0
$\begingroup$

I have some doubts about the way we apply time reversal to Dirac's Lagrangian in QFT. Looking for the transformed field, $\psi^t(x)$, I've found sources (see below) that claims:

$$\psi^t(x) = \gamma^1\gamma^3\psi(-t, \vec{x}),\quad x = (t, \vec{x}) \tag1$$

($\gamma^\mu$ in Weyl representation)

Nevertheless, I tried to prove it in the following way and find another result. The transformed field $\psi^t(x)$ is a fermion field so it has to satisfy Dirac's equation as $\psi(x)$ does, i.e.,

$$(i\gamma^\mu\partial_\mu - m)\psi(x)\ \Rightarrow\ (i\gamma^\mu\partial_\mu - m)\psi^t(x) \tag2$$

Also, due to left equation in (2), $\psi(-t, \vec{x})$ satisfies

$$(-i\gamma^0\partial_0 + i\vec{\gamma}·\vec{\partial} - m)\psi(-t, \vec{x})\ \Rightarrow\ (i\gamma^0\partial_0 - i\vec{\gamma}^{\ *}·\vec{\partial} - m)\psi^*(-t, \vec{x}) \tag3$$

(Star * denotes complex conjugation)

Now, in (2)'s right equation you can take $\psi^t(x) = M·\psi^*(-t, \vec{x})$ with $M$ unitary (the anti-unitary property of time reversal transformations is already applied over the field via complex conjugation) and such that

$$[M, \gamma^0] = 0, \quad \vec{\gamma}^{\ *}M = -M\vec{\gamma}\tag4$$

With this properties $M$ leads you from (4) to (3), so this $\psi^t(x) = M\psi^*(-t, \vec{x})$ is a fermionic field that satisfies the time reversal Dirac's equation. Moreover, the conditions in (4) force $M = \gamma^1\gamma^3$. Then, I obtain

$$\psi^t(x) = \gamma^1\gamma^3\psi^*(-t, \vec{x}) \tag5$$

(5) is different from (1), so am I wrong or the sources are?


References: http://homepages.ulb.ac.be/~boblak/Miscellaneous/Oblak_CPT.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.