# General formulation of time reversal symmetry action on fermions

I'm wondering about a general way to define the action of time reversal on a fermion field $$\psi$$. From a few sources I've read (e.g. appendix A of Witten's paper on fermion path integrals), it seems like in general, a reflection of the coordinate $$x^\mu$$ in space time corresponds to the transformation $$R_\mu:\psi(x) \mapsto \gamma_\mu \psi(R_\mu(x)),$$ where $$R_\mu(x^\nu) = -x^\nu$$ if $$\mu=\nu$$ and $$R_\mu(x^\nu)=x^\nu$$ otherwise. If this is true, we should be able to choose time reversal so that it acts on fermions as $$T: \psi(t,x) \mapsto \gamma_0 \psi(-t,x).$$ Indeed, I have seen papers (e.g. this paper by Cordova et al) where such a choice is made, at least for theories in 2+1 dimensions.

I'm curious if this is a satisfactory definition of time reversal in general (which seems to be implied by the references mentioned), and if it's not, whether there is a simple modification that makes it into a general definition.

One reason why I don't think it can be general is that it seems to be dependent on the signature we choose. For example, the Dirac mass in 2+1D is supposed to be $$T$$-odd. If we are in $$(-,+,+)$$ signature then we can choose e.g. $$\gamma_0 = i\sigma^y$$, in which case one checks that the Hermitian term $$im\bar\psi\psi$$ is indeed $$T$$-odd. However, if we are in $$(+,-,-)$$ signature where we can choose e.g. $$\gamma_0=\sigma^x$$, then the Hermitian term $$m\bar\psi\psi$$ is $$T$$-even. So clearly something with our definition of $$T$$ is not right.

## 1 Answer

1. A spinor, in the sense of transforming in a representation of $$\mathrm{Spin}(1,3)$$, the double cover of the proper orthochronous Lorentz transformations $$\mathrm{SO}^+(1,3)$$, does not have a transformation behaviour under parity or time reversal transformations, since neither parity nor time reversal are in $$\mathrm{SO}^+(1,3)$$ - they are not continuously connected to the identity and therefore can't be generated by exponentiating the Lorentz algebra. What we really need is the notion of a pinor, something that transforms in a representation of the pin group, the double cover of the full Lorentz group $$\mathrm{O}(1,3)$$.

2. Very annoyingly, while the spinor representations are generally insensitive to the metric sign convention of our spacetime (mostly + or mostly -) because $$\mathrm{Spin}(1,3)$$ and $$\mathrm{Spin}(3,1)$$ are isomorphic, $$\mathrm{Pin}(1,3)$$ and $$\mathrm{Pin}(3,1)$$ are not. It is therefore rather easy to get tangled in inconsistent sign conventions when talking about the action of parity or time reversal on fermions.

3. In the end, it turns out that time reversal acting on a standard Dirac fermion will come out to be proportional to $$\gamma^1 \gamma^3$$, but the sign depends on several sign choices along the way. In order to derive this for your particular conventions, you need to consider the basic solutions of the Dirac equation in terms of a Weyl spinor $$\xi$$ which will look something like $$u(\vec p) = \begin{pmatrix}\sqrt{p\cdot \sigma}\xi \\\sqrt{p\cdot\bar\sigma}\xi\end{pmatrix}$$ and now apply time reversal to this expression, then figure out what sign you need to give $$\gamma^1\gamma^3$$ so that $$\pm\gamma^1\gamma^3u(\vec p)$$ is the same as this time-reversed expresson. Then take the full mode expansion of the fermion field, consider that the creation/annihilation operators also flip momentum and spin, and work out what the factor of $$\gamma^1\gamma^3$$ is that ends up in front of the expansion. This method is applied e.g. in Blagoje's notes on CPT symmetry.

• Can you comment on / suggest references for seeing the non-isomorphism between ${\rm Pin}(1,3)$ and ${\rm Pin}(3,1)$? Also, how does this situation, and the action of time reversal as $\pm \gamma^1\gamma^3$, generalize to other dimensions (in particular, odd dimensions)? – user3521569 Feb 11 at 18:34