# Show that the vacuum polarization is transverse

Tags:
1. Jul 4, 2017

### leo.

1. The problem statement, all variables and given/known data
Show that the vacuum polarization $\Pi^{\mu\nu}_2(p)$ in 1-loop is transverse. Decide whether you want to use Ward's identity and prove this to be true in all orders or only prove for 1-loop.

2. Relevant equations
Ward's identity $q_\mu \mathcal{M}^{\mu}=0$ which must hold where $\mathcal{M}=\epsilon_\mu \mathcal{M}^\mu$ is the amplitude for a process with an external photon with momentum $q$ and polarization $\epsilon_\mu(q)$.

Also the form of $\Pi^{\mu\nu}_2(p)$ guessed by means of Lorentz invariance by Schwartz in his book "Quantum Field Theory and the Standard Model":
$$\Pi^{\mu\nu}_2(p)=\Delta_1(p^2)g^{\mu\nu}+\Delta_2(p^2)p^\mu p^\nu$$.

3. The attempt at a solution
If I apply Ward's identity to $\Pi^{\mu\nu}_2$ using the form guessed by Lorentz invariance the result is easily obtained. Actually we have

$$p_\mu \Pi^{\mu\nu}_2(p)=\Delta_1(p^2) p^\nu+\Delta_2(p^2)p^2 p^\nu=0$$

and this gives that $\Delta_2(p^2)=-\frac{1}{p^2}\Delta_1(p^2)$. Since the photon isn't on-shell this four-momentum is not zero, and we are left with

$$\Pi^{\mu\nu}_2(p)=\left(g^{\mu\nu}-\dfrac{p^\mu p^\nu}{p^2}\right)\Pi_2(p)$$

where I renamed $\Pi_2(p)=\Delta_1(p)$. This was my initial guess. There are some problems however. The first one, is that this is redundant. If I assume Ward's identity to be valid to $\Pi^{\mu\nu}_2$ then I'm already obviously assuming transversality.

The second one is that I can't see why Ward's identity can be used in this case. As I've stated in the relevant equations, Ward's identity is applied to the amplitude for a process with an external photon. Here as the books explain, the vacuum polarization is just a part of a diagram, where the external photons cannot be considered on-shell.

More than that, in Peskin's book, he says Ward's identity is valid to $\Pi^{\mu\nu}$ where this last one is the "sum of all 1-particle-irreducible insertions", being $\Pi^{\mu\nu}_2$ the second order contribution. By his writing I understand Ward's identity should be valid just for that sum, not for the individual contributions.

So I'm quite confused on how Ward's identity must be used.

Also, how would be the solution without using it, just for one-loop as the problem states that can be done?

Last edited: Jul 4, 2017
2. Jul 9, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.