# What does spontaneous symmetry breaking have to do with decoherence?

## Background

The question here by Prof. Wen, and the answers that follow point out that spontaneous symmetry breaking (SSB) has something to do with decoherence if I understand it crudely correctly.

But the usual reasoning why SSB does not occur in a quantum mechanical system (for example, a particle confined in a double-well potential) is that (due to the tunnelling effects) the ground state is a symmetric or antisymmetric linear superposition of the ground state wavefunctions localized around the classical minima of the potential which respects the symmetry of the Hamiltonian. It's only in field theory where one has infinite degrees of freedom and the tunnelling effects are shut down so that one can have SSB.

## Question

If decoherence were truly the reason of SSB then should one not expect SSB to happen even in quantum mechanics, and the system to go to a mixed state? But SSB doesn't happen in quantum mechanics.

I guess I wrongly understood the points explained there, and I would like to be clarified on this issue.

• "It's only in field theory where one has infinite degrees of freedom and the tunnelling effects are shut down" . This is putting the cart in front of the horse. Field theory is a necessary mathematical tool in order to study and calculate many body problems in quantum mechanics. It does not generate/replace quantum mechanics. It is based on the the postulates of quantum mechanicsand the free particle solutions of the basic quantum mechanical equations (Dirac, Klein Gordon, quantized Maxwell) and is a sophisticated mathematical tool to be able to calculate scattering and decay probabilities. Jan 1, 2018 at 5:37
• I agree. But a field theory is different from particle mechanics (classical or quantum) where in the former you have systems with a finite number of degrees of freedom and in the latter infinite. And as far as I understand, it's crucial to have infinite degrees of freedom to have SSB. When I say quantum mechanics, I mean systems with one or a finite number of particles where you cannot take a thermodynamic limit. @annav
– SRS
Jan 1, 2018 at 5:40
• Then "If decoherence were truly the reason of SSB then should one not expect SSB to happen even in quantum mechanics, and the system to go to a mixed state?" If you mean for "in quantum mechanics" in solutions of the pertinent equations with a potential, you will have to provide an example of what you mean. If the potential has two minima in the simple equation , there is a probability for the particle to go to the lower minimum. Probabilities are spontaneous. They appear with the throw of the dice. What symmetry would be broken? Jan 1, 2018 at 5:52
• The symmetry broken in SSB is that of having all the gauge bosons at zero mass. In a particle in a potential well the definition cannot apply as masses are not variable. Jan 1, 2018 at 5:54
• @annav In case of Ising model, the $\mathbb{Z}_2$ symmetry of the Hamiltonian is broken by the choice of one of the two possible ground states. The actual ground state is not a linear combination of two allowed ground states. But a single quantum particle confined to a double well potential, with two degenerate minima, certainty does not exhibit $\mathbb{Z}_2$ symmetry breaking because the ground state becomes a linear combination of two wavefunctions localized around the two minima of the potential $V(x)=A (x^2-a^2)^2$ at $x=\pm a$. The point is, in the first case you have no superposition.
– SRS
Jan 1, 2018 at 7:20

I don't think I have anything really new to say here, but saying it again in different words might have some value, so I'll give it a try.

Consider these two seemingly-contradictory statements, both of which are known to be true:

• An ordinary ferromagnet becomes spontaneously magnetized at sufficiently low temperatures, even though it does not have infinite volume.

• On the other hand, a mathematical model of a something like a ferromagnet — the simplest example being the Ising model — does not show SSB except in the infintie-volume limit.

The resolution of this paradox is that a real ferromagnet does not exist in isolation. Even if we put it in a vacuum chamber, it still has a magnetic field, and the stuff outside the vacuum chamber (and in the walls of the chamber) can still be influenced by that magnetic field. Also, in order for the ferromagnet to cool down from above the SSB transition temperature to below that temperature, it must release energy into the environment. In the real world, there is no such thing as an isolated system. (Well, except maybe the whole universe, whatever that means.)

An accurate model of an isolated finite-volume ferromagnet predicts that SSB cannot occur, because the ground state is something like $$|\text{all spins up}\rangle +|\text{all spins down}\rangle$$ rather than either term individually. But in the real world, because the ferromagnet isn't isolated, the state ends up being something like this: $$|\text{all spins up}\rangle\otimes |E_\text{up}\rangle +|\text{all spins down}\rangle\otimes |E_\text{down}\rangle$$ where $$E_\text{up/down}$$ represent different states of the ambient system which are almost exactly mutually orthogonal to each other. Most importantly, their mutual orthogonality cannot be undone by the action of any local operator of manageable complexity. (Quantifying "complexity" here involves an unanswered question, but the qualitative idea is good enough here.) This is decoherence.

When we consider a model of something like a ferromagnet (or other system exhibiting SSB) in isolation, we need the infinite-volume limit in order to get a mathematically strict version of SSB. But even in this case, there is a sense in which decoherence is still relevant. In effect, we're using the far-away parts of the ferromagnet itself as the "environment" with which the local part becomes entangled.

In the real world, the tunneling effects that would prevent a mathematically-strict version of SSB in finite volume become ineffective FAPP (for all practical purposes) when the volume is large enough, for the same reason that we can't observe superpositions of different measurement outcomes. In the wake of a high-quality measurement, the thing being measured has influenced its surroundings in a prolific way that we have no hope of ever reversing in practice. So... (wave hands here)... the superposition might as well have "collapsed" to just one of those terms. That's where I'll end this paragraph, because continuing any further down that path would take us straight into the philosophical quagmire through which no mortal physicist has ever successfully passed.

By the way, this is closely related to the motivation for using the cluster property to distinguish between the "real" SSB vacuum states and arbitrary superpositions of them. A vacuum state isn't just a lowest-energy state; it's a lowest-energy state that satisfies the cluster property. Arbitrary superpositions still have zero energy, but they don't satisfy the cluster property. Here are a couple of references that explain how the cluster property selects the "correct" vacuum states:

• In the context of spin systems (like the Ising model): Section 23.3, "Order Parameter and Cluster Properties", of Zinn-Justin's book Quantum Field Theory and Critical Phenomena.

• In the context of QFT: Section 19.1 in Weinberg, The Quantum Theory of Fields, Volume II.

• This is a good answer and I enjoyed reading it. I don't know much about QFT, but I do know that a semi-infinite transmission line (or other wave-carrying medium) can be modeled as a set of harmonic oscillators in the limit that the number of oscillators goes to infinity. interestingly, a semi-infinite transmission line has the same linear response function as a resistor. In other words, an infinite set of lossless (i.e. coherent) modes looks lossy. So we can model decoherence with an infinite set of closed systems. I wonder if the infinite limit of the magnet works the same way somehow. Dec 16, 2018 at 5:36
• Look up "Cadeira-Leggett model". I actually have my own version of it written up in TeX. Email me (you can find my email reasonably easily via my profile page) if you'd like me to send it to you. Dec 16, 2018 at 17:04