# How do I derive this Ward-type identity?

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.

• Interesting, I did not know about the first identity. Are you aware of any specific use it has?
–  Mio
Jun 14 at 9:00
• @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 Jun 15 at 1:47

• 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.
• 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 $\sigma_{\mu\nu}\equiv \frac{i}{2}\left[ \gamma_\mu,\gamma_\nu \right]$. Jun 14 at 7:11
• Correction, above that should be $$\left<\cdots \frac{\delta S}{\delta \bar \psi (x)}\right>=0$$ Jun 14 at 7:23