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I'm confused by the CPT theorem. It states (more or less) that a Lorentz invariant quantum field theory needs to be CPT invariant. But what does it actually mean for a QFT to be CPT invariant? It surely means that it's Lagrangian is. What about the Hamiltonian? Does the invariance get inherited by it as well? What about other observables that I can measure in an experiment? Are all of those also invariant under CPT (even though they might not be Lorentz invariant)?

Related to this: If I have a Lagrangian (density) that transforms in a specific way under C, P, and T, and I derive a low energy Hamiltonian from it, does this one necessarily inherit the same transformation properties? And if yes, is the reverse true? If a Hamiltonian (of an isolated system) is odd under C, P, or T, will this necessarily correspond to violations of the same symmetries at high energies?

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    $\begingroup$ P refers to inverting the sign of the position coordinates $q_i$. This also entails inverting the sign of the partial differentiations $\partial/\partial q_i$. T refers to inverting the sign of $t$ and $\partial/\partial_t$. C refers to inverting the sign of all charges (e.g. the electric charge.) If you do all this together and a Lagrangian or Hamiltonian remains the same then it's CPT invariant. $\endgroup$
    – Chad K
    Commented Nov 22, 2023 at 17:04
  • $\begingroup$ In physics.stackexchange.com/q/545672/385138 it is said, that C becomes ill defined in low energy quantum mechanics. If this is true, wouldn't this imply that CPT also looses its meaning? $\endgroup$
    – user
    Commented Nov 23, 2023 at 9:36
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    $\begingroup$ Right: CPT is a concept of relativistic QFT, not QM, as answered there. Thinking of NR QM as the low energy limit of QFT has its limits, and several features of QFT don't pass down naturally... $\endgroup$
    – Cosmas Zachos
    Commented Nov 23, 2023 at 14:57

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The WP statement is that

Any Lorentz invariant local quantum field theory with a Hermitian Hamiltonian must have CPT symmetry.

The theory involves the vacuum state, the Lagrangian, and the Hamiltonian.

What about other observables that I can measure in an experiment? Are all of those also invariant under CPT?

They don't need to, in principle: that's how you might measure a violation; but they turn out to be, confirming CPT is a sacrosanct symmetry.

Re: your second paragraph. P and T are violated by the weak interactions, the first maximally and the second (CP) by a little, e.g. in neutron decay. CPT is still conserved there, low energies, just as in higher ones! The weak QFT Hamiltonian preserves CPT at all energies!

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  • $\begingroup$ Thanks for your answer! For the second part of my question, I tried to restate it now in another way. I hope it is clearer like this $\endgroup$
    – user
    Commented Nov 23, 2023 at 9:33
  • $\begingroup$ I adjusted my answer. $\endgroup$
    – Cosmas Zachos
    Commented Nov 23, 2023 at 11:55

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