Time reversal operation on $\gamma$-matrices

Question

I know that time reversal is realized as an anti-linear operator. Nevertheless I am quite bewildered by the realization of the $T$ reversal on $\gamma$-matrices.

We assume here a Minkowski metric $\eta_{\mu\nu}=diag(1,-1,-1,-1)$.

According to N. Beisert's QFT script (ETH) for time reversal $S_T=\gamma^1\gamma^3$ the following identity holds:

$$\Lambda^\mu_\nu S(T) \gamma^{\nu\ast} S(T)^{-1} = - \gamma^\mu \tag{1}$$

whereas for transformations of the orthochrone Lorentz group $L^{\uparrow}$ the transformation rule is:

$$\Lambda^\mu_\nu S(L)\gamma^\nu S^{-1}(L) = \gamma^\mu \tag{2}$$

where $\Lambda^\mu_\nu$ symbolizes a Lorentz-transformation.

(2) is valid for parity transformations $P^\mu_\nu$ since we know according to the chosen metric $(\gamma^0)^2=1$:

$$ \gamma^{\dagger 1,2,3} = -\gamma^{1,2,3} \quad \text{and}\quad \gamma^{\dagger 0} =\gamma^{ 0}\quad \text{or shortly}\quad P^{\mu}_\nu\gamma^{\dagger\nu} = \gamma^{\mu}$$

$$ P^\mu_\nu \gamma^0\gamma^\nu(\gamma^0)^{-1} = P^\mu_\nu \gamma^0\gamma^{\nu} \gamma^0 = P^\mu_\nu \gamma^{\nu\dagger} = \gamma^\mu$$

where $P^\mu_\nu$ are the components of the parity Lorentz transformation.

Therefore we get $S(P) = e^{i\alpha} \gamma^0$ because an additional arbitrary phase factor can be admitted. For reasons which do not matter in this context a non-zero number $z$ as additional factor with $|z|\neq 1$ can be usally be excluded.

Similarly we have $(PT)^\mu_\nu = -\delta^\mu_\nu$ as Lorentz transformation of the 4-dimensional reflection:

$$(PT)^\mu_\nu\gamma^5 \gamma^\nu (\gamma^5)^{-1} = -\gamma^5 \gamma^\mu (\gamma^5)^{-1}=\gamma^\mu$$

which is valid because the $\gamma^5$ matrix anticommutes with the usual $\gamma$ matrices. Therefore $S(PT) = \gamma^5 e^{i\alpha}$. An additional arbitrary phase factor as above can be admitted again.

Actually in order concatenate a parity and a time-reversal transformation, naively I would assume:

$$\gamma^{\mu} = P^\mu_\sigma S(P) \gamma^{\sigma} S(P)^{-1} = P^\mu_\sigma T^\sigma_\nu S(P)S(T) \gamma^\nu S(T)^{-1} S(P)^{-1} = (PT)^\mu_\nu S(PT) \gamma^\nu S(PT)^{-1}$$

in order to make P-transformations compatible with PT-transformations. $T^\mu_\nu$ represents the Lorentz transformation of time reversal. However, according to N.Beisert the transformation law (1) the time-reversal is not

$$\gamma^{\mu}= T^\mu_\nu S(T) \gamma^\nu S(T)^{-1} \quad\text{but}\quad T^\mu_\nu S(T)(\gamma^\nu)^\ast S(T)^{-1} = -\gamma^\mu$$

Is it not a contradiction ?

Answer 1

I found the answer in the paper arXiv:math-ph/0012006: The Pin Groups in Physics C, P and T from M.Berg, C. DeWitt-Morette, S.Gwo, E.Kramer.

I won't develop here on Pin groups which is essential part of the paper, I will only touch upon it. As I fixed the metric to

$$\eta_{\mu\nu}=diag(1,-1,-1,-1)$$

I will use the formalism of the Pin(1,3) group.

The notation in the paper is that we have to translate:

$ \Lambda \rightarrow S$

$L\rightarrow \Lambda$

$\Gamma \rightarrow \gamma$

On page 30 of the mentioned paper it is said that for an anti-unitarian transformation A the new pinor $\psi'$ is related to the original one by

$$\psi'(Lx) ={\cal {A_L}}\psi^{\ast}(x)$$

with --- and here I will adapt Beisert transformation rule (1):

$${\cal {A_L}}\gamma^\ast_\alpha{\cal{A_L}}^{-1} = -\gamma_\beta \Lambda^\beta_\alpha$$

so in particular for the time reversal ${\cal{T}}$:

$${\cal {T}}\gamma^\ast_\alpha{\cal{T}}^{-1} = -\gamma_\beta T^\beta_\alpha$$

According to the definition of the charge conjugation operator of the mentioned paper page 8 ${\cal{C}}$ we have (according to the paper ${\cal{C}}=\pm \gamma^2$ -- usually we have: $C^{-1}\gamma^{\mu} C = -\gamma^T$, but this $C={\cal{C}}\gamma^0$):

$${\cal{C}} \gamma_\alpha^\ast {\cal{C}}^{-1} = -\gamma_\alpha$$

So we can write:

$$({\cal{T} C^{-1}})\gamma_\alpha ({\cal{C}T^{-1}})=\gamma_\beta T^\beta_\alpha$$

Now we can actually define according to page 30 of the mentioned paper:

$${\cal{T} C^{-1}} =:S(T)$$

So we get the desired relationship:

$$S(T)\gamma^\nu S(T)^{-1} = \gamma^\mu T^\nu_\mu$$

On the following page 31 the paper explains:

CPT invariance means invariance under the combined transformation of charge, parity and antiunitary time reversal. It follows from the above relation between unitary and antiunitary time reversal that CPT is simply invariance under S(P)S(T) where we emphasize that S(T) is unitary. The combination $S(P)S(T)$ covers the component PT of the full Lorentz group, which together with the component connected to unity constitutes the component of the orientation preserving transformation (determinant 1). Thus CPT invariance is invariance under orientation preserving Lorentz tranformations.

Another citation (page 30):

The reason for introducing the antiunitarian operation $A_T\equiv T$ performed by complex conjugation and the matrix ${\cal{T}}$ is the requirement, necessary in a theory free of negative energy states, that the fourth component of the energy-momentum vector does not change sign under time reversal.

Conclusion:

Beisert's definition is the antiunitarian one, whereas the one that reproduces best the $S(P)S(T) = S(PT)$ is ${\cal{T}C^{-1}} = S(T)$. I would call it the computational time reversal --- actually we have to deal with 2 types of time-reversals according to the context.

I am sure that the explanation here certainly still misses out many details, so I refer to the paper.