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

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March 24

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Pauli Matrices

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Defining a vector   for   where   is the   identity and the other elements are the pauli matrices. I believe there is a simple form for the rank-4 tensor given by   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,  , which appears to have the right symmetries. Cuzkatzimhut (talk) 13:38, 24 March 2015 (UTC)[reply]
I think this would be correct if the indices only cycle over  , but the inclusion of the identity as   makes things more complicated. — Preceding unsigned comment added by 128.40.61.82 (talk) 14:23, 24 March 2015 (UTC)[reply]
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)[reply]

In a related question if there is similarly a closed form for the orthogonal matrix   in terms of operations in the 4-space involving the vector   where   is a general invertible complex   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)[reply]

Hmm. I don't know, but I might know the right person to ping. YohanN7 (talk) 12:33, 24 March 2015 (UTC)[reply]
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   instead of  , and using the above expression,   orthogonal, all right. Cuzkatzimhut (talk) 14:22, 24 March 2015 (UTC)[reply]
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)[reply]
I was looking at similarity transformations on a system of spin-  particles, and so any spectrum preserving transformation is relevant. For what its worth, writing A= a0 I + i a⋅σ, I obtained  . Thanks anyway.

Logarithmic number

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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)[reply]

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? --NorwegianBlue talk 12:27, 25 March 2015 (UTC)[reply]
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. --NorwegianBlue talk 12:38, 25 March 2015 (UTC)[reply]
They're not the same. The mathworld definition only includes rational numbers. For an irrational example of Georgia guy's notion, consider  . It's irrational, and  .--80.109.80.31 (talk) 12:52, 25 March 2015 (UTC)[reply]
Thanks! --NorwegianBlue talk 13:20, 25 March 2015 (UTC)[reply]