As a simple example, take a constant magnetic field $\vec{B} = (0,0,B)$. This is invariant under rotations about the $z$ axis. However, we can express $\vec{B}$ as the curl of a vector potential $\vec{A}$:

$$\vec{A} = \begin{pmatrix} 0 \\Bx \\ 0 \end{pmatrix} + \nabla \lambda$$

for any scalar function $\lambda$. The vector potential is not invariant under rotations about the $z$ axis: if we rotate the gauge field by an angle $\theta$ about the $z$ axis the resultant $\vec{B}$ field is $(0, 0, B \cos \theta)$.

Though I've used a toy case, I'm interested in the case of a more general theory: is there any way to see the symmetry of the observables from the gauge field?


The relation of magnetic field and the vector potential is independent of the coordinate system. My first attempt I got the same result, however, soon I realized that the coordinates also have to undergo the same rotation. So if one has the following vector potential

$$ \mathbf{A}(x,y,z) = \left( \begin{array}{c} 0, & Bx, & 0 \end{array}\right)^T $$

so rotating it by the well-known rotation matrix gives: $$\mathbf{A'}(x,y,z)=\left(\begin{array}{c} Bx \sin\theta, & Bx \cos\theta, & 0 \end{array}\right)^T $$

For the curl computation, the curl has to be computed in the rotated coordinates. The rotated coordinates are:

$$ x' = \cos\theta x + \sin\theta y \quad \text{and} \quad y' = -\sin\theta x + \cos\theta y$$

Actually we need the inverse relation as we want to express the old coordinates by the new ones (fortunately we need it only for x):

$$ x = \cos\theta x' - \sin\theta y' $$

We now express the rotated vector potential by the new coordinates:

$$ A'(x',y',z') = ( \begin{array}{c} B(\sin\theta \cos\theta x' - \sin^2 \theta y'), & B(\cos^2\theta x' - \sin\theta\cos\theta y'), & 0 \end{array} )^T$$

If we take now the curl of $\mathbf{A}$ we get the desired result:

$$ \mathbf{B'}=\nabla' \times A'(x',y',z') = \left( \begin{array}{c} 0, & 0, & B(\cos^2\theta + \sin^2\theta )\end{array} \right)^T$$

So the raised question has nothing to do with gauge invariance, it is all about rotational invariance.

  • $\begingroup$ Ah yes, of course, I forgot that the curl operator transforms as well. Actually I think this answer gives a link to gauge invariance too---I'm adding a self-answer to add to yours. $\endgroup$ – DavidH Oct 6 '20 at 10:02

Following Frederic Thomas's answer pointing out the error in my example of not accounting for rotation of the coordinates, I am now able to answer the question of the constraint on the gauge field:

Under the symmetry transformation $x \to x'$, $A(x) \to A'(x')$ may be expressed $A(x') + \nabla \lambda(x')$.

Using Frederic Thomas's expression for $A'(x')$ we can see that

$$A'(x') = B(\sin \theta \cos \theta \, x' - \sin^2 \theta y', \cos^2 \theta x' - \sin \theta \cos \theta \, y', 0)^T = B(0, x', 0)^T + \nabla \lambda(x')$$


$$\lambda(x') = \sin^2 \theta \, x' y' - \sin \theta \cos \theta (x'^2 + y'^2).$$


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