Photon propagator in an arbitrary gauge

In summary: 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.
  • #1
kryshen
7
0
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:
[tex]
{\cal L}(x) = -\,\frac{1}{4}\,F_{\mu\nu}(x)F^{\mu\nu}(x) -
\,\frac{1}{2\xi}\,(\partial_{\mu}A^{\mu}(x))^2
[/tex]
I get the following equation of motion:
[tex]
\left(\Box g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) \partial_\mu \partial_\nu \right) A^\nu (x) = 0
[/tex]

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

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

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

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

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.
 
Last edited:
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  • #2
[tex]
\left(\Box g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) \partial_\mu \partial_\nu \right) A^\nu (x) = 0
[/tex]

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

[tex]
\int \frac{d^{4}k}{(2 \pi)^{4}}
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)
[/tex]

The quadratic differential operator becomes a (0 2) tensor [itex]M_{\mu \nu} \equiv \left(k^{\beta} k_{\beta} g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) k_\mu k_\nu \right)[/itex], which you can invert by linear algebra methods.
 
  • #3
I know this is an old post, but could anyone comment on how to invert the [tex] M_{\mu\nu} [/tex] matrix above by linear algebra methods?
 
  • #4
LAHLH said:
I know this is an old post, but could anyone comment on how to invert the [tex] M_{\mu\nu} [/tex] matrix above by linear algebra methods?
The inverse matrix has to be a linear combination of [tex]g^{\mu \nu}[/tex] and [tex]k^{\mu}k^{\nu}[/tex], so Ansatz [tex]Ag^{\mu \nu}+Bk^{\mu}k^{\nu}[/tex] should work.
 
  • #5
Thank you, that did work, how did you know it had to be a linear combination of those things?
 
  • #6
LAHLH said:
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.
 

1. What is a photon propagator in an arbitrary gauge?

A photon propagator is a mathematical representation of the probability amplitude for a photon to travel from one point to another in a given gauge. It takes into account the gauge fixing condition, which is a mathematical constraint used to simplify calculations in gauge theories.

2. Why is the photon propagator important in physics?

The photon propagator is important in physics because it allows us to calculate the probability of interactions between particles that involve the exchange of photons. This is crucial in understanding the behavior of fundamental particles and the forces that govern their interactions.

3. How is the photon propagator affected by the choice of gauge?

The photon propagator is affected by the choice of gauge through the inclusion of the gauge fixing condition. Different gauges will result in different forms of the propagator, but they will all yield the same physical predictions.

4. Can the photon propagator be experimentally verified?

Yes, the predictions of the photon propagator can be experimentally verified through high-energy particle collisions. The results of these experiments have confirmed the accuracy of the propagator and its role in describing the behavior of fundamental particles.

5. Are there any limitations to using the photon propagator in an arbitrary gauge?

One limitation of using the photon propagator in an arbitrary gauge is that it can become very complicated and difficult to calculate in certain gauges. In these cases, approximations and other mathematical techniques may need to be employed to simplify the calculations.

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