# Photon Path Integral and Lorenz Gauge

I am reading Srednicki's QFT book (http://web.physics.ucsb.edu/~mark/qft.html). In chapter 57, specifically in page 343, the book stated that there's a problem with the path integral because the projection operator cannot be inverted. In the first paragraph of page 344, it was explained that, since the issue is due to the fact that $$P^{\mu\nu}(k)k_{\nu}=0,\tag{57.7}$$ the problem can be solved by taking the path integral to be only span fields that satisfy $$k_{\mu}\tilde{A}^{\mu}(k)=0$$.

So, according to the textbook, the whole problem is solved by preventing the path integral to contain fields with $$\tilde{A}^{\mu}(k)\propto k^{\mu}$$, which can be done by requiring $$k_{\mu}\tilde{A}^{\mu}(k)=0$$. However, since $$k_{\mu}k^{\mu}=0$$, wouldn't any $$\tilde{A}^{\mu}(k)\propto k^{\mu}$$ still satisfy $$k_{\mu}\tilde{A}^{\mu}(k)=0$$?

• No, $k_\mu$ is off-shell – Nihar Karve Jan 4 at 3:13

You might have also come across this problem when trying to invert $$k^2P^{\mu\nu} \rightarrow (1/k^2)P_{\mu\nu}$$. The nullspace equation for the projector, $$P^{\mu\nu}(k)k_\nu = 0$$, is a linear algebra equation rather than something that stems from the equations of motion. Generally, whenever an equation involves a Fourier transform from position space to momentum space, the corresponding dummy integration variable is going to be off-shell, meaning it does not have to satisfy the "on-shell" condition for a real particle: $$k^2 = m^2$$ (the term "on-shell" comes from "on mass shell", since the equation constrains the momentum to lie on a hyperboloid fixed by the mass). The value of $$k^2$$ is unconstrained here - in essence, all momenta are actually off-shell to start with until we use the LSZ reduction formula to put the external legs on-shell (this is an artefact of the Feynman procedure, where definite on-shellness is sacrificed for momentum conservation).

The gauge condition $$k_\mu \tilde{A}^\mu(k)$$ involves restricting the domain of interest to the subspace orthogonal to $$k_\mu$$, in which the projection operator simply functions as the identity, allowing you to invert it (you can show that the eigenvalues of $$P_{\mu\nu}$$ are $$1, 1, 1, 0$$, with the $$0$$ corresponding to $$k^\mu$$).

So, taking $$\tilde{A}^\mu(k) = f(k)k^\mu$$ does not automatically satisfy $$k_\mu\tilde{A}^\mu(k)$$, since $$k_\mu\tilde{A}^\mu = f(k)(k_\mu k^\mu) \ne 0$$ as $$k^\mu$$ is off-shell. This momentum space propagator is actually divergent in the on-shell limit, but the LSZ procedure ensures that this divergence is exactly cancelled while calculating S-matrix amplitudes in terms of the correlators in which the divergence appears.

• I see... but the fourier-transform integral does pass through the point when the four-momentum is on-shell (k^2=m^2). How do we deal with the point when the integral pass through that point? – The Gypsy King Jan 4 at 5:16
• Ah I think I have an answer to the question I wrote in the comment: that point is only a small part of the integral so we can ignore it! – The Gypsy King Jan 4 at 5:26
• @SoldierofSteroids not exactly, see the edit – Nihar Karve Jan 25 at 12:59