Why this expression is time-reversal odd?

[Chenkb]

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?

 

[ChrisVer]

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

 

[ZapperZ]

... and please make a full citation of the paper.

Zz.

 

[George Jones]

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]

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

 

[Chenkb]

... 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]

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]

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

 

[Chenkb]

Doesn't that say that $A_{4}$ 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]

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.."
is your answer...

 

[Chenkb]

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.."
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]

Well, I think the $i$ in front is just for convenience.

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.

 

[Chenkb]

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!

 

[Bill_K]

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, σ_0i is even, σ_jk is odd, P⁰ and K⁰ are even, while P^j and K^j are odd. All the terms in the sum σ_μνP^μK^ν wind up transforming the same way, namely they are all odd.

 

[Chenkb]

No, σ₀₂ is even. In the sum, σ_0i is even, σ_jk is odd, P⁰ and K⁰ are even, while P^j and K^j are odd. All the terms in the sum σ_μνP^μK^ν 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]

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^-1γ^μD = γ'^μ where γ'⁰ = - γ⁰ and γ'^j = γ^j.

 

[Chenkb]

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^-1γ^μD = γ'^μ where γ'⁰ = - γ⁰ and γ'^j = γ^j.

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

 

[M. Bachmeier]

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

 

[Bill_K]

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

Which do you think has not already been given?

 

[Chenkb]

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

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