything inside of S, then S is central. Proof: Let x be outside of S, and s and t be in S. Then xs is not in S, so xst=(xs)t=t(xs)=(tx)s=xts. Cancelling x from the far left and right, you get st=ts. Thus everything in S commutes with everything outside and inside of S, and so it is central. (Note this did not use the hypothesis that S is closed under conjugation.)
 * I hope to be able to use the hypotheses given to show that everything in S commutes with everything outside of S! Rschwieb (talk) 20:00, 10 December 2012 (UTC)
 * I don't quite follow the chain of logic, particularly what you are referring to by "the hypotheses given". To show that everything outside of S commutes with everything inside it seems tricky, since it only true when there exists an element x∉S that is a member of a division ring containing S for which xS=Sx.  That is still a lot of necessary conditions – either I'm missing something, or there is a still a huge leap to be made.
 * Some other thoughts have occured to me. We start with the premise x∈X,s∈R⊆X,xs≠sx,xR=Rx, R and X rings.
 * If we can show that either x∈R or that a zero divisor (alternatively idempotent) exists in X, the required result follows.
 * We can easily show that α(x):S→S:s↦xsx−1 is a ring automorphism. Can this be used?
 * Somehow the zero divisors seem to me to be key. I'm not sure how to use the necessary condition x∉R. — Quondum 12:31, 11 December 2012 (UTC)

Yes, proving it commutes with things outside is 'hard' because it still contains the part of the problem we didn't grasp :) But of course, it no matter how hard it is, it's still a simplification from the old one!

If I frame what I was doing this way, I think you'll see what I mean:
 * Big problem: If S is a proper subdivision ring of division ring D such that S is closed under conjugation by nonzero elements of D, then all elements of D commute with all elements of S (="S is central").
 * Lemma: For any proper subdivision ring S of division ring D, if all elements of S commute with all elements of D\S, then S is central.
 * Task: Prove that in the big problem, the conditions for the lemma are met. Then the big problem is solved.

I'm not sure how zero divisors or idempotents could come into play for our case of division rings, because the only zero divisor is 0 and the only idempotents are 0 and 1. The inner automorphism you mentioned is also important, but I haven't seen that it fits into a solution yet.

You're right that the condition x ∉S is a puzzle to use :) At the very least we have: x+s∉S for any s∈S and xs∉S for any nonzero s∈S. I've been trying to use those, but no luck so far. Rschwieb (talk) 18:59, 11 December 2012 (UTC)


 * I understand and agree with your process; it works, only the "big problem" is pretty much as big as the original problem, and the lemma is pretty straightforward (at least when laid out for one). I do not agree with your bit about zero divisors/idempotents. Look at my premise: I did not assume that R was a division ring.  My idea is to do something along the lines of: we show ef=0, but because xs−sx≠0, e≠0 and f≠0, therefore e and f are zero divisors. This cannot happen because we are dealing with division ring. Hence the premise must be false: proof of the desired result by contradiction.
 * One thing is bugging me. My original quaternion proof critically relies on a whole continuous spectrum (in the sense of real numbers, or at least the underlying field) of values x (with the sandwich product of the automorphism), whereas it seems that from the premise we are only able to say that there exists at least one x that lies outside S. Not good. — Quondum 19:53, 11 December 2012 (UTC)
 * I did not expect the "big problem" to be trivial compared the original problem: I just expected it to focus my attention a little better on a smaller problem.
 * I saw that you were considering more general rings, but at the time I wasn't very clear on why that was the case. I think I probably read it too fast. Certainly if you show that xy-yx is a zero divisor, or an idempotent which isn't 1, then that would be a good way to prove x and y commute in a division ring.
 * We are a little better than just "there exists x outside of S": we have that xs and sx is outside of S for every s. Incidentally, xs-sx is either 0 or outside of S. I had been trying to prove for some time that it could not be outside of S.Rschwieb (talk) 16:00, 12 December 2012 (UTC)
 * I can see the membership. It is reasonably easy to show that in a division ring, if x∉S, then sx, s+x and sx−xs all fall outside S or are zero. In a division ring, if x∉S, and s≠0, then it is strictly outside S (i.e. the result is nonzero, even the last case). I think that I've been misleading us. Both conditions must be used: (a) ∃x∉S, and (b) ∀x:xsx−1∈S. One chain of reasoning will use these two conditions (and in particular the arbitrary choice of x in a way to eliminate all noncentral elements from S – i.e. that every s∈S is central. Which is to say, we must show that if any x∉S commutes with s∈S, then every y commutes with s, using the fact that every nonzero element has an inverse. Or perhaps in the form of: if x∉S commutes with s∈S and y∈D does not commute with s, then we can produce a zero divisor or else prove that x∈S, which would be a contradiction. It feels so close, just needs the right expression to prove it . — Quondum 08:59, 13 December 2012 (UTC)
 * Can we establish what can (and can't) be proved within general group theory, ignoring the addition operation of the (division) rings? Why I suggest this is that division rings are precisely rings for which the nonzero elements constitute a group.
 * With the omission of addition, the given problem seems to translate into: given that S is a proper normal subgroup of D, show that S is a subgroup of the center of D. There are other related concepts such as the commutator subgroup and the characteristic subgroup. The latter introduces the concept of an abelian group into the mix (possibly the biggest abelian subgroup?).
 * Being a proper normal subgroup does not imply being central: one can (presumably) find counterexamples. This presumably implies the existence of groups that cannot become (with the inclusion of zero), or be extended to become the multiplicative monoid of a ring. Any feeling about whether there might be anything to be learned from the group structure of the product? At a minimum, this shows that addition cannot be omitted from the argument. — Quondum 05:54, 15 December 2012 (UTC)

Yeah, you're right that there are certainly groups with noncentral normal subgroups (in particular that would mean the subgroup is abelian. It's very easy to see that the alternating group of the symmetric group with any n>3 is a normal subgroup that's not abelian.)

Definitely addition comes into play. The fact that addition is there, and the fact that it has the special relationship with multiplication (distributivity) is going to yield the result, somehow. The ring axioms are more "rigid" than just the additive group's axioms merged with the multiplicative axioms.

There is probably some identity that allows us to compute a product (xs-sx)(a±b)=0, where we have ensured a±b is nonzero. I'll be very surprised if it turns out to be deeper than that, although with the level of difficulty I have been experiencing, I could well be wrong :) Rschwieb (talk) 12:10, 15 December 2012 (UTC)
 * I'm afraid I peeked. The math.stackexchange solution seems to make step-wise sense and it seems to be complete, though I suspect it might simplify (or at least might be amenable to reframing so that the path of reasoning is more intuitive). I wouldn't consider it any more obvious than my tortuous quaternion version. I definitely would not have found it before developing considerably more facility with rings. Until you ask for further hints (e.g. whether there are any similarities with your or my arguments), I'll have to consider myself honour-bound to keep mum. — Quondum 14:06, 15 December 2012 (UTC)
 * Well I'm beginning to question myself the value of continuing to struggle with it. I'm not convinced it's going to teach me anything useful outside of this problem, so I may end up looking at it soon myself. Rschwieb (talk) 02:47, 16 December 2012 (UTC)
 * Before you do, perhaps you would like to ask about the form of the solution. Given the steps, in particular what gets used and what gets proved (significant results along the way), you could develop different detail. Alternately, you could look at the solution and see whether you can't find a more direct way of traversing each leg. The choice of expressions to obtain various quantities/properties is utterly obscure to me (but that could be due to my limited algebraic manipulation skills and lack of familiarity with commutators), so I don't know how much you'll like it either. Remember though, there is always the possibility of an incorrect step that I did not notice, which would invalidate the whole thing. In terms of teaching you anything, I suspect all you'll learn by doing it yourself is practicing a skill – making an intuitive algebraic leap given a context, but you'll probably learn much the same from looking at the solution. I've no idea what'll be more satisfying to you. — Quondum 05:07, 16 December 2012 (UTC)
 * Just to goad you a bit, I'll mention that I've reworked the proof into something quite neat. You're going to kick yourself (but not too hard, I hope: I wouldn't have known where to start). — Quondum 08:05, 17 December 2012 (UTC)


 * I will be entirely unsurprised but greatly frustrated if it is simple. Normally you eventually see simple things if you look long enough. And sometimes I guess you never see them :)
 * Before I bother with the solution on math.SE, I want to know if you can give me an insightful hint based on the solution you arrived at. If you think I should just go read the one you read, just let me know. Rschwieb (talk) 14:33, 17 December 2012 (UTC)
 * Firstly, your general approach is actually very close, and mine not so. In particular, you must show that elements commute as you wanted, and then you use your lemma to show that it extends to the whole of D. The insight that you might need is that it is not a product per se as we've been assuming that we show to be zero, but rather two expressions being equal, one of which we can show is in one set, the other of which is in another set, where the intersection of the two sets is {0}. It is indeed rather simple overall. It is just a matter of finding the right expression. The one on math.* does not exactly fit this description, but when cleaned up, it does, and is thus equivalent. — Quondum 20:37, 17 December 2012 (UTC)
 * I had that in mind the whole time too, though! Nevertheless I'll still meditate on it, since it's a very general hint which did not give anything away yet... I might be begging for more later. Thank you for keeping the pressure on! Rschwieb (talk) 21:21, 17 December 2012 (UTC)
 * You might not think that I gave anything away, but given this, I'm sure the next hint will get you there. You gotta get this one by Christmas. — Quondum 18:37, 21 December 2012 (UTC)

Sign of gravitational constant
On your user page, you have an equation which includes "1 = −1/(4πG)". This implies that G<0. But that contradicts what Gravitational constant says. JRSpriggs (talk) 05:51, 8 December 2012 (UTC)


 * Indeed it does. But then, so does the factor of 4π, if one insists on using the formula on that page as defined. Implicit in any discussion on natural units is the freedom to rewrite equations to suit the choice of units, as happens with Gaussian units. In this case, we would reference the force in the opposing direction, just as one does for the electrostatic force.  What I like about this is that in the static case, the gravitational and electrostatic formulations both treated classically as fields in flat space (coulomb/gravitational force as well as the energy density of the fields) are indistinguishable but for interchanging the constants and masses/charges, as well as getting rid of the annoying sign change when comparing the equations in Gravitoelectromagnetism. This, of course, becomes moot in the nonstatic case since the formulations diverge under special relativity. — Quondum 06:54, 8 December 2012 (UTC)


 * I like your convention! Maschen (talk) 20:28, 10 December 2012 (UTC)


 * Thanks. Since it appears to require explanation, I've added a note to the choice on my user page. Of course, there are other ways of handling the sign, such as by changing the sign in the gravitational force equation, but incorporating it in the constant itself seems to make everything more "regular". — Quondum 06:03, 11 December 2012 (UTC)

Hyperbolic quaternion
Not to distract you from our ongoing thread, but I wondered if you had ever looked at hyperbolic quaternion. I'm looking it over because I noticed several places that seemed to smack of armchair mathematician. I don't expect it to be as useful as quaternions, but then again, quaternions were basically dismissed in that manner initially, so I guess I shouldn't be carelessly dismissive! :) Rschwieb (talk) 03:22, 13 December 2012 (UTC)
 * The article comes across as semi-dismissive: it's focus is on history and comparisons, not on its mathematical properties and applications. I don't see quite what "places" you are referring to. The algebra is nonassociative, but it serves as a simple example of a nonassociative algebra over R (simpler than the octonions); it will have merit for that reason alone. Your comment about quaternions being dismissed is also insightful: quaternions were dismissed partly because the one-sided product did not model rotation, but with the two-sided product they come into their own. No indication is given of whether this insight has been applied to hyperbolic quaternions. The Lorentz transform is associative, the hyperbolic quaternions not. But still, the two sided-product may be far better behaved than the single-sided product.
 * To model a Lorentz transform, a more direct alternative approach should work. Just as the quaternions can be derived as an even subalgebra of a Clifford algebra over three Euclidean dimensions, one should be able to do the same with a 3D space with a Lorentzian signature. Another insight is that the quaternions are useful for three dimensions (not four, except under a curious 2-quaternion conjugation); it does not make sense to expect more from hyperbolic quaternions. Nevertheless, my hunch is that though all the results of quaternions may apply to some other algebra when moving to a Lorentzian signature, this algebra will be associative.
 * But since I think of myself as an armchair mathematician, I hardly feel qualified to comment. — Quondum 05:30, 13 December 2012 (UTC)


 * Well really by the time I posted you about this, I had edited the two most egregious things: the phrase "in the abstract algebra of algebras over a field" and a little nonsensical blurb to the effect of "structures = categories." I'm not sure exactly what to make of User:Rgdboer since I've never talked with him. However, it looks like he's contributed to several pages with this sort of flavor. That's ok... I enjoy hearing about new things, and if they're written by a mathematics aficianado then I have a good chance of being able to understand it! Rschwieb (talk) 19:22, 13 December 2012 (UTC)
 * My own impression is of a committed, fairly solid editor with whom I've had little interaction; I don't always agree with every detail of style (probably a different background, possibly informal), but that is unimportant when in the bulk an editor is contributing good material without fuss. Has a definite interest in the lower-dimensional hypercomplex numbers, which I share. I'm sure editing for style, clarity and correctness will not generate any offence. — Quondum 20:40, 13 December 2012 (UTC)


 * If it helps, and if I could cut in - I have recently been collaborating with User:Rgdboer - an expert mathematician, very friendly, on hyperbolic geometry articles... ^_^ Maschen (talk) 20:48, 13 December 2012 (UTC)
 * Thanks, yes. Friendly collaboration on WP is great. I find it stimulating and rewarding. I'm sure we'll see more of the same. — Quondum 05:57, 14 December 2012 (UTC)

Semantics
I appreciate the typo checking that you have been doing. One strange issue you ran into is that "semantics" is treated as a singular word in contemporary logic and philosophy. We define "a semantics" for a logical system or programming language, and each individual semantics has its own definition (not "each individual semantic"). Google has about 460,000 hits for me when I search for "a semantics". Confusingly, the plural of semantics is also semantics, but in many situations the plural doesn't make sense, for example "I want to compare Henkin semantics to full semantics, each of which is a semantics for second-order logic." &mdash; Carl (CBM · talk) 13:32, 19 December 2012 (UTC)
 * Oh, wow – this is new to me. It seems to occur only as a plural noun in general English (at least from consulting the online Oxford dictionary). However, unusual uses in specialized fields are not a surprise to me. Thanks for fixing it and for being so kind as to alert me to this subtlety. — Quondum 18:17, 19 December 2012 (UTC)

false witnesses subsection of multiplicative group of integers modulo n
Hi, can you help me convert the "square" in:...basic primality check, hence 341 = 11 ⋅ 31... to a "times" sign? Thanks76.218.104.120 (talk) 05:31, 24 December 2012 (UTC)


 * If you are referring to the dot as square, this suggests that your browser has very limited support for the default sans serif font that it uses. You should sort this out with your browser (select a more suitable font, or install a font that supports a more recent set of Unicode characters). Editing Wikipedia as a workaround is not generally the solution. In this instance, however, it seems neither here nor there from the perspective of the article, so I have replaced the two occurrences with a times (×) – I hope this at least displays correctly. — Quondum 06:20, 24 December 2012 (UTC)
 * Thanks. RPeterson63.196.195.108 (talk) 05:05, 25 December 2012 (UTC)
 * Pleasure. Please don't delete earlier conversation as you did in your last edit, as this confuses things. Clutter is managed by archiving, thus eventually removing it from this page. — Quondum 05:23, 25 December 2012 (UTC)
 * Didnt know i did that. wonder how it happened, oh well.63.196.195.108 (talk) 06:54, 25 December 2012 (UTC)

Create template: Clifford algebra?
Hi! Do you think this is a good idea? It could include articles on GA (APS and STA) and spinors; there doesn't seem to be a template:spinor, recently I may have overloaded the template:tensors with spinor-related/-biased links, although there is a Template:Algebra of Physical Space. I'm not sure if this has been discussed before (and haven’t had much chance to look..). M&and;Ŝc2ħεИτlk 13:18, 31 January 2013 (UTC)


 * It may well be. It feels to me like a more sensible template than a Template:Spinor. There seem to be two styles of template for this purpose: those that go at top right of an article (column-style, like Template:Tensors), those that and go at the foot of an article, like Template:Vitamins. We'd have to decide which form to use. Template:Algebra of Physical Space should be merged into it (I see no reason for a separate template for this). I tend to concur that spinors do not really belong in Template:Tensors, and although spinors naturally have a very close connection with Clifford algebras, some people might object that the concept does not live there in its entirety. But until they do object, I think spinors should be put there without creating a separate Template:Spinor. I don't recall a discussion relating to this. Others that have worked more in the area (e.g. Jheald) may have their own opinions to add. I'm not directly monitoring WP at the moment, so I'll possibly only respond to changes on my talk page (or email). — Quondum 14:30, 31 January 2013 (UTC)


 * Your response makes sense. I wrote to the lonely Template talk:Algebra of Physical Space saying that template could be merged into the Clifford algebra, and agree that spinors overlap between the tensor and Clifford algebra (if/when created) templates, and am neutral to a spinor template existing or not existing. About the format, perhaps just a vertical column box? As a head start we could just take the APS template and extend it.
 * I'm busy right now also, just a thought for spare time. Best, M&and;Ŝc2ħεИτlk 14:45, 31 January 2013 (UTC)


 * When you have some energy/time, this'll need attention drawn to it if you want others to pay attention, as I doubt many have it on their watchlist. The colummn format should work for now, and is more uniform in this context.  A change to the footnote template would really be part of a more general style change in this regard. In someways I prefer the footnote approach, but don't have strong feelings in this regard.
 * I was thinking rather that there are many who have worked with something called spinors in physics (as with the Dirac equation), but would not identify them immediately as belonging inherently in a Clifford algebra. Tensor algebra does not accommodate them, so I don't see the overlap there. — Quondum 17:33, 31 January 2013 (UTC)


 * No, tensor algebra doesn't as such, I was thinking about the notational similarities and "related abstractions" section in the tensor template.
 * I really don't mind what format to use, footer or column is fine, one can always be converted to the other anyway (with time and work...)
 * Something else to think about is the small Category:Clifford algebras, there are subcategories for GA, quaternions, and spinors which make the CA category large in all.... M&and;Ŝc2ħεИτlk 22:05, 31 January 2013 (UTC)


 * Hope you don't mind if I repspond here. I don't know a whole lot about creating templates... I don't have a feel for when to create them. Secondly, I've had a hard time finding a clear description of spinors, as well. I'll keep an eye out and jump in if I can, though. Rschwieb (talk) 22:14, 31 January 2013 (UTC)


 * You don't have to worry about making the template if desired; you (and others) would be valuable for helping with links. M&and;Ŝc2ħεИτlk 22:14, 1 February 2013 (UTC)


 * For the next few months I will probably not be editing, but thereafter can help look into this and other WP edits. I'm sort of assuming that you (Maschen) were looking for consensus to proceed, and I think it is a good call by you.  At the moment I'm monitoring only my own talk page.  The only links I would suggest would be from the articles and categories that you have already mentioned.  So, if you have the time/energy/inclination, I'm sure you also have the familiarity/skill to go ahead. — Quondum 06:21, 2 February 2013 (UTC)

Lorentz reps
Hi Q!

I finally made at least a minor edit on the article. I just couldn't stand seeing that so(3;1)=su(2)+su(2) any more ;)

Of course, it's nothing at all like what I wanted to do some time ago. I still on occasion work on that thing, and something will come out of it. By the way, I want to thank you and R (I'm sure he is reading this) for all the help and the good discussions back then. YohanN7 (talk) 20:13, 13 February 2013 (UTC)


 * Any fixing of a mathematical error is naturally good. I cannot comment at this stage on correctness; I have yet to make a proper study of Lie groups, Lie algebras, and especially representations. The level of abstractness is sufficient to require serious study.
 * Editing WP is quite a challenge, and I've found that actually going forward can be a little more convoluted than one would anticipate. Nevertheless, when a group of editors manage to communicate and collaborate constructively, as seems to have been the spirit here, I take great pleasure in the process. — Quondum 06:00, 14 February 2013 (UTC)


 * Yes, the reactions have been overwhelming. Both the BG19 bot and the DPL bot have had strong opinions ;) YohanN7 (talk) 06:26, 16 February 2013 (UTC)

Spacetime
On random thoughts: If spacetime exists, and if it is possible to model it mathematically as a set, then the easiest and most natural way to model spacetime is as a set of spacetime points called events? I don't think the uncertainty principle applies, since an event is not an observable as far as I can see. But true, the nature of spacetime is questioned, but the replacements are never simpler in any sense. If one abandons point particles (QFT version), then strings looks like the next simplest model. YohanN7 (talk) 11:00, 30 March 2013 (UTC)
 * I'm probably coming in in the middle of something, and may not respond to everything in this thread as I'm not on WP much at the moment, so please excuse apparent scattiness on my part. As a mathematical model, I see no problem with the model as a set of events, where a local (i.e. differential) metric is defined, and for using this as a domain on which to define fields in QFT.  This does not mean that the points (events) exist in any "real" sense, as long as the fields defined on this spacetime behave and interact as the QFT predicts; it certainly should not have a problem with the uncertainty principle.  Starting from this, one does of course have to add the whole extra structure of quantum superposition (Hilbert space) not embodied in the set-of-events picture.  Spacetime allows one to describe a field by its value at every point in spacetime, which is equivalent to defining it in terms of a basis consisting of Dirac impulses at every point in spacetime.  Seen from this perspective, it will be obvious that any other set of basis functions is equivalent, and that other sets (e.g. momentum space) are equivalent in describing a field.  I would argue that spacetime is therefore not the most natural, but ranked equally amongst many equivalent domains.  The choice of spacetime events as our domain is presumably thus motivated only by our specific way of perceiving the world, extrapolated into a domain that is beyond our perception. This may start coming to pieces when quantum gravity is introduced, though. — Quondum 14:28, 30 March 2013 (UTC)

Don't worry Quondum, anyone is always welcome no matter how random or rarely!! I don't think either that spacetime is an ideal "arena" for dynamics, and neither do some researchers. John Archibald Wheeler is one example who tried to popularize superspace as "dynamic 3-geometries", and even said once "down with points - up with geometries!" or something like that in a paper on the HJEE. Of course there is no problem with spacetime in SR and GR, and the idea of events as points is actually appealing in some way ("there and then"). But for a new way forward possibilities of spacetime singularities should be excluded - shouldn't they? String theory seems to do this by spreading infinities (can't remember of what...) over the length of the string.

Don't get the wrong idea(s)... About the uncertainty principle - I just meant that there is no way to tell the exact position and momentum of particles (etc.) simultaneously, since all "particles" (etc) are always moving. I'm not at all applying the HUP to spacetime itself, nor saying the HUPs are problematic; if anything they must be part of any physical theory incorporating quantum theory.

Although, still disagree with string theory. Always have and always will, unless it really is shown experimentally to be correct description of nature (which would probably take a very long time or be unlikely). M&and;Ŝc2ħεИτlk 18:04, 30 March 2013 (UTC)


 * All that is agreed upon (by the vast majority of researchers) is that QFT is not the final solution, it is only a pretty good approximation to low energy phenomena. String theory is not a QFT, but it is a theory of QM, and of SR, which are both built in at the outset. I have no idea what I believe, but string theory is fun because it is so goofy.
 * Regarding space/spacetime, are the HJE and the ADM formulation really different theories from anything else (GR)? I can't see from the Wikepedia articles that they change anything fundamental about the nature of spacetime. The ADM formulation assumes that the spacetime manifold is a foliation, which not all manifolds are but I'm only vaguely familiar with these ideas. Basically, it would say that spacetime is a disjoint union of "leaves" which all have a spacelike structure. GR is geometrical either way. When I first read the front page I got the impression that you didn't like the idea of points in either space or spacetime, but now I'm not sure.
 * Then I disagree slightly with Quondum about any other description (Hilbert space basis) being equally "natural". This is true only within mathematics where any basis would be as good as any other. But even there, there is asymmetry, because it wouldn't be easy to make e.g. the momentum formulation without having the coordinate basis at hand. (d/dx -> d/d(what?)) YohanN7 (talk) 21:56, 30 March 2013 (UTC)


 * I'll nail my colours to the mast on string theory: Although QFT clearly is incomplete ("not the final solution"), it feels like a partial solution. In contrast, string theory feels wrong on so many levels, IMO merely to "blur" the semi-problematic renormalization that may be an artefact of the choice of the event-based image of spacetime, by entrenching the event-based picture in the very definition of the theory to allow definition of the strings.  This is definitely a huge step backwards: it breaks an important symmetry of the theoretical underpinning – the very freedom of choice of basis that I mentioned above.  Decades of intense work by physics academia has not managed to bring it to the level of respectability/completeness of QFT, so I feel comfortable calling it unduly complicated and a fringe theory despite its mainstream popularity.  On YohanN7's final question: Every linear operator over a basis maps to a linear operator over any other basis.  −iℏ$d⁄dx$ → px· is the well-known operator correspondence when changing from an event basis to a momentum−energy basis.  Beware of "easiness": you should be aware of the importance of the concept of basis-independence.  This is a fundamentally important and powerful concept in geometry, and no less so in the description of fields over spacetime.  To break the symmetry by selecting a preferred basis (e.g. the event basis) would be like denying the fundamental postulate of special relativity.  Don't break a symmetry of the theory without any reason just because it "seems simpler".  Every basis is just as good as any other.  — Quondum 02:37, 31 March 2013 (UTC)
 * I still prefer to think that spacetime have physical existence (whatever that means) while momentum space is a purely mathematical equivalent set of points. Either is a "easy" as the other, but we live and breathe only in the more "natural" spacetime. YohanN7 (talk) 12:17, 31 March 2013 (UTC)


 * True, but "space and time" would be better. M&and;Ŝc2ħεИτlk 12:22, 31 March 2013 (UTC)


 * Obviously HJEE and ADM are just alternative formalisms of GR, not theories in themselves because they are GR (as far as I can tell), but they do interpret space and time differently. Again - I don't like the idea of points in space or spacetime (there's not that much difference is there?), for reasons already explained.


 * Yes, QFT is incomplete, and people (Roger Penrose is an example) do think that any form of quantum theory is incomplete, because there are open questions and awkward ideas, compared to SR/GR which appear to be more closed and logical (even though counterintuitive, it's still possible to understand things).


 * I'm not sure what the hype is about Hilbert spaces because they are additional abstract spaces for handling quantum states - not physical spaces themselves. Also a nicer formulation of QM is the phase space formulation, in which case you don't need to take sides with position and momentum operators and representations, using phase space ideas doesn't change any physics but offers new insight.


 * The worst part of string theory is right from the very beginning, is the assertion particles "are" strings etc. How can we ever know that experimentally? We can sit down speculating/guessing/modelling what particles "really are" (only to change again and again anyway in the future) a much as we like:
 * "let's pretend everything, even the "fabric" of space and time, is vibrating strings or springs or trampolines or twirling tops or spinning wheels or pendulums or twiddling knots/links or... then there are a number of fundamental normal modes/rotational frequencies/tensions of these ... and everything in the universe is derived from these fundamental things..."
 * It's weak - just making stuff up just for the sake of explaining physical phenomena, and yet the maths is ludicrously excessively complicated just so it works... CDT is not, at all, like that. Rather it takes a logical approach that spacetime is quantized at small scales and appears as smooth, curved spacetime at large scales, that timelines must agree and casualty is preserved, not just describing, but possibly explaining the very nature of space and time, and the interesting thing is it's automatic fractal nature. Perfectly simple (minimum number of assumptions) and extremely appealing. M&and;Ŝc2ħεИτlk 09:09, 31 March 2013 (UTC)


 * The number of abbreviations unknown to me is climbing – which is highlighting my lack of formal study in this area. But I am in agreement, and do not even bother examining string theory much. Equivalent formalisms are equivalent, the only difference being the difference in ease of use in different contexts, and immediacy of insights. The change of basis that I was referring to is however not even a change of formalism: |ψ⟩ is an abstract vector that can be expressed in components on time–position and energy–momentum bases resp. as |ψ⟩ = ∫VdVδ(r)ψ(r) = ∫VdVexp(p)ψ(p), the vectors being four-dimensional, and constants discarded. Both these bases are shown in event space, but one reformulate the bases and integrals in momentum space. Or one could use any of an infinite number of other bases – there's nothing special about either position or momentum (or the combination).  This suggests that spacetime itself is an emergent phenomenon, possibly inherently anthropocentric.
 * I would argue that QFT is no less complete than and GR: each of them contradicts experiment in some domain, but is complete in its own domain. Okay, so it is "obvious" that QFT is missing something: there are arbitrary choices and unexplained physics, and possibly further forces and particles.  GR is flawed in a more obvious way: it has inherent singularities, and also has an arbitrary choice of G.  SR does not have such flaws but is weakest, and is incorporated into both the others.
 * An interesting insight from the formulation of a wavefunction as an abstract vector on an arbitrary basis (effectively in any of infinitely many alternate 4-manifolds) is that we can be pretty confident that the superposition principle is exact within a full ToE. The GR picture breaks down under the superposition principle, and QFT breaks down in a GR background.  But there must exist a basis for the wavefunction that does not break down, even in the presence of GR singularities within some of the components of a superposition.  From this it seems that one should be able to build a QFT ToE in terms of a basis that is not defined over a 4-manifold. The challenge my be how to express the metric tensor mathematically as it applies to such a basis. — Quondum 12:21, 31 March 2013 (UTC)

Forgot to mention above to YohanN7, did you mean this:


 * $$\mathbf{\hat p} = -i \hbar\frac{\partial}{\partial \mathbf{r}}\,\rightleftharpoons\,\mathbf{\hat r} = +i \hbar\frac{\partial}{\partial \mathbf{p}}\,?$$


 * Something like that, but I meant nothing really specific. You'll have a hard time even defining classical momentum (or anything else) without space and time at hand. YohanN7 (talk) 19:18, 31 March 2013 (UTC)
 * The idea of classical anything in this context only confuses things: you have to think in functions over some (arbitrary) domain. And then defining momentum becomes easy, e.g. as an impulse in the momentum domain ;-). — Quondum 03:12, 1 April 2013 (UTC)
 * How would you define the momentum domain? (You are not allowed to use the spacetime manifold in the definition.)
 * Classical ideas about spacetime are still around. In QFT spacetime is classical (whether curved or not). String theory modifies spacetime, but it still occupies the central stage, and it is not quantized. I don't think that functions over an arbitrary domain are as "natural" as functions of spacetime. The momentum formulation is perhaps the second most "natural", but this naturality comes about only after the postulates of QM are in place. Quantum Mechanics is not "natural" to most people. It took thousands of brilliant people 100 years to come up with today's version of QM. By contrast, GR was the product of one man (brilliant, but still humanly so). YohanN7 (talk) 15:29, 1 April 2013 (UTC)
 * Yes, but: starting from your more "natural" position of defining spacetime, we run into a dead end. Hence my more abstract (i.e. as yet undefined) domain, while still retaining superposition. In answer to your question about defining the momentum space without defining spacetime: define it as a 4-d affine space, to use as the domain of a wavefunction. It really only works for flat momentum–energy space, but then using spacetime as the space of definition has the same problem (I don't think you can define a momentum operator over curved spacetime, but I have no real knowledge on this). Be careful of your use of "natural" – what is "natural to people" is pretty irrelevant (meaning useless).  What is mathematically "natural" is what is relevant here, which is to something that works in the broadest context with the fewest tweaks and conditions. Newton's mechanics are very "natural" (in both senses), but do not apply. Your argument seems to be that we must stick to something that is intuitively natural, even if it has no hope of taking us further, which is to say, throw QFT out of the window. I'd say the correct approach is find where our intuition has misled us, drop that part and proceed. I've identified the flat momentum-or-spacetime domain picture of QFT as broken, selected a part that evidently isn't broken (superposition over a domain – as yet unidentified), and saying let's work with that.  (And no, I think it took tens of brilliant people few decades to come up with the basic QFT we have today.  Investigating and refining this picture has taken the effort you describe.)  — Quondum 22:54, 1 April 2013 (UTC)


 * YohanN7, I can't tell if you're misunderstanding us or I'm misunderstanding you (probably the latter...).
 * "Momentum domain" as in space of all momentum vectors? Yes, momentum(-energy) spaces or other arbitrary spaces are just mathematical, and we live in space and time so of course it does seem more natural.
 * The difficulty in casting a physical theory without space and time is pretty obvious, since physics really is geometry one way or another; that's not the subject of discussion. All that's being said (mostly by Quondum and researchers) is that unified space and time ("spacetime") is not be the best way, the most obvious point is that spacetime singularities are problematic (excuse pun).
 * Notions of space and time can have subtleties no one has thought of, in the process leading to different insights into physics. Only in the past few decades (approx) has fractal geometry been realized to extend well beyond Euclidean geometry, and has revolutionized the way we look at anything from ferns to galaxies. OK - so I contradict myself by supporting CDT which applies spacetime and not just separated space and time, but I also said CDT is a logical approach - not "the best" approach.
 * M&and;Ŝc2ħεИτlk 08:16, 2 April 2013 (UTC)

Wavefunctions are not the only game in town - quasiprobability distributions in phase space formulation offer new insight. Phase space is nice since one slice of it is space(-time), the other is (energy-)momentum. It's easy to see how the "physical space(-time) we live in" compares with motion occurring in space(-time) (even though (energy-)momenta are elements of (energy-)momentum space).

Considering Quondum's quote:


 * "This suggests that spacetime itself is an emergent phenomenon, possibly inherently anthropocentric."

have you seen the biocentrism (theory of everything) article?

M&and;Ŝc2ħεИτlk 12:38, 31 March 2013 (UTC)


 * No, I hadn't seen that article. Nor do I feel that it relates in any real sense (it feels fringe); the anthropic principle is as close as I was getting to that.  Relating to Yohan's comment a bit above: if momentum and space operators can be shown to be entirely symmetric, then there should be an equivalent concept a locality (a metric tensor) in momentum space, and there should be creatures living in that space that perceive our momentum dimension as a physical space dimension and vice versa.  Which would make momentum–energy space every bit as physical as spacetime; we are then merely disadvantaged by our perspective. — Quondum 13:06, 31 March 2013 (UTC)


 * Interesting... I'm sure there is a phase space formulation in GR which is basically what you're saying. In the ADM formalism, the metric g and the momentum &pi; conjugate to the metric are analogues of the generalized coordinates q and momenta p in analytical mechanics.
 * Don't take that biology article too seriously - just thought to point it out. I think the ideas are of interest, but yes all natural sciences are emergent from physics. M&and;Ŝc2ħεИτlk 15:49, 31 March 2013 (UTC)
 * Nope, GR in any formulation is not what I'm saying. Different formulations may be mathematically equivalent. What I'm proposing is inequivalent to either GR or QFT, even though it shares a lot with QFT.  I cannot really comment on the ADM formalism, and do not know whether it is essentially equivalent to GR, and whether it could easily accommodate QFT superposition, though this does not seem to be the objective from the article.  — Quondum 03:12, 1 April 2013 (UTC)


 * OK I may have been carried away thinking immediately of GR as soon as you said "metric"... M&and;Ŝc2ħεИτlk 07:42, 1 April 2013 (UTC)


 * Yes, that's the challenge of a ToE: finding a way of formulating it so that it maintains QFT's superposition, an at the same time accommodates GR's metric in some sense within each component of a superposition, all the while being mathematically consistent. So it should generate GR (essentially exactly) in the classical limit.  And it should produce QFT in the low density limit.  But the picture of two black holes in offset locations linearly superposed on each other suggests that the spacetime is not a great way to frame the problem: we do not know how to form the superposition.  So essentially what I'm doing is to frame the whole thing in a way that does not rely on any definition of spacetime, but relies only on that superposition of fields appears to be exact. And in the process I'm saying: at a glance it looks as though the maths of wavefunctions looks exactly the same in momentum–energy space and in spacetime; is this in fact a true symmetry of the physics?  Possibly merely untrained speculations on my part.  The spaces do seem to be true duals (both are 4-d affine spaces containing complex fields, wavefunctions identically described over each, and fields on each translate via Fourier transform).  But I do not know how to describe multiple particles of different rest masses, and how their equations of motion (the Dirac equation of fermions, for example) look, how the rest masses and charges translate, and how the EM field translates.  Interacting particles must have overlapping wavefunctions in spacetime – is locality in momentum space the same?  This would in principle be a pretty elementary investigation.  — Quondum 14:52, 1 April 2013 (UTC)


 * "I would argue that QFT is no less complete than and GR: each of them contradicts experiment in some domain, but is complete in its own domain." Has any one of them been contradicted in experiment? Surely not, it would have been a sensation. They may contradict expectations, but that's another thing. Then the question about fields and superposition. The physical existence of fields is not certain at all. In QFT, no assumption about existence of fields is actually needed. What is needed is a set of abstract states. Wave functions of particles are under doubt. None of them are actually measurable. (You could argue that the EM fields are measurable, but what is actually measured lies closer to the classical EM fields) The abundance of bases supports the thought that no QFT field has physical existence. The existence of spacetime is under less doubt, as well as that of particles in one way or another. The QFT fields describe point particles, even though the "wave functions" make them appear to be spread out. In sorts, QFT then has built in "singularities". I'd have to defend string theory a bit here. If particles are not 0-dimensional, then the next simplest thing would be that they are 1-dimensional. It doesn't stop there, since the theory allows for structures of any dimension.
 * One probably needs to be a little bit careful about the superposition principle too. Mathematics is one thing, and physically realizable states is another thing. For instance, it is commonly believed that a superposition of a half integer spin state and an integer spin state does not exist in nature. Taking thoughts like this further, mathematically "possible" things might not be realized in nature. In short, I'm unsure about the physical existence of any field; if they do exist physically, then they certainly don't exist physically in every possible mathematical way.
 * Equations of motion for multiple interacting particles are difficult and quite useless. The trouble is that there really is no such thing as an interacting n-particle state. The best explored example is probably the QFT treatment (beyond Dirac) of the hydrogen atom, but here too one makes plenty of approximations. (Chapter 13 in Weinberg if you still have it Maschen). YohanN7 (talk) 09:45, 2 April 2013 (UTC)
 * I'm afraid I take a virtually diametrically opposite position. This sounds like arguing the de Broglie–Bohm theory versus the many-worlds interpretation.  How can you say the theories have not been experimentally contradicted?  It is well-known that they have limited domains of applicability: QFT and GR are mutually incompatible on theoretical grounds.  GR's spectacular experimental success on macroscopic scales is a sufficient contradiction of QFT, and vice versa.  It is only in the low-density, large scale limit that they are compatible (where neighbourhoods of spacetime can be treated as a flat background to QFT).
 * How the "states" of QFT that you mention evolve does require (AFAICT) those states to be treated as wavefunctions in some multidimensional "space", so they cannot simply be abstracted to states. In the solution to the Schrödinger equation, the evolution operator is determined by the Hamiltonian operator, which depends directly upon the background "space", whichever (by way of illustration only: the Schrödinger equation is non-relativistic and hence this treatment also has a limited domain of applicability). My point remains: without fields, the mathematical guts of QFT is ripped out. Beware of saying that particles are pointlike in QFT: this is presumably an artefact of the maths used (in particular, Feynman's formalism), no different from saying that a square pulse has infinite extent because each of its components under a Fourier transform are infinite in extent.
 * "Wave functions of particles are under doubt"? Just because they do not behave classically, does not put them in doubt – be careful of what you mean by "physical existence".  I start from the perspective that there seems to be some form of reality out there, and that we can try to characterize it, not that it has to conform beyond the domain of my experience to the concepts of reality that I have built up through personal experience.
 * The superposition of a half-integer and whole integer spin (of the same particle/field) is not possible in QFT with the permitted fields, so I'm not sure what you're getting at here. — Quondum 10:32, 2 April 2013 (UTC)
 * Last point first. Superposition of any states is possible in QFT. Some superpositions just aren't physically realizable. There is no formal "rule" prohibiting mixes of integer and half integer states, it is just thought that it is impossible to prepare such a state in an experiment.
 * Wave functions play a very little role in QFT. The ones that enter are free field wave functions in the approximation of no interactions. (Typically tensor products of exponentials times a coefficient vector with appropriate LT transformation properties.) Particles are pointlike in QFT, this is standard terminology. They are quantized classical point particles. Steven Weinberg (The Quantum theory of Fields) develops QFT from states. Using Lorentz invariance and causality (here is where spacetime enters), the free field equations (and their solutions) appear only as byproducts. Interactions are handled abstractly as well. The scattering matrix really only involves free trivial wave functions following from prescribed Lorentz transformation properties of the asymptotical fields. There is never even an attempt to follow the detailed course of events (or time evolution of a wave function) unless this is done in a classical approximation (potential function i.e. classical field).
 * The approach to QFT through wave functions is historical (Dirac had a huge influence), but what the fundamental order of nature is (are fields or particles - of some dimension and extent - the fundamental entities), is simply not known. The difference to me is that particles feel more certain to exist. It is another matter that fields describe nature very accurately. So does particles. An example: the Dirac wave function of an electron seems to be in no way measurable. How can we assign physical existence to something that is not measurable? My standpoint is that I simply don't know.
 * Can you name one experiment contradicting QFT or GR? They are believed to have limited domain of applicability, I believe that too, but there is no experimental evidence where any one of them break down. Moreover, QFT and GR have historically been though of as incompatible due to the fact that GR cannot be quantized as a renormalizable theory. These lines of thought are being reversed a bit. Nonrenormalizability does not automatically "destroy" a theory any longer. YohanN7 (talk) 12:07, 2 April 2013 (UTC)
 * I should mention that states and the particle content building up these states are defined as (elements of) representation spaces of the Lorentz group. Also, there are field equations, at least if there is a known Lagrangian density (which isn't a prerequisite for a QFT), but these are pretty much of no use if there are interacting particles. One wouldn't be able to tell how many of them there are, even given initial conditions. The true dynamics lies in the field operators. YohanN7 (talk) 20:10, 2 April 2013 (UTC)
 * When you speak of states, you are moving outside my area of familiarity (I can only interpret that in which there are clearcut eigenstates - like in the electron orbitals in an atom). I think of QFT as being phrased in the Dirac tradition: as fields in spacetime, extended via the Hilbert space concept, and excitations allow n-particle states.  I really only think of the Dirac electron wavefunction, which satisfies a multi-dimensional differential equation that includes a coupling with the electromagnetic field, which simultaneously satisfies a similar equation.  Ignoring for the moment the concept of a wavefunction collapse, there is nothing in this picture that corresponds to a point particle.  This picture is (so I've been led to believe) mathematically equivalent to the Feynman and other formalisms.  Ergo, the concept of point particles is a mathematical convenience, not a reality.  You can't truly have a point particle: its energy would be infinite.  Terminology be damned.  That does not mean the particles are pointlike, only (I guess) that you can treat a particle as a mathematical superposition of an infinite number of Dirac impulses.  Yet an isolated Dirac impulse cannot occur, not even mathematically as a normalizable wavefunction.  I really do not see how you can say particles (interpreted as isolated point particles) describe nature at all.
 * The electron wavefunction has spin ±$1⁄2$ per excitation level, and the photon field has spin ±1 per excitation level. Thus, you never can get a single-excitation electron field in a spin-1 state.  You can of course get a superposition of 1-exitation (one "particle") and a 2-exitation (2-"particle") electron fields – nothing wrong with that.
 * Think in the spacetime wave paradigm in Hilbert space. Produce a Schrödinger's black hole (in lieu of his cat): a particle reflects off or passes through a semi-silvered mirror, passing either side of a pico black hole, imparting a tiny jolt of momentum through gravitational interaction.  The two components of the black hole drift apart over time, creating incompatible spacetime geometries for the superposition.  QFT broken, at least inasmuch as it does not describe this superposition.  The fact that GR can be shown to produce curves spacetime is sufficient to experimentally invalidate (and hence falsify) QFT.
 * Sorry, I know most of what you've said is over my head, but the bits about QFT that I've internalized seem to contradict your conclusions. — Quondum 23:30, 2 April 2013 (UTC)
 * The Dirac equation as originally thought of by Dirac is essentially a relativistic version of the Scrödinger equation, i.e. it's RQM, not QFT, so in this sense it is a classical theory. In addition to n-paricle states you need the number of particles to be nonconstant. This forces the introduction of creation and annihilation operators. Out of these the QFT fields are built up. They are thus operators on the set of states in the QM Hilbert space (Fock space). In the old (now deprecated) terminology, one speaks of second quantization because the Scrödinger and Dirac fields, like the EM field, are "quantized once again". In fullblown QFT, the coupling of the electron must be done at the level of operators, not fields.
 * Quantum mechanical point particles are characterized by not having internal structure. The electron is certainly believed to be a point particle in this sense. A nucleus is composed of constituents (protons, neutrons) in turn composed of structureless particles (quarks). In this sense the elementary particles are believed to be physical point particles.
 * What you do in your spin calculation is that you are taking tensor products (a⊗b) and using a Clebsch-Gordan decomposition (perhaps without realizing it) to find a new basis in which the spin operator is diagonal. There is no rule against simply adding two state vectors (after all, it is a vector space), and normalizing the result to unity. This would be a state with spin mixes.
 * I am not sure what you mean by the experiments contradicting QFT or GR. As "experiments" I count experiments in a laboratory, or astronomical observations, I wouldn't count experiments of thought, good as they may be. Of course, QFT assuming Minkowski spacetime should break down in extremely curved spacetime, but this has not been observed experimentally. Then there is QFT in curved spacetime to perhaps take care of this. The same goes for two black holes drifting apart. YohanN7 (talk) 09:09, 3 April 2013 (UTC)

Okay, no argument on your description of Fock space etc. – I'm certainly getting an education (my concept of QFT is sorely lacking). But I still maintain that you should be thinking of "point particle" as being in inverted commas, meaning as you say: without internal structure, not that it in any sense exists only on one spacetime worldline. On experiments contradicting QFT, I think it is sufficient to experimentally show that spacetime is curved to contradict QFT, if you consider a wavefunction as being a superposition within spacetime (whether or not you then expand this to Fock space). The theoretical construction fails as soon as the geometry of spacetime is nonconstant, and depends on the component of the superposition (with which you may be entangled). This shows that there is a real domain beyond the scope of QFT as it stands, if you incorporate a classical definition of space. If we throw out a classical (GR) concept of space, QFT might still be valid, and this is the direction that I was suggesting. I think you are being too restrictive in what you will accept as an experiment; I think to prove that the theory inherently cannot apply to known experimental observations qualifies. Two black holes drifting apart? No: one black hole drifts in different directions, depending on an entangled particle that is in a superposition of two very different states. Leading to a quantum superposition of two spacetimes with incompatible geometries. It was this problem (not even with black holes, merely small masses) that led Roger Penrose to propose an objective collapse of a wavefunction. — Quondum 01:31, 4 April 2013 (UTC)
 * Ok, inverted commas might be deserved for the term "point particle". It's still quite clear though, because in string theory, one speaks of world sheets as the objects corresponding to world lines. They are very much present in that theory. Also, the probability interpretation in QM, and the scattering processes (calculation of S-matrix elements described by Feynman diagrams) in QFT presuppose that interactions, i.e. destruction of incoming particles together with creation of new ones take place at points.
 * I don't know anything about QFT in curved spacetime, but why would the geometry of spacetime prohibit superposition? This b t w would contradict QM, which is much more general than QFT. Given any GR spacetime (without singularities at least), there are tangent bundles, i.e. vector bundles, tensor bundles, spinor bundles and other bundles. These are (mathematically) the ranges of wave functions (or more general quantum operator fields) and there is no problem with superposition as far as I can see. The QM wave functions have no "amplitude" in physical spacetime. Besides, it is a postulate of GR that SR holds locally. I.e. spacetime is as flat as we wish provided we look at a sufficiently small volume. These small volumes are where the arenas of QFT with SR.
 * Quantum superposition of two spacetimes is beyond what I have encountered. Are we talking about some quantum theory of gravity here? I'll have to have a look at the links. YohanN7 (talk) 09:58, 4 April 2013 (UTC)
 * Penrose's interpretation of the wave function seems interesting, and as good as any other provided it is backed up by math. It is also another reason to be careful with the physical existence of wave functions. All interpretations of the "collapse" of them seem to me (to be honest) either a little goofy or mathematically undefined.
 * Is it the superposition of two metric tensor fields (or their quantized operator versions) that is problematic? If so, I could see more what you mean. YohanN7 (talk) 10:25, 4 April 2013 (UTC)
 * World sheets: I don't think that these necessarily have the same implied inverted commas, even in string theory, because as I understand it (which I don't), string theory inherits all the QM mechanisms.
 * I disagree that QFT presupposes real events happen at points. I'm convinced that what is meant by this terminology is that interactions between fields is local, meaning that the evolution of the system can be described by differential equations that relate only quantities at a spacetime event. In Feynman's path integral formulation, one actually starts thinking of the interaction occurring at zero-dimensional points, and is at risk of forgetting that there is an inherent required integration over these mathematically convenient points. One can mathematically describe any continuous function (wavefunction) as an integral over a basis of Dirac functions, but this does not mean that every (or even any) function is a Dirac function (point particle).  The de Broglie–Bohm theory attempts to postulate actual worldlines for point particles, but it also keeps the wavefunction as a real entity.
 * Yes, QFT in a curved spacetime determined by GR is not problematic. What I was pointing out is that this picture is impossible, because the gravitational effects are not independent, but in fact depend on the QFT result.  This leads to the problem of trying to form the superposition of incompatible geometries.  One could hypothesize the metric as a bundle that is a field in a non-metric (but I suppose topologically flat) spacetime, though this is rather ad hoc and unsatisfactory.  A superposition of metric tensor fields could be treated in the same way as any other field, I guess, but its modification of all other fields seems to me to be a problem.  But yes, I am talking about some form of quantum gravity – that's what I meant when I referred to a ToE earlier. QFT and GR are not contradicted in their own domains, but I've been arguing that those domains have limits, and hence experiments testing effects outside those domains can (and do) falsify the theories.
 * Penrose's interpretation is merely useful for illustrating the problem – I wouldn't take it seriously. As I understand it, there is no math behind it, and I think that it shows his classical prejudices showing through – a mistake IMO.  Too many brilliant people (such as Penrose) try to hang onto aspects of the classical picture that they hold too dear.  I agree that "collapse" theories are goofy, which is why I like the MWI. This does not introduce any problems with the physicality of the wavefunction, though. — Quondum 12:40, 4 April 2013 (UTC)
 * The emphasis given to wave functions probably differs greatly between curricula at different universities, and in the literature. If one takes the longer, more historical route to QFT, then RQM (Relativistic Quantum Mechanics, not full QFT) naturally gives great emphasis to wave functions and wave equations describing particles. In this picture, it is possible to consider e.g. the hydrogen atom in great detail, but this does break down without full QFT when going into detail. Histoirical example: The Lamb shift. In the QFT books I have, the wave functions have a minor role. The dynamics lie in the Heisenberg equations of motion for the field operators. These are the "master equations". So I really don't recognize the importance you seem to give wave functions, but I totally open to changing my mind.
 * The only reasonable interpretation of the Dyson series for the calculation of S-matrix elements is to consider particle creation and annihilation at spacetime points and then integrate over space to get total amplitudes for a process. The (boring, just complex exponentials, mostly yielding momentum conserving delta functions) wave functions that are there are free particle ones whose properties are fully governed by their LT properties. The more interesting objects are the propagators. These are interpreted as probability amplitudes for a particle propagating from event x to event y. Moreover, the interpretation of the action of the field operator Ψ(x) (rather, its "creation part") on Hilbert space is to tack on one more particle at x, since Ψ(x) is an integral over (creation operators) all definite momenta with equal weight. I am trying to convince you that QFT being about point particles is a standard interpretation. This does not mean that they are localized. So is the nonrelativistic QM of massive particles as well. But wave functions do accurately describe these point particles in RQM, I'll agree with that. YohanN7 (talk) 14:26, 4 April 2013 (UTC)
 * There you go again: "...is an integral over [all points]". You are describing system that satisfies some differential properties, definitely not of a point-like particle. I'll concede that I think in RQM, not QFT (though I don't see how adding a layer of operators and moving from Hilbert to Fock space changes the argument).  The typical emphasis given is not relevant (and I have done no formal study at all in QM/RQM/QFT) – the question is what is meant by "point particle", and whether the spacetime picture is (a) consistent, and (b) the most "natural" or in any way unique.  I'll give you a challenge: if you can show that locality has no equivalent in other any basis than the spacetime basis, I'll concede that it might be "natural".  For example, if the Dirac equation has no equivalent (in form, not direct mathematical equivalence) in the momentum basis, I might have it wrong.  How would the Dirac equation be written in the momentum basis? — Quondum 00:12, 5 April 2013 (UTC)

Spacetime 2
The section above became too big to be practical. Let's start a new one.

If you read that passage about the integration again, then you will see that the field operators operators do their job at events. It is another matter that you can Fourier transform the whole scattering problem and have creation and annihilation operators creating particles with definite momenta. They are points in momentum space. With extended object (like strings) you have this duality too, but it looks different because you have (essentially) a further variable (parameter on the world sheet)

I still don't see what the problem you see with RQM of, say, an electron in a potential, as a theory where particles are point particles. Note, they are quantum point particles, I'm not suggesting classical point particles. And yes, particle states are described by wave functions that do spread out in time. But if you encounter a particle, then you encounter it at a point. You can't encounter it experimentally at two places. That would be two particles.

At the QFT level, the above problem (electron in potential) is described differently. It would be described as an ongoing scattering process with constantly ongoing (in the bound case) more or less complicated exchanges of photons. This process is impossible to describe by one-particle or two particle wave functions because photons are present, and you don't know how many. So, yes, QFT brings in new ingredients.

I don't have any problem with nonlocalized point particles described by wave functions. (I might have had that initially a number of years ago, everybody has some problems with QM initially.) We are perhaps only arguing about terminology. If you want a mathematical distinction between the RQM point particles, and something that is not point particles, try to acquire this book: http://www.amazon.com/First-Course-String-Theory/dp/0521880327. It might be in the library if you have access to it.

If I put it this way: The old concept wave-particle duality should really be called the old wave-point particle duality. YohanN7 (talk) 09:25, 5 April 2013 (UTC)


 * "But if you encounter a particle, then you encounter it at a point." Sorry, that's nonsense.  That's only true if you use the position operator to observe the position of the particle; this is not the only type of operator measurement .  That's like saying that all Smarties are blue. — Quondum 22:37, 5 April 2013 (UTC)
 * Perhaps you should take the QM/SR/QFT/GR and even String Theory courses. Speculation about TOE becomes easier (and more correct) when the prerequisites are pinned down. I think we should end this particular discussion because none of us are an expert and we have clung on to ideas outside the scope of the other. YohanN7 (talk) 11:26, 6 April 2013 (UTC)
 * As an aside, the position operator is not used in QFT. Not only is there the HUP, there is also the possibility of more particles being present when a particle is very well localized. This is essentially the starting point for RQM, when its domain of validity is considered. It says the RQM is ok, but you really need QFT. YohanN7 (talk) 14:09, 6 April 2013 (UTC)


 * Yes, in QFT spatial position is "demoted" to a parameter like time so space and time are on equal footing, and ideal approach. Another appealing thing (in QFT, not just RQM) is the time-ordering symbol used to permute states so that casualty is preserved. M&and;Ŝc2ħεИτlk 16:07, 6 April 2013 (UTC)

If I could state the obvious before you two possibly finish this thread...

YohanN7: The probability density is always $&psi;(x, t)&psi;(x, t)*$. Integrating this over a small region of space gives the probability of finding the particle within that space, the probability of finding it "at" a point is always zero for every point (which makes sense since there are uncountably infinitely many points, and no experiment can say exactly which "point" the particle is "at").

Quondum: What does "use the position operator to observe the particle" even mean? If you mean to act the position operator on a state in position representation then that's not observing/measuring at all - only the possible positions (eigenvalues) are obtained. You may know already, but the action of any operator on any state in any representation does not correspond to measurement at all. Maybe you mean something else?

We can describe the state of the system by wavefunctions or field operators or quasiprobability distributions, place time dependence in states or operators, etc... but you two seem to be discussing wavefunctions vs field operators (given the RQM vs QFT). The idea of a state is the most important and completely general thing in anything "quantum", since from there we can calculate anything (probability densities, probabilities, averages, dispersions, root-mean-squares, density matrix, Von Neumann entropy, etc.) and do what we like (place time-dependence wherever), as far as I know.

For a ToE, we need a principle that decides where to place the space and time dependence with an appropriate evolution equation based on minimizing action (include the metric here) and maximizing entropy (determinism of GR and stochastic nature of quantum information), given the mass-energy distribution and initial and boundary conditions, with conservation laws and symmetries as automatic by-products read off from the Action/Lagrangian (density)?

M&and;Ŝc2ħεИτlk 12:21, 6 April 2013 (UTC)

P.S. By "state" I mean quantum state (from now on "QS") using Dirac notation and Linear algebra (although that may go without saying in this context, just clarifying). M&and;Ŝc2ħεИτlk 16:07, 6 April 2013 (UTC)


 * I'm mixing concepts and terminology that're over my head, and producing a bad result (the position operator can be used to determine an expected position ( eigenvalues? or the relative probabilities of specific position measurements (equivalently, of eigenfunctions of the position operator) ), not to measure an actual position). Which kind of makes me agree with Yohan: this is beyond my area of expertise.  Although I am taking (as of a few weeks ago) a QM course (it bills itself as for advanced undergrads/postgrads), I'm not likely to answer my own questions soon, as I cannot devote much time to study.  And yes, pinning down fundamentals (and shedding unsubstantiated preconceptions) is essential.
 * "we need a principle that decides where to place the space and time dependence": yes, this I think is the nub of my questioning, and what I felt should be deduced rather than postulated. But, I can only speculate on approaches. — Quondum 13:25, 6 April 2013 (UTC)


 * Could the time-ordering symbol be part of the solution? Maybe. M&and;Ŝc2ħεИτlk 16:11, 6 April 2013 (UTC)


 * To Maschen. You need to reintrepret $&psi;(x, t)&psi;(x, t)*$ as well, already at the RQM level. There is not always a conserved current so that the probability density interpretation isn't possible at all these times. YohanN7 (talk) 16:40, 6 April 2013 (UTC)
 * Yes, not every current has to be conserved, but how can probability not be conserved? The explicit expressions for the probability density $&rho;$ and probability current $j$ can vary, yet surely they are always components of a probability 4-current $J^{&mu;} = (&rho;, j)$ satisfying the continuity equation:
 * $$\partial_\mu J^\mu = 0 \, ?$$
 * M&and;Ŝc2ħεИτlk 17:11, 6 April 2013 (UTC)
 * You might not be able to find a positive definite density that is possible to interpret as a probability density. Example: The Klein-Gordon equation. A deeper argument: the Lorentz group has no unitary finite-dimensional representations. YohanN7 (talk) 17:36, 6 April 2013 (UTC)

It's true the probability density is not positive definite in RQM... but the probability current is still conserved. M&and;Ŝc2ħεИτlk 18:19, 6 April 2013 (UTC)
 * Except that it by definition is not a probability current anymore. In the case of the Klein-Gordon equation it may be interpreted as charge density if I recall correctly. YohanN7 (talk) 18:55, 6 April 2013 (UTC)
 * I'm playing catch-up here. From a mathematical perspective, if $&psi;$ can be represented as a (vertical) vector of complex values, then $&psi;^{∗}&psi;$ (interpreted as the dot product, or matrix product using the Hermitian transpose) must be positive-definite at every point in any space in which it can be expressed this way. $&psi;&psi;^{∗}$ is then a matrix, so the ordering or the factors seems wrong (except in GA, where it would make no difference).  The Klein–Gordon equation can be applied to more general wavefunctions (scalars etc.), but I don't think that this changes the argument.  I'm pretty sure other representations (e.g. GA) will give the same result.  What have I missed?  — Quondum 19:48, 6 April 2013 (UTC)


 * You have missed that you need a continuity equation. There is none for $ΨΨ*$ in the case of the Klein-Gordon equation (and many others). In case you find a continuity equation, the time-component of the corresponding current might not be positive definite, as required from a probability density. The probability postulates of QM in RQM and beyond are not expressed in terms of wave functions. You will both have to come to terms sooner or later with the fact that wave functions are not of much use except in non-relativistic QM, and as an approximation in RQM. I'll reiterate what I said way up. The wave functions enter only in the approximation of no interaction (in which case they are exp(ipx)×column vector depending on under which representation of the Lorentz group it transforms). Even in this case, the wave function is 100% determined by Lorentz invariance. When particles are interacting, then there is no state that can be well described by a, say 2-particle wave function, because the interaction falsifies the assumption of 2 particles. RQM is different from non-relativistic QM, and QFT is different from RQM. I am sorry about sounding grumpy. YohanN7 (talk) 20:16, 6 April 2013 (UTC)
 * Going further, by considering gauge theories like QED and QCD, the gauge invariance, if nothing else, makes it worse for any concept of wave function, especially one with physical existence. Again, sorry about sounding grumpy. YohanN7 (talk) 20:27, 6 April 2013 (UTC)
 * One further technical detail that might shed some light onto some issues in this thread. In QFT, the Heisenberg equation of motion of the operators and the corresponding Schrödinger picture equation are not on the same footing. The former is fundamental, and the latter is derived. See e.g. Field Quantization by Walter Greiner. YohanN7 (talk) 20:43, 6 April 2013 (UTC)
 * I can quite see how I'm off-base in QFT, what with particle creation and destruction. I gather then that you are not speaking of wavefunctions, presumably hence your use of capital psi.  Meaning that my question about wavefunctions did not apply.  Your comment about RQM suggests more than I had expected: that it is more than QM adjusted to be Lorentz invariant, so I'll have to brush up on at least that before embarrassing myself further.  — Quondum 21:29, 6 April 2013 (UTC)
 * RQM is QM with Lorentz invariance. I guess it would probably be correct to say so, but SR brings more than Lorentz invariance. When energy densities are high, pair creation becomes more and more of a problem. The Compton wavelength of a particle is traditionally taken as a distance where talking about 1,2, ..., n particles is useless. It is interesting that a huge part of QFT can be logically deduced from RQM and SR. For a logical (but not entirely rigorous) approach to QFT, see Weinberg, The Quantum Theory of Fields, vol 1. YohanN7 (talk) 11:16, 7 April 2013 (UTC)


 * It's understandable YohanN7 that you may be annoyed... I'll concede that I stretched the application of $&psi;^{∗}&psi;$ beyond non-RQM for nothing without thinking clearly, which just adds to the verbosity of the discussions. However I don't deny that wavefunctions and fields are on a different footing, nor that "wavefuntions" are only of use in non-RQM while QS vectors are completely general. We can agree on that much.


 * We seem to be on different pages - YohanN7 ahead, Quondum a little behind (no offense at all intended), and me way backwards... To make the most of this page why don't we list all the
 * principle features,
 * axiomatic postulates,
 * treatments of space and time,
 * successes/pros and failures/cons
 * of SR, GR, QM, RQM, QFT, SUSY (?), and if desired other formulations/theories (Yang-Mills fields, string/M theory, spin networks, CDT etc.) take it from there? In this way we don't have to bicker/re-explain between facts and our personal interpretations since the facts are summarized... M&and;Ŝc2ħεИτlk 21:45, 6 April 2013 (UTC)

These are quotes from Zweibach, A first Course in String Theory:

"There is a natural identification of the quantum states of a relativistic point particle of mass m with the one-particle states of the quantum theory of a scalar field of mass m."

"There is a correspondence between the relativistic quantum point particle wave functions and the classical scalar field, such that the Schrödinger equation for the quantum point particle wavefunctions becomes the classical field equations for the scalar fields."

"Because it provides a natural description of multi-particle states, the scalar field theory can be said to be a more complete theory."

I'd say it applies to most of our discussion here. Please note that the one-particle states of the scalar field are not the only states of it. Note too that we are talking about point particles. YohanN7 (talk) 11:33, 7 April 2013 (UTC)


 * Thanks. Quotes like this would help in a new article Introduction to quantum field theory, to parallel Introduction to quantum mechanics. M&and;Ŝc2ħεИτlk 11:53, 7 April 2013 (UTC)

Summary
Applying these to nature, we find...

Interesting/relevant articles
To save ourselves repeatedly typing in the search bar:


 * List of unsolved problems in physics, beyond the Standard Model
 * Relativistic heat conduction
 * Yang–Mills theory
 * Clifford algebra
 * Geometric algebra
 * Geometric calculus
 * Algebra of physical space, Dirac equation in the algebra of physical space
 * Space and time (Privileged character of 3+1 spacetime)


 * User:YohanN7/Representation theory of the Lorentz group - Good link to read (not used for a reference): it's a keeper!

Discussion
Ideally let's not label our signatures in the above summary section. Feel free to add/subtract content or delete/transform to other list/summary formats. Thanks, M&and;Ŝc2ħεИτlk 22:40, 6 April 2013 (UTC)
 * I see that relativistic quantum mechanics redirects straight to quantum field theory, which is a pity because I would have expected that it should be an extra entry in the table above between non-RQM and QFT, presumably referencing the Dirac picture. — Quondum 00:28, 7 April 2013 (UTC)


 * Yes, RQM is a redirect so didn't have the inclination to enter it. About CM as inconsistent: consider how it incorrectly explains the photoelectric effect, the model of electrons orbiting deterministically a nucleus (with the obvious and famous implication that atoms radiate EM waves and collapse), its lead to the ultraviolet catastrophe, and so on... M&and;Ŝc2ħεИτlk 07:30, 7 April 2013 (UTC)


 * Agree totally about the need for an article on RQM.


 * This table could develop into something nice.
 * "Probability densities are not positive definite." This needs to go, it is simply nonsense. You don't have the probability density interpretation unless it is positive definite.
 * "Negative energies are predicted (KG and Dirac equations)" No, but this is another potentially very lengthy discussion.
 * Why take computational difficulty as a drawback of any theory? I think SR is simplicity itself. YohanN7 (talk) 11:48, 7 April 2013 (UTC)


 * You two might see here - I took the RQM redirect issue to WP physics.
 * About drawbacks - delete them, just an intrinsic feature. M&and;Ŝc2ħεИτlk 11:53, 7 April 2013 (UTC)


 * On CM being "inconsistent for a closed theory": as a theory, it appears to be self-consistent, albeit not even addressing aspects such as the nature of matter at the microscopic level. It is obviously both incomplete and inconsistent as a theory for describing observed physics (not what I understood the wording to mean, though), but then this applies to all the theories in the table.
 * Oh, and just because WP has a silly redirect for RQM is no reason to omit RQM as a table entry (especially one privately exploring the distinctions). — Quondum 12:29, 7 April 2013 (UTC)
 * Yes, RQM should have an entry. But then you need to be careful (i.e. define) terms such as "inconsistent". Note that neither GR nor QFT are so far observed to be inconsistent, either internally as mathematical theories, or contradicted by experiment. You could argue that the existence of singularities in GR hints towards incompleteness, but hardly inconsistency. I have heard about inconsistencies in QED, but since QED most certainly isn't complete, it doesn't show inconsistency of QFT. Compare this with experimental breakdown of CM. This has been around for some time. Still, CM is considered internally consistent within it's domain of applicability.
 * Proven inconsistence and/or experimental failure of GR/QFT would be a sensation. But we have had this discussion. Incompleteness is more obvious since no fully accepted QFT includes gravity. Gravitons are very much believed to exist, in which case GR is incomplete, since a classical field theories have no particles. Only a experimentally proved breakdown of QM or SR would be a bigger bang than experimentally proved breakdown of QFT or GR. To quote Weinberg, A breakdown of QFT would be a sensation, a breakdown of QM or SR would be a cataclysm.
 * Remember that these theories have been around for some time now, and it is not like people haven't tried to dethrone them - theoretically or experimentally YohanN7 (talk) 14:22, 7 April 2013 (UTC)
 * It's interesting to note that the inconsistency of classical EM theory and classical mechanics when applied to classical electron orbits (or the self energy of a point charge) pointed out by Maschen really only gets worse in QFT, that is, the problem is present and even more divergent than in classical calculations before the so called renormalization. This is one of the major successes of QFT. If there is a divergent Feynman diagram, then, after renormalization, there is precisely one (or perhaps finitely many) Feynman diagram of higher order that will cancel the infinity in the original graph (provided the theory is renormalizable). GR is not renormalizable as a QFT. On the other hand, if you let all processes that are allowed by symmetries enter the theory, then it might turn out that nonrenormalizable theories are just as renormalizable as the renormalizable ones. This is one line of thought in effective field theories. YohanN7 (talk) 14:40, 7 April 2013 (UTC)

I actually object to having second quantization in the axiom column for QFT above. You don't need a preexisting field to have a QFT. This relates a bit to what we have been discussing. YohanN7 (talk) 15:41, 7 April 2013 (UTC) Then feel free to remove it but it would help to have that link somewhere...

Why did you remove "Postulate proper time $τ$ is Lorentz invariant"? If you assume this you can derive all 4-vectors, the Lorentz transformations, it is the Lorentz invariant inner product:
 * $$(c\tau)^2= dX_\mu dX^\mu \quad\Rightarrow\quad\tau=\frac{1}{c}\sqrt{dX_\mu dX^\mu}$$

Yes, what you replaced is what is usually in the textbooks/lecture courses, but we should still include proper time somewhere. (Actually I was taught that the invariance of proper time with the constancy of c leads to all of SR).M&and;Ŝc2ħεИτlk 16:11, 7 April 2013 (UTC)


 * I think we should take Einsteins own choice of postulates. I think the ones I gave were his final versions of them. I'll dig up a reference. I have never seen the other version before as a postulate, just as one of many consequences (and you will need two postulates).YohanN7 (talk) 16:33, 7 April 2013 (UTC)
 * I added a ref in the table. Happy? M&and;Ŝc2ħεИτlk 09:44, 8 April 2013 (UTC)
 * Besides, how do you define proper time? You need a world line for that, and you are then singling out one (actually a rotationally and translationally invariant class of them) reference frame. YohanN7 (talk) 16:40, 7 April 2013 (UTC)


 * Um... the formula above? Why should that be frame-dependent? M&and;Ŝc2ħεИτlk 16:45, 7 April 2013 (UTC)


 * What if


 * $$dX_\mu dX^\mu < 0?$$


 * It is just not a good postulate. "Intervals, dS2, are invariant" would be better (since it doesn't single out anything), but still not as nice as the usual formulation. B t w, the second postulate can be sharpened to "There is a universal limiting speed c" without reference to light. It makes the generality and applicability to all of physics clear. (See Jackson, Classical Electrodynamics.) YohanN7 (talk) 17:03, 7 April 2013 (UTC)


 * Invariants can be spacelike ( 0), timelike ( > 0 or < 0) or lightlike ( = 0), signs depending on the metric sign convention. M&and;Ŝc2ħεИτlk 22:08, 7 April 2013 (UTC)


 * Exactly. But invariance of proper time is bad in at least three different ways as a postulate. The first problem is that it isn't symmetric in space and time, singling out a class of reference frames. (Before you object again to this, yes, you do.) Secondly, not all intervals are timelike, thus proper time isn't defined for all intervals. A third problem is that you need more concepts to explain what you mean by proper time. The typical way is to talk about a clock attached to a massive particle. Proper time belongs one of the other columns. YohanN7 (talk) 09:39, 8 April 2013 (UTC)


 * Proper time has been moved.. M&and;Ŝc2ħεИτlk 10:17, 8 April 2013 (UTC)


 * There is a question mark in the table associated with "Phase space" in the SR and RQM entries. The phase space equations are just Fourier transforms of the spacetime equations. Also, momentum is a 4-vector; p->Λp under LT. YohanN7 (talk) 16:03, 7 April 2013 (UTC)

No, the FT relates x space to p space. In RQM do we have $φ(x, p, t)$? A phase space distribution would have that form. M&and;Ŝc2ħεИτlk 16:11, 7 April 2013 (UTC)

And yes there is a 4-momentum which we all know transforms under LTs. The question marks query whether or not phase space $(X, P) = (ct, x, E/c, p)$ is used in relativity or RQM. M&and;Ŝc2ħεИτlk 16:16, 7 April 2013 (UTC)


 * I think you got me there. It is not always as simple as a Fourier transform. At least a qualification is needed. If we work with generalized variables (or fields), some care needs to be taken. E.g. formulating EM in canonical fields (say A and Π) is tricky because of equations of constraint. YohanN7 (talk) 16:24, 7 April 2013 (UTC)


 * Before you raise it: move Fock space where you want to (QM or RQM or QFT?). M&and;Ŝc2ħεИτlk 22:08, 7 April 2013 (UTC)


 * Probably QFT is best. Multi-particle Hilbert spaces (hence Fock spaces) exist in QM, but the particular occupation number basis is usually heavily only in beginning QFT with variable particle content. I'll try to find time to work a little on the table tomorrow. It has gotten pretty far.


 * A question: Is entropy maximization not something for the axiom column? Statistical mechanics (not to mention thermodynamics) is something I know very little about. YohanN7 (talk) 09:39, 8 April 2013 (UTC)


 * Yes, and so should the least action principle (if you assume $δS = 0$ with a classical Lagrangian; all of CM can be derived). It's done now. M&and;Ŝc2ħεИτlk 09:54, 8 April 2013 (UTC)


 * Sounds good! All of QM, QFT, and even string theory can be derived. (The only drawback is that Lagrangians are always guessed or reverse engineered.)


 * GR: Position and time are coupled into the 4-position X = (ct, x). This is too restrictive. Any set of local coordinates will do. I don't have a sharp formulation in mind right now, but it could be taken from smooth manifold theory directly. YohanN7 (talk) 10:45, 8 April 2013 (UTC)


 * Fixed that too. I'll hold off on the table since I'm dominating it too much (and this page for that matter...)... M&and;Ŝc2ħεИτlk 00:02, 9 April 2013 (UTC)

Page alive?
I made, for fun, a couple of tweaks to the table.

I reread all of our discussion above. One thing I took notice of this time around was the problem of superposition in GR emphasized by Quondum. It is true that "adding two geometries" is problematic. It is, in fact, impossible in GR, since the equations of motion determining the metric are nonlinear. However, it is not at the level of "geometry" that a quantum theory of gravity requires the superposition principle. (Thus there is no way of adding gravitational fields from two gravitating bodies. One must solve the complete problem.) It is at the level of graviton states where superposition is required. There is nothing in principle that prohibits this because these states mathematically have their amplitudes (call them wave functions if you want) in a vector bundle derived from the tangent bundle of the spacetime manifold. This is (at each point in space time) a vector space, so, a priori, superposition cannot be ruled out. (If so, they would have thought of that many many years ago.)

By the same token, curved spacetime doesn't in any way invalidate QFT; Those amplitudes (fields and field operators) too reside in vector bundles or spinor bundles, which are vector spaces at each point.

But needless to say, the problem obviously isn't easily solved. Einstein wrote in a late paper (I think it was his very last paper - and one of the best the last 30 years - it's in the book "The Meaning of Relativity" (real stuff, not popular science even if the title hints otherwise).) that attempting to solve the problem with singularities by quantization would amount to trying to solve an inherently nonlinear problem using linear methods. YohanN7 (talk) 01:33, 17 July 2013 (UTC)


 * I don't entirely follow, but I guess I'm a bit behind on QFT "states". Superposition works in state space, so superposition of arbitrary states should be possible. But when you start speaking of tangent bundles of the spacetime manifold, I climb off the wagon, because different states have different manifolds (geometries).  One could embed QFT in a manifold to get a very workable theory, but one would have to make the invalid assumption that the spacetime manifold geometry is independent of the state.  So somewhere in the interpretation of superposition as applied to spacetime something has gone wrong. One does not have to resort to singularites to discover the problem. Thus I disagree with the statement 'However, it is not at the level of "geometry" that a quantum theory of gravity requires the superposition principle.' — Quondum 19:53, 17 July 2013 (UTC)


 * There is no difference whatsoever between QFT states and QM states as a concept.


 * Tangent bundles are basic objects in manifold theory. At each point in a manifold there is a tangent space, which is a vector space. Now take the union of these tangent spaces over all points in the manifold. That is the tangent bundle - as a set. From this, other bundles are defined. The Cotangent bundle, dual to the tangent bundle. For each point, form the tensor product of the dual space with itself, and then again form the union over all points in the manifold. Its elements are second order tensor (covariant) fields. It is here that the metric tensor of general relativity resides.


 * You are not entirely correct when you say that "different states have different manifolds". The superposition principle is a mathematical device. You can have any spacetime you want and put in any state of a somehow quantized gravitational field. Now, this total configuration may be such that it cannot emerge by itself, and may not be possible to prepare. Still, since gravitation interact with everyting, including its own gravitational field (gravitons), the total configuration would evolve in time from our artificial state in a way such that in the classical limit, the equations of classical GR hold.


 * This embedding of QFT (and other theories, like classical EM) that you mention is not anything like a unified theory. It is just the classical way of doing things in a GR background, so I don't see how it relates to the discussion here.


 * You are unclear about what you mean by "different states having different manifolds". I believe that there is no conflict whatsoever. Classically, the "same problem" is very much there. You cannot, as in EM theory (to a good approximation), have a background field and calculate the motion of charged particle. Likewise, in EM, you can take a fixed source and (to a good approximation) calculate the field it produces. The corresponding things aren't possible in GR, except for "test particles" of negligible mass. One must solve the complete problem. Very much related to this is the fact that you don't have superposition at the global level in GR. You cannot calculate the field of two gravitating bodies and them add them up. Thus, again, it is not at the level of geometries (solutions of the field equations for GR) you would want superposition.


 * I also don't see what you mean by "superposition as applied to spacetime". I have tried to explain where the superposition is supposed to take place mathematically. YohanN7 (talk) 15:02, 19 July 2013 (UTC)


 * Hmmm..., I think I know from where you get your reasoning about "adding geometries". Chapter 30 in Penrose's Road to Reality? That chapter seems highly speculative, and mixes concepts from an existing theory (classical GR) and a so far nonexistent quantum theory of gravity. By no means is he treating a combined theory and he hasn't proved that no such theory exist, and if no such theory exists, he hasn't proved which principles (one?, several, all) of QM and GR break down. Those Killing vectors are classical objects, and the quantum states inhabit a Hilbert space he hasn't really defined. More importantly, (as an objection to the conclusiveness of Pensrose's reasoning) a quantum theory of gravitation doesn't need to have a Schrödinger equation as a basic ingredient. Fundamental QFT does not need the artefact of Lagrangians, or wave equations and the like.


 * What Penrose really puts his finger on - as far as I am concerned - is what I quoted Einstain for above: "Attempting to quantize general relativity seems to me as an attempt to solve a nonlinear problem with linear methods". This problem, in other words, is not a new one.


 * I really don't argue for or against any speculative theory, but I think we should have kept the distinction between speculation established theories and other ones. It really confuses things when they are all treated as being equal. YohanN7 (talk) 18:35, 19 July 2013 (UTC)


 * IMO Penrose gets a couple of things badly wrong: he likes the idea of long-term macroscopic entanglement to explain how our brains work, and he likes to objectify the quantum collapse through gravitational disparities between states. We are however left with working out how to match the implicit geometries for different states for the purposes of determining outcomes of experiments based on superposition, which he does highlight.  The linear/nonlinear problem surely is not new: even an atom is a highly nonlinear system, but using a Fock/Hilbert space solves this perfectly.  — Quondum 19:43, 19 July 2013 (UTC)


 * I should say that I haven't read Penrose's book, I just got hold of it. I have just glanced through that chapter 30.


 * In what way do you mean that an atom is a nonlinear system? In the usual terminology, a system is linear if it's solutions obey the superposition principle, which happens if the underlying differential equations are linear. At the level of RQM, the equations are surely linear. This is true even for quarks. There is no reason that the composite equations (for the complete atom) should be nonlinear.


 * I would agree with you if you said that the sum of two one-electron states isn't a two-particle state, but this is not what is meant by non-linearity. What is your definition of nonlinearity? One should keep in mind that there is a subspace of Hilbert space that is called the physical Hilbert space, which is smaller than the full Hilbert space. (Do you recall an earlier post of mine talking about the impossibility of superimposing different states with integer and half-integer spin? Such a thing is an element of Hilbert space because it solves the equations. nonetheless, such states don't exist in nature.)


 * In QFT (which must be used for the complete system for best accuracy) it is less transparent what is meant by linearity, but suffice to say, that quantization using a Lagrangian yielding linear classical equations (e.g the Dirac equation) will not introduce nonlinearity. Of course, I have no idea what happens beyond QFT. YohanN7 (talk) 20:30, 19 July 2013 (UTC)


 * By an atom being nonlinear I mean that it is self-interacting. The electron and proton of an unexited  hydrogen atom will be found to be close together, even if the atom as a whole may have a very uncertain location.  Gravitational interaction is nonlinear in the same sense, even though there is self-interaction at the graviton level, just like the Higgs boson self-interacts.  I do not see why this should make the system nonlinear in the sense you mean specifically for gravitation and not for the Higgs boson, or for an atom: I would think that the superposition principle would apply equally across the board, and I've seen arguments in this direction.  The problem with gravitons probably only manifests if one insists on a shared background manifold, or at least a metric tensor field that is not permitted to exist in a state of superposition just like all the other fields in a Hilbert space.
 * I recall, but didn't pay much attention to the superposition of incompatible spins, mainly because I could not follow what you were getting at. It seemed to imply different fields (e.g. electrons and photons). Since spin-half photons etc. are not solutions in the first place... – anyway, you see where I am at on that. — Quondum 21:10, 19 July 2013 (UTC)

We had an edit conflict. I wrote the below stuff. Will respond shortly to your last post.


 * In QFT, when considering e.g. a two-particle system in an external potential and where the two particles interact among themselves one does obtain a nonlinear equation for the field operators. (It becomes a nonlinear partial integro-differential equation, see Walter Greiner, Field Quantization, Chapter 3) So, yes, the fundamental problem (that of the time evolution of the operators which is the fundamental one in QFT) is nonlinear. If this is projected down to Hilbert space (the space of states), it must however become linear - or the model of a Fock space would be simply wrong.
 * In other words:
 * Quantum states with fixed particle content are governed by linear laws of nature.
 * The operators on the space of states may be governed by nonlinear equations.
 * The latter equations are the fundamental ones.
 * Is this the way you perceive the atom to be a nonlinear system? in that case, I agree. YohanN7 (talk) 21:29, 19 July 2013 (UTC)


 * With superposition of states of different spins I mean something very simple. Let |ψ1> be a spin 1 state and let |ψ½> be a spin ½ state. Then |ψ> = 1/sqrt(2)(|ψ1> + |ψ½>) is a state which upon measurement would result in either spin 1/2 or spin 1 with equal probabilities according to the general postulates of QM. Such states are believed not to exist in nature, and to be impossible to prepare. (Weinberg, The Quantum Theory of Fields, Chapter 2). YohanN7 (talk) 21:47, 19 July 2013 (UTC)


 * I see better now what you mean, but your choice of terminology calls for plenty of misunderstanding in the first place. "Self-interaction" is one thing, nonlinear equations is another thing, both very well defined. For self-interaction in QFT you would need a cubic (or higher) term in the Lagrangian in the field, yielding a 3 particle vertex in the Feynman diagrams. (A quadratic term is not self-interaction because for such a vertex, only forward scattering is possible due to 4-momentum conservation - i.e. nothing "happens".) True enough, any viable quantum theory of gravitation must have self-interaction in this sense, because gravity interacts with everything possessing energy.
 * I still can't see how an atom in any precise sense is self-interacting. It's constituents interact - yes, but the atom as a whole with itself? YohanN7 (talk) 22:06, 19 July 2013 (UTC)
 * Apologies for terminology issues: these will not disappear, since I have no formal training. I suppose you are referring to the spin of an entire system of particles.  Probably not worth digging into here.
 * I mean effectively that the constituents of an atom interact. The same goes for gravity: a gravitating system's constituents interact.  The system as a whole cannot self-interact any more than an atom does.  Thus, a ToE must be linear in Fock space like any QFT.  And I'm afraid Lagrangians are beyond my ken, though I've seen them mentioned everywhere.  — Quondum 22:36, 19 July 2013 (UTC)

Lagrangians
Quondum: the simple idea of a Lagrangian makes its truly astonishing.
 * In general - Lagrangians summarize the dynamics for a system of particles, Lagrangian densities (Lagrangians per unit spatial volume) for fields. Similar to Hamiltonians, they "store" the dynamical information of the system - the physics is contained entirely in the Lagrangian, while the rest of the calculation is reduced to mathematics...
 * In classical Lagrangian mechanics: The Lagrangian is a function of the generalized coordinates and generalized velocities (time derivatives of generalized coordinates), and possibly time:
 * $$L(q_1,q_2,\ldots,\dot{q}_1,\dot{q}_2,\ldots,t)$$
 * Any generalized coordinate which doesn't enter L corresponds to a conserved generalized momentum, by definition:
 * $$p_i = \frac{\partial L}{\partial \dot{q}_i}$$
 * so it's easy to read off conserved quantities for the system. Still in classical mechanics, it is always the total kinetic energy (as a function of the generalized velocities) minus the total potential energy (as a function of the generalized coordinates, and possibly generalized velocities and time):
 * $$L(q_1,q_2,\ldots,\dot{q}_1,\dot{q}_2,\ldots,t) = T(\dot{q}_1,\dot{q}_2,\ldots) - V(q_1,q_2,\ldots,\dot{q}_1,\dot{q}_2,\ldots,t) $$
 * Substitution into the Euler-Lagrange equation:
 * $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) = \frac{\partial L}{\partial q_i} $$
 * generates the coupled equations of motion in terms of the generalized coordinates, where the generalized coordinates and velocities are treated as separate variables (take the partial derivatives wrt each separately).


 * In classical field theory: I'm less familiar/proficient with this, but the Lagrangian density is now a function of fields, space and time partial derivatives of fields, and time, instead of generalized coordinates. Each field is a function of space and time, $$\phi_i = \phi_i(\mathbf{r},t)$$. I think it's still the kinetic energy density minus the potential energy density, but not 100% sure:
 * $$\mathcal{L}(\phi_1,\phi_2,\ldots,\partial_\mu \phi_1,\partial_\mu \phi_2,\ldots,t) $$
 * the fields can be scalar, vector, tensor, spinor... anything. Substitution into the Euler-Lagrange equation for fields:
 * $$ \partial_\mu \left(\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_i)}\right) = \frac{\partial \mathcal{L}}{\partial \phi_i} $$
 * gives the field equation. This is definitely true for scalar fields, but more derivatives may be needed for multi-component objects.

Hope this helps, M&and;Ŝc2ħεИτlk 08:00, 20 July 2013 (UTC)
 * In GR and QFT: Same ideas apply, but in general it's certainly not so simple to write down the Lagrangian which has to be empirically constructed and guessed, and the Lagrangian is not necessarily the kinetic minus the potential energy (densities) anymore. Analogous to the partition function and grand partition function in statistical mechanics, the Lagrangian of a realistic system can become very complicated...


 * Yes, astonishing I guess it is, especially considering the breadth of its applicability. I like to understand things at a fairly deep intuitive level before I really accept them. If there is an unanswered "why" about it, I tend to resist learning much detail about it, and Lagrangians have fallen into this category for me.  They point to a very powerful underlying mathematical mechanism or truth, but don't seem to capture the essence of it.  I suppose Lagrangians have seemed too much like "black magic" (some cookbook tool that does the job) for me to think that they are fundamental.  Lagrangians have the curious feature that they take the difference of two types of energy and not the intuitively more comfortable total energy of the system. Lorentzian symmetry is also not evident in the formulation of a Lagrangian. Worst is that it seems that there is no obvious way to find the "correct" Lagrangian in every context.  At least the formulation using a Lagrangian density appears to have the correct symmetries.  The electromagnetic four-potential also seems to condense the information of a system in a similar way, simplifies expressions and even appears in the Dirac equation.  But to me there is something unsatisfying about it, as though there is a missing insight.  And, I guess, the entry-level effort is a bit more than I can muster.  An example of a a lifting of the mist that I liked was GA's reinterpretation of the Dirac matrices and Pauli matrices as vectors; they seemed to much like "just something that works". — Quondum 21:19, 21 July 2013 (UTC)

Geometric calculus
I made some rather major updates to geometric calculus. I noticed on its talk page that you seem interested in editing that article. You may wish to have a look and correct/improve further. Teply (talk) 23:38, 21 May 2013 (UTC)
 * I have been watching your work on the article with interest: it is on my watchlist and an interest area for me. Thank you for enhancing it so much – you took it a big step forwards.  I'll work on it as I am able, though this might be limited by a lack of knowledge, reference material.  — Quondum 23:57, 21 May 2013 (UTC)
 * Then you may also be interested in outermorphism. The theory of linear algebra embedded in geometric algebra isn't really covered in the current version of geometric algebra though I think it deserves some mention. Teply (talk) 01:39, 26 May 2013 (UTC)
 * Thanks for pointing it out. It probably belongs in the GA article rather than as separate article – both the term and the concept belong in the GA context.  The area of linear transformations in GA is perplexing me, so I still have a lot to learn on this.  I have figured out that general linear transformations on a GA/CA cannot readily be expressed neatly, but have not looked into outermorphisms, these being a very important subclass of linear transformation in the GA context.  It's really bugging me that a general outermorphism begs to be written in terms of an element of the algebra (as a rotor can), but seems to be resistant to this.  — Quondum 02:20, 26 May 2013 (UTC)


 * Keep outermorphism in a separate article, likely geometric algebra would become too big. M&and;Ŝc2ħεИτlk 07:20, 26 May 2013 (UTC)


 * Well, yes, I agree, but it should be treated as a sub-article of GA. Which it already is written as.  But Geometric algebra must give a nutshell definition and must link to it.  — Quondum 11:23, 26 May 2013 (UTC)

Recent article Symmetry in quantum mechanics
Judging by your previous discussions with Rschwieb, you have probably more experience in Lorentz, Lie, and Poincaré group theory, than myself, and would be helpful in any clarification of the article when you have inclination, time, and energy. As you'll be able to see, the article is far from finished and needs plenty more on gauge theories in QFTs and discrete symmetries in particle physics and numerous other things on that talk page, and may need several layers of rewriting. I plan to do this soon (today hopefully) but it may take time...

Could I please notify that you make no dash changes to this article (yet)? While you are hard working and altruistic in all articles on this matter, it's best to wait until this article reaches an asymptotically stable and complete form, else you may end up editing the entire thing several times over for no reason (except wasting your own time, which we don't want)... Also it'll give me a chance to get my own dashing correct.

Thanks, M&and;Ŝc2ħεИτlk 05:37, 8 June 2013 (UTC)


 * I don't agree with you about "more experience" – there are simply certain aspects that I have gained an intuitive perception of, combined with a tendency to express myself quite assertively. I tend to focus on the big picture and some of the details, both of which can be simple, whereas the workings of a topic will often elude me.  I've added a comment on the talk page.
 * Point taken about the dashes. This is an example of a cognitively mechanical activity in which I indulge, perhaps increasing general awareness of WP style and simply adding "safe" improvements, but where it becomes distracting or irritating it risks becoming counterproductive, as would be the case in any nascent article.  — Quondum 14:12, 8 June 2013 (UTC)

Finally: Spherical basis
I finally started it for now. Spherical harmonics are familiar from partial differential equations and quantum mechanics, but what we really need to be careful on is the notations/conventions various authors seem to use (and the errors in some pdfs floating around). It is a huge mess, incomplete and probably full of typos/inaccuracies, and I intend to finish it soon, though at least others can edit it now. Thanks, M&and;Ŝc2ħεИτlk 08:29, 12 June 2013 (UTC)
 * Great. Since the form is already well-developed, attention from others will be good to sort out details and improving consistency. And don't talk it down so much – you've clearly put a lot of careful work into it. — Quondum 11:02, 12 June 2013 (UTC)


 * You'll notice content split as explained at Talk:Spherical basis, Talk:Tensor operator, wikiproject mathematics, feel free to edit the two. Thanks, M&and;Ŝc2ħεИτlk 08:59, 13 June 2013 (UTC)

Anapole DM
I see you recently reverted a change to Majorana fermion that mentioned their role as dark matter candidates because of their electromagnetic properties. Perhaps my recent, sourced edits in that article and toroidal moment are to your liking? Teply (talk) 08:31, 13 June 2013 (UTC)
 * Yes, definitely much better. More comment at toroidal moment talk page.  — Quondum 11:57, 13 June 2013 (UTC)

Editing other's Talk page entries
I just want to be sure you're aware that it's generally considered poor form to edit another person's WP:Talk page entry, as you did in this edit. In this case, it's pretty clear that the link you provided was BB's intent, but it would have been better to say so in a follow-up comment rather than editing his post (which, after all, has BB's signature and time stamp, not yours). This is addressed in WP:TPO. -- Scray (talk) 03:13, 30 July 2013 (UTC)
 * Yes, I'm aware of it. This seems to be one example that is permitted:
 * Some examples of appropriately editing others' comments:
 * Fixing format errors that render material difficult to read. In this case, restrict the edits to formatting changes only and preserve the content as much as possible. Examples include fixing indentation levels, removing bullets from discussions that are not consensus polls or requests for comment (RfC), fixing list markup, using and other technical markup to fix code samples, and providing wikilinks if it helps in better navigation. [my added emphasis]
 * In this instance, I preferred to not visibly post additional material, as it is a thread that does not belong there and adding a follow-up comment partially counters that impression, but the OP was apparently inexperienced and the link may have made his/her barrier to entry lower. — Quondum 10:52, 30 July 2013 (UTC)
 * OK, you're aware. -- Scray (talk) 11:44, 30 July 2013 (UTC)

Thanks!
I really should have checked before assuming it was Lorentz, not Lorenz! Elroch (talk) 19:18, 7 September 2013 (UTC)
 * It's an easy mistake to make, probably because both the names and the concepts are so closely related. I think most of us make that assumption until we learn otherwise. In some articles, this change was being made so frequently that editors have taken to leaving edit notes in the wiki markup warning about it. Perhaps the same could be done here. — Quondum 19:27, 7 September 2013 (UTC)

Generators and submonoids
I would want to ask what reference did you use to write Generators and submonoids.

Having in mind that there is a unique identity element in a monoid and that a submonoid, by definition, has the identity element I think that in this sentence

A submonoid of a monoid M is a subset N of M that is closed under the monoid operation and contains an identity element, not necessarily that of M.

"not necessarily that of M" is a false statement.

I have searched several books that mention the need of being a submonoid when generated by a subset N of a monoid. A subset N of M is said to be a generator of M if M is the smallest set containing N that is closed under the monoid operation, or equivalently M is the result of applying the finitary closure operator to N. Not every set N will generate a monoid, as the generated structure may lack an identity element.

Therefore, it's impossible that "Not every set N will generate a monoid, as the generated structure may lack an identity element." — Preceding unsigned comment added by Ydus (talk • contribs) 06:48, 17 September 2013 (UTC)


 * I found the statement in the article to be ambiguous, and sought to clarify it. There are two feasible definitions, one as I have phrased it, and one that further imposes the condition that its identity element be that of the original monoid. The two both make sense; the only question is which is the standard definition.
 * The definitions that I found may not be notable, and references that require the two identity elements to be the same element would be welcome. What I found  makes it clear that at least as far as these sites are concerned, the two identities need not be the same.  An example of the definition requiring the identity elements to be the same also exist (e.g. ).
 * Now to address a few of your statements, which might indicate misconceptions:
 * "not necessarily that of M" is a false statement
 * I'm not sure of the basis you are proposing for it being false; it is only false if the definition requires the identity element of N to be the identity element of M. Most of my links do not corroborate your point of view. Unfortunately, I do not at present have Jacobson to hand. In case this might be confusing things, it is possible to have a proper subset N of a monoid M that is itself a monoid, but the identity element of N is not that of M. Take for example the multiplicative monoid of the integers Z, and form the Cartesian product of it with itself: M = Z × Z, with multiplication defined via component-wise multiplication: (a,b) · (c,d) = (ac,bd).  Its identity element is (1,1).  The set defined by N = Z × {0} is a subset of M, and under M 's multiplication has identity element (1,0) and is thus a monoid, and hence a submonoid of M by my definition, but not under the alternate definition that requires the identities to be the same.
 * it's impossible that "Not every set N will generate a monoid, as the generated structure may lack an identity element."
 * I don't see where you get this. Counterexample: Again take M = Z with multiplication, which is a monoid. Take a subset N = {2}.  This generates the semigroup {2,4,8,16,...} under multiplication, but as this lacks an identity element, it is not a monoid.
 * I suggest that we need to find which of the two definitions (possibly both) are notable, and indicate the result in the article with references. Finding notable references is apparently not easy though. — Quondum 12:10, 17 September 2013 (UTC)

First of all, I want to apologise for my English. I get your point, it's true that depends on where you get the definition.

Taking into account what it is written in Fundamental concepts of algebra - Chevalley:


 * "If we assume that B contains e and is stable, then it constitutes a monoid when equipped with the induced law of composition. In that case, B is called a submonoid of A".

(A is a monoid and e is the neutral element of A)

we come up with the idea that a submonoid has the same properties as the monoid operation.

I will remark that, if we use the previous defintion of submonoid, the definition that you get from Wolfram mathworld implicitly defines a submonoid with the same identity element as the monoid and the same properties as the monoid operation:


 * "A submonoid is a subset of the elements of a monoid that are themselves a monoid under the same monoid operation."

Quoting Jacobson in Basic Algebra I,


 * "If M is a monoid, a subset N of M is called a submonoid if N contains 1 and N is closed under the product in M."

It confuse me. In my understanding, there is no need for the operation to be associative. For the time being, I didn't find this definition anywhere else.

About the generator of a submonoid, quoting Chevalley,


 * "B is the smallest submonoid of A containing S (in the sense that it is contained in any submonoid which contains S). It is called the submonoid generated by S."

Again, it depends on the definition of the concepts used. In this case is talking about a submonoid.

Appart from what we are discussing, I want to mention that using N in the first paragraph as a submonoid and as a subset in the second one, it could mislead someone. — Preceding unsigned comment added by Ydus (talk • contribs) 15:55, 17 September 2013 (UTC)
 * By the way, please sign your posts. — Quondum 21:53, 17 September 2013 (UTC)


 * On your last point, I have renamed the second use of N. Good observation.
 * This leaves us with your other two observations: That there are effectively two definitions of a submonoid, and two definitions of generation of a monoid. My guess is that both definitions occur in each case (e.g. your quote of Jacobson requires the identity of a submonoid to be the same as the of the monoid).  We will have to include both, I think, with references.  — Quondum 21:48, 17 September 2013 (UTC)


 * Consider the natural numbers, N, with the operation being maximum. Then this is a commutative monoid with identity 0. And every non-empty subset of N constitutes a submonoid which is also commutative and has its least element as its identity. JRSpriggs (talk) 23:49, 17 September 2013 (UTC)


 * I'm not too sure what your point is. This depends on which of the two definitions of "submonoid" you are using. I agree that it is a subset and a monoid, but if the definition of "submonoid" also requires that the identity element be the same (as, for example, Jacobson's definition quoted above: "If M is a monoid, a subset N of M is called a submonoid if N contains 1 and N is closed under the product in M."), then the submonoids of the natural numbers under this operation are only those subsets that contain the identity (in this case 0).  — Quondum 04:07, 18 September 2013 (UTC)


 * It was not clear to me from the above discussion whether Ydus understood that such a situation was possible. Perhaps he thought that the identity element had to be the same regardless of the definition used. That is why I mentioned the example. JRSpriggs (talk) 01:13, 19 September 2013 (UTC)


 * I understood what Quondum was explaining. But, as we mentioned, it depends on the definition. Your example wouldn't be true with Jacobson's definition of submonoid. I will research about this topic. — Preceding unsigned comment added by 27.253.101.141 (talk) 02:37, 19 September 2013 (UTC)

I thought about the definition of submonoid. Normally, when we use the prefix sub- it is because we are dealing with a subset that accomplish the same properties of the object that it belongs to. In consequence, I think that the right definition of submonoid would be:

A submonoid of a monoid M is a subset N of M that is closed under the monoid operation and contains the identity element of the monoid.

I would say that after the confusion respect the definition, it is a good idea to emphazise the fact that it containts the identity element of the monoid.

We could add, afterall, that it also exists other monoids incluided in a given monoid with another identity elements. — Preceding unsigned comment added by 95.19.87.240 (talk) 23:38, 25 September 2013 (UTC)


 * I assume you mean by "the same properties of the object" that you mean the category-theoretic approach that treats the identity element as a nullary operator, and that all operators are shared. Thus, the identity element is shared.  I agree with this approach, so I would refer to a "monoid subsemigroup" if I did not mean that the identity element had to be the same. Yet, one should give due weight to other approaches, which means someone with familiarity with the area should decide whether we should present this as and alternative definition.  I also agree that it is important to point out that a subset of a monoid that is a monoid under the same binary operator can have a distinct identity element.  — Quondum 00:17, 26 September 2013 (UTC)


 * I have changed the article (Monoid) in line with this. See what you think. — Quondum 11:44, 26 September 2013 (UTC)

Gamma 5
Hi Q!

You removed my note on $γ_{5}$ forming a Clifford algebra together with the other gammas for five spacetime dimensions. Why?

I think it is relevant, and I can source the removed statement verbatim.

Cheers, YohanN7 (talk) 16:28, 6 November 2013 (UTC)


 * Hi. No offence intended.  For context, this is the bit I removed after you added it:
 * But it is a particularly appropriate name because the set ${γ^{0},γ^{1},γ^{2},γ^{3},γ_{5}}|undefined$ provide a set of gamma matrices in five spacetime dimensions.
 * Even as a quote from a source, it appears to be the opinion of the author you're quoting, and (if you'll forgive the wikilawyering) thus would have to be reported as such, even if said opinion was notable. However, quoting the opinions of individual authors is not what the article is about. And the article does not deal with generalizations to other dimensions at all, aside from the statement "It is also possible to define higher-dimensional gamma matrices" in the lead (duly linked to in another article). Its presentation was only a point of incidental interest. Relevance would only be shown if the name was actually notably used in this particular way, not simply some source saying that it forms a Clifford algebra.
 * The "appropriateness" of the name is that authors (Weinberg, The Quantum Theory of Fields vol 1) opinion of course (and should go out anyhow). But ${γ^{0},γ^{1},γ^{2},γ^{3},γ_{5}}|undefined$ giving gamma matrices in 5-d is just a simple fact, notable or not. I think we mention somewhere that the Pauli matrices are 3-d gamma matrices. Why not mention the 5-d? As for spacetime in various dimensions, it is treated in plenty of advanced QFT texts, not to mention string theory. The fact that the article doesn't treat arbitrary dimensions doesn't mean it shouldn't.
 * You'd also have had to indicate that it would be a real Clifford algebra. There is already enough confusion about whether the Dirac algebra is Cl1,3(R) or its complexification Cl4(C). This algebra is Cl2,3(R), which we shouldn't consider as being of a 5-dimensional "spacetime", since this term would normally be intended to mean having a Lorentzian metric. This is a slightly scatty reply. but I think you get the message that I think it is inappropriate to include this. — Quondum 06:39, 7 November 2013 (UTC)
 * I don't follow you exactly. I need to read up on this, but from where do you get that the algebra is Cl2,3(R)? I'd say the algebra is Cl4,1(R) (or Cl1,4(R)). YohanN7 (talk) 14:41, 7 November 2013 (UTC)
 * If the five matrices span the grade 1 space of a real Clifford algebra and also anti-commute, then their squares necessarily give the signature of the space. And here we have that two of the matrices square to +1, and three to −1.  As a check on this, we already know that the algebra is isomorphic to M4(C).  As a real Clifford algebra over five dimensions, Classification of Clifford algebras gives three candidates with this ring isomorphism: Cl4,1(R), Cl2,3(R), Cl0,5(R).  So while you could argue that though starting with a (1,n−1) convention, we've had to flip to a (n−1,1) convention to match your assertion. But to do so, we've had to replace at least two of the original matrices (e.g. by multiplying 2 or 4 of them by $i$), to get the signature to be (+ + + + −).  That is to say, the five matrices as listed do not span any grade 1 subspace of this Clifford algebra.  Which disqualifies them as what we mean by "gammas", even if we leave the rest as a hidden puzzle for the reader to disentangle.
 * But ignoring the maths, the statement still does not qualify for inclusion. You included it as a comment on the name γ5, which it fails to do in a coherent fashion. The mathematical argument is actually a diversion from this point. — Quondum 15:27, 7 November 2013 (UTC)
 * I included it mostly for its mathematical message, the name was just an odd twist. Forget that.
 * On the math: Note that it says ${γ^{0},γ^{1},γ^{2},γ^{3},γ_{5}}|undefined$, not ${γ^{0},γ^{1},γ^{2},γ^{3},γ^{5}}|undefined$. The Weinberg books are famous for containing extremely few typos, and of course no such gross errors as you hint above. YohanN7 (talk) 16:18, 7 November 2013 (UTC)
 * I noticed the subscript. But raising the index is equivalent to flipping the sign only if the corresponding element squares to −I.  But since (γ5)2 = +I, we have γ5 = +γ5, and it makes no difference whether the index is lower or upper.  C'mon, what do you read into the sequence $((γ^{0})^{2},(γ^{1})^{2},(γ^{2})^{2},(γ^{3})^{2},(γ_{5})^{2}) = (+I, –I, –I, –I, +I)$?squaring added — Quondum 02:14, 8 November 2013 (UTC)
 * Since I have now gone into details of the isomorphism, it is clear that I am not contradicting the statement that this can be used as a 5d "spacetime" Clifford algebra (by minor changes to the matrices it gives you a signature (+ + + + −), so depending on how it was presented, there is no deep contradiction. Starting with different gamma matrices that have a (− + + +) signature (just multiply each by $i$), you get a (− + + + +) signature with the exact construction you give. But it remains esoteric, since this construction is probably confined to only a few cases (i.e. choice of number of dimensions and signature). Add to this that at some level Clifford algebras of opposite signature are equivalent for this use, and you might find that Weinberg et al are simply glossing over this detail? Or perhaps they just use the opposite sign convention? — Quondum 19:13, 7 November 2013 (UTC)


 * You are right. Weinberg uses the (− + + +) convention. What we should say (if anything) is that ${γ^{0},γ^{1},γ^{2},γ^{3},iγ^{5}}|undefined$ gives a 5-d Clifford algebra. B t w, I checked another reference (that uses (+ - - -)): http://www.damtp.cam.ac.uk/user/dt281/qft/four.pdf. Respected author, freely available online, page 93. He states that the reason for the terminology is, in fact, that ${γ^{0},γ^{1},γ^{2},γ^{3},iγ^{5}}|undefined$ is a set of gamma matrices for five spacetime dimensions. YohanN7 (talk) 20:03, 7 November 2013 (UTC)


 * With the Weinberg sign convention, I guess someone might have used this as a "reason", but it doesn't quite work for me as a reason with the added $i$. The source you gave a reference to seems to do as you say: use a (+ − − −) convention, and then adds $iγ^{5}$ in an attempt to make a five dimensional (+ − − − −) Clifford algebra.  The anticommutation relations given in the source and required for a Clifford algebra do hold, but this is a crucial: they are not sufficient.  As you can check for yourself, ${γ^{0},γ^{1},γ^{2},γ^{3},iγ^{5}}|undefined$ does not work for five dimensions at all: we have the identity $iγ^{5} = −γ^{0}γ^{1}γ^{2}γ^{3}$}, which is not R-linearly independent of the space that is generated by the other matrices; it is already present in the Clifford algebra. This collapses the 5-d Clifford algebra to a 4-d one.  Weird, but it looks to me to be a genuine mistake to call this a "five-dimensional Clifford algebra".  — Quondum 20:47, 7 November 2013 (UTC)


 * There is nothing about linear independence of the gammas from everything generated by the gammas in the defining condition. In odd spacetime dimension $d$ the totally antisymmetric (AS) tensors of rank $n$ are linearly related to the AS tensors of rank $d - n$. [The AS tensors are obtained by antisymmetrizing the products of the gammas. This is something I'll put into the Dirac algebra article.] There is a formula expressing this in the Weinberg book (with a million of indices and subscripts/superscripts, too lazy to write it here). This formula implies the identity $iγ^{5} = −γ^{0}γ^{1}γ^{2}γ^{3}$, so everything is as it should be. YohanN7 (talk) 10:35, 8 November 2013 (UTC)


 * You seem to be mixing up the two cases. You cannot transfer Weinberg's mathematical details to the other case. Weinberg uses different gammas from the ones in the article (this must be the case to get the different signature). Also, take care not to confuse R-linear with C-linear. ${1, i}$ is R-linear independent, but C-linear dependent.  Remember that the two cases use different matrices throughout, so they must be dealt with independently.  For the moment I'm going to use another name ($γ^{4}$) for one of the matrices, to try to keep things straight.
 * Weinberg's $(− + + +)$ convention, and his proposal $(γ^{0},γ^{1},γ^{2},γ^{3},γ^{4})$ with $γ^{4} = γ^{5} = iγ^{0}γ^{1}γ^{2}γ^{3}$ as a 1-vector basis for the generating space of a real 5-dimensional Clifford algebra: Here we have the relation $γ^{4} = iγ^{0}γ^{1}γ^{2}γ^{3}$, so $γ^{4}$ is R-linearly independent of the product $γ^{0}γ^{1}γ^{2}γ^{3}$.  We also get the desired squaring to $(− + + + +)$.
 * The $(+ − − −)$ convention, and the proposal $(γ^{0},γ^{1},γ^{2},γ^{3},γ^{4})$ with $γ^{4} = iγ^{5} = −γ^{0}γ^{1}γ^{2}γ^{3}$ as a 1-vector basis for the generating space of a real Clifford algebra: Here we have the relation $γ^{4} = −γ^{0}γ^{1}γ^{2}γ^{3}$, so $γ^{4}$ is R-linearly dependent on the product $γ^{0}γ^{1}γ^{2}γ^{3}$.  We get an awkward squaring to $(+ − − − +)$ apparent squaring to $(+ − − − −)$, which is not the case for any Clifford algebra isomorphic to $M_{4}(C)$, denoted $C(4)$ in the classification article.
 * The confusion possibly comes in because previously we did not have a symbol for the fifth basis element, and were using the an expression in its place. One of the properties of a Clifford algebra of a vector space of dimension n over a field $F$ (often referred to as an $n$-dimensional Clifford algebra, even though it has a higher dimension over $F$ as a vector space in its own right) is that its dimension is $2^{n}$ over $F$. That is to say, a basis of the algebra has $2^{n}$ $F$-linearly independent elements. If this is not what we find, we know something went wrong.  — Quondum 15:34, 8 November 2013 (UTC)
 * Weinberg writes that there are $2^{n}$ independent elements in even spacetime dimensions, and $2^{n-1}$ in odd spacetime dimensions. Edit: No, he doesn't exactly. He is talking about antisymmetrized products, and I can't say immeduatly that it is the same thing. That aside, I find it hard to believe that the metric signature (exactly one + or exactly one -) has any real significance for anything we are discussing. YohanN7 (talk) 16:43, 8 November 2013 (UTC)
 * If we have a set of (linearly independent!) gamma matrices squaring and anticommuting the right way, I'd say we have a Clifford algebra. The article Clifford algebra says that the dimension of the algebra is $2^{n}$ (no exceptions mentioned), that's true. But the definition of a Clifford algebra doesn't say anything about dimension. There is a problem somewhere. YohanN7 (talk) 17:31, 8 November 2013 (UTC)
 * Perhaps we could continue the discussion here: User:YohanN7/Gamma matrices. I wrote a program to confirm anticommutation/squaring. The matrices with signature $(+ − − −)$ certainly behave as advertised. YohanN7 (talk) 19:14, 8 November 2013 (UTC)

(ec: posting this here, but will continue discussion on your page when I get back to it later.)

As I've said before, linear independence and the commutation relations of a basis for a vector space is not a sufficient condition for producing a full Clifford algebra. If this were the case, we could construct higher-dimensional Clifford algebras with very limited representations. Take for example the quaternions $H ≈ Cℓ_{0,2}(R)$, which is a "two-dimensional" Clifford algebra, with the algebra having $2^{2} = 4$ dimensions of the algebra. (If you want a strictly real full matrix representation example, we could use $Cℓ_{1,1}(R) ≈ M_{2}(R)$ instead, which can be used to illustrate the same point.) Now take the three elements $i, j, k$, which are $R$-linearly independent, and which satisfy the necessary commutation relations. Yet, no amount of finding products is going to build a Clifford algebra over three dimensions (i.e. is going to construct $Cℓ_{0,3}(R)$ with an algebra dimension of 8), for the simple reason that we have a predefined relation $ij = k$, which would normally be a new, linearly independent basis element of the algebra in the construction of a Clifford algebra. The relation $ij = k$ is an additional equivalence that collapses the attempted construction back to $H$. This is exactly the problem that is occurring in our example with gamma matrices. Or, to put it more succinctly, $M_{4}(C)$ (or any subalgebra of it) is not algebra isomorphic with $Cℓ_{1,4}(R)$; it simply cannot be used as a representation of the Clifford algebra $Cℓ_{1,4}(R)$, even though it is a representation of $Cℓ_{4,1}(R)$.

Your difficulty in accepting that switching the signature should have any effect is shared by many people, but it is a fact of Clifford algebras: this switch often produces a completely nonisomorphic algebra. However, in physics we are usually interested only in specific subspaces of the algebra, for example scalars, 1-vectors, bivectors, spinors/rotors, pseudovectors, etc., and the full algebra never comes into play. Because of this, the sign convention (and the fact that we would be working in nonequivalent algebras) has no effect: we always get the same result, regardless of the sign convention. Which will hopefully allay the feeling that this contradicts your intuition. But it does not change my assertion that the source got it wrong. — Quondum 19:24, 8 November 2013 (UTC)


 * If you two don't mind me cutting in, I just found a very nice pdf: on the Gamma matrices. It looks relevant and useful for RQM articles. Unfortunately, the pdf alone is not reliable as a source, and there could be errors in it, but at least it has a clear presentation style. M&and;Ŝc2ħεИτlk 20:45, 8 November 2013 (UTC)

Emphasis
Hi!

I often emphasize text a little bit to much. When I go back it might not read well. Here in Talk:Symmetry in quantum mechanics the phrase "Somebody knowledgeable of the subject might know" reads like an insult. It was unintentional and the italics is removed. YohanN7 (talk) 03:28, 12 November 2013 (UTC)


 * No sweat. It seems to have at least attracted an opinion from someone else, who may have more experience in the field. I welcome anyone with the knowledge and experience to talk with some confidence; I am guilty of speaking above my confidence level (even if that is just to express doubt). You at least have a number of years of study in the general field to go on.  —Quondum 05:53, 12 November 2013 (UTC)

Nontrivial idempotent
Hi Quondam,

At zero divisor, I think it is probably safer to say "idempotent $$e \ne 0,1$$" instead of "nontrivial idempotent e" since I think that some people might interpret "nontrivial" as meaning nonzero. In any case, the idempotent needs to be named e, for the sake of the equations at the end of the sentence. Anyway, this is a small matter, so I leave it up to you on how to edit it. Ebony Jackson (talk) 22:47, 16 November 2013 (UTC)


 * Thanks. I've edited it, settling on a reduced restriction (we do not need to exclude 0).
 * A different point: in the next edit did you not mean "nontrivial ring" where you put "nonzero ring"? (Zero ring has a different meaning, which makes "nonzero ring" really confusing.) —Quondum 23:05, 16 November 2013 (UTC)


 * You are right; no need to exclude 0! As for "nontrivial ring" vs. "nonzero ring": the standard name (in modern notable textbooks such as Artin, Atiyah-Macdonald, Bourbaki, Hartshorne, Lang, etc.) for the ring with 0=1 is the zero ring.  I'm going to try to persuade Wikipedia to reflect this. Ebony Jackson (talk) 23:16, 16 November 2013 (UTC)


 * I saw your suggestion at Trivial ring. A book search suggests that "trivial ring" occurs in more publications than "zero ring" does.  So determination of the "best" (most notable) term is liable to be tricky (or contested).  This is well outside my sphere of knowledge.
 * I see that "nonzero ring" is used (notwithstanding that I find it strange, but I guess one can get used to that). —Quondum 23:27, 16 November 2013 (UTC)

Pauli exclusion principle
I've added some of the reasons I think the article should state all electrons rather than all electrons in the same atom on the talk page here. I'd be grateful if you reviewed them regarding the wording of the article. Wolfmankurd (talk) 21:14, 24 November 2013 (UTC)


 * I've added a comment. This is a tricky point. The Pauli exclusion principle applies to all electrons, as you say. But the example should perhaps remain worded (for simplicity) to consider a system consisting of a single atom, in which all the electrons are in the atom's orbitals. As you left it, the wording could suggest that the state with the same quantum numbers in a different atom is the same state; this is not so, because the other atom is in a different place. An electron here is in a different state from an electron there, even if the energy levels are mathematically identical. —Quondum 00:44, 25 November 2013 (UTC)

Group structure and the axiom of choice
Hi Q!

Thanks for cleaning the article. (I didn't know what to do with it really, had it for a year or so, wrote it most for fun. User:Arthur Rubin had a look at it, said it was correct, but probably didn't belong in main space. Well, instead of tossing it, I submitted, expecting it to take ages for an assessment (it said it would). It took 2 hours;)) YohanN7 (talk) 12:13, 20 December 2013 (UTC)


 * Sure, just my OCD getting the better of me, and it was already pretty cleanly presented, notwithstanding that it is way over my head. Actually, it has the feel of being such a fundamental but somehow surprising result (theorem, whatever), that it "just belongs", even if the title might need a bit of thought. You've had some mathematicians make tweaks without much comment, which suggests that my uneducated gut feel is not way off.  It will be good to have it exposed to mathematician editors in main space to see what comes of it.  It might end up merged somewhere, but until then, no harm leaving it as its own article.  Perhaps Arthur meant that it does not deserve a main space article to itself, but that does not mean that it does not belong in some form in main space. Perhaps it should be linked under Axiom of choice? —Quondum 15:57, 20 December 2013 (UTC)
 * It is linked from there.
 * You are right about "fundamental but somehow surprising result", especially group -> choice. Over at mathworld mathoverflow, where I found most stuff to begin with, they were pretty enthusiastic about it. Follow the link from the articles talk page, it's worth a read. YohanN7 (talk) 16:18, 20 December 2013 (UTC)

Quasigroups and cancellative magmas
It seems to me that quasigroups and cancellative magmas are in one-to-one correspondence. The cancellative magmas are just the corresponding quasigroups stripped off of right and left division? YohanN7 (talk) 04:45, 9 February 2014 (UTC)
 * Consider the monoid of non-negative integers with addition. If $a + x = b$ and $a + y = b$, this implies that $x = y$, so this is a cancellative magma.  There is no element $a$ such that $a + 5 = 3$, thus it is not a quasigroup.  So we have an example of a cancellative magma that is not a quasigroup.  We can't even speak of a "corresponding quasigroup" to relate it to. Rather counterintuitive at first; I wrestled with this a bit. —Quondum 05:55, 9 February 2014 (UTC)
 * Yep, you are right. (See article talk page too, I think i blundered there as well and striked out some ). It is possible that we don't have a Latin square, but we have a subset of a Latin square, and that's what is important, so that $ax ≠ bx$ and $xa ≠ xb$ whenever $a ≠ b$. This fortunately follows directly from the cancellation properties. I'm to quick to draw conclusions, especially since a few hours ago, I had no clue about what a quasigroup or a cancellative magma is... YohanN7 (talk) 06:29, 9 February 2014 (UTC)
 * I've been doing some rapid learning about them too. Quasigroups I'd seen a while ago, but cancellation is new to me. I've been monkeying around with "groupes" (groups plus the empty groupe) defined as associative quasigroups, so that I can produce a consistent definition of the field with one element. And succeeded, I think. —Quondum 06:44, 9 February 2014 (UTC)

Etymology of magma
Hi Quondum, I've just seen that you have removed my addition to the etymology of the term magma in algebra. When we are referring to the etymology of a word we must give the first meaning, its root. I understand that the article want to give the meaning that the Bourdaki group wanted and what they might have in mind, but then the title should not be "Etymology" but something like "Meaning" or "History of the term" etc. Also i don't find it bad to add the origin of the word and coexist with all the other meanings, as many other articles in wikipedia do the same thing. Waiting for you opinion! Happy to discuss it!--Papxr (talk) 11:55, 2 March 2014 (UTC)
 * Any mention of the etymology in this context (mathematics) would be to aid us think about the use of the word, and why it came to be used in the sense being discussed, and then only if it differs from a readily available etymology, as the longer-term linguistic history (i.e. its evolution before any non-mathematical use) being of interest would be found in a dictionary. Remember that, unlike a dictionary, a WP article is about a concept, not about a word, and distinct meanings will not be discussed in the same article. You say that many other articles in WP do the same thing? Could you give examples? Preferably of mathematical terms, but the same principle applies to all articles. I would find it very strange if the original etymology of words such as ring, group etc. were so much as mentioned in the articles. I'm happy to get involved in a discussion, though at Wikipedia Talk:Wikiproject Mathematics you'd get a broader cross-section of opinions. —Quondum 21:20, 2 March 2014 (UTC)

Hi again, thanks for your reply. Well, an encyclopedia like WP is more than a dictionary. So it has all of a dictionary and much more. As I said before I got your point but I think the term "etymology" is misleading and I believe it should be changed. As far as the examples are concerned, take the word school --> Etymology: The word school derives from Greek σχολή (scholē), originally meaning "leisure" and also "that in which leisure is employed", but later "a group to whom lectures were given, school". So we have the original meaning which is different from the current use. Also you can take a look at the etymology of the word "Mathematics" in WP. In addition, many terms that didn't exist in the antiquity and formed later come from greek words. For example take the term Homomorphism. We surely will not say that the one who introduced the term was thinking about the meaning of the word morph in english which means shape but we just say that morph comes from ancient greek μορφή (morphe) meaning "shape"(see homomorphism). So in the same way we shall say that magma comes from greek and its metaphorical meaning is a mixture of various things. Lastly (that's beside the point), do you know why they had chosen the name "Burbaki" which is a Greek name? Is there any "tale" or it was completely random. Happy again to have this conversation! --Papxr (talk) 01:02, 3 March 2014 (UTC)
 * I disagree on "it has all of a dictionary and much more". It is an explicit principle to avoid the functions of a dictionary, although this is not always implemented strictly. The use of the term "etymology" might be better restricted to a dictionary, and yes, I think it should not be used in WP except where it is of historical relevance to the subject, and in general a paragraph under History could be devoted to it, not a top-level section on itself. Your example of the etymology of school has some relevance to the history of the concept, but I'll stick with what I've just said, and I don't feel that even there a separate section is appropriate.  In the article Mathematics, it is a subsection as I'd expect.  In both cases, the aspects of the etymology mentioned relate to the origin of the word in the meaning of the article, at least as a progression; this does not appear to apply to the "magma" and numerous other mathematical terms.  Further, in your examples, the topics are fairly broad, so the articles are descriptive rather than acting as references for the detail of the topic, and can include more of general interest and in particular history. You are presumably aware that mathematics co-opts words for various uses, a bit like we use variable names: the general meaning of the word and its mathematical use are often only tenuously connected, so to draw attention to its other uses prior to it having been (nearly arbitrarily) co-opted is simply a distraction.  There are instances where compounds are formed, so homo-, auto-, morphism etc. have specific meanings, which are worth knowing about, as this helps to quickly understand new compounds that incorporate from the same roots and modifiers.  Again, origins are to be considered in the historical context.  (As an aside, one of my pet hates is the way the same word may be used in several sub-disciplines of mathematics, but often with confusingly overlapping but distinct meaning; what's worse is that there will be many arguments precipitated where this is not recognized.)
 * I have no idea why Nicolas Bourbaki was chosen as a pseudonym for the group (I'm not well-read, so I don't know whether this is even discussed in the literature); the reason for using a pseudonym at all would be more interesting than the inspiration for the choice of name. Perhaps the members put names into a hat and that was drawn? —Quondum 01:52, 3 March 2014 (UTC)

It is interesting to note that the removed passage seems to be undisputable facts (unsourced), and what remains is disputable speculation (It is likely that ..., equally unsourced). Sorry Q, couldn't resist. YohanN7 (talk) 03:36, 3 March 2014 (UTC)
 * Yeah, I was tempted to remove that as well, and do think that it should go (it is even phrased as speculation); I just felt less confident removing that as it would have some relevance. However, now that you point it out... —Quondum 05:17, 3 March 2014 (UTC)

Well I was a little bit exaggerating when I said "it has all of a dictionary and much more"... wanted to say that it has same common ground like the basics of an etymology and so it's not completely "foreign". I agree with what you said and I am happy that we came to a conclusion and deposited our thoughts! --Papxr (talk) 12:23, 3 March 2014 (UTC)


 * Regarding that reference desk question just now, as we know, magmas have peculiar implications. The presence of a cancellative one implies that the underlying set is wellorderable, the presence of one on every set implies the axiom of choice. I think that is kind of cute (and remarkable). YohanN7 (talk) 22:22, 11 April 2014 (UTC)


 * But no, that is only partly correct. The presence of a magma doesn't mean the set is wellorderable. When I thought about what I wrote in the post above, I realized that if it is true, then I, in passing, had solved the second part of Hilbert's first problem (can the reals be wellordered?, CH was the first part). :D YohanN7 (talk) 22:33, 11 April 2014 (UTC)


 * Yeah, this is dabbling in things that stretch my mind. I'm liable to tie myself in knots on the subtleties of this. —Quondum 04:26, 12 April 2014 (UTC)

Interval (mathematics)
Thank you for your recent correction! While I do not strongly object the correction itself, you might be interested to know that the term "remarkable" is used in mathematical literature to denote exactly the properties that are peculiar and deserve being noticed. Whether such properties are "useful" is a separate issue. The term "useful" is more applicable in the context of a theorem proof, where certain properties can assist with the task at hand. Without a context of application, any peculiarity is just "remarkable", not "useful". Nyq (talk) 15:25, 31 March 2014 (UTC)
 * Thanks for the info. Terminology usage is always a challenge as it varies from context to context. Here it is doubly challenging, since an encyclopaedia does not qualify (IMO) as "mathematical literature" due to the different audience: the reader cannot be assumed to be familiar with quirky usage. To avoid potential subtleties of this nature, I've rephrased it again to avoid the term "useful" as well, hopefully without detracting from the reference value in any way. —Quondum 16:33, 31 March 2014 (UTC)