quantum field theory - Is it possible to have renormalisable C-, P-, or T-violating terms in QED?

In a recent paper (1710.01791), Witten claims that there are no renormalisable $\mathrm C,\mathrm P,\mathrm T$-violating terms that can be included in the QED Lagrangian:

What does this picture say about $\mathrm C,\mathrm P$ and $\mathrm T$? One basic question is why these symmetries are conserved by the strong and electromagnetic forces, given that they are not full symmetries of nature. In the case of electromagnetism, the answer is clear. Large symmetry violation would have to be induced by a renormalizable operator, that is one of dimension $\le4$; unrenormalizable operators with a mass scale characteristic of new physics beyond the strong and electromagnetic interactions produce small effects, as above. But there is no way to perturb Quantum Electrodynamics (QED) by an operator of dimension $\le4$ that violates any of its global symmetries, including the ones we have mentioned and some, such as strangeness, that we have not.

I'm not sure I agree with this claim. I believe there are several quadratic terms one can add to the QED Lagrangian without breaking gauge invariance, Lorentz invariance and (power-counting) renormalisability. For example, one could consider the kinetic terms $$z_5\bar\psi\gamma_5\!\!\not\!\partial\psi+z_c\psi C\gamma^0\!\!\not\!\partial\psi$$ and the mass terms $$m_5\bar\psi\gamma_5 \psi+m_c\psi^T C\gamma^0\psi$$ where $\gamma_5=\gamma^0\gamma^1\gamma^2\gamma^3$ and $C=i\gamma^2$? Surely all these terms break $\mathrm P$ and $\mathrm C$, but they respect the rest of symmetries, contradicting Witten's claim.

Needless to say, I am wrong and Witten is right. What am I missing here? Did I misunderstand Witten's claim?

It is interesting to note that Srednicki, chapter 62, lists $\mathrm C$ and $\mathrm P$ symmetries as imposed rather than derived. In other words, it appears that in his construction of QED from first principles, he lists non $\mathrm P$-symmetric terms as consistent with the rest of symmetries, and discards them as invalid only on grounds of $\mathrm P$ conservation. This is consistent with my claims above, but not with Witten's. I really don't know what to think.

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^{Remark: the $C$ terms above are Lorentz invariant (LI) because the matrix $C=i\gamma^2$ is LI. To see this, note that this matrix is nothing but $\epsilon^{ab}\oplus\epsilon^{\dot a\dot b}$, which is the metric in spinor space. Under the Lorentz transformation $\psi\to S\psi$, the contravariant spinor $\psi^TC\gamma^0$ transforms as $$\psi^TC\gamma^0\equiv\bar\psi^c\to\bar\psi^c\bar S$$ so that $\bar\psi^c\psi$ is a scalar. In the previous version of this post I wrote $C=\gamma^0\gamma^2$ instead of $C=i\gamma^2$, which is the charge conjugation matrix in the Dirac basis instead of the Weyl basis. This made this analysis less transparent, which made some people suggest my $\mathrm C$-breaking term was not LI. I hope these matters are more clear now.}

^{Even in the case where the $C$-terms are actually non-Lorentz covariant (should I have made some algebraic mistake somewhere), there are two $\mathrm P$-breaking terms that cannot be rotated away at the same time, so my criticism of the quote stands.}

Answer

As has been noted both in the comments, any $\mathrm C$-breaking term does also break gauge invariance. The reason is simple: gauge invariance requires the fermion fields to appear in the combination $\sim\psi^\dagger\psi$ (which is independent of the phase of $\psi$), while any $\mathrm C$-breaking term requires the introduction of $\psi^c$ in the form $\sim\psi^T\psi$, which depends on the phase of $\psi$. Therefore, charge conjugation is out of the question. Parity, on the other hand, is not so simple. It appears, as in the OP, that one can indeed introduce $\mathrm P$-breaking terms in a manner that is consistent with the rest of symmetries.

It turns out that all the apparently $\mathrm P$-breaking terms can actually be rotated in such a way that the Lagrangian becomes manifestly $\mathrm P$-invariant. In fact, we can prove a slightly stronger result: QED conserves flavour as well. Indeed, we have

Theorem: The most general gauge- and Lorentz-invariant, and power-counting renormalisable Lagrangian constructed out of a vector field and a set of bispinor fields is necessarily $\mathrm P,\mathrm C,\mathrm T$ and flavour conserving.

The detailed proof can be found in Weinberg's QFT, section 12.5. We summarise the essential points here.

The most general gauge and Lorentz invariant, and power-counting renormalisable Lagrangian for a vector field $A$ and a set of bispinor fields $\psi_i$ reads $$\mathcal L=-\frac14Z_3F^2-\sum_{ij}\left(Z_{Lij}\bar\psi_{Li}(\not\!\partial+ie\not A)\psi_{Lj}-M_{ij}\bar\psi_{Li}\psi_{Rj}\right)+(L\leftrightarrow R)$$ where $\psi_{L,R}:=\frac12(1+\pm\gamma_5)\psi$.

This is, in principle, not $\mathrm P$ invariant, nor does this conserve flavour. Nevertheless, there exists a similarity transformation $$\psi_{L,R}\to S_{L,R}\psi_{L,R}$$ such that $Z_L=Z_R=1$, and such that $M$ becomes diagonal. When we do this, we get $$\mathcal L=-\frac14Z_3F^2-\sum_{i}\bar\psi_{i}(\not\!\partial+ie\not A)\psi_{i}-m_i\bar\psi_{i}\psi_{i}$$ which is clearly $\mathrm P,\mathrm C,\mathrm T$ and flavour conserving. QED (pun intended). Surprise, surprise, Witten was right.