Spontaneous symmetry breaking in classical mechanics, quantum mechanics and quantum field theory

Question

I wondered if someone could help me understand spontaneous symmetry breaking (SSB) in classical mechanics, quantum mechanics and quantum field theory. Consider a Higgs-like potential, with a local maximum surrounded by a degenerate ground state - a pencil balanced on its point, for example.

Classical Mechanics (CM) exhibits spontaneously symmetry breaking if and only if the system is perturbed.

Quantum mechanics (QM) exhibits no symmetry breaking, because the ground state is a superposition of the degenerate vacua.

Quantum field theory (QFT) At infinite volume, spontaneous symmetry breaking occurs. Because the degenerate vacua are orthogonal, $$ \langle \theta^\prime | \theta \rangle = \delta(\theta^\prime-\theta), $$ a ground state is chosen.

Q1 Is it true that QM never exhibits SSB? Some sources suggest otherwise. But I can't see a way round the basic argument.

Q2 In QFT, is it correct that a conceptual difference with CM is that the system needn't be perturbed? I guess this is the case, because we simply look at the expectation of the field $\langle 0 | \phi | 0 \rangle$. But how can I convince someone that the field cannot just sit at the local maxima?

Q3 I find it strange that SSB disappears from QM to CM, then reappears in QFT. Are there other phenomena that have this feature? Is there a nice way to understand this?

Answer 1

The third question Q3 is basically the subject of a very recent work by N.P. Landsman.

In quantum theory, spontaneous symmetry breaking requires the system to be infinite dimensional. When the number of degrees of freedom is finite, spontaneous symmetry breaking does not take place. Consider for example a particle in one dimension moving in the potential of a double well, tunneling takes place between the two degenerate states, corresponding to the minima of the potential, resulting a unique linear superposition ground state. In the infinite number of degrees of freedom limit. The transition probabilities between the degenerate states vanish, thus partitioning the Hilbert space into mutually inaccessible sectors built up over each ground state.

It is well known that in classical systems with finite numbers of degrees of freedom spontaneous symmetry breaking is possible as emphasized also in the following review by Narnhofer and Thirring. Given that a (pure) state in a classical system is a point in phase space (a mixed state is a probability distribution over the phase space); then classical spontaneous symmetry breaking means that there are initial conditions leading to time invariant solutions which are not invariant under the symmetry group. For example, in the double well placing the particle in one well without enough energy to get out describes a spontaneously broken state.

There are many other examples of finite classical systems exhibiting spontaneous symmetry breaking, the most known one is, may be, the buckling of bars, another example is the Bead, Hoop, and Spring system .

Now, as Landsman emphasizes, Large $N$ quantum systems are analogous to classical systems in the sense that quantum correlations vanish as $\frac{1}{N}$, which leads to the posed question that for a finite $N$ no matter how large it is, spontaneous symmetry breaking is not allowed whereas in the thermodynamic limit the system becomes infinite and the spontaneous symmetry breaking is allowed. The same question can be asked for $\hbar \rightarrow 0$.

Landsman explanation is that when N becomes very large, the system becomes exponentially unstable to a symmetry breaking perturbation which drives the system to one of the degenerate states already at a very large but finite $N$. Landsman performs the analysis by means of algebraic quantum mechanics and a full comprehension of the article needs acquaintance with his previous work.

Answer 2

I think you are running together two different kinds of "symmetry breaking". The usual notion of spontaneous symmetry breaking in condensed matter occurs in the thermodynamic limit. This happens in both classical and quantum system. In this scenario, different ground states become infinitely separated from each other. So in a model with spins on a lattice - quantum or classical - if the spins become aligned, it would take an infinite amount of energy to align them in a different direction by local fluctuations.

It is frequently understood that spontaneous symmetry breaking can only occur in the thermodynamic limit. For example, the partition function must be analytic at finite particle number and therefore there cannot be a phase transition, and therefore no SSB. (I freely admit that I do not entirely why facts about the infinite size limit apply so well to large but finite physical systems in the quantum case.)

What you have listed as the first two cases is a distinction between classical and quantum finite systems, viz. the classical system can have symmetry breaking ground states, whereas a finite quantum system cannot. This is qualitatively different from the usual SSB. It is not spontaneous.

Take your Higgs-like potential with the classical particle starting at the top. If you do nothing it sits at the top, symmetry is maintained. If you tap it once it will waddle back and forth between the two minima, by energy conservation, so symmetry is maintained on average. If you try to dissipate away the energy by coupling your particle to noise, then you will spend a lot of time at the bottom of one the minima. But there's a finite probability the noise will fluctuate and kick your particle over to the other well. So again if you look long enough symmetry is maintained on average. You need the infinite potential barrier to have real spontaneous symmetry breaking.

Answer 3

In the case of symmetric double well potential (the Hamiltonian is even under parity) tunneling happens between two states localized at the two minima provided the barrier is finite. Those two states are superposition of the ground and 1st excited states of the Hamiltonian thus they are not eigenstates of the Hamiltonian. If we tune the barrier height as a parameter towards infinite the energy difference between ground and 1st excited states decreases and at infinite height it disappears thus forming a degenerate ground state one of which is antisymmetric under parity (though the Hamiltonian is symmetric under parity). This is spontaneous symmetry breaking.

To observe a phase transition we have to reach this symmetry broken ground state starting from initial ground state i.e we have to go beyond this gapless (degenerate) condition to a negative gap. As we increase the barrier height and drive the system towards phase transition the procedure gets stopped just coming at the critical point. Infortunately there is no way to go beyond this point and observe the phase transition in this system.