# Virtual particle propagators in QFT

1. Jul 15, 2011

### Matthaeus

I am reading a nice book (Quarks and Leptons, by Halzen and Martin) about particle physics. It states that the general form of the propagator of a virtual particle is:
$$\dfrac{i\sum_{\text{spins}}}{p^2 - m^2}$$

I see that this is the case for the Dirac propagator:

$$\dfrac{i(\displaystyle{\not}{p} + m)}{p^2 - m^2} = \dfrac{i\sum_{s}u_s(p)\bar{u}_s(p)}{p^2 - m^2}$$

but how can I prove that always holds?

2. Jul 15, 2011

### Bill_K

Matthaeus , I agree, Halzen and Martin is a very well written book, and highly useful. Although it does tend to be a bit informal, and you are not going to catch them "proving" anything. So in the same spirit, here is not a proof but an argument.

Let's say you have an operator L, which is generally a differential operator, and in particular will be the Klein-Gordon operator. You want to find the field |φ> at one point produced by a source |S> at another. The equation is L|φ> = |S>. The solution can be written formally as |φ> = L-1|S>. Expand L in terms of its eigenstates, namely L = ∑|n>Ln<n| where |n> are the eigenstates and Ln the eigenvalues. Then L-1 = ∑|n>Ln-1<n|. Thus |φ> = G|S> where G is the Green's function or propagator, G = ∑|n>Ln-1<n|.

For the Klein-Gordon operator the eigenstates are the plane wave solutions, Ln = k2 + m2, and the numerator ∑|n><n| is the sum over spins. Halzen and Martin go on to write this out in more detail for the only cases that are physically realistic: spin 0, 1/2 and 1.

3. Jul 16, 2011

### Matthaeus

Yes, I was also thinking about something along these lines.
Thanks.