16
$\begingroup$

This question is about how the ground state is chosen in a spontaneous symmetry breaking process. Say we have a Mexican Hat potential (e.g. the one for the Higgs field) and are sitting at the unstable equilibrium in the middle. My lecturer said that then the "state would decay into" a random one of the stable equilibria that are located on a circle around the origin. My question is now how exactly this "random" direction the symmetry is broken into is determined.

To illustrate my problem: For the equivalent process in a superconductor I understand at least conceptually how that could happen. We have some thermal motion that is a statistical process, so the random choice of the direction of symmetry breaking really just comes from the chaotic dependence on the thermal initial conditions. What I don't understand is where such a random process could come from in the case of particle physics since I thought of the fields as fundamental.

But maybe I am completely misunderstanding and such a decay from an unstable equilibrium to a Ground State never actually happens.

EDIT

From the comments knzhou seems to think that the breaking is an actual process while seeking_infinity says that it is just that you have to "expand your Taylor series" about a minimum. The latter would resolve the problem completely, but is it true that this only matters when we do perturbation theory?

$\endgroup$
4
+50
$\begingroup$

In relativistic QFT, this cannot be a process in time. The unstable initial state does not exist at all: An unstable ground state is impossible in relativistic QFT at temperature T=0 (i.e., the textbook theory in which scattering calculations are done) since it would be a tachyonic state with imaginary mass, while the Kallen-Lehmann formulas require $m^2\ge 0$. (In particular, supercooling the symmetric phase down until $T=0$ is impossible.)

The unstable state can be defined only with a cutoff, but diverges into meaninglessness as the cutoff is removed. After renormalization, only the state of broken symmetry makes sense. All these are equivalent, so describe exactly the same physics. The ground state is equally described by any of these - they all produce equivalent physics. (Whenever a gauge symmetry is broken, the symmetry is reflected in the broken phase by the impossibility to distinguish between the different broken symmetry ground states. These form a single orbit under the gauge group, and hence any choice of gauge is equally valid.)

In the bending of a bar under a sufficiently strong longitudinal force - which bends in a random direction determined by the unmodeled details in the bar, the surrounding is fixed, hence the direction makes a physical difference. But the standard model doesn't float in a fixed ether - so there is nothing at all which could describe the difference between the various ground states. This means that the person doing explicit computations can choose the symmetry-breaking direction arbitrarily, for convenience, which is what is indeed done in the textbooks.

If the QFT is described by causal perturbation theory (the conceptually most - though not fully - rigorous construction), one never sees the unbroken symmetry.

At sufficiently high temperature $T$, the symmetry may sometimes be restored - but the particle interpretation is then quite different as we have instead only quasiparticles. In this case, lowering the temperature below the phase transition temperature $T_c$ breaks the symmetry, and essentially the same happens as in the case of a superconductor - the nature of the quasiparticles changes at the phase transition.

Dynamically, what happens when the temperature drops slightly below $T_c$ the physical state moves away from the previously stable and now unstable symmetric ground state to move towards the just barely emerged ring of (equivalent and indistinguishable) new ground states with broken symmetry. While it is moving it is out of [local] equilibrium and hence not easily described - in the equilibrium formulation this is visible as a discontinuity in the response functions. Once [local] equilibrium is recovered at a new broken symmetry ground state, the changes when lowering the temperature further are smooth again, as the ring of global minimizers changes smoothly. In the limit of zero temperature, it will have converged to the textbook theory with broken symmetry.

$\endgroup$
  • 1
    $\begingroup$ There is a difference between out of equilibrium states (of which there are many) and unbroken symmetry states (which do not exist in the zero temperature case).. Note also that baryogenesis is at high temperature, so the last paragraph of my answer applies there. $\endgroup$ – Arnold Neumaier Apr 20 '16 at 10:28
  • 2
    $\begingroup$ A tachyonic state is impossible in relativistic QFT at T=0 since the Kallen-Lehman formulas require m^2>0. $\endgroup$ – Arnold Neumaier Apr 20 '16 at 16:21
  • 1
    $\begingroup$ @ArnoldNeumaier You say that 'All these are equivalent, so describe exactly the same physics'. Then how come only one of them is chosen? They are all as likely to be realised and there is nothing that pushes the system towards one particular ground state. A particular direction is chosen spontaneously even though the quantum fluctuations happen in an exactly symmetric way. $\endgroup$ – Steven Mathey Apr 22 '16 at 10:39
  • 2
    $\begingroup$ @StevenMathey: Nothing is chosen. The ground state is equally described by any of these - they all produce equivalent physics. In the bending of a bar under a sufficiently strong longitudinal force - which bends in a random direction determined by the unmodelled details in the bar, the surrounding is fixed, hence the direction makes a physical difference. But the standard model doesn't float in a fixed ether - so there is nothing at all which could describe the difference between the various ground states. $\endgroup$ – Arnold Neumaier Apr 22 '16 at 11:45
  • 1
    $\begingroup$ This means that the person doing explicit computations can choose the symmetry-breaking direction arbitrarily, which is what is indeed done in the textbooks. $\endgroup$ – Arnold Neumaier Apr 22 '16 at 11:45
1
$\begingroup$

In fact in some sense the full state of the quantum system shouldn't break any symmetry. It's just that the overlap of the different vacuum states (in field theory) vanishes.

So in other words the full quantum state is a superposition of all the different vacua, but when we make observations we "collapse the wave function" (feel free to insert your favorite interpretation of quantum measurement between the quote marks) and that picks out a specific state.

But there is something subtle about field theory vs qm. In qm after "collapsing the wave function" the wave function will spread out again to become a superposition of all vacua. In field theory the wave function doesn't spread. That's related to the fact that in field theory there are infinite number of degrees of freedom.

$\endgroup$
  • $\begingroup$ In exactly solvable 2D models, the different groud states form different superselection sectors, hence they cannot be superimposed. Thus one expects the same in general. $\endgroup$ – Arnold Neumaier Apr 22 '16 at 10:12
1
$\begingroup$

Many answers discuss the transition from the unstable state to the stable one. Let me thus discuss the issue of choosing the ground state itself.

I will suppose a two dimensional Mexican hat potential. As Numrock realised, it has degeneracy. There is nothing which lift this degeneracy in principle. Then you can change the ground state without energy. This kind of mechanism is usually referred to the Goldstone phenomenon : fluctuations can change the ground state with linear dispersion. To fix the ground state you need the Higgs mechanism.

To illustrate, suppose a Ginzburg-Landau functional. Without gauge potential, it reads

$$F=\int dx\left[g\left|\nabla\Psi\right|^{2}+\alpha\left|\Psi\right|^{2}+\beta\left|\Psi\right|^{4}\right]$$

and its minimum is at

$$\dfrac{\delta F}{\delta\Psi}=0\Rightarrow\left|\Psi\right|^{2}=\dfrac{-\alpha}{2\beta}$$

whereas $\nabla\Psi=0$ and so any phase is allowed: $\Psi\rightarrow\Psi e^{\mathbf{i}\varphi}$ makes no harm to the ground state, i.e. you can change the ground state without energy. If you add a gauge potential

$$F=\int dx\left[g\left|\left(\nabla-\mathbf{i}A\right)\Psi\right|^{2}+\alpha\left|\Psi\right|^{2}+\beta\left|\Psi\right|^{4}+\kappa\left(\nabla\times A\right)^{2}\right]$$

the phase is somehow fixed $\left(\nabla-\mathbf{i}A\right)\Psi=0\Rightarrow\nabla\varphi=A$ when one chooses the Ansatz $\left|\Psi\right|e^{\mathbf{i}\varphi\left(x\right)}$ with only the phase depending on position. In principle $\left|\Psi\right|=\pm\sqrt{\dfrac{-\alpha}{2\beta}}$ is already fixed. Varying carefully with respect to $A$ also give you the Meißner effect.

Note the above argument is really sketchy (and contains sign errors certainly). You can find more details in many questions on this website (check for Goldstone and Higgs). I particularly appreciate the account given in a book by Rubakov, V. (2002). Classical theory of gauge fields. Princeton University Press. An other important account is in Weinberg, S. (1995). The Quantum Theory of Fields (Volume 2). Cambridge University Press.

$\endgroup$
  • $\begingroup$ thank you for your answer, I particularly appreciate the references. You have my +1 $\endgroup$ – Wolpertinger Apr 22 '16 at 20:10
1
$\begingroup$

I think the question was already answered here, look at either my answer of the one by Adam.

The punch line is that that the expectation value of a field (such as the field $\phi$ at the bottom of the mexican hat potential) is fixed by the way the source $j$ that couples to $\phi$ in the path-integral for the functional generator is sent to zero. As there is spontaneous symmetry breaking the functional generator $Z[j]$ is not an analytic functions of the source and the results of the correlators, including the 1-point functions of $\phi$, depend on the way the limit of vanishing sources is performed. In other words, it depends on the details of the history of the system. But more practically, one trades such a freedom for the value of $\langle\phi\rangle$ itself that becomes hence a free parameter that defines the phase of the theory you are in.

Nicely enough, for internal spontaneously broken symmetries there is no physical effect that can distinguish one vacuum from the other. For spontaneously broken spacetime symmetries the story is somewhat different. For example, breaking spontaneously dilatations in a scale invariant theory generates a scale that controls e.g. the masses of the particles which have clearly a physically distinct impact.

$\endgroup$
  • $\begingroup$ thank you for the link to the other detailed answer. +1 on both $\endgroup$ – Wolpertinger Apr 24 '16 at 9:05
0
$\begingroup$

Physically, you look for the ground state of your theory in order to make a correct predictive calculus around the ground state. The "random" choice of the ground state depends on the dimension of the theory, in fact, you can obtain a "hat" shape or a simple 2-dimensional shape with just 2 possibilities for the ground state (look the scalar quantum electrodynamics formalism for an example of it). Surely, if you have an even more higher dimensional theory the potential has a more complicated topology and then the possibilities of choose the ground state will be higher. The important thing here is that does not matter the ground state you choose, the physics will be independent of that choice.

$\endgroup$
  • $\begingroup$ If the ground states are randomly distributed throughout space, then how can cosmic strings develop (or cosmic textures, of wich the proven nonexistence seems strange to me)? $\endgroup$ – descheleschilder May 14 '16 at 9:59

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.