No SSB in finite systems

Consider a system interacting with a heat bath at inverse temperature $\beta$, with the resultant dynamics of the system described by a Liouvillian superoperator $\mathcal{L}$. If this system is finite, then under fairly general conditions on $\mathcal{L}$, we expect the equilibrium state $\rho$, meaning $\mathcal{L}(\rho) = 0$, to be uniquely given by the Gibbs state

$$ \rho_{\mathrm{gibbs}} = \dfrac{e^{-\beta H}}{\mathrm{Tr}\left[ e^{-\beta H} \right]}, $$

where $H$ is the Hamiltonian of the system.

Let $\mathcal{G}$ be the symmetry group of the Hamiltonian, meaning there is a unitary representation $U$ of $\mathcal{G}$ such that $[H, U(g)] = 0$ for all $g \in \mathcal{G}$. It is clear that we also have $[\rho_{\mathrm{gibbs}}, U(g)] = 0$, so the Gibbs state preserves all the symmetries of the Hamiltonian. In this sense, it seems that there cannot be spontaneous symmetry breaking (SSB) in finite systems.

SSB in infinite systems (and KMS states)

The typical narrative then proceeds to say that, in fact, SSB can occur, but only in infinite systems. Here there is no guarantee that $e^{-\beta H}$ is trace-class, so in general the Gibbs state is not well defined. To extend the notion of a "thermal" state to infinite systems, one usually defines the so-called KMS states. These are the states $\phi$ which satisfy the KMS condition, which can (informally) be stated as

$$ \langle A (t) B \rangle_{\phi} = \langle B(t + i\beta) A \rangle_{\phi}, $$

for all operators $A$ and $B$ in the operator algebra, where $\langle \cdot \rangle_{\phi}$ indicates an expectation value with respect to the state $\phi$. (I omit all $C^{*}$-algebraic details here for brevity.)

There is a large body of literature showing that KMS states preserve the properties that we consider key to the definition of a thermal state, such as being equilibrium states, but remain well-defined for infinite systems.

For finite systems, I believe the KMS condition uniquely specifies a state: the Gibbs state. However, for infinite systems this is not necessarily the case, and, roughly speaking, SSB occurs when there are multiple KMS states, each of which is not preserved by the symmetry group of the Hamiltonian.


Both experiments and numerical simulations show systems with behaviour that seems very similar to that of SSB (ferromagnets exist!). However, these real-world systems are clearly finite, so the above arguments would suggest that they cannot truly display SSB. What is the explanation for this discrepancy?

Thoughts on an answer

Though finite, real-world experiments can often by fairly effectively described by taking the infinite size limit. If this is appropriate, then perhaps the dynamics of these large finite systems can be well approximated by infinite systems, at least up to some large timescale $\tau$ which presumably grows quickly with system size. Then we might expect these finite systems to display signatures of SSB over the timescale $\tau$, after which they will decay to the Gibbs state and the symmetry will be restored. If this is along the right lines, can any of this be made precise?

  • $\begingroup$ Some quick thoughts: 1) the absolute ground state is a macroscopic superposition of the symmetry breaking state. 2). The tinyest amount of coupling will destroy coherence for this cat state 3) my guess is the smallest temperature gibs state is a mixed state of the symmetry broken states, so over all its symmetric. This is consistent with experiments where you don’t know how the symmetry is broken before hand. I don’t know how to prove this. It might just be physically the macroscopic tunneling time diverges and there is never a true gibs state thermalization $\endgroup$ Jul 30, 2018 at 22:08
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    $\begingroup$ This sounds like excess mathematical formalism clouding a very simple physical point. Forget all about Liouvillian superoperators, KMS states, and $C^*$ algebras. The point is that a big system has degenerate ground states with high energy barriers between them. In the statistical mechanical case, the timescale for thermal fluctuations between them is very small. In the quantum case, a superposition of them is not stable, and decoheres immediately. So in either case you see the system in a single, symmetry broken state. $\endgroup$
    – knzhou
    Jul 31, 2018 at 9:54
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    $\begingroup$ @OliverLunt No, there's no qualitative difference between finite and infinite systems, because every physical property of an infinite system should be approximated arbitrarily well by a sufficiently large finite one. If you did find a quantity that didn't obey this rule (i.e. where the infinite case would look completely different from any finite case, no matter how large), then it would be irrelevant for all physical purposes because no infinite systems exist. $\endgroup$
    – knzhou
    Jul 31, 2018 at 10:19
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    $\begingroup$ In the classical case, you can't mix the ground states; the magnet points either this way or that way. In the quantum case any superposition of the two will immediately decohere upon interaction with just about anything else. (Or, said another way, any charged particle passing by will "measure" the direction and hence collapse the superposition.) At finite temperature tunneling between the ground states by thermal fluctuations becomes possible, but that doesn't change the conclusions much. $\endgroup$
    – knzhou
    Jul 31, 2018 at 10:22
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    $\begingroup$ While it makes little difference to physical system, mathematically infinite and finite systems are very different. For finite systems, dechorence doesn’t exist, there is always a recurrence time where the original state reappears. This time goes to infinity as system size increases and subsystems can appear to thermalize as they do for infinite systems. But infinite systems the information about the initial state spreads to infinity and the system never recurs. $\endgroup$ Aug 1, 2018 at 23:22

2 Answers 2


This question exactly is addressed and rigorously treated by N.P. Landsman. The explanation is that in the large $N$ limit the symmetric ground state becomes exponentially sensitive to asymmetric perturbations, while the first excited states, although unstable, become very close in energy to the symmetric state and decay exponentially slowly in any direction, thus the system dynamically finds itself in a symmetry broken state already at a finite by very large $N$.


Not sure if this matches your puzzle, but: when you write something like $[\rho, \mathcal{G}]$, you are thinking about measuring the symmetry existence by looking at the expectation value of the operator itself. However, in completely quantum sense, you should define the symmetry with the expectation of correlators other than operators themselves.

You might find the note of Lec.1 here helpful, which used transverse Ising model as an example, and the SB definition is mentioned on Pg.9


  • $\begingroup$ Hi, I know this is a while ago but would you mind sharing the lecture note you had? I couldn't access this myself... Thanks in advance. $\endgroup$ Sep 13, 2019 at 22:10

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