Can the vacuum $| 0 \rangle$ of some QFT break C, P, T symmetry?

  1. It could be that the QFT break C, P, T symmetry explicitly. Are there good examples?

p.s. The weak interaction breaks P explicitly and maximally, I believe.

  1. It could be that the QFT does break C, P, T symmetry explicitly. But the vacuum $| 0 \rangle$ of some QFT break C, P, T symmetry spontaneously. Are there good examples?

p.s. We can also change the parity P to the reflection symmetry. Follow people's advice.

  • $\begingroup$ I don't understand this question. What is your definition of "vacuum" where the vacuum is not by definition invariant under the full Poincaré group, and hence at least $P$ and $T$? $\endgroup$
    – ACuriousMind
    Commented May 5, 2021 at 16:20
  • $\begingroup$ maybe take the weak interaction sector as an example, it does break $P$. And in our standard model, the vacuum $|0>$ includes the weak interaction breaks P explicitly. Agree? $\endgroup$ Commented May 5, 2021 at 16:28
  • $\begingroup$ I don't agree that the weak interaction not being parity invariant would mean that the vacuum is not parity invariant. You would have to spell out an actual argument for that. $\endgroup$
    – ACuriousMind
    Commented May 5, 2021 at 16:29
  • $\begingroup$ The |0⟩ vacuum just means the lowest energy ground state of the QFT without any particle or operator insertions. I thought the weak interaction QFT partition function (without any particle or operator insertions), has the action $S_{weak}$ and the partition function $Z = \sum_{gauge} \exp(i S_{weak})$ already violates the P. This also means the vacuum ground state |0⟩ also breaks the P. This is my view. Thanks for challenging. $\endgroup$ Commented May 5, 2021 at 16:35
  • $\begingroup$ @ACuriousMind That isn't correct, is it? We only strictly need to demand invariance under the connected part of the Poincare group. If the vacuum were still parity invariant, then any parity odd currents would need to have vanishing expectation. But wasn't the point of the kaon decay discovery that such amplitudes are non-zero (for CP rather than just P, but the argument is the same formally)? $\endgroup$ Commented May 5, 2021 at 18:57

1 Answer 1


I am writing the answer based on my conversations with Prof. Michael Peskin.

We both agree that there are QFTs that do not break C, P, T symmetry explicitly. But the vacuum |0⟩ of some QFT break C, P, T symmetry spontaneously.

There are many theories in which the weak interactions are fundamentally P and C symmetric, but P and C are spontaneously broken. One of the original papers is

G. Senjanovic and R. N. Mohapatra Phys. Rev. D 12, 1502 (1975)

Consider an $SU(2) \times SU(2)$ left-right symmetric theory of weak interactions, in which $P$ exchanges the two $SU(2)$'s. Postulate a set of Higgs fields in the representation $(1,0) + (0,1)$ where the numbers are the isospin under the corresponding $SU(2)$. It is easy to write a potential energy in which one Higgs field acquires a vacuum expectation value but both cannot. The the parity symmetry is spontaneously broken. The vacuum in which the first $SU(2)$ is broken is equivalent to the vacuum in which the second $SU(2)$ is broken. In both cases, the unbroken symmetry is $SU(2) \times U(1)$, which is just right.

Prof. Michael Peskin: "Some people writing in the stackexchange post seem to be going around in mathematical circles. They do not seem to have picked up a textbook to find out what spontaneous symmetry breaking is or what the actual properties of the weak interaction are."

Prof. Michael Peskin generously suggested: "So how can I help them?" Look at Prof. Michael Peskin's textbook Concepts of Elementary Particle Physics. Spontaneous symmetry breaking is discussed in a physical way.

  • $\begingroup$ Prof. Michael Peskin does not have an account so we decide that I post an answer for him. $\endgroup$ Commented May 6, 2021 at 16:42

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