Consider the theory with Lagrangian

$$ \mathcal{L} = -\frac{1}{4}F_{\mu \nu} F^{\mu \nu} + (D_\mu \phi)^* (D^\mu \phi) - U(\phi) \,, $$ where $U(\phi)$ breaks the $U(1)$ symmetry of the system. If we were working in an ungauged theory, in which the $U(1)$ symmetry is global, we would have no problem claiming that there are infinitely many distinct vacuum states, all with the same energy, and labelled by an angle $\theta$.

My question is simple:

Is the same true in the gauged theory? Are there states $|\theta \rangle$ such that $\langle \theta' | \theta \rangle = 0$ for $\theta \neq \theta'$, with $\langle \psi |H|\psi\rangle \geq \langle \theta |H|\theta\rangle$ with equality iff $|\psi\rangle = |\theta'\rangle$?

This question seems essentially equivalent, to me at least, to the question of whether the global transformations in a gauged theory remain legitimate physical transformations, as they are in the ungauged theory, or instead join the local transformations as "redundancies of description".

On the one hand, I want to argue that global transformations should remain "physical" in the gauged theory, and so the vacuum should be degenerate. For instance, if the symmetry weren't spontaneously broken, we would take $\phi$ (being complex) to have two degrees of freedom, with the theory being symmetric under rotation of these degrees of freedom amongst themselves.

On the other hand, we know that there are no massless excitations in the spontaneously broken theory, and perhaps this is suggesting to us that the vacuum state isn't connected to any other states of the same energy. So in fact the vacuum is unique.

PS: if anybody could suggest a resource that discusses the subtleties of gauge transformations, in terms of which are physical and which are not, how gauge transformations must behave at infinity, and how the picture differs in the classical and quantum theory, it would be much appreciated!

  • 1
    $\begingroup$ I was just thinking about this yesterday! My impression is: yes, there are infinitely many degenerate vacua, and choosing one breaks the global $U(1)$ symmetry. However we retain the gauge $U(1)$ symmetry which only applies for gauge parameters $\alpha \to 0$ at infinity. $\endgroup$ – knzhou Mar 1 '18 at 14:16
  • $\begingroup$ Quantitatively let $j^\mu(x)$ be the conserved current of the global $U(1)$ symmetry. The total charge $Q$ still generates the global symmetry and moves between vacua. Goldstone bosons would instead be created by acting with $j^\mu(x)$ (or a wavepacket made from it, etc.) but since this is a local operation, it is a do-nothing transformation by the gauge symmetry. $\endgroup$ – knzhou Mar 1 '18 at 14:18
  • $\begingroup$ @knzhou Thanks for the comment. Could you explain a bit further what you mean by "acting with $j^\mu(x)$ is a local operation, and is hence a do-nothing operation"? $\endgroup$ – gj255 Mar 1 '18 at 15:32
  • $\begingroup$ On the level of classical field theory, $j^\mu(x)$ acts on the field by Poisson bracket and has the effect of doing a local $U(1)$ rotation at $x$. But this is exactly what a gauge transformation does, so the physical state remains the same; that's why there isn't a Goldstone boson. $\endgroup$ – knzhou Mar 1 '18 at 15:55
  • $\begingroup$ This article discuss this question (from a cond-mat point of view, but that's pretty much the same problem) : arxiv.org/abs/cond-mat/0503400v1 $\endgroup$ – Adam May 3 '18 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.