What is spontaneous symmetry breaking in quantum systems?

Question

Most descriptions of spontaneous symmetry breaking, even for spontaneous symmetry breaking in quantum systems, actually only give a classical picture. According to the classical picture, spontaneous symmetry breaking can only happen for non-linear systems. Classical linear systems, such as harmonic oscillators, can never have spontaneous symmetry breaking (here "linear" means that the equation of motion is linear).

But the real quantum systems are always linear since the Schrodinger equation is always linear. So how can a linear quantum system have spontaneous symmetry breaking? Do we have a simple intuitive understanding for spontaneous symmetry breaking within quantum mechanics (without using the classical picture, such a Mexican hat -- the logo of physics.stackexchange)?

The Mexican hat does give us an intuitive and pictorial understanding of spontaneous symmetry breaking in classical systems. Do we have an intuitive and pictorial understanding of spontaneous symmetry breaking in quantum systems?

Answer 1

I just discovered this very interesting website through Prof Wen's homepage. Thanks Prof Wen for the very interesting question. Here is my tentative "answer":

The spontaneous symmetry breaking in the ground state of a quantum system can be defined as the long range entanglement between any two far-separated points in this system, in any ground state that preserves the global symmetries of the system.

To be more precise, denote $G$ as the symmetry group of the system and $|\Psi\rangle$ a ground state that carries a 1d representation of $G$. For an Ising ferromagnet, the ground state will be $|\Psi_\pm\rangle =\frac{1}{\sqrt{2}}\left(|\text{all up}\rangle \pm |\text{all down}\rangle\right)$. Then consider two points 1 and 2 separated by distance $R$ in the space, and two small balls around points 1 and 2 with radius $r\ll R$, denoted by $B_1$ and $B_2$. Define $\rho_1$, $\rho_2$ and $\rho_{12}$ as the reduced density matrices of the region $B_1$, $B_2$ and $B_1+B_2$, and correspondingly the entropy $S_{1}=-tr(\rho_1\log \rho_1)$ (and similarly for $2$ and $12$). The mutual information between the two regions is defined as $I_{12}=S_1+S_2-S_{12}$. If $I_{12}> 0$ in the $R\rightarrow \infty$ limit for all symmetric ground states, the system is considered as in a spontaneous symmetry breaking state.

In the example of Ising FM, $S_{12}=\log 2$ for both ground states $|\Psi_\pm\rangle$.

I am afraid it's just a rephrasing of ODLRO but it might be an alternative way to look at spontaneous symmetry breaking.

Answer 2

This question posted by Prof. Wen is so profound that I had hasitated to response. However motivated by Jimmy's insightful answer, I eventually decided to join the discussion, and share my immature ideas.

1) Quantum SSB is a non-linear quantum dynamics beyond the description of Schordinger's equation.

Regarding the transverse field Ising model mentioned in the comments of the question, with a small B field, the ground state is a Schordinger's cat state. Asking how does the SSB happen in the $B\to 0$ limit is the same as asking how does the cat state collapses to a definite state of live or death. Quantum decoherence plays the key role here. However quantum decoherence is an irriversible dynamics with entropy production, which, I believe, can not be described by the linear dynamics of quantum mechanics that preserves the entropy. To understand quantum SSB, we may have to understand the dynamics of quantum decoherence first.

2) Quantum SSB is a result of information renormization, which may be described by the tensor network RG.

The key of understanding quantum decoherence is to understand how entropy was produced. It had been a mystery for a long time that what is the origin of entropy? Until Shannon related entropy to information, we started to realize that entropy is produced due to the lost of information. Information is lost in the experiments inevitably because we can only collect and process finite amount of data. Because all experiements are conducted under a finite energy and information (or entropy) scale, so only the low energy and low information effective theory is meaningful to physicists. Renormalization group (RG) technique had been developed to obtained the low energy effective theory successfully. Now we need to develop the informational RG to obtain the low information effective theory. DMRG and tensor network RG developed in recent years are indeed examples of informational RG. Quantum information is lost through the truncation of density matrix, and entropy is produced at the same time, which makes quantum decoherence and quantum SSB possible. In fact, quantum SSB can been observed in both DMRG and tensor network RG as I know. Along this line of thought, quantum SSB is not a final state of time evolution under linear quantum dynamics, but a fixed point of informational RG of quantum many-body state, which is non-linear and beyond our current text-book understanding of quantum mechanics.

Answer 3

Bei Zeng and I wrote a paper http://arxiv.org/abs/1406.5090 , which addresses this question:

A symmetry breaking phase for finite group G is a gLU equivalent class formed by symmetric many-body states that have GHZ entanglement.

In other words, a symmetry breaking phase is a set of

1. symmetric states $U_g \Psi = \Psi$ up to a phase, $g \in G$, and
2. those symmetric states have the same GHZ entanglement $\Psi = \sum_\alpha \Psi_\alpha ,\ \ \alpha \in G/H,\ \ H\ \subset G$, where $\Psi_\alpha$'s are locally distinguishable.

We say those symmetric states are equivalent. The set of equivalent symmetric states is a symmetry breaking phase.

So symmetry breaking = GHZ entanglement which are classified by pairs $(G , H),\ H \subset G$.

More precisely:

1. A symmetric many-body state has spontaneous symmetry breaking implies that the state has a GHZ entanglement.

2. One can detect spontaneous symmetry breaking in a symmetric many-body state even without knowing the group and/or order parameter of the symmetry. One can detect spontaneous symmetry breaking in a symmetric many-body state using only probes that respect the symmetry.

3. The symmetric exact ground state of a generic symmetric Hamiltonian has spontaneous symmetry breaking iff it has GHZ entanglement.

Answer 4

I'm sure Prof. Wen understands this question very well and is posting this just to inspire some discussions. So I'm just gonna go ahead and give my 2 cents.

A classical spontaneous symmetry breaking happens when the classical ground state breaks the symmetry of the Hamiltonian. For example, for a classical Ising model in 1D, spontaneous magnetization in a particular direction happens at low T, which breaks the $S\rightarrow-S$ symmetry of the Hamiltonian.

A quantum spontaneous symmetry breaking doesn't necessarily mean the quantum ground state breaks the symmetry of the Hamiltonian; instead, it's signatured by the splitting of the ground state degeneracy. Say in the case of transverse Ising model, $H=-\sum{S_i^z S_j^z}-B\sum{S_i^x}$. The ground state of the Hamiltonian for very small $B$ is the superposition of all spin up and all spin down, which still has the $S_z\rightarrow -S_z$ symmetry; but now the ground state degeneracy is lost---the ground state is now unique, rather than having a 2-fold degeneracy.

This is just a preliminary answer, so please feel free to correct me/improve the answer.

Answer 5

I think that one way to visualize spontaneous symmetry breaking in quantum systems is as follows:

The theory's Hilbert space is infinite dimensional. Given a Hamiltonian, one method to seek approximate solutions of its spectrum is by formulating a variational principle with respect to a finite dimensional Hilbert space of trial functions.

In many cases when there is a continuous symmetry group $G$ of the Hamiltonian, the manifold of trial functions can be chosen as a homogeneous symplectic $G$-space, which implies that the (Lie algebra of ) the symmetry group generates all observables and the approximate Hamiltonian is some element in the universal enveloping algebra.

On these types of manifolds the quantum and classical dynamics are greatly similar and offer a simple relation between the classical and the quantum picture of the spontaneous symmetry breaking;

Explicitely, When, the (approximate) classical Hamiltonian on the trial function manifold acquires a minimum at a nonvanishing expectation value of some generator, the vacuum of the quantum Hamiltonian on the quantization of this manifold becomes degenrate.

Answer 6

One possible understanding of SSB in quantum systems may be the following: we all know that classically there is a ground state manifold and one can choose to locate the ground state on one point which breaks the symmetry. However, in quantum systems, due to superposition principle one can form linear combinations that restore the symmetry. However, SSB means that for the low-energy states, there are a certain basis (which are the "classical" states), such that, if one looks at the matrix elements of local physical operators(operators with local support) between different basis states they always vanish in the thermodynamic limit. This may provide a quantum characterization of SSB, although I'm not fully confident that this is sufficient and necessary. Finite size effect may be included by considering how the matrix elements scales with the system size.

Obviously there are some hand-wavingness in the above definition, since we are talking about "basis" for only low-energy states. But I still find it a useful way of understanding SSB.

Answer 7

A way to study the quantum system that closely parallels the discussion in classical physics is to use the (quantum) effective action: Compute the partition function $Z[B]$ as a function of the external field. Then $\beta\log(Z)$ is the free energy $F$ and $\partial F/\partial B$ is the magnetization $m$. Now perform a Legendre transform to get the quantum effective action $\Gamma[m]$. Then we look for an effective action that has the shape of the physics stackexchange logo (with the usual caveat that strictly speaking, the effective action is always convex).

Answer 8

The best answer I´ve come up with is arXiv:1205.4773v1

Spontaneous symmetry breakdown in non-relativistic quantum mechanics

R. Munoz, A. Garcia-Quiroz, Ernesto Lopez-Chavez, Encarnacion Salinas-Hernandez

The advantages and disadvantages of some pedagogical non-relativistic
quantum-mechanical models, used to illustrate spontaneous symmetry breakdown,
are discussed. A simple quantum-mechanical toy model (a spinor on the line,
subject to a magnetostatic interaction) is presented, that exhibits the
spontaneous breakdown of an internal symmetry. 

Comments:19 pages, 5 figures. arXiv admin note: substantial text overlap with arXiv:1111.1213

Answer 9

Insofar as SSB causes or corresponds to the existence of arbitrarily long range order at space-like separation, it may be understandable in terms of violation of cluster decomposition. As such, SSB corresponds to the existence of a set of vacuum vectors in the Hilbert space that is invariant under the action of the field operators (within the Wightman axiomatic approach, part of the proof of the Wightman reconstruction theorem is to show that cluster decomposition, a property of the VEVs, is equivalent to the reducibility of the Hilbert space).

Whenever the observables of a theory are a nontrivial subset of the set of operators that can be constructed from the field operators, typically because the observables are required to be invariant under the action of some symmetry, the vacuum state will be reducible under the action of the observables, and there will be violation of cluster decomposition.

Cluster decomposition is largely restored by the introduction of gauge fields (which I take not to be part of SSB, although one could of course take SSB to include the introduction of gauge fields). To me it's not clear whether cluster decomposition is completely restored by the introduction of gauge fields.

EDIT: This is to me moderately intuitive, but, focusing on your last paragraph, I guess it won't seem pictorial to most people --- and it's only a little pictorial to me. I take it mostly to depend on algebraic intuition.

Answer 10

The analog is superselection sectors. If a symmetry transformation acting upon a quantum states puts it in a different superselection state, we say that symmetry in question is spontaneously broken.

Answer 11

The answer is in decoherence. for classical systems, if a subsystem breaks a symmetry, the system as a whole also breaks the symmetry. not so in quantum mechanics because of entanglement. here lies the complication.

think of zurek's pointer states. there lies the clue. i may give you a many body quantum state which literally is invariant under the symmetry in question, but if it decomposes into decoherent pointer states which aren't invariant, feel free to say the symmetry is spontaneously broken? but zurek's analysis only works for open systems.

can this work for finite closed systems? unfortunately no because of poincare recurrences. we might naively think a symmetry is spontaneously broken, but wait long enough and the slight (or not so slight) energy differences between the various energy eigenvalues corresponding to different irreps will lead to a washout in phase differences in energy eigenstates carry info on symmetry breaking.

what are zurek's pointer states? those which preserve information longest in time while minimizing dynamical generation of entanglement with the environment. sometimes, a pointer state invariant under a symmetry will generate more entanglement with the environment than one not invariant.

complications abound. take a collection of helium-4 atoms at a low temperature. superfluid phase. u(1) symmetry corresponding to number of he-4 atoms. put the atoms in a very sealed box where not even a single he-4 atom can pass but info can pass. idealized, yes, but bear with me. quantum state with a fixed specific value for number of he-4 atoms. invariant under u(1)? what are the pointer states? unfortunately, not condensate states with a superposition in number of he-4 atoms? but the dynamical generation of enviroentanglement remains small in either case anyway: fixed atom num and condensate. just that over very long periods of time, fixed atom num has slightly more entanglement. because dynamical processes sensitive to total num of he-4 atoms will dominate but only because of absolute suppression of permeability. unrealistic, no?

but loosen up. make box slightly permeable. just let only one or two he-4 atoms pass after relatively long time. voila? pointer state changes favoring condensates? confused yet? the number of he-4 atoms in the environment is in a superposition entangled with the num of he-4 atoms in the box. THE ENVIRONMENT!!! the symmetry has to be broken in the environment, not the system.

but what about the universe as a whole? it has no external environment. aah, but there are no global symmetries in quantum gravity. ok, what about gauge symmetries then. oh boy, another huge can of worms. What is spontaneous symmetry breaking in QUANTUM GAUGE systems? that is worth another s.e. question.

Answer 12

This question seems to be based on a false premise, namely, that systems that are nonlinear classically are linear when quantized. Really, the opposite tends to be the case. E.g., Maxwell's equations in vacuum are exactly linear but in QED there is nonlinearity due to interactions mediated by electron loops.

Quantum mechanics is linear at the level of the Schrödinger equation. If you interpret the wave function as a quantum analog of a probability distribution, then classical mechanics is also linear at this level. E.g., if a system is in state $A_i$ with probability $p_i$ for all $i$, and the chance it will evolve from $A_i$ to $B$ is $q_i$, then the chance it will end up in state $B$ is $\sum_i p_i q_i$. Classically, there can be no nonlinear interaction between the alternatives because only one of them actually happens. Quantum mechanics preserves that linearity, though the justification of it in terms of an underlying classical reality doesn't work any more.

Classically, if you start with a uniform (or at least symmetric) distribution over microstates of a fluid, and let it crystallize, you'll end up with a symmetric distribution over all possible orientations of the resulting crystal. In some sense the system is still symmetric, if you take "the system" to be this probability distribution, but no one in the world can see that symmetry; they only see one particular orientation of the crystal. In QM, the same thing happens.