I've been doing some reading about chiral symmetry breaking since it was not touched in my particle physics course

I found these slides

As explained in the above link, if we take $|\psi \rangle$ as the energy eigenstate of our Hamiltonian, commutation of the massless QCD Hamiltonian with conserved axial and vector charges imply degeneracy:

$$H_0 |\psi \rangle = E|\psi \rangle \rightarrow H_0Q^{\alpha}_{V,A}|\psi \rangle = Q^{\alpha}_{V,A}H_0|\psi \rangle = E Q^{\alpha}_{V,A}|\psi \rangle$$

If $|\psi \rangle$ is a parity eigenstate, axial charge will flip the parity of $|\psi \rangle$ state due to $\gamma_5$ so $|\psi \rangle$ state will be degenerate with $Q^{\alpha}_{A}|\psi \rangle$ state with opposite parity.

If $SU(3) \times SU(3)$ is an exact symmetry and unbroken, we would expect to see $|\psi \rangle$ state degenerate with $Q^{\alpha}_{A}|\psi \rangle$ state with opposite parity.

Since no such parity doubling has been observed, we conclude that the symmetry must be broken, hence we get $Q^{\alpha}_{A}|0 \rangle \neq 0$ from which one can see that axial charges create 8 massless Goldstone bosons. I feel like I am missing something obvious but I don't see how this implies the absence of parity doubling.


1 Answer 1


Yes, it is obvious, but you ignored the key point your source makes in the lower page 13.

If $|\psi \rangle$ is a parity eigenstate, any axial charge will flip the parity of the $|\psi \rangle$ state due to the $\gamma_5$, so the $|\psi \rangle$ state will be degenerate with the $Q^{\alpha}_{A}|\psi \rangle$ state with opposite parity.

All true, but as he points out, $Q^{\alpha}_{A}|\psi \rangle$ is not a one particle state! Instead, it is a superposition of the parity doublet partner state you'd have, plus the original particles state with a Goldstone boson with the quantum numbers of the axial charge, etc... so your output state is mush and you have no right to interpret it as just the parity doublet particle with the same energy/mass as the original one!

The point should be clear if you kept reading, but the pedagogical sadism in your notes (probably unconscious, and self-inflicted first on the author himself) prevents you from focussing on it, as you should.

Consider a sloppy/schematic cartoon version of the axial charge $Q_A^+\sim \int\!\!d^3x ~\bar p \gamma_5 \gamma_0 \partial_0 n $, plus irrelevant terms, in the (linear, non-SB) Wigner-Weyl mode: it converts a neutron to a parity-doublet "proton", $$ Q_A^+ |n\rangle= |\tilde p\rangle. $$ The way this works is by $$ Q_A^+ n^\dagger |0\rangle = ([Q_A^+, n^\dagger] +n^\dagger Q_A^+)|0\rangle = \tilde p ^\dagger |0\rangle, $$ where it is crucial that the vacuum is a singlet under axial rotations, i.e. $Q_A^+ |0\rangle$. So the degeneracy of the parity doublet $\tilde p$-n is predicated on the properties of the vacuum, the foundation on which you built your spectrum.

In the Nambu-Goldstone mode, however, something truly dramatic happens. Crudely, the charge (skipping lots of fine print, since it basically does not exist!) does not parity-isorotate all fields. Its most important, extra, leading, term is a linear term in the pions (pseudoscalar octet goldstons) which pumps pions out of the vacuum (nonlinearly, then), $$ Q_A^+|0\rangle \sim f \pi^{+\dagger} |0\rangle=f|\pi^+\rangle \neq 0. $$ So, in the above argument, the second term does not vanish, but amounts to $$ f|n \pi^+\rangle, $$ whence your transform is a multiparticle superposition state with the opposite parity to n, but, who cares? The pairing of parity doublets broke down, and $\tilde p$ need not be degenerate to n.

  • $\begingroup$ So if I chose $|\psi \rangle$ as parity eigenstate, $Q^{\alpha}_A |\psi \rangle$ state will have opposite parity thus I expect to see parity doubling. However, in Goldstone mode, since axial charge creates Goldstone bosons $Q^{\alpha}_A |\psi \rangle = |\psi, NG_{1}, ... \rangle$, I can not interpret the final state as the opposite parity state because it's not a single particle state? $\endgroup$
    – Monopole
    Feb 3, 2022 at 18:05
  • 1
    $\begingroup$ It is an opposite parity state, but it is a jumble of states, $\tilde p$, $n\pi^+$, etc... and you really don't care about silly states: you only care about the one particle state spectrum, and $\tilde p$ need not be degenerate to n ! $\endgroup$ Feb 3, 2022 at 18:10
  • $\begingroup$ Could you clarify what you meant by silly states? Because I was expecting those Goldstone bosons to be pions and etc. $\endgroup$
    – Monopole
    Feb 3, 2022 at 19:28
  • $\begingroup$ "Silly state" = superposition of one, two, ..... n-particle states, what you original one particle state has axially transformed to. The energy of this superposition tells you little about the mass of the $\tilde p$, which you determined in the Wigner-Weyl mode. The $|\psi,NG1, ...\rangle$ state is completely useless ("silly") and uninformative for your purposes! $\endgroup$ Feb 3, 2022 at 19:50
  • 1
    $\begingroup$ This is what I say, isn't it? The parity doubles exist, but they are non-degenerate. N(1535) with P=-. $\endgroup$ Feb 3, 2022 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.