# Why this expression is time-reversal odd?

1. ### Chenkb

28
P and k are four-momentum of two particles.
I read in a paper which said that
$[\slashed{P},\slashed{k}]=\slashed{P}\slashed{k} - \slashed{k}\slashed{P}$
is time-reversal odd.
Why?

2. ### ChrisVer

could you please fix the Latex?
What is P and k? momenta?

3. ### ZapperZ

29,656
Staff Emeritus
... and please make a full citation of the paper.

Zz.

4. ### George Jones

6,357
Staff Emeritus
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##.

Or just use (the possible more useful in this case) ##\gamma^\mu P_\mu## and ##\gamma^\nu k_\nu##.[/edit]

Last edited: Dec 13, 2013
5. ### dauto

Try [t e x]/ \!\!\!\! P[/ t e x]: $/ \!\!\!\! P$
and [t e x]/ \!\!\! k[/ t e x]: $/ \!\!\! k$

28
7. ### Chenkb

28
Thanks a lot for the texing.

8. ### ChrisVer

Well, the reason why there is T-oddness I guess is because there is a $i$ in front, making that term complex..

Last edited: Dec 13, 2013
9. ### Chenkb

28
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!

10. ### ChrisVer

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 $i$ in front, making that term complex.."

11. ### Chenkb

28
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]##?

12. ### Bill_K

4,160
No, it's not there just for convenience. It's there because i[γ·P, γ·K] = 2 σμν Pμ Kν. And if you look at the table on p.71 of Peskin, you'll see how σμν transforms.

13. ### Chenkb

28
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!

14. ### Bill_K

4,160
No, σ02 is even. In the sum, σ0i is even, σjk is odd, P0 and K0 are even, while Pj and Kj are odd. All the terms in the sum σμνPμKν wind up transforming the same way, namely they are all odd.

15. ### Chenkb

28
Could you please explain why ##\sigma_{0i}## is T-even? isn't it ##-(-1)^{\mu}(-1)^{\nu}##?

16. ### Bill_K

4,160
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 γ1γ3, but the value actually depends on your choice of representation for the gamma matrices. He uses the Weyl representation Eq.(3.25), in which γ2 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-1γμD = γ'μ where γ'0 = - γ0 and γ'j = γj.

17. ### Chenkb

28
WOW!:thumbs: Many thanks, I will check it.

18. ### M. Bachmeier

184
Your looking for a transformation of the equation... yes? or, an explanation?

19. ### Bill_K

4,160
Which do you think has not already been given?

20. ### Chenkb

28
The transformation, why it's T-odd. And I think it is solved already.