1
$\begingroup$

I'm going over Greiner's field quantization book.

In chapter 7 quantization of a massless spin 1 field, the book says in the Lorenz gauge, we have equation 7.26:

$$\hat{A}^\mu(x)=\int\frac{d^3k}{\sqrt{2\omega_k(2\pi)^3}}\sum_{\lambda=0}^3\big(\hat{a}_{\boldsymbol{k}\lambda}\epsilon^\mu(\boldsymbol{k},\lambda)e^{-ik\cdot x}+\hat{a}_{\boldsymbol{k}\lambda}^\dagger\epsilon^\mu(\boldsymbol{k},\lambda)e^{ik\cdot x}\big).\tag{7.26}$$

here $\epsilon^\mu(\boldsymbol{k},\lambda)$ are certain chosen polarization vectors.

In the Coulomb gauge, we have $A_0(\boldsymbol{x},t)=0$ and equation 7.105

$$\hat{\boldsymbol{A}}(\boldsymbol{x},t)=\int\frac{d^3k}{\sqrt{2\omega_k(2\pi)^3}}\sum_{\lambda=1}^2\boldsymbol{\epsilon}(k,\lambda)(\hat{a}_{\boldsymbol{k}\lambda}e^{-ik\cdot x}+\hat{a}_{\boldsymbol{k}\lambda}^\dagger e^{ik\cdot x}\big).\tag{7.105}$$

Here $\epsilon^\mu(\boldsymbol{k},\lambda)$ are another set of polarization vectors.

My questions are: which one should I use in general? How do we see that they give the same answer?

$\endgroup$

1 Answer 1

3
$\begingroup$

One may show that gauge-invariant physical observables of a gauge theory (in this case QED) does not depend on the specific gauge-fixing condition (e.g. Lorenz gauge, Coulomb gauge, etc). See e.g. my related Phys.SE answer here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.