Assume that I have the usual $U(1)$ gauge field $A_{\mu}$. We know that observable quantities are invariant under global transformations of the form $A_{\mu}\rightarrow A_{\mu}'=A_{\mu}+\partial_{\mu}\lambda$. One particular example of such an observable quantity is the electric and the magnetic field, right? So, $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ is invariant. This can be proved as well.

However, if I choose the very specific choice of $A_{x}=0$, for example, wouldn't that make my field strength components $F_{xt}$, for example equal to $F_{xt}=\partial_xA_t-\partial_tA_x=\partial_xA_t$? Suppose that I now perform a gauge transformation. What happens to my $F_{xt}$ component of the field strength? It will be transformed like that $$F_{xt}\rightarrow F_{xt}'=\partial_xA_t+\partial_x\partial_t\lambda-\partial_t\partial_x\lambda=F_{xt}$$ hence remaining the same as it was before performing the transformation. What if I was making a gauge choice of the form $A_t=0\Rightarrow F_{xt}=-\partial_t A_x$? Wouldn't that change the physics? Do the physics depend on my choice of gauge? What does the gauge choice has to do with gauge invariance? In the same way I have chosen $A_{\mu}=0$ for $\mu=x$, could I choose whatever form I wished for my $x$-component of the gauge field? Similarly, could I do that for every component? What is stopping me from choosing a gauge in which all the components are zero? How much freedom do I have in choosing a gauge and what are some gauge choices that are usually chosen?

Any help will be appreciated?


1 Answer 1


By postulating gauge invariance, you start by assuming that the physics does not depend on the choice of gauge. From there, the physical observables are constructed such as to be gauge invariant. Typically, you try to do it using gauge invariant objects like $F_{\mu\nu}$, but sometimes it is easier to use the potential and then check for gauge invariance.

The choice of the gauge is merely for mathematical convenience. The liberty of the gauge is determined by your equation of gauge transformation. For example, if you start from a potential $A$, you want to know whether you can fix $A_x = 0$. This amounts to choosing $\lambda$ st: $$ A_x+\partial_x\lambda = 0 $$ So any $\lambda$ of the form: $$ \lambda = -\int dxA_x+\lambda' $$ with $\lambda'$ independent of $x$. Therefore, your gauge fixing is possible, but does not fix the potential entirely.

If you try to fix the gauge by setting two components to $0$, then you'll be in trouble already. Say you want $A_x=A_t=0$. You therefore want: $$ A_x+\partial_x\lambda = A_y+\partial_y\lambda = 0 $$ However, this is only possible for the special fields satisfying: $$ \partial_x A_y = \partial_yA_x $$ A counter example for this would be $A=(0,y,-x,0)$. This gauge fixing is therefore not possible for all fields, it's invalid. A fortiori, fixing all the components to $0$ is not admissible either.

Two usual gauge fixing are the Lorentz gauge: $$ \partial\cdot A = 0 $$ which has the advantage of staying gauge invariant and transforms EM into uncoupled wave equations. However, there is a "leftover degree of freedom" in some sense, and makes quantisation trickier. The other is the Coulomb gauge: $$ \nabla\cdot \vec A = 0 $$ It is easier to quantise and the potential satisfies the same equations of electrostatic. However you lose relativistic invariance and the vector potential is harder to calculate.

A good way to build intuition is to Fourier space. Gauge invariance says that you can add something proportional to the four-wave vector. This is why in EM, you can restrict to the case of transverse waves.

Btw, classically, the full gauge invariance is rather: $$ A\to A+B $$ with $\partial_\mu B_\nu = \partial_\nu B_\mu$. Locally (in a simply connected neighbourhood), you can always find a $\lambda$ such that $B = \partial \lambda$, but this is not true globally. In contrast, gauge invariance in quantum mechanics is more restrictive as is is rather: $$ A\to A+\partial\lambda $$ with $\lambda$ defined globally up to an additive integer multiple of the inverse charge.

Hope this helps.

Answer to comment

You can view gauge freedom when you have a finite number of degrees of freedom. Say you want to describe a particle on a sphere subject only to the constraining force. You have two degrees of freedom, which you can keep track of by using an atlas of the sphere (spherical coordinates etc.).

Another way of approaching the problem, which is analogous to gauge freedom, is rather to say that the particle actually lives in 3D. However, since the real particle lives on the sphere you introduced an extra, non-physical degree of freedom (radial direction). All physical quantities will not depend on this extra degree of freedom by definition. However, looking at your new system, its symmetries are enhanced, which gives you more freedom to choose appropriate coordinates.

The analogue of when your gauge fixing is too restrictive would be to define a projection that is not defined for every point of the sphere. You would need to introduce more projections in order to cover the sphere completely.

  • $\begingroup$ Hi @lpz and thanks for the answer. You say at some point "This gauge fixing is therefore not possible for all fields, it's invalid". Why does the gauge fixing not being possible for all fields make the gauge choice invalid? This is what you are trying to say, right? Am I somehow allowed to fix a gauge, but at the same time I should be cautious to choose a gauge that it is appropriate for every single field (like $A_x=0$) ? And if so, why? $\endgroup$
    – schris38
    Mar 24, 2023 at 5:12
  • $\begingroup$ Remember that the end goal is to represent conveniently your field in all its possible configurations. Gauge fixing gets rid of superfluous degrees of freedom making them more economical. However, if they are too restrictive, you need to precise what to do when they are not applicable. In theory, this is not a problem since it is like assigning an atlas to a manifold. In practice though, it is too cumbersome (especially for quantisation). You should focus on some concrete examples where gauge fixing simplifies the math in order to see what is desirable. $\endgroup$
    – LPZ
    Mar 24, 2023 at 9:32
  • $\begingroup$ Hi and thanks again for the reply. Am I lead to understand that what we are usually after is making a gauge choice that fixes the "redundant" degrees of freedom, but not more than those degrees of freedom? And the sweet spot is the one that allows you to fix all the redundant ones (and not less of them)? And if I try to be naughty and I wish to fix more than the redundant degrees of freedom, it is not that it is not strictly allowed, but I have to pay the cost of not having gauge symmetry anymore? Is this the point of your comment? And I guess this generalizes to all gauge symmetric theories? $\endgroup$
    – schris38
    Mar 24, 2023 at 9:38
  • $\begingroup$ No, perhaps if I try to get rid of more degrees of freedom than the redundant ones, then I will have restricted my field to a certain form and the gauge symmetry will still hold (I think?). However, how is this a problem for quantization, or in any other case? $\endgroup$
    – schris38
    Mar 24, 2023 at 9:51
  • 1
    $\begingroup$ Going between the two gauges is not restrictive. To compare it with the analogy, this time say you describe the particle in a sphere with a 4D space. Each of the gauge fixing would correspond to restricting your attention to a 3d hyperplane. Your other gauge fixing $A_x=A_t=0$ corresponds to restricting to a 2D plane which does not cover all the equivalence classes representing a point of the sphere. $\endgroup$
    – LPZ
    Mar 24, 2023 at 12:55

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