Wikipedia:Reference desk/Archives/Mathematics/2015 March 24

= March 24 =

Pauli Matrices
Defining a vector $$\sigma_i$$ for $$i=0,1,2,3$$ where $$\sigma_0$$ is the $$2 \times 2$$ identity and the other elements are the pauli matrices. I believe there is a simple form for the rank-4 tensor given by $$T_{ijkl} = \frac{1}{2} \mathrm{tr} \left[ \sigma_i \sigma_j \sigma_k \sigma_l \right]$$ but I can't seem to find it. If anybody is able to point me in the right direction I would be quite grateful.
 * Since pinged, I could point out the first boxed eqn of Pauli matrices leads directly to, I think, $$T_{ijkl} =\delta_{ij} \delta_{kl} -\delta_{ik} \delta_{jl}+ \delta_{il} \delta_{jk} $$, which appears to have the right symmetries. Cuzkatzimhut (talk) 13:38, 24 March 2015 (UTC)
 * I think this would be correct if the indices only cycle over $$i=1,2,3$$, but the inclusion of the identity as $$\sigma_0$$ makes things more complicated. — Preceding unsigned comment added by 128.40.61.82 (talk) 14:23, 24 March 2015 (UTC)
 * Apologies, yes, I took only indices 1,2,3. If 3 indices are 0, the expression vanishes. If two, say k,l, it is a delta of the other two. If one, l=0, it is i εijk. Cuzkatzimhut (talk) 15:13, 24 March 2015 (UTC)

In a related question if there is similarly a closed form for the orthogonal matrix $$O_{ij} = \frac{1}{2} \mathrm{tr} \left[ \sigma_i A \sigma_j A^{-1} \right]$$ in terms of operations in the 4-space involving the vector $$ a_i = \frac{1}{2} \mathrm{tr} \left[ \sigma_i A \right]$$ where $$A$$ is a general invertible complex $$2 \times 2$$ matrix it would be very useful to know.

Thank you. — Preceding unsigned comment added by 128.40.61.82 (talk) 11:42, 24 March 2015 (UTC)


 * Hmm. I don't know, but I might know the right person to ping. YohanN7 (talk) 12:33, 24 March 2015 (UTC)


 * So, utilizing the above, you should first take out the inert determinant of $A$ killed by the inverse $A$ in your expression, and supplant it with −|$a$| the length of your 3-vector a, and the determinant of your now normalized $A$, which is thus now reduced to the mere Pauli vector $a⋅σ$. The inverse of $A$ is just the similar Pauli vector with $$ a_i/a^2$$ instead of $$ a_i$$, and using the above expression, $$O_{ij}= (-\delta_{ij} + 2 a_i a_j /a^2)$$ orthogonal, all right. Cuzkatzimhut (talk) 14:22, 24 March 2015 (UTC)


 * This I think is correct if we restrict only to the case that $A$ is traceless.


 * Of course! take the trace out as an identity piece, if you like. Otherwise, you have to do the calculation with a $A= bI + i a⋅σ$, invert it, etc... In "customary" applications, $A$ is not arbitrary, but a unitary, rotation matrix, whose form is then substantially constrained. Cuzkatzimhut (talk) 15:09, 24 March 2015 (UTC)


 * I was looking at similarity transformations on a system of spin-$$\tfrac{1}{2}$$ particles, and so any spectrum preserving transformation is relevant. For what its worth, writing $A= a_{0} I + i a⋅σ$, I obtained $$O_{ij} = \frac{1}{a_0^2-a^2} \left[ \left( a_0^2+a^2 \right) \delta_{ij} - 2 a_i a_j + 2 i a_0 \epsilon_{ijk} a_k \right]$$. Thanks anyway.

Logarithmic number
Has there been any proof that pi is not a logarithmic number?? (This means a number that can be written as b in a^b = c where a and c are natural numbers and a > 1.) It is known that all non-negative rational numbers are logarithmic, as are many irrational numbers. Georgia guy (talk) 19:37, 24 March 2015 (UTC)
 * Since no-one yet has answered your question, and I'm uncertain if I've understood it correctly, could you (or someone else) give an example of an irrational number b and two natural numbers a and c that satisfy the equation a^b = c? --NorwegianBluetalk 12:27, 25 March 2015 (UTC)
 * There is a definition of "logaritmic number" at mathworld.wolfram.com. I'm having trouble seeing that that definition and Georgia guy's definition are the same. --NorwegianBluetalk 12:38, 25 March 2015 (UTC)
 * They're not the same. The mathworld definition only includes rational numbers.  For an irrational example of Georgia guy's notion, consider $$\log_2 3$$.  It's irrational, and $$2^{\log_2 3} = 3$$.--80.109.80.31 (talk) 12:52, 25 March 2015 (UTC)
 * Thanks! --NorwegianBluetalk 13:20, 25 March 2015 (UTC)