- #1

kryshen

- 7

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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.

[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.

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