P and k are four-momentum of two particles. I read in a paper which said that [itex][\slashed{P},\slashed{k}]=\slashed{P}\slashed{k} - \slashed{k}\slashed{P}[/itex] is time-reversal odd. Why?
Unfortunately, MathJax does not support the "slashed" command from the amsmath package, which is used to implement Feynman's slash notation. Try $$\left[\gamma \cdot P , \gamma \cdot k \right] = \gamma \cdot P \ \gamma \cdot k − \gamma \cdot k \ \gamma \cdot P, $$ where ## \gamma \cdot P = \gamma^\mu P_\mu##. [edit]Or just use (the possible more useful in this case) ##\gamma^\mu P_\mu## and ##\gamma^\nu k_\nu##.[/edit]
Try [t e x]/ \!\!\!\! P[/ t e x]: [itex]/ \!\!\!\! P[/itex] and [t e x]/ \!\!\! k[/ t e x]: [itex]/ \!\!\! k[/itex]
*wrong answer* Well, the reason why there is T-oddness I guess is because there is a [itex]i[/itex] in front, making that term complex..
But ##\Phi(P,k,S|n)## on the left of the Eq. must be T-even according to it's physical meaning, so, if ##A_4## is T-odd, ##[\gamma \cdot P, \gamma \cdot k]## must be T-odd. Actually, I think it is ##[P, k]## be T-odd that infers ##A_4## be T-odd, not the opposite, because ##A_4## is an unknown coefficient function. So I wonder why. Thanks a lot!
lol, I changed my initial post because it was wrong, I guess that: "Well, the reason why there is T-oddness I guess is because there is a [itex]i[/itex] in front, making that term complex.." is your answer...
Well, I think the ##i## in front is just for convenience. And the gamma matrices are not all real, for example, in Dirac representation, ##\gamma^2## is complex. So, I think the problem is how does ##\left[\gamma \cdot P , \gamma \cdot k \right]##change under time reversal operation. I know that, under time reversal, ##P=(P_0,\vec{P})##becomes ##(P_0,-\vec{P})##, and in QFT course I've learned that ##\gamma^{\mu}##becomes ##(-1)^{\mu}\gamma^{\mu}##(Peskin's book, Page71). But what about ##\left[\gamma \cdot P , \gamma \cdot k \right]##?
No, it's not there just for convenience. It's there because i[γ·P, γ·K] = 2 σ_{μν} P^{μ} K^{ν}. And if you look at the table on p.71 of Peskin, you'll see how σ_{μν} transforms.
Year, but for the terms in ##\sigma_{\mu\nu}P^{\mu}k^{\nu}## like ##\sigma_{02}P_0k_2##, is T-even, because ##\sigma_{02}## odd, ##P_0## even, and ##k_2## odd. Regards!
No, σ_{02} is even. In the sum, σ_{0i} is even, σ_{jk} is odd, P^{0} and K^{0} are even, while P^{j} and K^{j} are odd. All the terms in the sum σ_{μν}P^{μ}K^{ν} wind up transforming the same way, namely they are all odd.
Peskin uses a very strange notation here! He says "(-1)^{μ}" is supposed to mean 1 for μ=0 and -1 for μ=1, 2, 3. In the table, for T we're given - (-1)^{0}(-1)^{i}, which is interpreted to mean (-1)(1)(-1) which is +1. His derivation of the results in this table could use a little cleanup work too. In addition to complex conjugation ψ → ψ*, the time reversal operation includes a matrix acting on ψ. He says this matrix is γ^{1}γ^{3}, but the value actually depends on your choice of representation for the gamma matrices. He uses the Weyl representation Eq.(3.25), in which γ^{2} is imaginary and all the others real. In general, time reversal can be written ψ → Dψ* where D is a matrix having the property D^{-1}γ^{μ}D = γ'^{μ} where γ'^{0} = - γ^{0} and γ'^{j} = γ^{j}.