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How would you prove that $T^{-1}\gamma_\mu T=\gamma_\mu$? Being $T$ the time reversal operator defined as $T=\gamma_1\gamma_3 K$ with $K$ the complex conjugate operator and $\gamma$ the Dirac gamma matrices?

By substitution I wrote $K^{-1}\gamma_3 \gamma_1 \gamma_\mu \gamma_1 \gamma_3 K$ but then I don't know if the inverse of complex conjugator is equal to itself, if so I've tried to proceed as follows:

$K\gamma_3\gamma_1\gamma_\mu\gamma_1\gamma_3K$ and then I have to consider whether $\mu$ is a time index or a spatial index.

If $\mu=4$ then I have: $K\gamma_3\gamma_1\gamma_4\gamma_1\gamma_3K=-K\gamma_3\gamma_4\gamma_1\gamma_1\gamma_3K=-K\gamma_3\gamma_4\gamma_3K=K\gamma_3\gamma_3\gamma_4K=K\gamma_4K$ so I have to prove that $K\gamma_4K=\gamma_4$ which implies $[K, \gamma_4]=0$ and I don't know if this is true;

if $\mu=1$ I have $K\gamma_3\gamma_1\gamma_1\gamma_1\gamma_3K=K\gamma_3\gamma_1\gamma_3K=-K\gamma_3\gamma_3\gamma_1K=-K\gamma_1K$ and I'm left with $-K\gamma_1K=\gamma_1$

if $\mu=2$ then I have $K\gamma_3\gamma_1\gamma_2\gamma_1\gamma_3K=-K\gamma_3\gamma_1\gamma_1\gamma_2\gamma_3K=-K\gamma_3\gamma_2\gamma_3K=K\gamma_2K$

if $\mu=3$ I have $K\gamma_3\gamma_1\gamma_3\gamma_1\gamma_3K=-K\gamma_3K$

so it's a complete mess and I don't know how to proceed, maybe it is way simpler than what I did but I don't know how. It seems to me that I did something right since I get a minus sign only with $\gamma_1$ and $\gamma_3$ which are the only two matrices with imaginary numbers in them. Maybe this is the answer, a matrix filled with real numbers commutes with $K$ operator while a matrix filled with imaginary numbers anticommute with it?

I've used this relation to compute this: $\{\gamma_\mu, \gamma_\nu\}=2\delta_{\mu\nu}$

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    $\begingroup$ Well, what is $K^2=KK$? If you can answer this, then it should be clear what $K^{-1}$ is... $\endgroup$ Commented Dec 17, 2023 at 19:28
  • $\begingroup$ @TobiasFünke Thank you, I think now that $K$=$K^{-1}$ but then I'm still not able to compute the demonstration. $\endgroup$
    – Salmon
    Commented Dec 17, 2023 at 19:46

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It's best to avoid basis dependent formulae such as the one you use. The time reversal matrix is defined in all space-time signatures by $$ T \gamma^\mu T^{-1}= {(\gamma^\mu)}^T $$ where $(\gamma^\mu)^T$ is the transpose.

In the mostly minus metric, the antilinear time reversal operator is defined by its action on the Dirac field opertors by $$ {\mathcal T}^{-1}\psi(x,t) {\mathcal T}= \eta_T T\psi(x,-t)\\ {\mathcal T}^{-1}\bar\psi(x,t) {\mathcal T}= \eta^*_T \bar \psi(x,-t)T^{-1}. $$ Here $\eta_T$ is a phase. (Note that the field does not get conjugated). Perhaps you are confusing $T$ and $\mathcal T$?

Also note that in a vector space over the complex numbers a vector may have real components in one basis, but complex components in another basis. A object like $K$ is basis-dependendent therefore and best avoided.

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  • $\begingroup$ Why sandwiching $\gamma^\mu$ with the time reversal transformation should give the transpose ? According to wikipedia $S(T)\gamma^\mu S(T)^{-1} = (\Lambda(T)^{-1})^\mu_\nu\gamma^\nu$ So I'd guess $\Lambda(T)^{-1} = diag(-1,1,1,1)$ (which yields not the transpose). Actually I know that it is apparently not like this. But I've never understood. $\endgroup$ Commented Dec 18, 2023 at 16:13
  • $\begingroup$ @ Frederick Thomas Where in Wikipedia? You seem to be confusing a disconnected part of the Lorentz group with time reversal. They are not the same. Time reversal is an antilinear action on the field theory quantum Hilbert space. The $t$ matrix is poart of this. It is not directly rekated to any Lorentz transform. $\endgroup$
    – mike stone
    Commented Dec 19, 2023 at 0:02
  • $\begingroup$ wiki: en.wikipedia.org/wiki/Gamma_matrices In your post I cannot see any antilinear behaviour of the time reversal, neither on the gamma-matrices, nor on the field operator. I really would like to understand why the antilinear (+other usual) behaviour of $T$ transforms a gamma-matrix in its transpose. $\endgroup$ Commented Dec 19, 2023 at 9:32
  • $\begingroup$ You can try reading my article arXiv:2009.00518. It has references and a fair bit of pedagogical material. $\endgroup$
    – mike stone
    Commented Dec 19, 2023 at 13:20
  • $\begingroup$ I should also add the discussion after the question that lead to my paper: physics.stackexchange.com/questions/519265/… $\endgroup$
    – mike stone
    Commented Dec 19, 2023 at 13:31

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