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Ward Identity and Gupta-Bleuler condition

Physics Asked on November 4, 2020

Reading David Tong notes on QFT, he mentions about Gupta-Bleuler condition $$partial^{mu}A_{mu}^{+}|Psirangle=0tag{6.54},$$ which makes sure that matrix elements vanish,$$langle Psi|partial_{mu}A^{mu}|Psirangle=0.tag{6.55}$$ 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?

One Answer

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.

Answered by JulianDeV on November 4, 2020

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