Talk:Bra–ket notation/Archive 1

Clarification needed, or possible Error
In the section "The unit operator" shouldn't the formula be
 * $$ \langle v | w \rangle = \langle v | \sum_{i \in \mathbb{N}} | e_i \rangle \langle e_i | w \rangle = \langle v | \sum_{i \in \mathbb{N}} | e_i \rangle \langle e_i | \sum_{j \in \mathbb{N}} | e_j \rangle \langle e_j | w \rangle = \langle v | e_i \rangle \langle e_i | e_j \rangle \langle e_j | w \rangle $$

rather than
 * $$ \langle v | w \rangle = \langle v | \sum_{i \in \mathbb{N}} | e_i \rangle \langle e_i | w \rangle = \langle v | \sum_{i \in \mathbb{N}} | e_i \rangle \langle e_i | \sum_{j \in \mathbb{N}} | e_j \rangle \langle e_j | w \rangle = \langle w | e_i \rangle \langle e_i | e_j \rangle \langle e_j | v \rangle $$

? (v and w exchanged in the last part)


 * I don't see why. $$\langle w | e_i \rangle \langle e_i | e_j \rangle \langle e_j | v \rangle = \langle w | v \rangle$$, which is the complex conjugate of $$ \langle v | w \rangle $$, not in general equal.  That bit of the article is just trying to demonstrate how you can insert identity operators rather than any specific application of that technique.  &mdash; Laura Scudder &#9742; 22:11, 14 June 2007 (UTC)


 * That would be my mistake (see this section) although I find it very odd that myself nor anyone else has noticed this before. There are two possible fixes: putting a conjugation around the entire expression on the right or swapping the v and w. Since they are actually the same (interchanging the order of the scalars and renaming i ←→ j) I chose the latter. Thanks for (finally) noticing. --CompuChip 09:40, 16 June 2007 (UTC)

Unheadered stuff at top
I have changed the (unfinished) last paragraph about Hermitian operators of the section "Linear Operators" to include some physical meaning and to prevent operators from jumping onto the bras'/kets' label. I think this notation is unsuitable for beginners and not very useful for experts. Y!qtr9f 18:11, 30 November 2006 (UTC)

what does "is dual to" mean?


 * See Dual space! Y!qtr9f 18:11, 30 November 2006 (UTC)

Should we really use & rang ; and & lang ;? My Mozilla on Windows and IE both don't render it. (For mozilla this is reported as a bug in bugzilla: http://bugzilla.mozilla.org/show_bug.cgi?id=15731 ) Is there actually a browser that does? -- Jan Hidders 12:19 Mar 5, 2003 (UTC)

What's wrong with th ordinary angle brackets ? or & lt ; and & gt ; ? Theresa knott


 * Mozilla renders it fine for me. rang and lang are HTML 4.0 character entity references specifically for "bras" and "kets", so it's more correct to use them. It also looks more legible; make the bras and kets somewhat more difficult to read.


 * IE 4 should be able to display the characters (see here). Are you using IE 3? Do &#9001; and &#9002; (generated from the numeric codes) work for you? -- CYD

I use IE 6. I can see everything on Special characters and on http://www.unicode.org/iuc/iuc10/x-utf8.html, but neither &#9001; nor &lang; - Patrick 21:01 Mar 5, 2003 (UTC)

Can you see the characters on http://www.htmlhelp.com/reference/html40/entities/symbols.html ? -- CYD


 * I couldn't. I was also using IE6 under W'98 and there it didn't render. IE6 under XP also doesn't render it. Mozilla 1.2 under W'98 doesn't render it either, but Mozilla 1.3 under Linux and XP do. I'll see what happens if I upgrade to Mozialla 1.3 under W'98. If that works, than I'm happy with lang and rang, although strictly that would not be enough for the official policy on special characters. -- Jan Hidders 21:22 Mar 5, 2003 (UTC)
 * Is it a problem with the browser or simply the correct font that is missing?


 * Then maybe we should use an image, similar to what is done for [[Image:Del.svg]] ( [[Image:Del.gif]] ). Out of curiosity, what "official policy" are you referring to? -- CYD


 * You could, but images are really the last resort. They don't scale and are sometimes positioned awkwardly by different browsers. So I would suggest using . If you really don't like those then my next choice would be to use lang and rang anyway, just as long as we can tell people that if they want to see the page in its full glory they have to install the latest Mozilla. You might even want to plead for a change of policy on the mailing list. In that case you have my vote. :-) The "official policy" is more or less implicit in the page on special characters. From what I remember from previous discussions on the mailing list the main argument is always that we should keep Wikipedia as accessible as possible and therefore only use special characters if we really need them. -- Jan Hidders 21:35 Mar 5, 2003 (UTC)


 * I would say that having to use one particular browser is much worse than either an image or a regular <. - Patrick 22:11 Mar 5, 2003 (UTC)


 * Absolutely, however, where Mozilla goes so do the browsers that are based on the Gecko rendering engine, and since it follows the standards the KHTML-based browsers (Konquerer et al.) and other open source browsers are usually not far behind and even IE will probably catch up if it has not already. Besides, Mozilla is pretty easy to install these days and availiable on many platforms. -- Jan Hidders 22:40 Mar 5, 2003 (UTC)

Using & #9001 ; &lang ; I can see them here: &#9001; &lang;, in accordance with http://www.alanwood.net/unicode/miscellaneous_technical.html, which says "LEFT-POINTING ANGLE BRACKET (present in WGL4 and in Symbol font)" - Patrick 21:58 Mar 5, 2003 (UTC)


 * Ah I can see them now [IE 5 windows 2000]. Is it safe to assume that symbol is a pretty much unversal font? Theresa knott 14:20 Mar 6, 2003 (UTC)

Alternatively, we can use TeX all the time for these brackets, they work fine also. - Patrick 22:06 Mar 5, 2003 (UTC)


 * In-line TeX is usually discouraged. See WikiProject_Mathematics. -- Jan Hidders 22:40 Mar 5, 2003 (UTC)


 * comparison: the "correct" in-line symbol (?), the ordinary less-than, and the TeX symbol: &lang;<$$\langle$$. I see more difference between the correct in-line symbol and the TeX symbol than between the correct in-line symbol and the ordinary less-than! So using the ordinary seems best. - Patrick 04:32 Mar 6, 2003 (UTC)


 * That's strange. The TeX symbol looks exactly like &lang; to me. The difference between the correct symbol and an ordinary < looks to me like the difference between $$\langle$$ and $$<\;$$. -- CYD


 * Perhaps Patrick is referring to the size and not the shape? -- Jan Hidders 10:36 Mar 7, 2003 (UTC)


 * No, to the shape. In my case &lang; and < have a much smaller angle than $$\langle$$. - Patrick 12:40 Mar 7, 2003 (UTC)


 * Yup. I am now looking at it with my IE6 under W'98 and also there the shapes differ. The symbol font, byt the way, works for me in IE6 and Mozilla under W'98, XP and Mozilla under Linux, but seems a bit cumbersome to type. Does anybody know if there is some font that I could install under Windows to see lang and rang? -- Jan Hidders 15:51 Mar 7, 2003 (UTC)
 * Better to be cumbersome to type than impossible to read. I've changed the page accordingly. Could people check for mistakes please. Can anyone still not read the text ?Theresa knott 09:30 Mar 11, 2003 (UTC)

Regarding
 * Given any ket |&psi;&rsaquo;, bras &lsaquo;&phi;1| and &lsaquo;&phi;2|, and complex numbers c1 and c2, then, by the definition of addition and scalar multiplication of linear functionals,
 * $$(c_1 \langle\phi_1| + c_2 \langle\phi_2|)|\psi\rangle = c_1 \langle\phi_1|\psi\rangle + c_2\langle\phi_2|\psi\rangle. $$

As far as I know, it should be c1* and c2* since an inner product over the complex field is sesqi-linear in first component, or hermitian.

Proof: let A and B be vectors and c a complex number. Then, from hermitian property
 *  = < B|cA>* ,

but from linear property
 * < B|cA> = c< B|A>, hence  = (c< B|A>)* ,

hence
 *  = c*< B|A>* = c*.

Q.E.D. MathKnight 21:10, 28 Mar 2004 (UTC)

Regarding


 *  = < B|cA>* ,

This should instead be


 *  = < B|c*A>* ,

Your proof is therefore erroneous. -- CYD

The passage
 * <cA|B> = < B|cA>*

is correct, since we mark D = cA and then from Conjugate Symmetry,
 * $$ \lang D | B \rang = { \lang B | D \rang }^* $$

and subsitute again we get
 * <cA|B> = < B|cA>*

Notice that c (a scalar) multiplies A.

I can also show it directly, define bra-ket inner product as:
 * $$ \lang F | G \rang = \int{ F^* \cdot G dx } $$

you immidietly see that
 * $$ \lang c \cdot F | G \rang = \int{ (cF)^* \cdot G dx } = c^* \int{ F^* \cdot G dx } = c^* \cdot \lang F | G \rang $$

MathKnight 09:44, 15 May 2004 (UTC)

Okay, I see the confusion now. Well, since the relevant line in the article clearly refers to the addition and scalar multiplication of linear functionals, your original objection is obviously invalid. In any case, the issue has already addressed; look at the third point of that section:



c_1|\psi_1\rangle + c_2|\psi_2\rangle \;\; \hbox{is dual to} \;\; c_1^* \langle\psi_1| + c_2^* \langle\psi_2|. $$

The concept of "multiplying inside the ket" is redundant in b reprenstation of the Delta function around x we get
 * $$ \lang x | A | \psi \rang = \frac{1}{i} \psi ' (x) $$

as desired. The calculation is long and exausting, but since many useful operators are just combination of x and p = (h/i) * d/dx the common way is to skip the calculation just solving differential equation for the ket, where we search the ket expressed as a function of x (it later can represent as a function of p by the Fourier transform ).

MathKnight 09:47, 3 Aug 2004 (UTC)

Revision: I want later to enter this answer to the article as an explanation about "bra-ket and concrete representations". MathKnight 15:18, 3 Aug 2004 (UTC)

Hi MK, thanks for your explanation. I suspected the statement may have been refering to a basis representation of the operator, but am concerned that this subtlety might be overlooked by readers not previously familiar with the material. The simple fact remains that d/dx|&psi;&rang; is identically zero, unless you understand that d/dx is a label meant to imply that A is the operator defined so that &lang;x|A|&psi;&rang;=d/dx &lang;x|&psi;&rang;.

You are abeslutly right, therefore I think that this article should handle the representation issue. I think we can rely on the explaination here with few edits such as general intro for the representation problem. I want to add it to the article and I'll work on it. Also, I get white squres in in-text math-symbols (such as &lang; = &lang;. I prefer to work with LaTex for math, such as the previous thing you wrote: $$ \lang x | A | \psi \rang = \frac{d}{dx} \lang x | \psi \rang = \frac{d}{dx} \psi (x) $$ . MathKnight 20:10, 4 Aug 2004 (UTC)

Plane Waves
I'm not sure if this is just a different notation, but I believe the "plane waves" eigenstates of the momentum operator are denoted &lang;x|p&rang; rather than &lang;x|x&rang; which would be some strange delta function concotion.


 * You're right, of course. In any case, I've deleted the section on the momentum eigenstates, because it's not relevant to the article, which is about bra-ket notation not wavefunctions. -- CYD


 * You're right. The confusion was created by abuse of notation, since |x> is a reserved notation of a eigenstate of x operator. You are right, <x|p> is the proper notation. MathKnight 17:30, 16 Oct 2004 (UTC)

Linear Operators
I think the defenition Operators can also act on bras. Applying the operator A to the bra $$\langle\phi|$$ results in the bra ($$\langle\phi|$$A), defined as a linear functional on H by the rule


 * $$(\langle\phi|A) \; |\psi\rangle = \langle\phi| \; (A|\psi\rangle)$$.

This expression is commonly written as


 * $$\langle\phi|A|\psi\rangle.$$

is a little bit misleading and need clarification. For some reason it ignores the conjugation that A must go through when it swtiches sides in the bra-kets. MathKnight 10:52, 27 Nov 2004 (UTC)


 * The statement is obviously correct. I think you're getting confused by the "multiplication inside a ket" issue again. -- CYD


 * Take the operator to be the momentum operator, then the above statement is not true!? —Preceding unsigned comment added by Miobrad (talk • contribs) 22:26, 6 March 2008 (UTC)
 * The momentum operator is not a hermitian operator. but above property follows by exactly this. User:ibotty too busy to log in —Preceding unsigned comment added by 91.67.59.57 (talk) 17:25, 17 December 2008 (UTC)


 * The momentum operator is Hermitian. You can find a mathematical proof of this fact here.


 * I also disagree with MathKnight, I believe the definition given is a complete and mathematically correct definition of how an operator acts on a bra. (That doesn't necessarily mean it's phrased in a clear and pedagogical way.) --Steve (talk) 19:22, 17 December 2008 (UTC)

symbol changed midway through
"   \langle\phi|\psi\rangle.

In quantum mechanics, this is the probability amplitude for the state &#968; to collapse into the state &#966;."

Is that right, or did some &#968;'s get changed to phi's at the end of this section?


 * It is currently correct as is. <&phi;|&psi;> is the projection of state |&psi;> onto state |&phi;>, which can be interpreted as the probability for |&psi;> to change into |&phi;>.  --Laura Scudder | Talk 21:26, 16 May 2005 (UTC)

C Number Omission
Hello all: I've dared to put my remarks at the top of the page, as I'm unaware (as yet) of any convention concerning where to post newer remarks, other than below for responses. I'm sure I'll get well smacked around if I'm being presumptious about this.

I've noted that while Dirac's Bra and Ket notation is addressed, there is both a spelling omission and Dirac's "c number", which accounts for the omitted letter referenced.

A complete bra and ket notation discussion actually should reference a number c which goes between the bra and ket.

I'll include more about this when I access my copy of Dirac's 4th edition, which at the moment is probably in my storage unit at home. Doug Reiss 15:08 January 27, 2006 (EDST)

Quantum Computing
The beginning of the article says: "It has recently become popular in quantum computing." Has it ever not been popular in quantum computing? It seemed to be popular in 1999, which I don't consider "recent". I don't remember reading any quantum computer papers without bra-ket notation. A5 19:52, 27 February 2006 (UTC)

Banach Spaces
I have a question about these statements:

1. "Consider a continuous basis and a Dirac delta function or a sine or cosine wave as a wave function. Such functions are not square integrable and therefore it arises that there are bras that exist with no corresponding ket. This does not hinder quantum mechanics because all physically realistic wave functions are square integrable."

What exactly is the bra here? Shouldn't it say "therefore it arises that there are kets without any bra"?

2. "Bra-ket notation can be used even if the vector space is not a Hilbert space."

Is the space mentioned in the statement 1 a Hilbert space? It shouldn't be, because otherwise the Riesz theorem would hold, right? Is it a Banach space? If it is, shouldn't it be moved to the following paragraph with statement 2 as a (useful) example? A5 20:41, 27 February 2006 (UTC)


 * So far as I know, the two statements should be equivalent, but I'm no expert.
 * I believe that the current ordering is attempting a transition from the physically motivated to theory and spaces. It's a wiki so be bold if you don't like it as is. &mdash; Laura Scudder &#9742; 21:12, 27 February 2006 (UTC)

You can use a rigged Hilbert space for the first thing. You can use any reflexive Banach space for the second thing. Charles Matthews 16:19, 3 March 2006 (UTC)

yeah, how often does one find a piece of literature on Banach spaces, or functional analysis in general for that matter, that uses the bra ket notation? never. it's essentially a physical notation, and as such perhaps lend itself more to physical interpretation. doesn't really pay to try to pin down its rigorous meaning. although Charles is right in what he said. In the finite dim case, of coure, rigorization can be done trivially. Mct mht 20:05, 22 April 2006 (UTC)


 * The statement quoted is typical physical language, can be rigorized using spectral theory for self adjoint operators on Hilbert spaces. Or, like Charles said, rigged Hilbert spaces. Mct mht 20:13, 22 April 2006 (UTC)

Riesz representation
small mistake in article. the Riesz representation is not quite an isomorphism; it's conjugate linear. i didn't change it because, IMHO, it's rather pointless to justify the notation rigorously. one does not use a piece of notation to speak of "continuous basis" on a separable Hilbert space, and use that same notation to discuss, say, Hahn-Banach. Mct mht

what does bra-ket notation mean?
Can bra-ket notation be converted to normal mathematics? I'm pretty sure it can, because Schrödinger's equation can be expressed both with bra-ket notation and derivatives. Fresheneesz 10:46, 25 April 2006 (UTC)


 * From what I can glean from a pervious discussion on this page: does this pattern describe how bra-ket notation is always used?
 * $$ \lang independant \,\, variable | operator | function \rang = operator \lang independant \,\, variable | \psi \rang = operator \psi (independant \,\, variable ) $$

or is it more general like this?:
 * $$ \lang independant \,\, variable | a | function \rang = a * \lang independant \,\, variable | \psi \rang = a * \psi (independant \,\, variable ) $$


 * ??? Fresheneesz 20:48, 25 April 2006 (UTC)


 * I think the most concrete example is
 * $$ \lang \psi | \hat O | \psi \rang = \int_{-\infty}^{+\infty} \psi^* \hat O \psi \, dx $$
 * - mako 22:53, 25 April 2006 (UTC)


 * Should we put that somewhere in the introduction, because in this article, its very unclear how bra-ket notation is translated into something meaningful. This article is rather on the properties of manipulating and using the notation. 68.6.112.70 02:14, 26 April 2006 (UTC)


 * Right now I think you're supposed to get that from the third equation when it says $$\lang \phi | \psi \rang$$ is an inner product, but it could be made more explicit. The problem with trying to tie it in to wavefunction mechanics with examples is that you don't want to accidentally imply that bra-ket notation is just some kind of shorthand for doing wavefunctions in some particular representations.  The above example, for instance, technically does this middle step with the identity matrix $$| x \rang \lang x |$$:
 * $$ \lang \psi | \hat O | \psi \rang = \lang \psi | x \rang \lang x | \hat O | x \rang \lang x | \psi \rang = \int_{-\infty}^{+\infty} \psi^*(x) \hat O_x \psi(x) \, dx $$
 * There's nothing special about doing the inner product in position space (could just as easily use any other identity), but that could easily be misunderstood. &mdash; Laura Scudder &#9742; 05:24, 26 April 2006 (UTC)


 * Also, so far as being meaningful, there's some interesting situations where with bra-ket notation you can find out a lot of useful information about a system for which you can't solve for the wavefunctions, so it's a useful standalone technique. I think the article could convey that better.  &mdash; Laura Scudder &#9742; 05:37, 26 April 2006 (UTC)


 * So can you correct/fill-in-the-blanks for me here?:
 * $$ \lang a(x) | \hat O | b(x) \rang = \int_{-\infty}^{+\infty} a(x)^* \hat O_x b(x) \, dx $$
 * where
 * $$ a(x) $$ is a function of x.
 * $$ \hat O $$ is ______?
 * $$ b(x) $$ is a second function of x.


 * Fresheneesz 06:11, 30 April 2006 (UTC)


 * $$ \hat O $$ is an operator (something like $$ \frac{d}{dx} $$ ) Trewornan 03:41, 2 June 2006 (UTC)


 * To be picky, I usually prefer not to write $$\lang a(x) | $$ as it's not perfectly clear what's meant. The bra $$ \lang a | $$ is a state that's not in any particular representation, while the projection of that bra onto a basis set is for instance $$ \lang x | a \rang = a(x) $$.  I would usually interpret $$\lang a(x) | $$ then as $$ \lang a | x \rang \lang x |$$.
 * But to answer your question about $$ \hat{O}$$,  in my equations above I meant the operator in no particular representation.  So for instance, in that notation I wouldn't be able to write down what the 1-D momentum operator $$\hat{p}$$ was because it will be different in position and momentum spaces and I haven't yet determined what I'm working in.  What I do know how to write down is $$\lang x | \hat p | x \rang = - i \hbar \frac{\partial}{\partial x} $$.  Projecting the abstract onto a representation basis set is what allows me to write things down in a traditional way.
 * So a concrete example of what this means:
 * $$\lang \phi | \hat{p} | \psi \rang =  \lang \phi | x \rang \lang x | \hat p | x \rang \lang x | \psi \rang = \int \phi^*(x) \left( - i \hbar \frac{\partial}{\partial x}  \right) \psi(x) dx $$
 * or equivalently
 * $$\lang \phi | \hat{p} | \psi \rang =  \lang \phi | p \rang \lang p | \hat p | p \rang \lang p | \psi \rang = \int \phi^*(p) p \psi(p) dp $$
 * which would be a much easier calculations in some situtations. &mdash; Laura Scudder &#9742; 18:20, 30 April 2006 (UTC)
 * Laura's explanation is awesome! It would do everybody a great favor is someone included it on the page's text. I read the page like 3 times over a 6 months period and didn't understand anything; with this explantion I got it (well, kinda, at least I got what the ket thing is about) :) --70.137.159.152 02:32, 25 July 2006 (UTC)
 * Laura's explanation is indeed awesome ;-) However, I have an issue... The article says that $$\lang \phi | \psi \rang$$ evaluates to a complex number.  Then it gives the example $$\langle x|\psi\rangle = \psi(x)\ = c e^{- ikx}$$, in which $$\langle x|\psi\rangle$$ fits the form $$\lang \phi | \psi \rang$$ but it is resolving to a function mapping the reals into the complex.  I presume that $$\langle x|$$ is to be understood as a variable representing the bra for each real value of x, but the dichotomy between the example showing the value of $$\lang bra|ket \rang$$ to be a function and the text stating that the value of $$\lang bra|ket \rang$$ is a complex number threw me off for many readings &amp; re-readings of the article.  Someone who knows more than me on the topic should probably remark about the meaning of the $$\langle x|\psi\rangle$$ example, or perhaps I'm still misunderstanding and you could clarify here.  Thanks! 76.186.139.122 (talk) 17:25, 18 April 2009 (UTC)
 * To me, bra-ket notation is really, really simple, whereas all the above explanations make it sound really, really complicated. It's possible that I've misunderstood completely, but here's what I thought.  Bra-ket notation is just a special way of writing vectors, where the column vector $$|x\rangle$$ is simply the conjugate transpose of the row vector $$\langle x|$$.  That's it.  When you put them right next to each other, the $$||$$ becomes $$|$$ just to make it look prettier.  Other than that syntax, they aren't special, they're just normal vectors.  So $$\langle x|A|x\rangle$$ just means a vector times a matrix times another vector, where the two vectors happen to be the conjugate transpose of one another.  Because of the angle brackets, it looks like some kind of magical expression, but really it's just 3 variables side by side, indicating multiplication.  Then, assuming linear operators, "vector" is another word for "function" and "matrix" is another word for "operator". Xezlec (talk) 04:56, 10 August 2010 (UTC)
 * Nope. I suggest you read some Dirac.  There's a reason this notation exists, and why you can derive Schrodinger from it but not the other way around.  —Preceding unsigned comment added by 199.89.180.254 (talk) 00:40, 11 February 2011 (UTC)

reason for deleting section
section on comparison with "long-S" notatation has been deleted. first there is no such thing as a "long-S notation". a ket is an element of a Hilbert space, whether the Hilbert space $$\mathbb{C}^n$$ or $$L^2$$. the inner product is just the inner product on those spaces. to say it is so by convention, as claimed in the section is awkward and unnecessary. it's not the notation that changes, but the underlying Hilbert space. while awkwardness is not sufficient for deletion, incorrectness is. Mct mht 21:25, 29 April 2006 (UTC)


 * More precisely, since the state space is almost always $$L^2(\mu)$$ for some measure $$\mu$$. it is the measure that changes, not the notation. Mct mht 21:29, 29 April 2006 (UTC)


 * Well firstly, there is such a thing as "long-S" notation, it was invented by Leibnitz (see Integral). Secondly, as it stands this article is opaque to anybody who doesn't already have an understanding of the notation, which defeats the point. However if my contribution offends your pedantry so be it - I really can't be bothered. Trewornan


 * i have no objection for further elucidation in article, as long as what is said is correct. the very claim that the notation actually changes shows conceptual deficiency. citing irrelevant material from high school calculus texts doesn't really do it. Mct mht 17:09, 30 April 2006 (UTC)


 * I agree with Mct mht, this article serves 100% *no* purpose for me, or any users I can imagine would want to look at the page. It *needs* explanation as to what it *means*. Laura Scudder gave some examples aboves, but I still find it impossible to generalize that into how to actually interpret bra-ket notation. Fresheneesz 07:32, 6 May 2006 (UTC)


 * It's actually not that complicated - if you've got a function like f(x)=g(x)+ih(x) and its conjugate f*(x)=g(x)-ih(x) in bra-ket notation you write f(x) as |f> and f*(x) as <f| the trick bit is that when you write both together like this <f|f> you mean the integral $$\int f \cdot f^* dx$$ in general (particularly when dealing with quantum mechanics) that's all there is to it. The problem certain individuals have is that the same notation is used for vector spaces as for function spaces (actually that's "conceptually deficient") so if you were talking about vectors like the x and y axes rather than functions the product <x|y> isn't an integral like above just a standard vector product. It's the easy way to explain but it's not absolutely correct in some ways.Trewornan

H*
In the last section, H* is defined to be the space of linear operators on H. Was "linear functionals" intended? The $$\phi_g$$ defined in the article seems to map h into the field, not back into H.

Notation used by mathematicians
Whoever wrote this -- THANK YOU. I think it would be a great idea to have a math translation of all the physics talk found across many of the quantum mechanics articles. I have not seen quantum physics before and I spent an hour pondering about what exactly is happening (and why certain notations are chosen) as I slowed chewed through the early sections. However, since I do have some math background, the last section resolved the entire article in a breeze -- and I wondered about whether I was abusing substances in the previous hour ... --yiliu60 01:40, 6 October 2006 (UTC)


 * actually that stuff is not quite right, e.g. 1. as pointed out above by unsigned, the dual space H* is not the same as the bounded linear operators on H, 2. H is actually conjugate isomorphic to H*, etc. Mct mht 09:40, 17 October 2006 (UTC)


 * That's ironic, as I'm an undergrad physics student and I don't have the faintest idea what this article is trying to tell me. Nobody who teaches our course uses much in the way of this notation until later on, so I don't have a clue what it's for, even though i'm pretty good with QM. 86.135.99.158 12:26, 10 December 2006 (UTC)

Small "errors" / omissions
In the Bras and kets section the bras and kets are written in (a1, a2, ...) respresentation. This is correct for finite and countable-dimensional vectors, but seems a little awkward for functions in a Hilbert space. Perhaps a remark should be added that this notation is an example.

Also, I'm missing the unit operator |ei><ei| (in the basis {|ei>}) (which is important probably, though perhaps not necessarily a property of the bra-ket notation but more an example of bra-ket notation applied to a mathematical identity (namely, that the sum over e_i* e_i is unity) and the equivalence between for example <v|w>, <v|e><e|w>, <v|e><e|e><e|e><e|w> and <v||||w> (which may not be important at all but still, it's a nice notational trick). 213.84.168.13 20:54, 6 November 2006 (UTC)


 * That's a pretty neat trick, and (yet) another example of how e turns up everywhere. Please be bold and edit the page to include this - you'll do a better job than me :-D --h2g2bob 21:23, 6 November 2006 (UTC)
 * I've been thinking about this, and I think it would work for any base, ie:
 * $$ | a^{ib} >< a^{ib} | = 1 $$
 * For any real a and b, as you end up with
 * $$ ( a^{ib} )( a^{ib} )^* = ( a^{ib} )( a^{-ib} ) = a^{ib} / a^{ib} = 1 $$
 * I suppose it's no surprise that the answer is real when squaring complex variables, but the result is very nice. I think this (with exponentials) is probably a quite useful trick. --h2g2bob 13:19, 7 November 2006 (UTC)
 * I heavily edited the first section for the new information to fit in, removing a (in my opinion useless and bad-quality) reference. Regarding the unit operator: I wanted to fit it in the Properties section at first, but it became quite a lot of text to fit in the enumeration so I split it into a separate section. Please review. CompuChip 14:08, 11 November 2006 (UTC) (formerly 213.84.168.13)

Rewrite?
There are some basic errors and yet there is some really good stuff here. Some of the errors are so nearly correct that fixing them would be difficult without just deleting them. I think it would be great if someone started over, and avoided reading too much into the bra-ket notation. Also, it's difficult to address the bra-ket notation without considering inner product spaces and bilinear forms. The article should be more explicit about the relation to those things.Mathchem271828, 19:30, 12 December 2006
 * Actually, I think the article should start with the physical use (like in section 1.1) of the bra-ket notation, avoiding mathematical preciosities (while still being correct of course) and stressing the ease of using it (for example, being able to just glue bars together, write $$\langle\phi | \psi\rangle$$ etc.) and some mathematical background. The talk about inner product spaces, bilinear forms, uniqueness, ring multiplication (?) and so on, should go in a separate section discussing the underlying mathematics (incorporating for example section 1.2). This would make it easy for (beginning) physicists to find out what they're actually doing without cluttering up the text with advanced mathematics, while still giving more interested readers access to the mathematical background of the notation. --CompuChip 11:43, 6 January 2007 (UTC)


 * Agreed. Leading with the mathematical background is very scary and leaves the physics student asking, "Do I really need to understand all this to use these?" and, "Why should I bother using these complicated things?"  Start with what they're used for and how easy they make things, then lay out all the details &mdash; as if writing an extended lead section.  &mdash; Laura Scudder &#9742; 12:05, 6 January 2007 (UTC)


 * Most students find logical inconsistencies more daunting than math itself. The purpose of the article isn't to help physics students, per se, anyhow.  It's most important that the article be rigorously correct and I think that should come before readability, at least initially.  Mathchem271828 14:13, 8 January 2007 (UTC)


 * I'm not saying the article should contain logical inconsistencies or not be rigorously correct. Just that the mathematical background should have its own section rather than being integrated in the text. Though its main purpose may not be helping physics students, I think it's very likely many readers will in fact be students and - along with everyone else - they should be properly served. As for readability, I very much disagree with you. I think readability is very important in any encyclopedic article, even more than rigorous correctness (mind the adverb, of course it should never be incorrect at all). Perhaps the article could start, after the lead, with a section describing how the notation entered quantum mechanics, a section with some properties (whose proofs could be given there, deferred to the mathematics section or even on a Proofs subpage) and then a section with the more general mathematical information (like what's actually behind it, why it's well-defined, etc). --CompuChip 17:00, 10 January 2007 (UTC)

Outer product example?
In section 'Linear operators':


 * If $$\langle\phi|$$ is a bra and $$|\psi\rangle$$ is a ket, the outer product $$ |\phi\rang \lang \psi| $$ denotes the rank one operator that...

Can someone give an example about the outer product? I mean if we let


 * $$\langle\phi|=(a_1^*,a_2^*,a_3^*)$$


 * and


 * $$|\psi\rangle=(b_1,b_2,b_3)^T$$


 * where $$a_1,a_2,a_3,b_1,b_2,b_3\in\mathbb{C}$$, the inner product should be


 * $$\langle\phi|\psi\rangle=a_1^*b_1+a_2^*b_2+a_3^*b_3$$

What will it be for the outer product?


 * $$ |\phi\rang \lang \psi|=???$$

Thanks! Justin545 01:56, 14 October 2007 (UTC)


 * I assume $$ |\phi\rang \lang \psi|$$ becomes the $$3 \times 3$$ matrix
 * $$\left( \begin{matrix}

b_1 a_1^* & b_1 a_2^* & b_1 a_3^* \\ b_2 a_1^* & b_2 a_2^* & b_2 a_3^* \\ b_3 a_1^* & b_3 a_2^* & b_3 a_3^* \end{matrix} \right)$$
 * if I ain't mistaking.
 * Jens Persson (193.10.251.182 08:29, 18 October 2007 (UTC))


 * I just found a website :
 * According to the site, it seems that the outer product is just a position exchange of the bra and ket such that if we right-multiply $$|\phi\rang \lang \psi|$$ by the general ket $$ |C\rang$$. It becomes:


 * $$|\phi\rang \lang \psi|C\rang=\lang\psi|C\rang|\phi\rang$$


 * which is just another ket since $$\lang\psi|C\rang$$ is a (complex) number.


 * Justin545 00:23, 20 October 2007 (UTC)

Relation to Riesz isomorphism
Is it correct to say that (if we are in a Hilbert space) the bra is the image of the ket under the Riesz isomorphism? I.e. $$\langle\phi| = J(|\phi\rangle)$$? Eriatarka (talk) 15:00, 14 June 2008 (UTC)


 * This is true, and mentioned in the article, although not prominently. You're welcome to try to emphasize it more if you think it would be appropriate. :-) --Steve (talk) 17:21, 14 June 2008 (UTC)


 * Thank you. For me as a mathematician, this helps to clarify matters a lot. I did read the short paragraph which mentioned the relation to the Riesz isomorphism, but I couldn't find any mention of this explicit mapping. I added it; please review, if you want. Eriatarka (talk) 10:49, 16 June 2008 (UTC)

"Unit Ray"
What is a "unit ray"? That sounds like a contradiction since a ray, as far as I know, has no size. Xezlec (talk) 05:02, 17 June 2008 (UTC)


 * Agreed. I changed it to just "ray". :-) --Steve (talk) 06:04, 17 June 2008 (UTC)

Bra-ket as dot product
In the following equation:

$$ \langle\psi| : \mathcal H \to \mathbb{C}: \langle \psi | \left( |\rho\rangle \right) = \operatorname{IP}\left( |\psi\rangle \;,\; |\rho\rangle \right) for all kets |\rho\rangle $$

In the bit which states: $$\operatorname{IP}\left( |\psi\rangle \;,\; |\rho\rangle \right)$$ Should it not be: $$\operatorname{IP}\left( \langle\psi| \;,\; |\rho\rangle \right)$$ ? —Preceding unsigned comment added by Etanol (talk • contribs) 16:38, 17 March 2009 (UTC)


 * No, I don't think so. This is defining $$\langle\psi|$$ in terms of $$|\psi\rangle$$, it would be a circular definition the way you're proposing. That said, I really don't like that line...things that can be said just as well in English should be said in English, not symbols, since wikipedia is intended to be read by humans, not robots. --Steve (talk) 17:03, 17 March 2009 (UTC)


 * the inner product is a function with domain H^2 (= H x H). bras are elements from the dual space. this is exactly what is said there. reread some linear algebra. this is in any book (dual space, a base of the dual space and so on). -- 91.15.144.196 (talk) 21:23, 2 April 2009 (UTC)

More general uses
"Bra-ket notation can be used even if the vector space is not a Hilbert space. In any Banach space B, the vectors may be notated by kets and the continuous linear functionals by bras. Over any vector space without topology, we may also notate the vectors by kets and the linear functionals by bras. In these more general contexts, the bracket does not have the meaning of an inner product, because the Riesz representation theorem does not apply." relevance? non-existent. trivial. i can denote any vector by a ket. i can denote any OBJECT by a ket. —Preceding unsigned comment added by 91.15.144.196 (talk) 21:18, 2 April 2009 (UTC)

Comment
This article, as written, is math rather than physics. The Dirac notation has some nice practical applications as described by Feynman and other authors. However, with comments like

"(cur) (prev) 19:22, 17 October 2009 Sbyrnes321 (talk | contribs) m (20,364 bytes) (Undid revision 320439307 by 128.151.39.155 (talk) silly to describe two examples of a totally ubiquitous thing) (undo)"

this article will most likely remain as a math piece. Sbyrness321 deleted the short applications paragraph because it gave a only few practical examples (not just two) when the article right now has NONE. Very interesting. —Preceding unsigned comment added by 128.151.39.161 (talk) 20:22, 18 October 2009 (UTC)


 * Sorry I didn't explain myself better. What I meant was, essentially every quantum mechanical calculation ever done in the last fifty years uses bra-ket notation. That would be millions of papers and tens of thousands of books, probably. It's misleading to list out a few examples, because even a hundred examples wouldn't begin to do justice to the applications of bra-ket notation, which is the same as the applications of quantum mechanics. It's like having an article on the English language, and saying a few applications of the English language are: The movie "Titatic" uses the English language, and William Shakespeare uses the English language, and Barack Obama uses the English language. You see what I mean? :-)


 * It's also misleading because bra-ket notation is a notation. Anything that can be figured out using bra-ket notation could also be figured out just as well using a different notation, for example the linear-algebra notation common among mathematicians.


 * Anyway, the "applications" of bra-ket notation, such as they are, are the same as the applications of quantum mechanics. I'll add a sentence or two to the top of the article explaining that. See what you think. :-) --Steve (talk) 21:47, 18 October 2009 (UTC)

Rendering of brackets
3 months ago, changed the rendering in the lead from the standard LaTeX usage, $$\langle\phi|\psi\rangle$$, to a nowiki style, <Φ|Ψ>, to avoid potential confusion between ( and < for some resolutions. While I agree that $$\langle$$ can appear similar to $$(\!$$ at first glance, it is the standard used throughout Wikipedia and most other sources. I don't see anything unusual about the MediaWiki rendering compared to other sources using LaTeX. I think we should use the same style for this lead, especially considering the large number of incoming links from articles that use the LaTeX style. I experimented with another style that renders phi and psi better, $$<\phi|\psi>$$, but I ultimately decided to go back to the standard used by the scientific community. —UncleDouggie (talk) 13:58, 24 January 2011 (UTC)
 * I have no problem with following a standard (for the field, and for consistency in the articles). But I have a huge problem the intro for an article being immediately and intentionally unclear to anyone who's not already well-versed in the field or at least able to overlook this unclarity. This particular topic is relevant to down to mid-level college students, who are just trying to get started with the symbology itself--they really may not know what symbol to use and really do need to learn that detail. DMacks (talk) 15:32, 24 January 2011 (UTC)
 * Since this is an article on the notation itself, perhaps a paragraph on the symbology is in order. We could even describe the use of \langle and \rangle in LaTeX. Readers shouldn't be forced to view the wikitext if they are curious about how to generate the brackets. —UncleDouggie (talk) 10:10, 25 January 2011 (UTC)

Abuse of Notation
Why is the differential operator applied to a ket considered to be an abuse of notation? It seems to me like the derivative of a vector is well defined, even in the absence of a chosen basis.Hiiiiiiiiiiiiiiiiiiiii (talk) 00:57, 30 March 2011 (UTC)


 * This is trickier than it seems. See.


 * You can indeed define the operator d/dx (a linear operator on the vector space) by:
 * $$(d/dx)|y\rangle = \lim_{\delta\rightarrow0} \frac{|y+\delta\rangle-|y\rangle}{\delta}$$
 * where $$|y\rangle$$ is an element of the position basis (a delta-function wavefunction). This limit doesn't really exist (it's a sequence of vectors that doesn't converge to a vector), but we'll keep going anyway....
 * Now we apply this operator to a general ket $$|\psi\rangle$$ (not necessarily an element of the position basis):
 * $$(d/dx)|\psi\rangle = \int_y (d/dx)|y\rangle\langle y | \psi \rangle = \int_y \frac{|y+\delta\rangle\langle y | \psi \rangle-|y\rangle\langle y | \psi \rangle}{\delta}$$
 * $$ = \int_y \frac{|y\rangle\langle y - \delta | \psi \rangle-|y\rangle\langle y | \psi \rangle}{\delta}$$
 * $$ = \int_y |y\rangle \frac{\psi(y-\delta)-\psi(y)}{\delta}$$
 * $$ = -\int_y |y\rangle \langle y | \psi'\rangle$$
 * $$ = -\psi'(x)$$
 * So actually the operator d/dx properly interpreted gives the negative derivative of the wavefunction. So there's a sign-error difference based on whether d/dx is interpreted as a vector derivative or as "an abuse of notation" wavefunction derivative. That's why you usually see the momentum defined as $$p=-i\hbar (d/dx)$$ (wavefunction derivative), but you also occasionally see $$p=+i\hbar (d/dx)$$ (vector derivative)! --Steve (talk) 07:44, 30 March 2011 (UTC)


 * I hardly understand your definition of
 * $$(d/dx)|y\rangle = \lim_{\delta\rightarrow0} \frac{|y+\delta\rangle-|y\rangle}{\delta}.$$
 * What is y and x? Shouldn't they be the same?
 * And $$\langle y | \psi \rangle$$ is a number, which commutes with $$|y\rangle$$, so why can't you just say
 * $$\frac{d}{dx}|\psi\rangle=\frac{d}{dx}\int\langle x|\psi\rangle|x\rangle=\int\frac{d}{dx}\langle x|\psi\rangle|x\rangle=\int\psi'(x)|x\rangle=|\psi'\rangle?$$--Netheril96 (talk) 08:15, 30 March 2011 (UTC)


 * d/dx is a specific linear operator on the Hilbert space. $$|y\rangle$$ is any element of the Hilbert space in the position basis (a delta-function wavefunction centered at the coordinate y). I used different symbols "x" and "y" to emphasize that it's just one operator, but can be applied to many different kets. Maybe it would be clearer if you mentally replace d/dx with $$\nabla$$. :-)
 * Since (d/dx) is a linear operator, it satisfies the linearity condition: (d/dx)(scalar * vector) = scalar * ((d/dx) vector). Therefore,
 * $$\frac{d}{dx}\langle x|\psi\rangle|x\rangle=\langle x|\psi\rangle(\frac{d}{dx}|x\rangle)$$
 * You can't apply a linear operator to just a scalar like $$\langle x|\psi\rangle$$. You also can't apply (d/dx) to a bra like $$\langle x|$$, because (d/dx) is an operator on the vector space of kets, and bras are not part of the vector space of kets. (They're in a different vector space. You can apply (d/dx) to a bra on the left as a function composition, but not on the right.) :-) --Steve (talk) 17:40, 30 March 2011 (UTC)


 * Well, if you must define d/dx like that, please tell me how to define d/dt.--Netheril96 (talk) 00:53, 31 March 2011 (UTC)


 * You could have written like this
 * $$\langle x|\frac{d}{dx}|\psi\rangle =$$
 * $$=\int\langle x|\frac{d}{dx}|x'\rangle\langle x'|\psi\rangle dx'$$
 * $$=\int\left(\lim_{\Delta x'\to0}\frac{\langle x|x'+\Delta x'\rangle-\langle x|x'\rangle}{\Delta x'}\right)\langle x'|\psi\rangle dx'$$
 * $$=\int\left(\lim_{\Delta x'\to0}\frac{\delta(x'+\Delta x'-x)-\delta(x'-x)}{\Delta x'}\right)\langle x'|\psi\rangle dx'$$
 * $$=\int\delta'(x-x')\psi(x')dx'$$
 * $$=-\psi'(x)$$
 * --Netheril96 (talk) 02:01, 31 March 2011 (UTC)


 * In the Schrodinger picture, any ket will evolve as a function of time according to the Schrodinger equation. So I would define (d/dt) in the obvious way:
 * $$(d/dt)_{t=t_0}|\psi\rangle = \frac{|\psi(t_0+\delta)\rangle - |\psi(t_0)\rangle}{\delta}$$
 * Did you have something else in mind? The definitions of d/dx and d/dt cannot be exactly analogous, because space and time are treated differently in non-relativistic quantum mechanics...there is a "position basis" but there is not a "time basis", etc. :-) [or consider: $$|\psi(t_1)\rangle$$ is one ket, $$|\psi(t_2)\rangle$$ is a different ket...but you can't say that $$|\psi(x_1)\rangle$$ is one ket and $$|\psi(x_2)\rangle$$ is a different ket!]
 * I agree, that's a clearer way to write the derivation. :-) --Steve (talk) 02:41, 31 March 2011 (UTC)


 * Well, I wanted to say that Schrodinger equation
 * $$\hat{H}|\psi\rangle=i\hbar\frac{d}{dt}|\psi\rangle$$
 * is not an abuse of notation.--Netheril96 (talk) 04:48, 31 March 2011 (UTC)


 * Yea, I would say $$|\psi\rangle$$ is a function of t but not of x, so it is not an abuse of notation to differentiate with respect to t but is for x. More specifically, if you give me one t, I can give you the corresponding $$|\psi(t)\rangle$$, and if you give me a different t', I can give you the different corresponding $$|\psi(t')\rangle$$. But that statement doesn't work if you replace t by x. $$|\psi\rangle$$ corresponds to a function of x (namely $$\psi(x)$$, but is not itself a function of x, it's just a single vector not a function. :-) That's my understanding. [I didn't myself write that section.] --Steve (talk) 08:10, 31 March 2011 (UTC)

Please, somebody, add a short "physical meaning of these things" section
I chased wikipedia pages all over the place to try to understand this stuff. My complaint is rather like one of the othe commenters here - I could't see any connection between hamiltonian spaces and lasers or electrons. I finally had to view the Oxford University lecture videos to "get" something which is probably obvious to everyone else. Here's kinda/sorta what I wish someone had written here on wikipedia for interested amateurs such as myself. If someone could express the following more clearly and correctly, that would be great.

The basis vectors of a ket vector correspond to "classical"states. For instance, in QM the energy of a particle may be one of an infinite series of discrete values - E0, E1, E2 and so on. Each position in the ket corresponds (loosely speaking) to its corresponding energy level. The value psi_1 is the probability amplitude for the system having  energy level e1, and so on. Each value is a complex number, and the sum of the moduli of the squares of those numbers ( ie: sum of (psi_i. conj psi_i) ) must be 1.

The key thing, the really important thing, is that the system is not "in" state E0, e1, e2 and so on - its the whole of that ket vector that is its physical state, and its the whole of the ket vector that changes as the system propagates forward in time, or however. For this reason, it often makes sense to express the ket as a function rather than as a list of numbers.

For any physical system, there will be several ways that you can describe it completely. For instance, it's often the case that describing the *position* of a particle comlpetely also describes the *moment* of the particle completely. This means that the position ket can be transformed into a momentum ket by choosing different basis vectors and doing some sort of transform. A takeaway here is that sometimes you want to think of the ket as being "the state of the system" without being specific about how that state is expressed. It's still a meaningful thing to do.

A bra can be thought of as an observation that you would like to make on a ket. < bra | ket > - the projection of the ket on the bra - gives us the physical probability of making that observation, given that physical state. But "observation" is a term-of-art. < bra | ket > sort of corresponds to when you do something to a quantum system. It's an action that you can perform. It has *physical* meaning.

For any ket, we can construct a corresponding bra. This doing < foo | bar >, where we understand foo to be a bra-ified ket, gives us a probability - the probability that if the system is in the state bar, that it will look like it is in state foo should we observe it.

Note that < foo | foo > always sums to 1. The physical meaning of this is that if you know that a system is in some state, then the probability of observing it in that state is 1. Duh. This feature is mathematically equivalent to the rule "modulus squared of all the values in any ket must sum to one".

Wwe can construct a set of basis ket vectors


 * 1, 0, 0, 0 ... >
 * 0, 1, 0, 0 ... >
 * 0, 0, 1, 0 ... >

The process of converting these into bras and applying them to some ket does the job of inquiring about the value for each position in the ket. It's equivalent to simply reading the square of the modulus of the ket at a position.

A bra or ket does not need to have an infinite number of values. For instance, in a two slit experiment there are two states "photonn goes through the left slit" and "photon goes through the right slit". Another on is spin. Rather than e0, e1, e2, e3 and so on, if you test the spin of a particle you will get one of two possible answers: up or down. It can be represented by a ket with two values | spin up, spin down >. When we take a simple "spin up or down" observation on the system - ie, when we do a < 1, 0 | or <0, 1| on it, then we get a simple probability back. But the system actually has three degrees of freedom. The question "spin left or right?" is effectively a matter of choosing a different set of basis vectors and performing a coordinate transform on the ket to express its spin in terms of those new vectors.

The other thing I would like to know is - what's the story with reducing the size of a ket? When we perform a simple "yes or no" question of a complex system, obviously we are somehow mapping a system with many dimensions ont a systems with fewer. For instance, we could map | e0, e1, e2, e3 ...> onto a two-state | E lt e3, e ge E3 >. We can't do that with a simple dot-product - we have to use a matrix or something. What do you call that, and how does it fit into the notation?

Paul Murray (talk) 05:39, 22 June 2011 (UTC)


 * I intend to...-- F = q(E + v × B) 11:37, 22 January 2012 (UTC)


 * I am strongly against going too far down this path. Bra-ket notation is a notation for linear algebra, nothing more. Questions about quantum mechanics are best answered in articles about quantum mechanics, like wavefunction and quantum state and observable etc. I don't object to a few words as long as the topic is part of "why bra-ket notation is especially convenient for QM" rather than "I will now explain a QM concept using bra-ket notation". --Steve (talk) 14:43, 23 January 2012 (UTC)


 * By my previous statement - I meant the incorporating material on linear/abstract algebra below, not the above.-- F = q(E + v × B) 21:49, 23 January 2012 (UTC)

New lead section
I intend to re-write, not delete or remove content, but add and adjust content in the beginning, to make this far more approachable. It charges into so much mathematical terminology and notation which simply couldn't be understood by any layperson. This topic can be introduced in analogy with ordinary Euclidean vectors, bases, and the dot product, paralleling with the basis kets, dual vectors and the inner product - simultaneously explaining the generality of the Dirac formalism and its abstract scope beyond the physical, geometric Euclidean vectors from the elementary vector calculus lots of people can understand fairly well, and the reason pointing to the wavefunction article on its powerful application.-- F = q(E + v × B) 11:37, 22 January 2012 (UTC)

DIAGRAMS!!!
I have produced a number of diagrams to accompany the above quote: illustrate the parallels between "usual" vectors and bra-ket vectors, though more will be added soon to illistrate the abstractness - these are intended for the lead.







Opinions? --Maschen (talk) 17:51, 22 January 2012 (UTC)


 * Sorry - didn't notice all this time, they're pretty good! Be bold and add them, given that no-one has objected. This is another important page which people have, and still, repeatedly complain about being not understandable (which is perfectly reasonable, given the way it was written before), so these may help. =) -- F = q(E + v × B) 12:19, 23 January 2012 (UTC)08:21, 23 January 2012 (UTC)


 * Lets add them soon...-- F = q(E + v × B) 12:19, 23 January 2012 (UTC)
 * I had to break the multi-image into two seperate ones, they're too clumsy together. 4/5 have been added, the other can be added later, and maybe wavefunction also.-- F = q(E + v × B) 13:50, 23 January 2012 (UTC)


 * I suggest the following changes:
 * In ket_vector_components, change |x> to |e_x>, etc. This makes the connection to mathematician-notation clearer, and also avoids the possibility of conflating 3D vectors with wavefunction-position-space basis vectors very common in intro QM. Also, add the equations A_x = <e_x|A>, A_y = <e_y|A>, A_z = <e_z|A> to the pic or caption in order to get rid of Cartesian_and_ket_vector_components_projection, see below. Also, it's not "ket vector", it's just "ket". "ket" means "vector written in a certain notation", so "ket vector" is redundant.
 * Delete Cartesian_vector_components from the article. This is exactly the same as ket_vector_components, but in very slightly different notation. What's the point? If it really requires further explanation, the caption of ket_vector_components could be amended with something like "If we were not using bra-ket notation, we would write e_x instead of |e_x>, e_x dot A instead of <e_x|A>, etc."
 * Delete Cartesian_and_ket_vector_components_projection from the article, after adding the three equations A_x = <e_x|A>, A_y = <e_y|A>, A_z = <e_z|A> to ket_vector_components or its caption. Those three equations are the only thing it adds to ket_vector_components, right?
 * Delete bra_ket_inner_product from the article. The shadow length is not, in general, equal to the inner product. Various other problems with the figure and caption too, e.g. it should be |phi> not <phi|. Even if all problems were corrected, the geometric definition of inner product is not necessarily a helpful thing to have in mind when thinking about the infinite-dimensional complex vector spaces which are ubiquitous in intro QM.
 * Don't add position_spin_state_vector to the article. It seems to confuse a vector in 3D space with a wavefunction vector in 3D position space, which is actually a vector in infinite-dimensional space. It also gets spin wrong.
 * That's just my votes, but I'm not the boss :-P --Steve (talk) 14:24, 23 January 2012 (UTC)

I would naturally disagree, but thought as much someone would say this... user:F=q(E+v^B) you were 100% completley wrong, which disapoints me after all that excitement... =( The Cartesian_and_ket_vector_components_projection fig is to show that the shadow length is not equal to the inner product, just the projection of magnitudes. All else as suggested will be modified/annihilated now that its all wrong... yet another failure (suprise suprise). Actually Steve, in a sense out of us three you are "the boss", given that you know all this inside out, and we are still learning, which is why i'll do what you reccomend...--Maschen (talk) 20:13, 23 January 2012 (UTC)


 * Next time user:F=q(E+v^B) (if so), please leave these (phenomenally fuck-failed - so they result every time) images on the talk page until consensus is reached - I hadn't even realized you added them till now. They could have been error indicated here and repaired/obliterated sooner...--Maschen (talk) 20:21, 23 January 2012 (UTC)
 * One more thing - I think there is an error in the wavefunction example for the 1s overlapp wavefunctions, shouldn't it just be 1s, without the 2? Thats the shell where the electrons come from...--Maschen (talk) 20:41, 23 January 2012 (UTC)

All previous statements have been unified into here. The current diagrams in the article are no more (and never will).







--Maschen (talk) 21:17, 23 January 2012 (UTC)


 * All - sorry once again. Every now and then I seem to cause a "tidal wave" of editing problems, and this is the baddest one. Maschen: the reason why I accepted the images was because that’s exactly how I interpret kets, just as you do. Ever noticed how rare it is to find diagrams like the ones you drew in other sources? That’s why I appreciate your patience and effort in actually striving to create them. So - I am completely wrong according to the many statements above, and will take full responsibility for wrecking the article, but next time don't get excited just because I may say something which is +ve, but is wrong... Suppose now someone will revert back to the last "sane" version, before I extended the lead. That’s fine... It was hoped the current version would be clearer but obviously not. I just looked at the new diagram, so for now, I’ll match the current bra-ket formulae to the bases e_x etc and correct the typo about the 1s orbital overlap you pointed out above. The new diagram is certainly no worse than the previous set so if no-one opposes it can be added (sorry for using your images without permission also). =( -- F = q(E + v × B) 21:49, 23 January 2012 (UTC)


 * Don't worry. Thanks for your appreciation. In turn, irrelevant of other opinions, I think the new section you have added is far easier and clearer than before. Even if its wrong, the style is definitely there, compared to the original flood of jargon. Fingers crossed about the new image, assuming anyone else will reply of course.........--Maschen (talk) 22:26, 23 January 2012 (UTC)
 * UPDATE: I made a slight adjustment for clarity between the basis vector, originally the shadow (which was wrong anyway) and the x-component of the vector. It’s likely that only the right-hand side may be suggested for inclusion, so I chopped the image down the middle and have inserted the dissected sides here. The left side can be added elsewhere if not here, to (say) unit vector, or cartesian coordinate system, etc so it’s not all a complete waste (I hope). --Maschen (talk) 23:38, 23 January 2012 (UTC)


 * Jolly good. =) -- F = q(E + v × B) 22:25, 25 January 2012 (UTC)

So no-one (else) has anything to say...
Right now, the article is a slight mess so I’ll have a go at making it more coherent. I think the new images are fine Maschen - they do match Steve's recommendations above, so by all means add them when you see fit. -- F = q(E + v × B) 22:25, 25 January 2012 (UTC)
 * Yep - I'll do it now. --Maschen (talk) 10:44, 26 January 2012 (UTC)

spherical and cylindrical coordinates with basis vectors?
I don't understand the equation:
 * $$\mathbf{A} = A_\rho \mathbf{e}_\rho + A_\theta \mathbf{e}_\theta + A_\phi \mathbf{e}_\phi$$

I can't get it to work. For example, let's say A is (1,1,1) is Cartesian coordinates, (sqrt(3),54°, 45°) in spherical coordinates. What direction does $$\mathbf{e}_\rho$$ point? I would have guessed, parallel to A, so $$\mathbf{e}_\rho =$$ (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)) in Cartesian coordinates. But no, that's not right, because then $$\mathbf{A} = A_\rho \mathbf{e}_\rho$$ with no other terms. So what is $$\mathbf{e}_\rho$$? What are the other e 's? --Steve (talk) 15:54, 27 January 2012 (UTC)


 * Well, I can't argue with what you say here (except for the typo of 54° instead of 45°). Though you say this representation is mystifying? It shouldn't be. One example is the velocity in 2d plane polars:


 * $$\mathbf{\dot{r}} = \dot{r}\mathbf{e}_r + \omega r \mathbf{e}_\theta $$


 * in which the components of the vector are radial and tangential to the circular arc defined in the coordinate system. So in 3d the coordinates Aθ, Aφ are just the arc lengths as the radial vector is rotated first through by θ, then though by φ respectivley, though this has never been clear in any source I've seen. They just refer to "component of vector in the θ and φ directions". Also you asked what the e vectors are - is there any need to tell you? You know them, and the coord systems are linked so readers can look them up.


 * To summarize - take it you want to remove these also, since its going to confuse readers and is not clearly accessed in the literature? -- F = q(E + v × B) 00:31, 28 January 2012 (UTC)
 * To be honest these may as well be deleted not just because of the reasons said - it lengthens the lead of the article too much. The use of spherical and cylindrical polars can be simply linked. I'll eliminate them, so the only vector expressions are Cartesian and an arbitrary orthonormal system. Yes, ok? -- F = q(E + v × B) 08:50, 28 January 2012 (UTC)


 * Well, certainly removing them solves the problem.
 * I suggest you spend a few minutes writing down the numerical Cartesian coordinates of $$\mathbf{A}, \mathbf{e}_\rho, \mathbf{e}_\theta , \mathbf{e}_\phi$$ for the example of (1,1,1) in Cartesian coordinates. (Sorry for the bad rounding, I should have written 55° not 54° above.) You will be surprised to discover that
 * $$\mathbf{A} \neq A_\rho \mathbf{e}_\rho + A_\theta \mathbf{e}_\theta + A_\phi \mathbf{e}_\phi$$!!!
 * If you're too lazy, you can merely check that:
 * $$|\mathbf{A}|^2 \neq |A_\rho|^2 + |A_\theta|^2 + |A_\phi|^2$$
 * :-) --Steve (talk) 14:02, 29 January 2012 (UTC)

Yes, it is indeed incorrect, and I didn't even think to check that:
 * $$\theta = \cos^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right) \,\!$$

gives approx. 54.73561(...)° for x = y = z = 1... which is really bad... =(

I've seen that representation for spherical polars before, so decided to included it. (Most charming of you to say lazy - but then I am sometimes!...) -- F = q(E + v × B) 23:41, 29 January 2012 (UTC)

More embarresingly + trivial
In section Notation used by mathematicians, it was quite amusing/odd to read:


 * "Moreover (and more embarrassingly, although this is essentially trivial), mathematicians usually write the dual entity not at the first place, as the physicists do, but at the second one..."

I don't remember reading that from recently when the page was re-written, because that section was left alone and almost everything else it was re-written... Anyway is there any reason for embarrassment? F = q(E+v×B) ⇄ ∑ici 18:23, 6 June 2012 (UTC)

New templates introduced
See langle and rangle. Hope people like the change. F = q(E+v×B) ⇄ ∑ici 16:44, 12 June 2012 (UTC)


 * For now, in spite of extensive labour over these templates, there seems to be a slight bug in them. I'll remove them later, then maybe reinstate them. Its so effing annoying after all that time and energy...... =( F = q(E+v×B) ⇄ ∑ici 23:53, 12 June 2012 (UTC)


 * The templates have been reverted/tweaked back to the older version that actually worked. F = q(E+v×B) ⇄ ∑ici 08:59, 13 June 2012 (UTC)

Pronunciation
How do you say <A|B> when speaking? Is it standardized, or are there both lengthy and brief ways to read it? This should be included in the lede, just like the pronunciation of bra and ket. 130.60.6.54 (talk) 20:25, 10 June 2013 (UTC)


 * The last sentence in the lede actually already gives the longwinded way of saying it: "the probability amplitude for the state A to collapse into the state B." A lecturer who's just introduced bra-ket notation will describe every inner product this way until everyone gets used to it.


 * Later on, the shorthand you'll hear, especially when people are talking as they write, is simply "A B". Laura Scudder &#124; talk 20:59, 10 June 2013 (UTC)


 * These two extremes of pronunciation (I agree they exist) make the formulas in the "Unit operator" section for example either difficult to say or difficult to understand.


 * Is there a concise yet clear standard form for purely spoken usage, when there is not a shared visual formula to point to? For example, how would one say (on the phone, to another person familiar with quantum mechanics notation) the expression $$\langle v | e_i \rangle \langle e_i | e_j \rangle \langle e_j | w \rangle$$?  For the case of one careful physicist speaking to another physicist, or giving a formal spoken talk to physicists, I would expect something like "The amplitude of $$v$$ from $$e_i$$, times the amplitude of $$e_i$$ from $$e_j$$, times the amplitude of $$e_j$$ from $$w$$."  Is there standard terminology at this level (e.g. "the amplitude of $$v$$ from $$w$$", or "the amplitude from $$w$$ to $$v$$", or "the amplitude of $$v$$ given $$w$$", or "the amplitude for $$v$$ from $$w$$", etc.) that is concise yet clear, or, as the evidence below suggests, does every physicist "roll their own"?


 * The same question goes for standalone bra and ket vectors appearing in formulas.


 * Feynman speaks of "the amplitude for [some event]", but I do not have an example of him saying "$$\langle A | B \rangle$$". Oxford professor Binney in his youtube video (lecture 2) at 9:30 says that "$$\langle f|$$" should be read as "bra f".  He often says "ket $$\psi$$" similarly.  However, he is less consistent for the full "$$\langle j | i \rangle$$", which he often reads as "j of i" (10:56, 11:04, 11:07, 15:26), presumably due to his emphasis (e.g. 8:00-10:00) that the bra is a function acting on the ket, which is slightly different from this article's expositional approach.  He also uses "the probability amplitude of j given i" (38:40).  Of course, he is always also using a visual reference, so the spoken form does not need to be so clear, and his audience is non-expert, so he is expressing things in multiple ways to try to reinforce the meaning. In later lectures he gravitates towards the simple form you mention, "j i", relying on the accompanying written form for clarity. Stanford professor Susskind on the other hand reads it backwards as "i j" (video, 1:17:50, 1:18:00) or, also in later lectures, "the inner product of i with j".


 * From these sources, it is not clear that there is any standardized way of speaking the full bra-ket form. 130.60.6.54 (talk) 17:40, 11 June 2013 (UTC)

Another question is the pronunciation of '$$\langle$$'. The article currently states it as /brɑː/ (brah), but is it more logical that it is /bræ/ (bra, short 'a'), as in the pronunciation of the first syllable of bracket? I've never heard it spoken, so I don't know the answer. — Loadmaster (talk) 17:34, 18 December 2015 (UTC)

Vector spaces over ℂ
In the section introducing vector spaces over ℂ I amended the phrase 'though the coordinates and vectors are now all complex-valued' to read instead simply 'though the coordinates are now all complex-valued'. Although I see what the original author meant, it's not necessary for the vectors in a vector space over ℂ to in any way 'contain' complex numbers - it's simply necessary that the vectors (whatever they are) to be able to be scaled by elements in ℂ. For example, you can give ℝ2 the structure of a vector space over ℂ [when you scale (x,y) by a+ib you end up with (xa-yb,xb+ya)] but in this case the vectors (elements of ℝ2) don't 'contain' complex numbers. Felix116 (talk) 21:31, 14 October 2013 (UTC)

Bracket characters
I reluctantly reverted the edit of User:131.231.242.221 today, who complained: "The characters "〈" and "〉" appear on my computer as rectangles with little numbers in. Is there an alternative?" Indeed, older systems lack one or both pairs of html characters. Every few months, there a complaint of this sort somewhere or elsewhere in WP. There is also the alternate set of ⟨ψ|z⟩, I strongly suspect equally invisible/missing-character'd to him/her. Supplanting with <.> (lessthan-greater than) as a workaround gives the text an amateur look, and TeXing in text, $$\langle. \rangle$$ gives that distinctly ransom note look in those very systems. I have no good ideas, not even on how to advise this user to reset his/her preferrences to render formulas better. Cuzkatzimhut (talk) 11:17, 20 October 2013 (UTC)


 * I wrote on the IP's talk page here asking if some alternatives may work: the template:bra, template:ket, template:bra-ket, or just the angular brackets in the lower editing toolbar (the one just below the edit window): ⟨ ⟩ . M&and;Ŝc2ħεИτlk 14:29, 20 October 2013 (UTC)


 * Indeed, the templates use these characters, most likely the ones associated with z and ψ above. I can't speak for him/her, but, in fact, it is these brackets that are missing-boxed in early systems, about half a dozen years old. (The characters he actually complained about,  appeared more robust to me. Check these explicit HTML-character numbered &#9001; and  &#9002;,   & # 9 0 0 1 ; and  & # 9 0 0 2 ;  in the edit source.) Ultimately, such complaints will cease with computers modernizing, but I fear it is too much to punish everybody with in-text TeX, just  to assist a few readers missing characters. In my ignorance, I wonder if there were a diagnostic of the "do you see this well" type associated with one setting their preferences−−assuming they were registered to do that. Cuzkatzimhut (talk) 15:54, 20 October 2013 (UTC)

If it helps I saw bra-ket symbol issue viewing the page with google chrome browser, but not with firefox or mirosoft internet explorer. — Preceding unsigned comment added by 67.249.102.221 (talk) 21:30, 16 November 2013 (UTC)

Background is wrong.
This section has a number of errors (or simplifications to the point of excessive inaccuracy). A R3 point is NOT a vector. You do not add points. You do not multiply points. You don't take their dot (inner) products. A vector might be a entity with only magnitude and direction, but it is common in physics to also locate the vector in 2 or 3-d space. (ie. there are two distinct technical meanings for the term, (I'm not sure if vector fields can be used as equivalent to a located vector?)). Only vectors originating from the origin can be described with three scalars in 3-d space (n scalars in n-space). That is only vectors which only have magnitude and direction (not position) are the subject of this section.

ALSO: The coordinates are NOT "the number of basis vectors in each direction". Who wrote this??!!216.96.79.61 (talk) 22:19, 21 January 2014 (UTC)

The initial paragraphs of this section need a rewrite.

notation
kets and the vectors that represent kets are not equal! A ket is related to a vector that represents the ket, like a poster of a star is related to the star. A "coordinate vector" needs the reference to a basis, whereas the ket does not. See Modern Quantum Mechanics Revised Revision, Sakurai, p. 20. Therefore the equal sing between a ket and a vector needs to be changed to another symbol since the mathematical objects on both sides are not equal. For example instead of $$ | B \rangle = \begin{pmatrix} B_1 \\ B_2 \\ \vdots \\ B_N \end{pmatrix}$$ something like $$ | B \rangle \overset{.}{=} \begin{pmatrix} B_1 \\ B_2 \\ \vdots \\ B_N \end{pmatrix}$$ should be used (with $$\overset{.}{=}$$ like Sakurai for "is represented by" or $$\leftrightarrow$$ like Shankar) --Biggerj1 (talk) 12:27, 2 February 2014 (UTC)
 * I agree. This kind of shorthand should be avoided in any reference material for two reasons: it requires clarification before use, and it potentially leads a wide audience to believe that they are the same thing. This kind of misidentification is quite common in linear algebra (vectors, tensors etc.), but unfortunately hinders a more natural and complete understanding. Cleaning this up would be a definite improvement. —Quondum 17:35, 2 February 2014 (UTC)
 * I tried to present the problem. --Biggerj1 (talk) 19:20, 17 February 2014 (UTC)

✅

Square boxes
Any idea why the kets are showing up as square boxes? Having the issue on multiple computers. Sam Walton (talk) 14:38, 2 June 2014 (UTC)
 * See the thread Bracket characters above. This is generally a result of the browser using a font that does not support these characters. Some browsers allow a different font to be selected, which can solve the problem. But as indicated, older systems may have problems because fewer/none of their loaded fonts support these characters. As you may glean, it is problematic trying to prevent this on WP. —Quondum 23:14, 13 June 2014 (UTC)

Inner products and bras
In the article it was mentioned:
 * $$ \langle A | B \rangle = \text{the inner product of ket } | A \rangle \text{ with ket } | B \rangle$$

Shouldn't it have been
 * $$ \langle A | B \rangle $$ = the inner product of "bra-A" with "ket-B"?--Almuhammedi (talk) 17:44, 24 October 2014 (UTC)


 * The easy way to remember it is <bra|ket>. Pronounce the denoted terms as a bracket. Think of a ket like a named vector. Think of a bra- like a named adjoint (it's like a vector as well, but more general).
 * In other words, the article is OK right now. Not an error. --Ancheta Wis   (talk  &#124; contribs) 19:13, 24 October 2014 (UTC)