Photon propagator in an arbitrary gauge

[kryshen]

My aim is to derive the photon propagator in an arbitrary gauge. I follow Itzykson-Zuber Quantum Field Theory and start from the Lagrangian with gauge-fixing term:
$${\cal L}(x) = -\,\frac{1}{4}\,F_{\mu\nu}(x)F^{\mu\nu}(x) -<br /> \,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu}(x))^2$$
I get the following equation of motion:
$$\left(\Box g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) \partial_\mu \partial_\nu \right) A^\nu (x) = 0$$

We can write the solution as a superposition of plane waves:
$$A^{\mu}(x) = \sum_{\lambda}\int \frac{d^3k}{(2\pi)^3 2\omega(\vec{k}\,)}\,<br /> \Big[c(\vec{k},\lambda)\,e^{\mu}( \vec{k},\lambda)\,e^{-ik\cdot x}<br /> + c^{\dagger}(\vec{k},\lambda)\,e^{*\mu}( \vec{k},\lambda)\,<br /> e^{+ik\cdot x}\Big],$$
where $$e^{\mu}$$ is a polarization vector.

Then I use the following commutation relations for the creation and annihilation operators to quantize the field:
$$[c(\vec{k},\lambda),\,c^{\dagger}(\vec{k}\,',\lambda\,')] <br /> = (2 \pi)^3\, 2 \omega (\vec{k})\, \delta^{(3)} (\vec{k}-\vec{k}\,')\,\delta_{\lambda \lambda\,'}$$
$$[c(\vec{k},\lambda),\,c(\vec{k}\,',\lambda\,')] <br /> = [c^{\dagger}(\vec{k},\lambda),\,c^{\dagger}(\vec{k}\,',\lambda\,')] = 0$$

I want to calculate the Green function as a vacuum expectation value of the time-ordered product:
$$-\,i\,D^{\mu\nu}(x - y) = \langle 0|{\rm T}(A^{\mu}(x)A^{\nu}(y))|0\rangle$$

To do this I need to derive the expression:
$$\sum_\lambda e^{\mu}(\vec k, \lambda) e^{*\nu} (\vec k, \lambda) = <br /> \left(g^{\mu\nu} - (1-\xi) \frac{k^\mu k^\nu}{k^2}\right)$$

I am novice at QFT. Please, could you advise on the methodology, how I could derive the expression for the sum of the products of polarisation vector components.

 

[lonelyphysicist]

$$\left(\Box g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) \partial_\mu \partial_\nu \right) A^\nu (x) = 0$$

You simply need to invert the quadratic differential operator above in momentum space (aka Fourier space).

In momentum space the photon kinetic term looks like

$$\int \frac{d^{4}k}{(2 \pi)^{4}}<br /> A^{\mu} (k) \left(k^{\beta} k_{\beta} g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) k_\mu k_\nu \right) A^\nu (-k)$$

The quadratic differential operator becomes a (0 2) tensor $M_{\mu \nu} \equiv \left(k^{\beta} k_{\beta} g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) k_\mu k_\nu \right)$, which you can invert by linear algebra methods.

 

[LAHLH]

I know this is an old post, but could anyone comment on how to invert the $$M_{\mu\nu}$$ matrix above by linear algebra methods?

 

[saaskis]

I know this is an old post, but could anyone comment on how to invert the $$M_{\mu\nu}$$ matrix above by linear algebra methods?

The inverse matrix has to be a linear combination of $$g^{\mu \nu}$$ and $$k^{\mu}k^{\nu}$$, so Ansatz $$Ag^{\mu \nu}+Bk^{\mu}k^{\nu}$$ should work.

 

[LAHLH]

Thank you, that did work, how did you know it had to be a linear combination of those things?

 

[saaskis]

Thank you, that did work, how did you know it had to be a linear combination of those things?

Well, what else could it be? :) To construct a Lorentz tensor with two indices, there really is no other choice. I cannot justify it any better than by noticing that it works.