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Why this expression is time-reversal odd?

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Chenkb
#1
Dec13-13, 02:22 AM
P: 27
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?
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ChrisVer
#2
Dec13-13, 01:16 PM
P: 754
could you please fix the Latex?
What is P and k? momenta?
ZapperZ
#3
Dec13-13, 01:47 PM
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... and please make a full citation of the paper.

Zz.

George Jones
#4
Dec13-13, 02:05 PM
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Why this expression is time-reversal odd?

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]
dauto
#5
Dec13-13, 04:02 PM
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Try [t e x]/ \!\!\!\! P[/ t e x]: [itex]/ \!\!\!\! P[/itex]
and [t e x]/ \!\!\! k[/ t e x]: [itex]/ \!\!\! k[/itex]
Chenkb
#6
Dec13-13, 08:07 PM
P: 27
Quote Quote by ZapperZ View Post
... and please make a full citation of the paper.

Zz.
http://arxiv.org/abs/hep-ph/0504130
Eq.(7), the ## A_4 ## term.

Regards
Chenkb
#7
Dec13-13, 08:08 PM
P: 27
Quote Quote by George Jones View Post
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]
Thanks a lot for the texing.
ChrisVer
#8
Dec13-13, 08:15 PM
P: 754
*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..
Chenkb
#9
Dec13-13, 08:31 PM
P: 27
Quote Quote by ChrisVer View Post
Doesn't that say that [itex]A_{4}[/itex] is T-odd? no the [P,k]
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!
ChrisVer
#10
Dec14-13, 05:12 AM
P: 754
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...
Chenkb
#11
Dec14-13, 06:10 AM
P: 27
Quote Quote by ChrisVer View Post
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]##?
Bill_K
#12
Dec14-13, 09:32 AM
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Quote Quote by Chenkb View Post
Well, I think the ##i## in front is just for convenience.
No, it's not there just for convenience. It's there because i[γĚP, γĚK] = 2 σμν Pμ Kν. And if you look at the table on p.71 of Peskin, you'll see how σμν transforms.
Chenkb
#13
Dec14-13, 09:50 AM
P: 27
Quote Quote by Bill_K View Post
No, it's not there just for convenience. It's there because i[γĚP, γĚK] = 2 σμν Pμ Kν. And if you look at the table on p.71 of Peskin, you'll see how σμν 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!
Bill_K
#14
Dec14-13, 09:58 AM
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Quote Quote by Chenkb View Post
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, 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.
Chenkb
#15
Dec14-13, 10:06 AM
P: 27
Quote Quote by Bill_K View Post
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.
Could you please explain why ##\sigma_{0i}## is T-even? isn't it ##-(-1)^{\mu}(-1)^{\nu}##?
Bill_K
#16
Dec14-13, 11:39 AM
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Quote Quote by Chenkb View Post
Could you please explain why ##\sigma_{0i}## is T-even? isn't it ##-(-1)^{\mu}(-1)^{\nu}##?
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.
Chenkb
#17
Dec14-13, 12:07 PM
P: 27
Quote Quote by Bill_K View Post
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.
WOW! Many thanks, I will check it.
M. Bachmeier
#18
Dec31-13, 01:54 PM
P: 184
Your looking for a transformation of the equation... yes? or, an explanation?


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