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?

The linearity is at the wavefunction level, not on the operator level. I fail to see the problem exactly--- what is confusing exactly about quantum spontaneous symmetry breaking? If you have a probability distribution on the states a nonlinear system, the equation of motion for the probability distribution is also linear, but you still have spontaneous breaking of symmetry. — Ron Maimon, Jun 01 '12 at 06:03
Consider the ground state of transverse Ising model $H=-\sum S^z_iS^z_j + B \sum S^x_i$ of $N$ spins. For small $B$, the exact ground state still do not break the $S^z\to -S^z$ symmetry. So it is non-trivial to see the $S^z \to -S^z$ symmetry breaking for small $B$. — Xiao-Gang Wen, Jun 01 '12 at 06:43
I'm not sure the distinction between classical and quantum is very useful here. I mean, one talks about the ground state of the system, but one still has quantum fluctuations around it. — WIMP, Jun 01 '12 at 08:04
Classical states are points in phase space and quantum states are vectors in Hilbert space. So they are very different. Spontaneous symmetry breaking in classical systems does mean that the classical ground state (represented by a point in phase space) breaks the symmetry. However, spontaneous symmetry breaking in quantum systems may not mean that the quantum ground state (represented by a vector in Hilbert space) breaks the symmetry. — Xiao-Gang Wen, Jun 01 '12 at 12:24
With apologies that it's a little off topic, one can give a Koopman-von Neumann presentation of a classical statistical theory, in which case the states are vectors in a Hilbert space. In the relativistic field context, one can work with random fields instead of with quantum fields. Such fields do not satisfy the requirement that the spectrum of the energy-momentum lies in the forward light-cone, however, and one typically ends up with alternatives such as Stochastic ElectroDynamics (SED), which on the whole have not been productive. — Peter Morgan, Jun 01 '12 at 13:21
If the question is "Find an example that is intuitive to the person asking." Then this is arguably a guessing game. Nobody but the OP knows how that intuition works. — dmckee --- ex-moderator kitten, Jun 01 '12 at 14:06
fwiw I was trying to find a vector analogue with a rotating pendulum, the vector being the direction of the rotating bob from the center.Also looked at gyroscopes, but have not concluded on a good analogue. — anna v, Jun 01 '12 at 14:59
@dmckee: There is a standard understanding of SSB in quantum systems. If you are happy with it, then one does not need to go further. I ask the question since I myself is not happy with the standard understanding of SSB in quantum systems. So I try to see if there are alternative ways to understand SSB. Maybe this will lead to an understanding that is deeper, and that I find more satisfying. I do feel that there should be a deeper and better understanding of SSB in QUANTUM systems. — Xiao-Gang Wen, Jun 02 '12 at 03:00
@Xiao-GangWen: you are undoubtedly aware of Tony Leggett's work on this topic. For him, the motivation lies in the clearly finite size of real experimentally realisable condensates, especially in the early years. He defines a somewhat ad-hoc measure in the eigenvalues of the single-particle density matrix. Personally, I think it is the result of entanglement with the observer, and subsequent amplification due to quantum "locking". Or: though the system + observer is symmetric, once one traces out the observer it is equivalent to introducing a classical symmetry-breaking term. — genneth, Jun 02 '12 at 14:16
@dmckee: There is a meta-question I would like to raise: the OP is not asking about the kind of understanding that one finds in textbooks; after all, he wrote some and is undoubtedly well-read on this sort of thing. There are a couple of answers below which essentially miss the point of the question, even though they would be perfect for something not tagged with "research-level". Is it appropriate to down-vote them? Or even flag for removal? (This is the sort of thing which the loss of TP.SE makes tricky...) — genneth, Jun 02 '12 at 14:32
The introduction by Jimmy Liu of "very small B" takes us out of SSB, inasmuch as the symmetry is explicitly broken by the "B" term. That restores vacuum uniqueness and cluster decomposition at the cost of invariance. @Xiao-Gang Wen: do you intend this question mostly in terms of the condensed-matter tag, in which case I take the loss of invariance to be unproblematic? In the QFT context, allowing nonunique vacua introduces many new possible states over free fields, it's outside the Wightman field context, but perhaps we have to keep to exact Lorentz invariance. — Peter Morgan, Jun 02 '12 at 21:06
A finite $B$ does not break $S^z\to -S^z$ symmetry. Also there is Lorentz symmetry. — Xiao-Gang Wen, Jun 02 '12 at 21:52
@Xiao-Gang It's helpful if you put a name at the beginning of a comment. Nonetheless, the B term is not innocuous, in that it converts the mixed vacuum state of the zero-B model into the pure vacuum state of the non-zero-B model, which removes the long-range order of being in one or another state, replacing it with the long-range order of being in the pure state /enforced/ by the non-zero B term (which could be in the y or some other direction, not only in the x direction). Gauge symmetries seem to me harder than the SSB aspect. — Peter Morgan, Jun 05 '12 at 17:28
@Peter: Indeed, a non-zero $B$ in $x$-direction will lead to a unique ground state for any finite systems. — Xiao-Gang Wen, Jun 07 '12 at 10:47
@all, can someone explain clearly why "A finite does not break →− symmetry. Also there is Lorentz symmetry."?@Xiao-Gang Wen said, what does this mean here? — ann marie cœur, Sep 28 '20 at 20:44
@Peter Morgan, can you explain clearly why "A finite does not break →− symmetry. Also there is Lorentz symmetry."?@Xiao-Gang Wen said, what does this mean here to your comment on Jimmy Liu of "very small B"? — ann marie cœur, Sep 28 '20 at 20:44
@Peter Morgan, can you explain clearly "B converts the mixed vacuum state of the zero-B model into the pure vacuum state of the non-zero-B model, which removes the long-range order of being in one or another state, replacing it with the long-range order of being in the pure state /enforced/ by the non-zero B term (which could be in the y or some other direction, not only in the x direction). Gauge symmetries seem to me harder than the SSB aspect." ? — ann marie cœur, Sep 28 '20 at 21:04
Particularly I am confused by " long-range order of being in the pure state /enforced/ by the non-zero B term (which could be in the y or some other direction, not only in the x direction). Gauge symmetries seem to me harder than the SSB aspect." ? — ann marie cœur, Sep 28 '20 at 21:06

score 35 · Answer 1 · answered Sep 24 '12 at 21:45

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.

I like your answer. Rephrasing ODLRO in term of entanglement may be helpful. The motivation of my question is trying to have an understanding of symmetry breaking and topological order with the same framework. Thinking entanglements may allow us to do it. Hope you can be active in physics.stackexchange to make it more useful for physics graduate students. I find mathoverflow is very helpful at graduate level. — Xiao-Gang Wen, Sep 27 '12 at 00:09
Thanks very much. I think it is very interesting to express both conventional order and topological order in the same framework of long range entanglement. I will try to be active and I think this website will be very helpful not only for graduate students but also for myself. — Phynics, Sep 28 '12 at 05:53
@Phynics Welcome to physics stackexchange! May I ask what the physical interpretation of the following three cases are: (i) $I_{12}>0$, (ii) $I_{12}=0$ and (iii) $I_{12}<0$ in the $R\rightarrow\infty$ limit? — Heidar, Oct 14 '12 at 12:34
@Heidar: For the case I considered with two disconnected regions 1 and 2, I think $\I_{12}$ is always non-negative. $I_{12}=0$ means these two regions are completely independent, i.e., $\rho_{12}=\rho_1\otimes \rho_2$. $I_{12}>0$ means there is at least some correlation function between 1 and 2 which is non-vanished. — Phynics, Nov 05 '12 at 00:10

Everett You · Answer 2 · 2012-06-03T02:57:58.620

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.

Here, Everett brought out a very good point. Without SSB, a ground state remain to be a pure state even with a little quantum decoherence. However, with SSB, even a very small quantum decoherence may put the system into a mixed state of the nearly degenerate ground state. This may lead to a understanding of SSB in quantum system, if this idea can be made more quantitative. — Xiao-Gang Wen, Jun 08 '12 at 02:55
@Xiao-GangWen: I've been thinking about exactly this quantitative example. I think the easiest model is to take a Heisenberg model with $N$ $1/2$-spins, couple one of these to a measurement spin, consider the ground state and explicitly perform the trace explicitly. Textbook answer is that all spins point "down", but clearly there are actually $2N+1$ degenerate states, which hopefully would just automatically fall out of the calculation. — genneth, Jun 17 '12 at 16:46
I'm not sure I agree that "entropy production ... cannot be described by the linear dynamics of quantum mechanics that preserves the entropy". Certainly the "external" entanglement remains the same under unitary time evolution, but the "internal" entanglement within the system can increase. For example, if we start in a pure product state of spin-1/2s pointing in random directions and evolve under the Heisenberg model Hamiltonian, the different spins will quickly become entangled even as the total state remains pure. Isn't this "internal" entanglement enough to explain SSB via decoherence? — tparker, Feb 11 '18 at 18:47

score 12 · Accepted Answer · edited Nov 12 '20 at 03:41

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

symmetric states $U_g \Psi = \Psi$ up to a phase, $g \in G$, and
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:

A symmetric many-body state has spontaneous symmetry breaking implies that the state has a GHZ entanglement.
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.
The symmetric exact ground state of a generic symmetric Hamiltonian has spontaneous symmetry breaking iff it has GHZ entanglement.

I wonder whether the first condition should be modified to 1. $ U_g \Psi \propto \Psi$ up to the symmetric charge? also the second condition should be modified by 2. $\Psi_\alpha$ may NOT be locally distinguishable? — wonderich, Nov 29 '16 at 06:19
I made a correction on a typo to "$H \subset G$." Do you have comments about what I wrote on Nov 29 '16? — wonderich, Sep 28 '20 at 21:56

score 10 · Answer 4 · answered Jun 02 '12 at 14:20

10

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.

answered Jun 02 '12 at 14:20

Jimmy Liu

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The problem is actually more subtle than that. In your example, the issue is that the split only occurs because $B$ is classical, and introduced by hand in order to effect a symmetry breaking. In reality, the universe is isotropic, so it becomes circular to ask where that terms comes from --- after all, you only generate such fields by having some broken symmetry state somewhere! Nevertheless, I think there is a reasonable description available, which in the appropriate limits gives the various textbook answers. – genneth Jun 02 '12 at 14:27
1

@Jimmy: I only understand the standard description of SSB. Here I am asking for a different understanding of SSB. It is nice that you bring in the issue of nearly degenerate ground states. For d-dimensional systems, the splitting between nearly degenerate ground states from discrete symmetry breaking scales like $\Delta \sim e^{-L^d/\xi^d}$ with the linear size $L$ of the system. For topo. order from string condensation, the splitting between nearly degenerate ground states scales like $\Delta \sim e^{-L/\xi}$. So the nearly degenerate ground states may lead to a deeper understanding of SSB. – Xiao-Gang Wen Jun 02 '12 at 17:38
It is interesting to note that the two $\Delta$ scale in the same way in d=1 dimension. This hints that there is no topological order in 1D. – Xiao-Gang Wen Jun 02 '12 at 17:39
2

Hi Jimmy, I think Prof Wen is actually asking about what happens as B goes to absolute zero, that the symmetry is suddenly broken. I feel that this is the same sort of question as how does Schordinger's cat settle down into a definite state of alive or dead. This is about the dynamic of quantum decoherence, which, I believe, is beyond the scope of our current understanding of quantum mechanics. – Everett You Jun 02 '12 at 17:44

score 10 · Answer 5 · answered Jun 05 '12 at 14:25

10

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.

answered Jun 05 '12 at 14:25

David Bar Moshe

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Indeed, variational calculation will make a linear quantum problem into a nonlinear classical problem (of minimizing a non-quadratic function). In practice, this is very helpful trick. But can we have to deeper understanding of SSB WITHIN the linear quantum theory? – Xiao-Gang Wen Jun 06 '12 at 02:44
1

May be it was not emphasized enough in the answer, but it was supposed to be the main point. Here, spontaneous symmetry breaking can be tested on the finite dimensional (effective) Hilbert space obtained from the quantization (in the sense of geometric quantization) of the trial function manifold. This is the Hilbert space of nonequivalent vaccua. The mere existince of a nontrivial Hilbert space after quantization is the indication of spontaneous symmetry breaking. – David Bar Moshe Jun 06 '12 at 08:31
Let us first consider a transverse-field Ising model with only two spins. In that case, do we have the finite dimensional (effective) Hilbert space obtained from the quantization of the trial function manifold? If yes, do you mean there is a SSB for even for a two spin system? (We know that a two spin system not suppose to have SSB.) If no, we can consider a three spin system, a four spin system, etc, and ask, for which $N$, do we have SSB? [ie for which $N$, do we have the finite dimensional (effective) Hilbert space obtained from the quantization, which is the indication of SSB] – Xiao-Gang Wen Jun 07 '12 at 11:06
@DavidBarMoshe So i understand that the ground states of spontaneously broken continuous $G \rightarrow H$ symmetries are parametrized by the homogeneous space $G/H$, but in which sense do we get an approximate classical dynamics on it? – Lorenz Mayer Oct 23 '18 at 17:19
@Lorenz Mayer, sorry for the late response. We get classical dynamics in the following sense: The couplings in the effective Nambu-Goldstone theory on $G/H$ can be shown to result from a summation over many loops in the microscopic theory. One example the chiral current anomalous non-conservation in the microscopic theory reveals itself at the tree level in the effective theory through the Wess-Zumino-Witten term. – David Bar Moshe Oct 29 '18 at 07:51

score 8 · Answer 6 · answered Jun 13 '12 at 01:00

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.

Could you clarify "the matrix elements of local physical operators (operators with local support) between different basis states they always vanish or not vanish in the thermodynamic limit" for the correspondence to SSB or not SSB --- I am confused which way it should be...? Do you have typos here? — wonderich, Sep 28 '20 at 22:43

score 7 · Answer 7 · answered Jun 01 '12 at 16:17

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).

answered Jun 01 '12 at 16:17

Thomas

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This is the standard description of SSB, and it works, but in a quite complicated way (for the quantum case). To see why it is complicate we can perform the described calculation for two spins (exactly). But after we get the quantum effective action, we find that there is no phase transition and no SSB. We also do the calculation for three spins, again no SSB. Magically, for large $N$ spins (or when $N=\infty$), phase transition and SSB do appear. The above understanding is certainly correct, but I wonder if there is a more direct and a deeper way to see SSB. – Xiao-Gang Wen Jun 02 '12 at 02:49
1

This has nothing to with quantum mechanics. A classical system of N (finite) spins does not have a phase transition either. – Thomas Jun 02 '12 at 15:39
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+1, This is also how I also imagined SSB (i.e., for @Xiao-GangWen's example of the 1D quantum Ising model, the partition function is equivalent to the classical 2D Ising model and the SSB is inherited). At the level of wild conjecture, this seems to suggest that SSB in a model that does not have a classical (sign-free) effective action requires a radically different explanation, though. – wsc Jun 06 '12 at 17:50
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@wsc: You have a very good point. A quantum system correspond to a statistical system only if the path integral description of the quantum system is a path integral over a positive definite functional. If the path integral is not positive definite, the quantum system has no statistical analogue, but it still has SSB. – Xiao-Gang Wen Jun 07 '12 at 10:52
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@Thomas: Here, we are talking about quantum phase transition or SSB phase transition at ZERO temperature. At zero temperature, a classical system of $N$ (finite) spins does have a phase transition as we change parameters, such as $B$ in the transverse-field Ising model. – Xiao-Gang Wen Jun 07 '12 at 10:56

score 5 · Answer 8 · edited Jun 17 '12 at 16:27

5

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

edited Jun 17 '12 at 16:27

Peter Morgan

9,922

answered Jun 17 '12 at 04:18

Rod Munoz

51

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Link only (especially when the link isn't even a hyperlink!) answers are generally discouraged and you may be setting yourself up for a bunch of downvotes. It is better to summarize the contexts of the link. In the case of a paper, excerpting the abstract may be sufficient. – dmckee --- ex-moderator kitten Jun 17 '12 at 15:26

Peter Morgan · Answer 9 · 2012-06-01T13:40:59.633

4

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.

edited Jun 01 '12 at 13:40

answered Jun 01 '12 at 13:12

Peter Morgan

9,922

Viewing SSB "in terms of violation of cluster decomposition" may be a very interesting direction. I need to learn cluster decomposition to see what is it. – Xiao-Gang Wen Jun 06 '12 at 02:47
@Xiao-Gang That a state satisfies cluster decomposition might be no more than to say that the quantum state is local (in the specific QT sense of cluster decomposition) as well as the algebra of observables being local (in the specific QT sense of microcausality). Neither idea of locality is precisely reproduced in classical dynamics, both being largely stochastic. In contrast, but at some level similarly, for SSB in a classical field theory the state picks out a global direction (in some space), whereas no direction is picked out by the dynamics. – Peter Morgan Jun 06 '12 at 13:07
@Xiao-Gang Perhaps the paper http://arxiv.org/abs/1201.5459v1, "Spontaneous Symmetry Breaking in Quantum Systems. A review for Scholarpedia", author F. Strocchi, might be interesting. – Peter Morgan Jun 23 '12 at 21:19
But if spontaneous symmetry breaking occurs, the Hilbert space decomposes into superselection sectors labeled by the charge of the symmetry broken vacuum state. Within each sector, the cluster decompostition principle holds. – Lorenz Mayer Nov 27 '18 at 12:01

score 3 · Answer 10 · answered Jun 02 '12 at 12:07

3

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.

answered Jun 02 '12 at 12:07

Joseph K.

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Cart before horse. Superselection sectors can be a convenient description of SSB, but not a definition. After all, superselection completely fails to describe the process where SSB emerges from finite systems as the thermodynamic limit is taken, and thus does not account for finite-size corrections. – genneth Jun 02 '12 at 14:35
@Joseph: Indeed, for finite $B$ in the transverse field Ising model with a finite spins, there is no superselection sectors. – Xiao-Gang Wen Jun 07 '12 at 11:15

score 1 · Answer 11 · edited Apr 13 '17 at 12:39

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.

score 1 · Answer 12 · answered Apr 07 '21 at 21:44

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.

What is spontaneous symmetry breaking in quantum systems?

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