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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, \begin{equation} \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 \end{equation} where $\Psi$ and $\Psi'$ are in and out states. Can anyone provide a mathematical derivation and prove if they are related?

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The Ward identity and Gupta-Bleuler condition are two completely different concepts.

The Ward identity is the quantum mechanical analogue of the Noether Theorem. A standard derivation can be found in any textbook (Schwartz). Intuitively, it can be understood as follows: classical conservation laws hold quantum mechanically inside correlation function up to contact terms.
The most general Ward identity one can write is
\begin{equation} \partial_\mu <J^\mu(y) O_1(x_1)...O_2(x_2)> = \sum_i \delta(y-x_i) <O_1...\delta O_i...O_n> \end{equation} 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.

The Gupta-Bleuler condition is something else and can be interpreted as the physical state condition. Essentially it's an operator equation such that only the ones with physical polarisations remain. Hence it also prevents negative-norm states.

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  • $\begingroup$ That seems to clear things up. One follow up question is that both Gupta Bleular and Ward's identity seem to preserve only physical polarization right? Ward identity does it by removing the unphysical state(∝pμ) from appearing in matrix elements and Gupta bleular removes unphysical polarization by coupling them so that they don't contribute to Observables. Is this correct? $\endgroup$ – MadScientist Sep 8 '20 at 17:55
  • $\begingroup$ No, only the Gupta-Bleuler condition removes the unphysical polarisations. The Ward identity is just current conservation and leads to conservation of electric charge in QED processes. $\endgroup$ – JulianDeV Sep 8 '20 at 22:40

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