# 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.

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.

• Could you please double-check whether $\bar{\psi}C\psi$ is a valid mass term, i.e., that it anticommutes with momentum-dependent terms of the Dirac Hamiltonian? – higgsss Oct 6 '17 at 11:15
• @higgsss I have a Lagrangian description in mind here -- no Hamiltonians, not momentum-dependent terms. My Lagrangian is $\mathcal L=i\bar\psi\gamma^\mu\partial_\mu\psi+m\bar\psi\psi+m_5\bar\psi\gamma_5\psi+m_c\bar\psi C\psi+e\bar\psi A^\mu\gamma_\mu\psi-\frac14 F^2$. – AccidentalFourierTransform Oct 6 '17 at 11:17
• Even if you started with a Lagrangian, there would be a corresponding Hamiltonian. My point is that with a term proportional to $\bar{\psi}C\psi$ in the Lagrangian (or Hamiltonian), the spectrum of your Hamiltonian won't be of the form $\pm\sqrt{\textbf{p}^2 + m^2}$. – higgsss Oct 6 '17 at 11:46
• So Witten's argument would accept $\bar{\psi}C\psi$ even though it is not Lorentz-invariant? – user154997 Oct 6 '17 at 22:07
• I believe that the second kinetic term and the second mass term you wrote down break the $U(1)$ symmetry. – higgsss Oct 16 '17 at 14:49

Although I'm not sure about a general proof of the $$C$$, $$P$$, and $$T$$ invariance of QED, I can explain why OP's examples don't invalidate Witten's argument.

(1) $$m\bar{\psi} \psi + m_5 \bar{\psi} \gamma_5\psi$$ is unitarily equivalent to $$\sqrt{m^2+m_5^2}\,\bar{\psi} \psi$$, where the relevant unitary operation is of the form $$e^{\theta\gamma_5}$$. Here, $$\theta$$ is a real parameter, and notice that this operation leaves the kinetic term invariant. The parity and charge conjugation symmetries hold in modified forms, that is, the corresponding symmetry operations are $$P_{\mathrm{new}} := e^{\theta\gamma_5} P e^{-\theta\gamma_5}$$ and $$C_{\mathrm{new}} := e^{\theta\gamma_5} C e^{-\theta\gamma_5}$$.

(Note that OP's definition of $$\gamma_5$$ is missing a factor of $$i$$ from the most widely used one, i.e., $$i\gamma_0\gamma_1\gamma_2\gamma_3$$. As a consequence, $$\gamma_5$$ is an anti-Hermitian matrix here.)

Below are no longer valid. OP meant $$m_c\psi^T C\gamma^0 \psi$$ as a possible mass term, but in the first version of the post, the term was written as $$m_c \bar{\psi}C\psi$$. Discussions below are referring to the first version of the post.

(2) The term $$m_c \bar{\psi}C\psi$$ is not a mass term. To see this, consider the following Lagrangian: $$$$\mathcal{L} = i\bar{\psi}\gamma^\mu\partial_\mu \psi + m\bar{\psi}\psi + im_c \bar{\psi}C\psi,$$$$ where $$m$$ and $$m_c$$ are real parameters. The corresponding Hamiltonian is $$$$\mathcal{H} = -i\psi^\dagger\gamma^0\gamma^i \partial_i\psi - m\psi^\dagger\gamma^0\psi - im_c \psi^\dagger\gamma^2\psi.$$$$ (Be reminded that $$\bar{\psi} = \psi^\dagger\gamma^0$$.) Here, the usual mass term $$m\psi^\dagger\gamma^0\psi$$ anticommutes with the kinetic term $$-i\psi^\dagger\gamma^0\gamma^i \partial_i\psi$$, which is the origin of the Dirac spectrum $$\pm\sqrt{p^2 + m^2}$$.

However, $$im_c \psi^\dagger\gamma^2\psi$$ anticommutes with the conventional mass term and the $$i=2$$ part of the kinetic term, while commuting with the rest of the kinetic term. As a result, the spectrum of the above Hamiltonian is given by $$$$E(\textbf{p}) = \pm\sqrt{\Big(\sqrt{p_1^2 + p_3^2}\pm m_c\Big)^2 + p_2^2 + m^2}.$$$$ Here, $$m_c$$ obviously doesn't play the role of a mass gap, and even more, it makes the $$y$$ direction inequivalent from the $$x$$ and $$z$$ directions, breaking Lorentz invariance.

Update: In part (2), writing down the explicit spectrum certainly wasn't the best way to demonstrate why $$im_c \psi^\dagger\gamma^2\psi$$ cannot be a valid term in a QED Lagrangian. A better way would be to note that this term, unlike other terms in the Hamiltonian, does not commute with all three rotation generators $$$$J^{ij} = -i(x^i\partial_j - x^j\partial_i) + \frac{i}{4}[\gamma^i , \gamma^j].$$$$ To be specific, $$im_c \psi^\dagger\gamma^2\psi$$ commutes with $$J^{31}$$ but does not with $$J^{12}$$ or $$J^{23}$$. Therefore, the Hamiltonian is invariant under a rotation in the 3-1 plane but not in the 1-2 or 2-3 plane. This fact, indeed, is reflected in the above spectrum.

• +1 Thanks, this is indeed nice. I agree with 1). As for 2), I do not: you argue that my term is not a mass term, which is correct but essentially irrelevant: don't call it a mass term if you don't want to, but it is (unless I'm missing something else) a valid term. You could even regard it as an interaction if you want to. Following Witten's line of thought, such a term could appear as a low-energy effective term coming from a C-breaking interaction. The fact that it doesn't suggest that there is something else going on. – AccidentalFourierTransform Oct 6 '17 at 19:31
• also, your spectrum is not Lorentz covariant. The Lagrangian term $\bar\psi C\psi$ is LI, so something smells fishy here. Everything should be covariant, unless my C-breaking term is wrong for some reason. Do we at least agree that it is LI? – AccidentalFourierTransform Oct 6 '17 at 19:33
• @AccidentalFourierTransform You are probably right. I think I got the spectrum wrong. Let me rethink that part. – higgsss Oct 6 '17 at 19:38
• @AccidentalFourierTransform I checked again, and I now believe that the spectrum I wrote above is indeed correct. I will update the answer to give a simpler argument on why $\bar{\psi}C\psi$ is not a valid term in QED Lagrangian. – higgsss Oct 6 '17 at 20:04
• "...However, $im_c \psi^\dagger\gamma^2\psi$ anticommutes with the conventional mass term and the $i=2$ part of the kinetic term, while commuting with the rest of the kinetic term..." As far as I understand, this is at least a very Dirac matrix basis dependent statement. – Name YYY Jul 2 at 23:26

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.