# Gauge fixing for spin 1 field

Consider the Lagrangian for a spin 1 massless field with a gauge fixing term: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}^2-\frac{1}{2\xi}\left(\partial_\mu A^\mu\right)^2-J_\mu A^\mu,$$ where $\xi$ acts as a Lagrange multiplier. Writing down the equation of motion for $A^\mu$ and inverting the corresponding operator leads to the following expression for the photon propagator: $$i\Pi^{\mu\nu}(p)=\frac{-i}{p^2+i\epsilon}\left[g^{\mu\nu}-\left(1-\xi\right)\frac{p^\mu p^\nu}{p^2}\right].$$ Here, each different choice for $\xi$ corresponds to a different gauge.

On the other hand, going back to the Lagrangian, calculating the equation of motion for $\xi$ and excluding the case $\xi\rightarrow\infty$ gives $$\partial_\mu A^\mu=0,$$ which is the Lorenz gauge condition.

So it seems that, while I can choose different gauges for the propagator, I am enforcing a specific gauge condition with the equations of motion. How to solve this apparent contradiction? Is it because the fields are not always on shell, i.e. the classical equations of motion are not always satisfied?

Thanks.

• – Qmechanic Mar 9 '17 at 21:43

The parameter $\xi$ is not a Lagrange multiplier, despite the fact that many books insist that it is. It behaves similarly to a Lagrange multiplier, but if it actually were one, then it would enforce $\partial\cdot A\equiv0$ as an operator equation, something we certainly do not want. $\xi$ is not one of the phase space variables of the theory, which means that you cannot write down equations of motion for it - you cannot vary the action with respect to $\xi$. It is a fixed parameter, not a degree of freedom.

The parameter $\epsilon$ doesn't correspond to the choice of a gauge. It is just arbitrary parameter on which our theory is completely independent. Your gauge fixing condition is set by the functional in the front of $\frac{1}{\epsilon}$.

• What do you mean? To my understanding, the $\epsilon$ parameter is just a mathematical trick that specifies the prescription for the time ordering, while each different value of $\xi$ corresponds to a different gauge. – gabo_18 Mar 9 '17 at 22:13

It is a historical reason to call the different choice of $\xi$ a choice of gauge. Indeed the different $\xi$ corresponds to the different gauge parameter $\lambda$ for

$$\partial_\mu A^\mu = \lambda$$

R$_\xi$ gauge.
• This is wrong. $\partial\cdot A = \lambda$ does not hold as an operator equation (the correct equation is $\xi^{-1}\partial^2(\partial\cdot A)=0$), and it neither holds in the Faddeev-Popov gauge fixing (the correct constraint is $\partial\cdot A=f(x)$, where $f$ is an arbitrary function over which you integrate). – AccidentalFourierTransform Mar 10 '17 at 12:05