I am trying to derive a Ward-type identity between amplitudes involving $\bar\psi \sigma_{\mu\nu}\gamma_5\psi$, $\bar \psi \gamma_\mu \gamma_5 \psi$, and $\bar \psi \gamma_5 \psi$ in QCD (diagonal quark current-mass matrix). It should be the following:

$$\partial^\nu \langle \bar \psi (0) \sigma_{\mu\nu} \gamma_5 \psi(x)\rangle=-\partial_\mu \langle \bar \psi (0) i\gamma_5 \psi(x)\rangle+m\langle\bar\psi(0)\gamma_\mu\gamma_5 \psi(x)\rangle \tag{1}$$

How do I derive this? This is different than the normal chiral Ward identity, which is:

$$\partial^\nu\langle\bar\psi(x)\gamma_\nu\gamma_5\psi(x)\rangle=2m\langle\bar\psi(x)\gamma_5\psi(x)\rangle-\underset{\textrm{anomaly}}{\underbrace{\frac{N_f}{8\pi^2}\langle F(x) \tilde F (x)\rangle}}\tag{2}$$

I understand how to derive equation (2) - simply apply a (diagonal) chiral flavor transformation to the generating functional of QCD, then expand to first order (and take into account the anomaly, for example using the Fujikawa method). However I don't see how to do that for equation (1). The LHS of (2) is $\partial^\mu J_\mu^5(x)$ where $J_\mu^5(x)$ is the Noether current associated with the diagonal chiral flavor transformation. However on the LHS of (1), the divergence involves $\sigma_{\mu\nu}$ which is not the Noether current of any global symmetry.

Also, equation (2) involves field operators evaluated at a single point $x$, whereas equation (1) is point-split.

  • $\begingroup$ Interesting, I did not know about the first identity. Are you aware of any specific use it has? $\endgroup$
    –  Mio
    Commented Jun 14, 2021 at 9:00
  • 1
    $\begingroup$ @Mio Yeah it connects the two independent twist-3 two-particle (pseudoscalar) distribution amplitudes, the pseudoscalar and pseudotensor amplitudes. The overall normalization constants for these distributions are the same because of this identity - supposedly, since I haven't seen the explicit justification of this statment. See page 4 of this article: PhysRevD.71.014015 $\endgroup$ Commented Jun 15, 2021 at 1:47

1 Answer 1



  • The non-commuting nature of the operators is not relevant here, because the points are split and all terms have the same operators in the same order.

  • Start with the Dirac equation: $\big\langle\bar\psi(0)\gamma_\mu \gamma_5 (i\gamma^\nu\partial_\nu+m)\psi(x)\big\rangle=0$.

From here, it's just $\gamma$-matrix identities.

  • $\begingroup$ Thanks I forgot about using the quantum-version of the equations of motion - that $$\left< \cdots \frac{\delta S}{\delta \psi (x)}\right>=0$$ where $\cdots$ represent other $\bar\psi, \psi$ fields. In the ultimate equation we define $\sigma_{\mu\nu}\equiv \frac{i}{2}\left[ \gamma_\mu,\gamma_\nu \right]$. $\endgroup$ Commented Jun 14, 2021 at 7:11
  • $\begingroup$ Correction, above that should be $$\left<\cdots \frac{\delta S}{\delta \bar \psi (x)}\right>=0$$ $\endgroup$ Commented Jun 14, 2021 at 7:23

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