# Ward Identity and Gupta-Bleuler condition

Reading David Tong notes on QFT, he mentions about Gupta-Bleuler condition $$\partial^{\mu}A_{\mu}^{+}|\Psi\rangle=0\tag{6.54},$$ which makes sure that matrix elements vanish,$$\langle \Psi|\partial_{\mu}A^{\mu}|\Psi\rangle=0.\tag{6.55}$$ I also came across Ward identity $$p_{\mu}\mathcal{M}^{\mu}=0$$ in Schwartz QFT. At first sight, they look equivalent—i.e., it looks like they are related by Fourier transform. I roughly thought of a way to derive, but it wasn't rigorous and satisfactory. It goes something as follows, $$$$\langle{\Psi'}|\partial_{\mu}A^{\mu}|{\Psi}\rangle = 0\\ p_{\mu}\langle{\Psi'}|A^{\mu}|{\Psi}\rangle = 0\\ \approx p_{\mu}\mathcal{M^{\mu}}=0$$$$ where $$\Psi$$ and $$\Psi'$$ are in and out states. Can anyone provide a mathematical derivation and prove if they are related?

• – Qmechanic Sep 8 '20 at 17:16

$$$$\partial_\mu = \sum_i \delta(y-x_i) $$$$ Contact terms arise whenever $$y=x_i$$. $$J^\mu$$ is the current associated to the symmetry $$\delta O_i$$.
The $$p_\mu \mathcal{M}^\mu= 0$$ is indeed a Fourier transform, but not of the Gupta-Bleuler condition, it's rather from $$\partial_\mu J^\mu = 0$$ and only holds if the external particles are on-shell and the gauge field couples to this conserved current. In QED this is the case. If the external lines are not on-shell, the equations of motion will appear in the RHS instead.