Spontanous symmetry breaking in the Heisenberg model? Spontaneous symmetry breaking can be defined as follows (Teresa & Antonio, 1996; pg89):

A physical system has a symmetry that is spontaneously broken if the interactions governing the dynamics of the system possess such a symmetry
  but the ground state of this system does not.

I am slightly confused about the meaning of 'ground state' in this expression in the context of condensed matter. In the case of the Heisenberg model the ground state can be take as:
$$\rho\propto \sum_n \left|GS_n\right>\left<GS_n\right|$$
where $\left|GS_n\right>$ is the $n$th ground state. Now for such a ground state density matrix the $SU(2)$ symmetry of the Hamiltonian is preserved. Only in the case where we have a spontaneous symmetry breaking field do we get a density matrix  that does not preserve the $SU(2)$ symmetry of the Hamiltonian. But in such a case the Hamiltonian strictly speaking does not have this symmetry anyway. 
Does this system therefore strictly speaking have spontanous symmetry breaking? and what is meant by the term 'ground state' in the above definition does it refer to individual $\left| GS_n\right>$ or to the density matrix $\rho$? Does it refer to before or after symmetry breaking fields are applied?
 A: I would like to offer a contrasting answer to that of @tparker. I want to emphasize the fact that it is actually not necessary to introduce any symmetry-breaking field, provided you use the proper setup. (Of course, you can use symmetry-breaking fields, or suitable boundary conditions to achieve that, but I believe that it is conceptually interesting that you don't have to.)
I'll only discuss the case of classical systems, as I am much more familiar with the relevant mathematical framework.
The goal is to define a Gibbs measure for an infinite system. This is necessary, as only truly infinite systems undergo genuine phase transitions (of course, large finite systems can display approximate, smoothed-out "phase transitions"). The usual definition relying on the Boltzmann weight $e^{-\beta H}$ is useless in this case, as the energy of an infinite system is usually undefined.
The most efficient framework to solve this problem, the so-called Dobrushin-Lanford-Ruelle theory, works as follows. I discuss the case of the Ising model on $\mathbb{Z}^2$ for simplicity, but the approach is completely general. One says that a probability measure $\mu$ on the set of infinite configurations $\{-1,1\}^{\mathbb{Z}^2}$ is an infinite-volume Gibbs measure if, for any finite set $\Lambda\subset\mathbb{Z}^2$ and any configuration $\eta\in\{-1,1\}^{\mathbb{Z}^2\setminus\Lambda}$, the conditional probability of seeing a configuration $\sigma$ inside $\Lambda$, given that the configuration outside $\Lambda$ is given by $\eta$, is given by the finite-volume Gibbs measure in $\Lambda$ with boundary condition $\eta$. The latter is well-defined (through the usual Boltzmann weight), since $\Lambda$ is finite.
For models with compact spins (such as the Ising model or the Heisenberg model), existence of infinite-volume Gibbs measures is guaranteed. Uniqueness, however, does not hold in general. For the Ising model on $\mathbb{Z}^2$, for example, there exists a critical value $\beta_{\rm c}\in(0,\infty)$ of the inverse temperature such that there is a unique infinite-volume measure when $\beta\leq \beta_{\rm c}$, while there are infinitely-many infinite-volume Gibbs measures when $\beta>\beta_{\rm c}$. It turns out that any infinite-volume Gibbs measure $\mu$ can be expressed as a convex combination of two of them: $\mu = \alpha \mu^+_\beta + (1-\alpha)\mu^-_\beta$, for some $0\leq\alpha\leq 1$.  These two measures $\mu^+_\beta$ and $\mu^-_\beta$ thus contain all the relevant physics, and there are good reasons to consider them as the physically truly relevant ones. It turns out that the average magnetization under $\mu^+_\beta$ is equal to $m^*(\beta)>0$, the spontaneous magnetization (that is the value of the magnetization that you would get, had you first added a magnetic field $h>0$ and then let $h$ decrease to $0$). Under $\mu^-_\beta$, on the other hand, the average magnetization is $-m^*(\beta)<0$. In this precise sense, there is spontaneous symmetry breaking, even though the procedure described above does not explicitly break the symmetry at any step.  
Let me now briefly link the above approach (still for the Ising model on $\mathbb{Z}^2$) with the one using an explicit symmetry breaking. A standard way to proceed in this case is to consider an increasing sequence of finite subsets $\Lambda_n\subset\mathbb{Z}^2$. We then consider the Gibbs measure $\mu^+_{\Lambda_n;\beta}$ associated to the Ising model in $\Lambda_n$, with $+$ boundary condition. One can then consider the (weak) limit of the probability measures $\mu^+_{\Lambda_N;\beta}$ as $\Lambda_n$ grows to cover $\mathbb{Z}^2$. It turns out that the limiting measure coincides with the infinite-volume Gibbs measure $\mu_\beta^+$ derived above. Of course, one recovers the measure $\mu^-_\beta$ by using a sequence with $-$ boundary condition.
So the two approached yield the same result, but I insist that the former does not require explicit symmetry breaking. (Moreover, it provides a much more powerful framework.)
A: Three comments:


*

*It's true that at some point you need a tiny symmetry-breaking field. But it doesn't need to act on every site uniformly - even a field that acts on a single site is enough to break the symmetry. Realistically, you can't be expected to be able to able keep track of these tiny fields in a real system. So philosophically I suppose you could say that "symmetry breaking" doesn't happen in real life because the symmetry is never exact - but the point is that the system is unstable to tiny asymmetries that you can't realistically keep track of.

*You are assuming that the Hamiltonian doesn't change in time. But the point of SSB is that even a momentary symmetry-breaking field is enough to permanently break the symmetry. Once the system falls into a symmetry-breaking configuration, it gets "stuck" and can't "climb out" back into the symmetric configuration.

*It's true that the Gibbs ensemble is always symmetric. But the point of SSB is that in the symmetry-broken phase, the system isn't described by the Gibbs ensemble, but only by an asymmetric subset of the ensemble. So you only find yourself in one of the ground states not in an equally weighted mixture of all of them. This is called "ergodicity breaking".
A: The book  (Altland and Simons, 2010; pg258) (or pg263 in the 2006 edition) states the following (verbatim quote from 2006 ed):

In spite of the undeniable existence of solids, magnets, and Bose condensates of definite phase, the notion of a ground state that does not share the fully symmetry of the theory may appear paradoxical, or at least "unnatural". For example, even if any particular ground state of the "Mexican hat" potential shown in the figure above "breaks" the rotational symmetry, shouldn't all these states enter the partition sum with equal weight, such that the net outcome of the theory is again symmetric?

The book then goes onto discuss how symmetry breaking fields produce symmetry breaking as an observable phenomenon. From this discussion I therefore gather the following two things:


*

*Spontaneous symmetry breaking refers to the ground states $\left| GS_n \right>$ not possessing the same symmetry as the dynamics.

*The reason we see spontaneous symmetry breaking is due to imperfections.

