For $\mathbb{Z}_2$ symmetry breaking in a Classical Field Theory described by a potential $$V(\phi)=\lambda(\phi^2-v^2)^2,\tag{1}$$ there is a finite energy barrier of height $\epsilon=\lambda v^4$ that exists between two minima $\phi=\pm v$ of $V(\phi)$ separated by a maximum at $\phi=0$. This can be seen clearly from the picture of a double-well potential. This finite potential barrier when multiplied with the infinite volume of space gives rise to infinite (potential) energy barrier. To be explicit, the infinite energy barrier comes from $$[V(0)-V(\pm v)]\int d^3x=\epsilon\int d^3x\to \infty!\tag{2}$$

But for $U(1)$ symmetry breaking, the potential energy barrier cannot be seen between two neighbouring vacua from the picture of Mexican hat potential. According to (1), this would mean by the potential barrier between two minima can be finite! But probably changing the vacuum costs infinite energy. If so, does it mean that the "Mexican hat" potential give a misleading impression? Why does the figure incapable of showing the actual barrier?

  • 2
    $\begingroup$ Why would it give a "misleading impression"? There's nothing misleading about stating that there is no potential barrier between the vacua of the Mexican hat. $\endgroup$ – ACuriousMind Oct 27 '18 at 14:56
  • 1
    $\begingroup$ It would have been good if you had made that thought explicit in your question (and if you would also make explicit why you expect "infinite energy cost, and for what. I mean, I can probably guess what you mean, but please don't assume that all readers can immediately follow such an implicit train of thought). In that case, physics.stackexchange.com/q/331024/50583 seems very much related. $\endgroup$ – ACuriousMind Oct 27 '18 at 15:01
  • 3
    $\begingroup$ Zero times an infinity is an undefined term and can give various answers depending on the limits $\endgroup$ – Triatticus Oct 27 '18 at 15:46
  • 1
    $\begingroup$ I'm just saying that was an incorrect statement is all. $\endgroup$ – Triatticus Oct 27 '18 at 15:52
  • 1
    $\begingroup$ @ACuriousMind I don't think that question is really related. This question is about spontaneously breaking continuous global symmetries, and that one is about spontaneously breaking continuous gauge symmetries, which is conceptually a very different issue. In fact, in my experience most experts tend to now think that it's misleading to think of gauge symmetries as ever being "broken", e.g. because of Elitzur's theorem, while global symmetries can inarguably be broken. $\endgroup$ – tparker Oct 28 '18 at 17:11

Good question. You are correct that, in principle, it is possible to rigidly, uniformly, and adiabatically rotate the entire field configuration by an identical angle in such a way that you always stay exactly inside the degenerate ground-state manifold. This gives the misleading impression that you can tunnel from one ground state to another with zero energy cost, as you suggest in your question.

But in practice, this isn't really realistically physically possible. That's because of the local nature of the thermal or quantum fluctuations. For concreteness, imagine the $XY$ model $$H = J \sum_{\langle i, j\rangle} \cos(\theta_i - \theta_j)$$ on a hypercubic lattice in $D \geq 3$ spacetime dimensions, in which the global rotational symmetry gets simultaneously broken. You can't realistically set up a situation in which every spin in the entire lattice rotates from pointing toward angle $\theta_i$ to angle $\theta_f$ exactly simultaneously, in one giant nonlocal perturbation. (For one thing, such a "perturbation" would require an infinite amount of time to set up, since the "planning message" about what angle to rotate by would need to propagate arbitrarily far before the process actually begins.) In practice, all you can do is first rotate a small little local cluster of spins from $\theta_i$ to $\theta_j$, then rotate that cluster's neighboring spins, then the next neighboring spins, etc. That's because by assumption, in a local QFT (or lattice system), the microscopic couplings are purely local, so nonlocal "update moves" (in Monte Carlo language) are not allowed.

Eventually, the effect of the initial perturbation will have propagated out far enough that a huge region of spacetime will have rotated its angle, so that to a good approximation you're locally in a new ground state. But the point is that as you do this process, at any given instant there will always be a boundary between the old and new ground states (kind of like a domain wall, but less sharply localized). That boundary will need to move and hit every spin in order for the entire system to rotate. So even if the rotation angle is infinitesimal and each individual spin's rotation only costs energy $\epsilon J$, the total energy of the entire tunneling process will still get multiplied by the volume of spacetime and will be a huge $\epsilon J V$.

Thought of another way: you can create a Goldstone mode with arbitrarily low energy by delocalizing it enough, but in practice it's impossible to create a Goldstone mode with exactly zero energy. The difference is subtle, but conceptually important.

  • $\begingroup$ In your spin model, I understand that each spin flip causes finite energy cost and multiplying the volume of space costs infinite energy cost. In a continuum field theory, described by a Mexican hat potential where do you get this infinite energy from? For the discrete $Z_2$ case, I've explained that a finite potential barrier at each space(time) point give rise to infinite energy barrier when multiplied by space(time) volume. But in the continuous $U(1)$ case, the barrier need not be infinite because we are multiplying zero and infinity (instead of a finite barrier and infinity!) @tparker $\endgroup$ – SRS Oct 28 '18 at 3:32
  • $\begingroup$ @SRS Please read my answer. $\endgroup$ – tparker Oct 28 '18 at 3:39
  • $\begingroup$ I read your answer. Hence the question regarding the 'third paragraph' of your answer. Please see Eq. (2) of my question. In the continuum case, $\epsilon=0$. That, when multiplied by infinite spacetime volume, need not give infinite (potential) energy cost but may give a finite value only. Another source of energy cost can be gradient energy which can be arbitrarily small for a long wavelength Goldstone mode. @tparker $\endgroup$ – SRS Oct 28 '18 at 3:50
  • $\begingroup$ I don't disagree with your answer. It seems very much plausible. In fact, I completely understand and agree with the first and second paragraph of your answer. I'm concerned about what causes infinite energy cost in the latter case. For the discrete case, it is the finite potential barrier multiplied by infinite spacetime volume giving rise to infinite energy cost. @tparker $\endgroup$ – SRS Oct 28 '18 at 4:06
  • 1
    $\begingroup$ @SRS The gradient energy cost prevents the total energy from being exactly zero. That's what I was getting at with my discussion of moving boundaries. In order to not have any gradient energy, you would need to simultaneously shift the entire field value identically over all of spacetime, which is a nonlocal and therefore physically impossible operation. The gradient cost can be made arbitrarily small by delocalizing the Goldstone mode more and more, but it can never be made exactly zero using local operations. So $\epsilon$ is very small but is not equal to zero. $\endgroup$ – tparker Oct 28 '18 at 17:02

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.