# Mandl & Shaw QFT problem 14.1

1. Aug 2, 2013

### Vic Sandler

The problem is on pages 323 and 324 of the second edition.

1. The problem statement, all variables and given/known data
Given the lagrangian
$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}(x)F^{\mu\nu}(x) - \frac{1}{2\alpha}(\partial_{\mu}A^{\mu})^2$$
show that the momentum space photon propoagator is given by
$$D_F^{\mu\nu}(k) = \frac{-g^{\mu\nu} + \delta k^{\mu}k^{\nu}/k^2}{k^2 + i\epsilon}$$

2. Relevant equations
$$\delta = 1 - \alpha^{-1}$$

3. The attempt at a solution
I can solve this problem if I set
$$\delta = 1 - \alpha$$
but not with the delta stated in the book.

My question is this:

Should the book say $\delta = 1 - \alpha$ and not $\delta = 1 - \alpha^{-1}$?

This question and this question only. The meat of the answer will be one word.

2. Sep 3, 2013

### Vic Sandler

Yes, the text is a typo. It should say $\delta = 1 - \alpha$. This can be seen by referencing eqn (8.40) on page 154 in the second edition of the book "A First Book of Quantum Theory", by Lahiri & Pal. Eqn (8.40) is
$$D_{\mu\nu}(k) = -\frac{1}{k^2 + i\epsilon}[g_{\mu\nu} - (1 - \xi)\frac{k_{\mu}k_{\nu}}{k^2}]$$
and is the photon propagator when the lagrangian is given by eqn (8.11) on page 148 together with eqn (8.26) on page 152 to get
$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2\xi}(\partial_{\mu}A^{\mu})^2$$
which is the same as the eqn at the bottom of page 323 in Mandl & Shaw.

Last edited: Sep 3, 2013
3. Sep 11, 2013

### Bryson

I suppose you should give thanks to yourself!