7
$\begingroup$

Given any electric and magnetic field (or $F_{\mu\nu}$), there is always some freedom in defining what $A_\mu(x)$ should be. In fact, there are infinite choices for $A_\mu(x)$. This is because for an arbitrary function $\theta(x)$ $$A_\mu^\prime=A_\mu+\partial_\mu\theta(x)$$ is also a possible choice of $A_\mu$. The gauge condition $\partial^\mu A_\mu=0$, however, selects a $A_\mu(x)$ from an infinite set of $A_\mu(x)$. This is what I think should be called the gauge-fixing.

But in quantizing a radiation field, we add a term $-\frac{1}{2\xi}(\partial_\mu A^\mu)^2$ to the Lagrangian. This is called adding a gauge-fixing term.

  • Please help me understand what does this step really mean. Are we just adding zero to the Lagrangian (because the added term is zero in the Lorenz gauge)? If so, why? What's the point in adding a zero to the Lagrangian?

  • If we are adding this as a nonzero term, to begin with, what gives us the right to add this term in the first place which is not a divergence (but the square of a divergence) term?

As expected, this does change Maxwell's equation with the gauge fixing term $-\frac{1}{2\xi}(\partial_\mu A^\mu)^2$ added to the Lagrangian in the limit of arbitrary but finite $\xi$. The equation for $A_\mu$ and $F_{\mu\nu}$ is changed from $$\Box A^\nu=j^\nu,~~\partial_\mu F^{\mu\nu}=j^\nu$$ to $$\Box A^\nu-(1-\frac{1}{\xi})\partial^\nu(\partial_\mu A^\mu)=j^\nu,~~\partial_\mu F^{\mu\nu}-(1-\frac{1}{\xi})\partial^\nu(\partial_\mu A^\mu)=j^\nu.$$

$\endgroup$
1
3
$\begingroup$

There is a trick to derive this in path-integral formulation of QED by introducing an auxilliary field $\pi(x)$ (See Schwartz's QFT 14.5 on p. 267 for detailed explanation).


Consider following path-integral: $$ f = \int {\cal{D}} \pi \; exp \left\{ -i \int d^4x \; \frac{1}{2 \xi} \left( \Box \pi \right)^2 \right\}. $$ As with other integrals we can shift the integrated variable without affecting the value of the integral (since measure doesn't change under the shifts): $$ \pi \rightarrow \pi - \frac{1}{\Box} \left( \partial_\mu A^\mu \right). $$ Even though after this transformation $f$ becomes a functional of $A^\mu$: $$ f\left[ A \right] = \int {\cal{D}} \pi \; exp \left\{ -i \int d^4x \; \frac{1}{2 \xi} \left( \Box \pi - \left(\partial_\mu A^\mu\right) \right)^2 \right\}, $$ its value is evidently still independent of $A^\mu$.

For the next step we can take the QED path-integral correlation function: $$ \langle \cdots \rangle = \frac{1}{Z[0]} \int {\cal{D}} A^\mu \; exp \left\{ i \int d^4x \; {\cal{L}}(A)\right\} \cdots $$ and multiply and divide the integrand by $f\left[A\right]$ without affecting the result (since the value of $f$ is a constant independent of $A$): $$ \langle \cdots \rangle = \frac{1}{Z[0]}\frac{1}{f} \int {\cal{D}} A^\mu {\cal{D}} \pi \; exp \left\{ i \int d^4x \left[ {\cal{L}}(A) - \frac{1}{2\xi} \left( \Box \pi - \left( \partial_\mu A^\mu \right) \right)^2 \right]\right\} \cdots. $$

Then we can apply a gauge transformation $A^\mu \rightarrow A^\mu + \partial^\mu \pi$ to get rid of the $\pi$ in the integrand: $$ \langle \cdots \rangle = \frac{1}{Z[0]} \frac{1}{f} \left( \int {\cal{D}} \pi \right) \int {\cal{D}} A^\mu \; exp \left\{ i \int d^4x \left[ {\cal{L}}(A) - \frac{1}{2\xi} \left( \partial_\mu A^\mu \right)^2 \right]\right\} \cdots. $$

Note that $\frac{1}{f} \left( \int {\cal{D}} \pi \right)$ is a constant (albeit an infinite one) and does not affect the value of $\langle \cdots \rangle$ because the very same constant has to be present in $Z[0]$. Hence ${\cal{L}}$ and ${\cal{L}} - \frac{1}{2 \xi} \left(\partial_\mu A^\mu\right)^2$ are physically equivalent.


What we have done is sometimes called "integrating out the gauge orbit". The idea is that the gauge invariance reduces the number of degrees of freedom of the theory. When we integrate over all possible field configurations - we inevitably count "twice" configurations that are related to each other by gauge transformations. There is an infinite number of equivalent configurations, hence we get an infinite multiplier: $\int {\cal{D}} \pi$ in front of the path-integral.

$\endgroup$
1
$\begingroup$

Classically first: The Langrangian is a starting point for a variational principle. Formally, when you take the variational derivatives that yield the equations of motion, you're taking it over states that may or may not obey the constraint equations.

If you've got a proper Lagrangian, the constraint equations fall out of this process when you take the variations with respect to the Lagrange multipliers. That's an output of the theory, even though in many cases you put terms into your Langrangian just to make this happen. (Not always - in general relativity, some elements of the 4-metric make this happen in a way that I don't think would have been obvious from the Lagrangian.)

As you noted, you are effectively adding a term that equals zero to the Lagrangian. On one hand, that's why you're allowed to do it. On the other hand, that makes is seem strange. The reason this matters is that this term can be non-zero for variations in directions that don't obey the constraints. In other words, you're using prior knowledge of what the constraint will be to add this term, but formally you don't get to know that answer until after you've done the variational derivatives.

Note that even classically this can matter for things like numerical stability of solutions. Numerical errors don't necessarily obey the constraint equations, and so your approximate solution can depend on the non-physical, constraint-violating modes that are influenced by these "zeros". (Remember, they don't have to be zero once you stray from the constrained portion of the space.)

I have way less insight into exactly why the term that you suggested is added in the quantum context that brought it to you attention, but I imagine that it could at least partially be the same. Quantum theory may let you jump off of the constraint-hypersurface of the classical theory for short times, so the form of this term may matter to the theory there. In addition, it may serve some practical role with how and at what point in the derivation you quantize the theory.

$\endgroup$
1
$\begingroup$

OP seems mainly concerned about what happens classically at the level of equations of motion when we alter the E&M Lagrangian density $$ {\cal L}_0~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \mp j^{\mu}A_{\mu} \tag{A}$$ with a gauge-fixing term$^1$ $$ {\cal L}~=~{\cal L}_0 - \frac{1}{2\xi} (d_{\mu} A^{\mu})^2. \tag{B}$$ Here we are using the Minkowski sign convention $(\pm,\mp,\mp,\mp)$. It should be stressed that the gauge-fixing term is generically non-zero, so OP's question is non-trivial. Also we should emphasize that the gauge-fixed Lagrangian density ${\cal L}$ is no longer gauge invariant.

OP already mentions that the Euler-Lagrange (EL) equations is$^2$ $$ \pm j^{\nu}~\approx~d_{\mu} F^{\mu\nu}+\frac{1}{\xi}d^{\nu}d_{\mu}A^{\mu} ~=~ \Box A^{\nu}+\left(\frac{1}{\xi}-1\right)d^{\nu}d_{\mu}A^{\mu}. \tag{C}$$ The gauge-fixing means that the EL eqs. (C) are no longer gauge invariant with a gauge-orbit worth of $A_{\mu}$ solutions. We can formally solve for $A_{\mu}$ (ignoring the kernel of the d'Alembertian $\Box$). The formal classical solution reads $$ \pm A_{\mu}~\approx~ \Box^{-1}j_{\mu} + (\xi-1) d_{\mu}\Box^{-2}d_{\nu}j^{\nu}.\tag{D}$$ Now in order for the original Lagrangian density ${\cal L}_0$ to be gauge invariant (up to a total space-time derivative term), we must impose that the matter current $j^{\nu}$ satisfies a continuity equation $$ d_{\nu}j^{\nu} ~=~0,\tag{E}$$ so that the second term in the solution (D) is zero. (Even if the second term is non-zero, it is of the form of a gauge transformation.) We conclude that the presence of the gauge-fixing parameter $\xi$ does not physically change the the classical solution (D).

References:

  1. H. Motohashi, T. Suyama & K. Takahashi, Phys. Rev. D 94 (2016) 124021, arXiv:1608.00071.

--

$^1$ Note that the Lorenz gauge-fixing function does not fix massless modes, i.e. modes in the kernel of the d'Alembertian $\Box$.

$^2$ Here the $\approx$ symbol means equality modulo eqs. of motion.

$\endgroup$
1
  • $\begingroup$ Correction to the answer (v3): the the should be the. $\endgroup$ – Qmechanic Jan 30 '20 at 6:24
-2
$\begingroup$

First note the Lorenz gauge does not completely fix the potential, but enough to make the equation of motion invertible.

Gauge fixing of the lagrangian is necessary because a gauge invariant equation of motion can not be inverted to yield a propagator.

$\endgroup$
0

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.