My question concerns the gauge fixing in classical v.s. quantum $U(1)$ gauge theory. I will ask about the gauging fixing in quantum $U(1)$ gauge theory in a separated Phys-SE post.

For the classical $U(1)$ gauge theory,

we have the electric $\vec E$ and magnetic $\vec B$ written as the scalar $\phi $ and vector $\vec A$ potentials: $$ \vec E = - \vec \nabla \phi -\frac{\partial}{\partial t} \vec A. $$ $$ \vec B = \vec \nabla \times \vec A. $$

My understanding about the gauge fixing for this gauge theory is that we can chose for example, $(\phi, \vec A)$ to give a set of $\vec E$ and $\vec B$ fields.

  1. But we can also shift to $$(\phi, \vec A) \mapsto (\phi + C_0, \vec A + \vec C)$$ where arbitrary choices of $(C_0, \vec C)$ still give the same solutions of $\vec E$ and $\vec B$. So a certain but arbitrary choice of $(C_0, \vec C)$ can be regarded as a way of gauge fixing? correct?

  2. Furthermore, we can also shift to $$(\phi, \vec A) \mapsto (\phi + \phi_0, \vec A + \vec A')$$ such that the followings are satisfied: $$ - \vec \nabla \phi_0 -\frac{\partial}{\partial t} \vec A'=0 $$ $$ \vec \nabla \times \vec A'=0 $$ Then we have a choice of $(\phi, \vec A) \mapsto (\phi + \phi_0, \vec A + \vec A')$ such that any choice is a way of gauge fixing? correct?

  3. In a classical differential equation, we have $$ * d * F = J $$ $$ dF=0 $$ so $F=dA$ with $* d * dA = J$ is a solution. Say $F=dA'$ with $* d * dA' = J$ is also a solution. And the gauge fixing implies a different solutions of $F=dA$ and $F=dA'$ where $A$ and $A'$ are both valid solutions. What are the differential equation constraints then? Is my understanding complete to include ONLY: $$ * d * d (A-A')=0 $$ or do we need more constraints to do gauge fixing?

Am I correct to say that $A$ and $A'$ are in the same gauge profile thus should be regarded as the same gauge equivalent classes. Choose $A$ or choose $A'$ is simply a choice of gauge fixing?


1 Answer 1

  • A gauge transformation is any transformation on $A = (\phi, \vec{A})$ that does not change any physical observable, c.q. $F$. So any transformation $A \to A + d \lambda$ with $\lambda$ any scalar field.
  • A gauge fix is, within this gauge freedom, a particular choice for $A$. This choice can be either complete ($A$ completely fixed, no further gauge transformations allowed) or not (there is still some leftover freedom).

A gauge fix is usually applied to facilitate certain computations (you have to then make sure the end result is nonetheless gauge-invariant). Therefore, one is not actually interested in the exact particular form of $A$, but rather in certain gauge-fixing conditions that $A$ satisfies. Given any gauge potential $A$, one can apply a gauge transformation $A \to A_g = A + d \lambda$ so that $A_g$ satisfies the gauge-fixing condition.

Some examples are

  • In the Lorenz gauge fix, the condition $d \star A_g = (1/c^2)\partial_t \phi_g + \nabla \cdot \vec{A}_g = 0$ is satisfied.
  • In the Coulomb gauge fix, the condition $\nabla \cdot \vec{A}_g = 0$ is satisfied.

The Lorenz gauge fix is not complete, because instead of using $\lambda$ to arrive at the gauge fix, we could have used any $\lambda + \psi$ as long as $\psi$ satisfies $\Box \psi= 0$. In other words, instead of $A_g$ we can have $A_g + d \psi$ and still satisfy the gauge-fixing condition. On the other hand, the Coulomb gauge fix is complete.

In your list, the transformation $C$ in item 1. is not entirely free to choose. Instead, it must be of the form $C = d \lambda$ for any smooth scalar field $\lambda$. Then, items 1. and 2. are actually the same. I would use a different wording than what you have written down: $C$ is a gauge transformation. A particular transformation $C$ to arrive at $A$ satisfying a gauge-fixing condition can be called a gauge-fix.

I don't understand your question item 3. If $F = d A$ and $F = d A'$ then $A$ and $A'$ differ by some gauge transformation. A gauge fix is some condition on $A$, but it cannot affect any physical observable, in particular not $F$. So $d (A - A') = 0$ is always satisfied. Moreover, physical observables like $F$ and $J$ can tell you nothing about the gauge fix. Instead, a gauge-fixing condition (which is a choice) can be for instance $d \star A = 0$ (Lorenz gauge).

  • $\begingroup$ Edit in my post: "so $F=dA$ with $* d * dA = J$ is a solution. Say $F=dA'$ with $* d * dA' = J$ is also a solution." $\endgroup$ Commented Apr 15, 2021 at 3:29
  • $\begingroup$ I still don't understand the question, but I've tried to emphasize that gauge transformations do not and cannot (by definition) affect any observable properties. If $\star d \star d(A - A') =0$ but $d(A-A') \neq 0$ then $A$ and $A'$ are not gauge-equivalent. $\endgroup$ Commented Apr 15, 2021 at 3:54

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