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

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?

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For me all the definitions are different aspects of the same mechanism, called spontaneously broken symmetry: they all discuss a system, the ground state of which no more exhibits the symmetry the Lagrangian / Hamiltonian had. –  FraSchelle Jun 26 '13 at 19:23
Not sure I understand. For me, there is a single definition of SSB, that which you give. I then wonder whether and how SSB can be realized in CM, QM and QFT, and want to understand the differences in each case. –  innisfree Jun 26 '13 at 19:33
More on SSB and quantum systems: physics.stackexchange.com/q/29311/2451 –  Qmechanic Jun 26 '13 at 19:34
–  ramanujan_dirac Jun 27 '13 at 6:11
@innisfree : The previously given reference is very interesting. For instance, it it said (Chapter III, page 7) that "A theorem of elementary quantum mechanics tells us that one-dimensional hamiltonians with lower bounded continuous potentials are non-degenerate. Thus, no one-dimensional continuous lower bounded potential (sombrero or otherwise) can exhibit spontaneous symmetry breakdown". So you have to take discontinuous potentials as infinite double well, to get SSB. –  Trimok Jun 27 '13 at 12:31

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.

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I'm confused by your sentence "When the number of degrees of freedom is finite for example in the case of the double well potential in one dimension". Whether the dof is finite depends on the nature of the field, not the potential? E.g. a a real scalar field in QFT might have that potential, and it would be infinite dimensional. Thanks a lot of the last two paragraphs. They look helpful, but I need to do some more reading to make sense of it fully. –  innisfree Jun 29 '13 at 20:10
Of course, I changed the wording to hopefully avoid confusion. –  David Bar Moshe Jun 30 '13 at 2:55
"classical spontaneous symmetry breaking means that there are initial conditions leading to solutions which are not invariant under the symmetry group". Is this not the general case? For example, any non-trivial solution of the Klein-Gordon equation is not invariant under a Lorentz transformation. The transformation connects solutions with different boundary/initial conditions. –  drake Jul 5 '13 at 6:34
@drake I have changed "invariant solutions", to "time invariant solutions". This means that the classical distribution on the phase space does not change in time. This definition avoids the mention of vacuum. Actually, the Klein-Gordon example is appropriate here, because the invariant solution $\phi = 0$ indicates that the Lorentz invariance is not broken, however it is not a "vacuum", since the Hamiltonian is not bounded from below. –  David Bar Moshe Jul 7 '13 at 13:52
@DavidBarMoshe: In what sense does a buckling bar a finite system? I mean it is finite in physical length, but aren't there an infinite number of degrees of freedom, one for each fourier mode? –  BebopButUnsteady Jul 8 '13 at 0:55