I get that the lightest mesons are generically the (approximate) Goldstone bosons of spontaneous chiral symmetry breaking in the quark sector. I understand that they are spin-0, but why are they necessarily pseudoscalars (odd-parity)? You can see it in this PDG listing of known $\bar q q$-mesons — all the lightest entries are $J^{\pi}=0^-$.

Here is what I know. The generators of axial flavor transformations is given in spin-space (Weyl basis):

$$T_a^A=\begin{pmatrix}T_a & 0 \\ 0 & -T_a\end{pmatrix}=\gamma^5 T_a$$

where $T_a$ is the $N_f\times N_f$ flavor generator. The CCWZ construction tells us that the correct way to incorporate these Goldstone bosons is via:

$$U(x)=\exp \left(\frac{i}{f_\pi}\Pi^a T_a^A\right)$$

where under chiral rotations we have $U\rightarrow R U L^\dagger$ with $R,L\in SU(N_f)$. But how does this transform under parity transformations, or any other discrete transformations for that matter? Anyway, I know that $\bar\psi \gamma^5\psi$ is a pseudoscalar quantity, and therefore $\bar\psi T_a^A \Pi^a\psi$ is a scalar so long as $\Pi^a$ is a pseudoscalar.

How do I piece all these together to show that indeed $\Pi^a$ must be a pseudoscalar? What am I missing?


First, recall from your favorite QFT text that $P\psi(t,x)P=\gamma^0\psi(t,-x)$.

The SSBroken generators you wrote are odd under parity, by dint of the $\gamma^5$, effectively $$ P T_a^A P = -T^A_a ~; $$ and ditto for the charge you make out of them, integration being invariant under parity, $$ Q^A_a=\int dx^3 ~~ J_{0~~a}^A (x) = i\int dx^3 ~~ \bar \psi \gamma^0 T_a \gamma^5 \psi, \\ PQ^A_a P = -Q^A_a. $$

  • But the SSBroken currents are always linear in the gradient of the goldstons (essentially by definition of the nonlinear realization involved: $\langle \Pi_a(p)| J_{\mu~~b}^A(x)|0\rangle\sim f_\pi \delta_{ab} e^{ip\cdot x} p_\mu$), $$ \bbox[yellow]{ J_{0~~a}^A (x) \sim f_\pi \partial_0 \Pi^A_a (x)+ ...} \\ P~J_{0~~a}^A (x)~P =i \bar \psi (-x)\gamma^0 \gamma^0T_a \gamma^5 \gamma^0 \psi (-x) = - J_{0~~a}^A (-x) , $$

So, likewise for the goldstons: they are parity odd.

You already saw that, above, for their interpolating field operators, $$ \bar \psi (x) T_a P\gamma^5P \psi (x)=- \bar \psi (-x) T_a \gamma^5 \psi(-x). $$

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  • $\begingroup$ Stupid question... how do you prove line 1? I prove by showing that $P\left( \delta(\theta^a) \psi\right) P=\left( P \delta(\theta^a) P\right)\left( P\psi P\right)=\delta(-\theta^a) \gamma^0\psi$, but this doesn't say that the generators themselves "transform under parity". In my mind, the generators only make sense when considering an infinitesmial chiral rotation of a spinor $\delta\left(\theta^a\right)=i\theta^aT_a^A$. On the other hand, I know that the $T_a^A$ generators have a $\gamma^5$ which carries its spin indices, and a parity transformation must flip left with right handed! $\endgroup$ – Arturo don Juan Apr 26 at 4:23
  • $\begingroup$ And second question, how did you prove the line $P J_{0,a}^{A}(x)P=-J_{0,a}^{A}(-x)$? I thought the key to showing $Q_{a}^{A}$ was parity-odd was using $d^3x\rightarrow -d^3x$. But $J_{0,a}^{A}(x)$ doesn't have this luxury, so it would seem that $J_{0,a}^{A}(x)$ should actually be parity-even! $\endgroup$ – Arturo don Juan Apr 26 at 4:32
  • $\begingroup$ Go back (run) to your QFT text, or here. Effectively, $\gamma^5\to \gamma^0\gamma^5\gamma^0=-\gamma^5$. Integration preserves parity, since the differential and the limits of integration flips cancel each other. So the charge density flips overall sign and argument sign, just like the pions. $\endgroup$ – Cosmas Zachos Apr 26 at 8:56
  • $\begingroup$ Thanks, I can't believe I forgot that bit about integration. But regarding your first remark, I am aware of that, but my issue is that if I take the relation $PT_{a}^AP=-T_{a}^A$ at face value, I derive the incorrect transformation rule for $J_{0,a}^A$. Here's what I mean: [method #1, correct result] $P[\bar\psi(x) \gamma^0 T_{a}^A\psi(x)]P=\bar\psi(-x) \gamma^0 T_{a}^A\psi(-x)=-\bar\psi(x) \gamma^0 T_{a}^A\psi(x)$ where the last equality was obtained using $\psi(-x)=\gamma^0\psi(x)$ and shuffling the gammas around. $\endgroup$ – Arturo don Juan Apr 26 at 17:15
  • 1
    $\begingroup$ The matrix element of the current relation is, indeed, what provides PCAC upon taking the divergence of, for pseudogoldstons. The most important essence of the concept, however, is the nonlinear realization in the following formula, which Ill highlight. This is what the Nambu-Goldstone nonlinear realization really, really, means. $\endgroup$ – Cosmas Zachos Apr 26 at 18:29

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