In the book Condensed Matter Field Theory by Altland, on page 5, it is given that

$$H[\pi, \phi]=\int d x\left(\frac{\pi^{2}}{2 m}+\frac{k_{\mathrm{s}} a^{2}}{2}\left(\partial_{x} \phi\right)^{2}\right)$$ The Hamiltonian of an atomic chain is invariant under simultaneous translation of all atom coordinates by a fixed increment: $\phi_{I} \rightarrow \phi_{I}+\delta$, where $\delta$ is constant. This expresses the fact that a global translation of the solid as a whole does not affect the internal energy. Now, the ground state of any specific realization of the solid will be defined through a static array of atoms, each located at a fixed coordinate $R_{I}=I a \Rightarrow \phi_{I}=0 .$ We say that the translational symmetry is "spontaneously broken," i.e. the solid has to decide where exactly it wants to rest. However, spontaneous breakdown of a symmetry does not imply that the symmetry disappeared. On the contrary, infinite-wavelength deviations from the pre-assigned ground state come close to global translations of (macroscopically large portions of) the solid and, therefore, cost a vanishingly small amount of energy. This is the reason for the vanishing of the sound wave energy in the limit $\partial_{x} \phi \rightarrow 0 .$ It is also our first encounter with the aforementioned phenomenon that symmetries lead to the formation of soft, i.e. low-energy, excitations.

However, it is wrong to say that the ground state of the system will be given by $\phi (x) = 0$ Because the Hamiltonian depends on $\partial_x \phi$ not $\phi$ itself, so I don't understand how one can argue that "the solid has to decide where exactly it wants to rest".

I feel like there is a missing sentence before "We say that the translational symmetry is "spontaneously broken" because I can't understand why the translation symmetry is broken and under what conditions.


3 Answers 3


The spontaneous symmetry break happen only in thermodynamic limit(that is, sufficiently large number of particles or system sizes). In the example you gave there are indeed many degenerate ground state -- for any constant $c$ $\phi(x)=c$ is a ground state,since $H$ depends on $\partial_x \phi$ instead of $\phi$ itself, just as what you've mentioned. However, when we take thermodynamic limit, namely, let N (numbers of atoms here) go to infinity, we would finally get a specific $c$ instead of a linear combination of different $c$. In QFT of Weinberg you could see some formal description about this: any local operator will have zero matrix element between different state when the system size is large enough, so we could always consider one of the degenerate states -- in fact you could find a lot about SSB in Physic stack exchange, usually about spin model. For your example, we could consider the center of mass Hamiltonian $H=\frac{P_c^2}{2m}$. We set a small quadratic potential $V(r_c)=\frac{1}{2}\omega^2 x_c^2$ to confine it. The wavefunction will be $\psi(x_c)=\left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}\exp(-\frac{m\omega}{2\hbar}x_c^2)$, we will find that if we first take m to infinity (the string is long enough) the wave function will be a delta function and then we set $\omega$ to zero. This correspond to a certain choice of $c$. You will imediately find that if we change the sequence of taking limit the result will be change -- this is something usual appear in SSB. Naively you could think as this way: there will always be some small perturbation, and since $m$ is sufficiently large, any small perturbation (e.g quadratic potential centered in some point $X_c$ ) will lead to $\langle x_c\rangle=X_c$ and a specific choice among the degenerate ground states.


I think the key is this:

This expresses the fact that a global translation of the solid as a whole does not affect the internal energy. Now, the ground state of any specific realization of the solid will be defined through a static array of atoms each located at a fixed coordinate RI=Ia⇒ϕI=0.

It's contrasting moving all of the atoms together, versus how each atom individually acts.

Consider a sequence of masses connected by springs, arranged in a line along the x-axis. Let $U$ be the total potential energy stored in all the springs put together. Let $x_1$ be the x-coordinate of the first mass, $x_2$ the x-coordinate of the second, and so on. The choice of the origin is arbitrary. If we were to choose a different coordinate system with horizontal dimension $x'$, with $x'_1 = x_1+h$, $x'_2 = x_2+h$, etc., the physics would not change. $\frac {\delta U}{\delta h} =0$; translating the system as a whole doesn't affect the energy. However, $\frac {\delta U}{\delta x_i} \neq 0$ for any individual $i$; moving any mass individually change $U$, even though moving all of them simultaneously doesn't. Squishing a ball changes its internal energy, because its molecules are moving relative to each other, but moving it from one place to another doesn't.


The answer by Black Monolith is pretty good, so let me discuss the issue in a different, more general way.

We are taught that systems typically find themselves in their ground state. The implicit reasoning is that, if the system is coupled to some environment/bath/a bunch of other systems, then they can lose energy via that coupling, and by the second law of thermodynamics they typically will. You see that there are actually quite a lot of assumptions going on here.

Furthermore, you can consider the situation that there are many degenerate or almost-degenerate states. Then in which state, or in which combination of states, will the system end up?

The answer is that, in equilibrium, systems will be in stable states. A state is stable if its fluctuations (e.g. thermal fluctuations) are small. If you have some observable $\hat{A}$, then the fluctuations of $\hat{A}$ are quantified by its variance:

$$ \rm{Var} \hat{A} = \langle \hat{A} \hat{A} \rangle - \langle \hat{A} \rangle\langle \hat{A} \rangle$$

(Think about why that is, or look it up in a thermodynamics book.)

Now, spontaneous symmetry breaking leads to ordered states, which are macroscopic. The criterion for stability is very intuitive: the fluctuations of extensive quantities should not be extensive. (An extensive quantity is one whose magnitude scales with the volume or equivalently number of particles $N$ of the system.) Indeed, if fluctuations where of size $\sim N$, then clearly the state is not in equilibrium.

So we demand that

$$ \lim_{N \to \infty} \frac{\sqrt{ \rm{Var} \hat{A} }}{ \langle \hat{A} \rangle} \to 0,$$

which implies $\rm{Var} \hat{A} < \mathcal{O}(N^2)$.

So, in general, in equilibrium systems, we can demand that the fluctuations of any extensive quantity be small in the sense above.

In systems with spontaneous symmetry breaking, this holds in particular for the order parameter, which is the quantity that distinguishes the ordered from the disordered state. The above criterion leads to the fact that systems with SSB find themselves in eigenstates of their order parameter operator. But in almost all cases, the order parameter operator does not commute with the Hamiltonian, so their eigenstates are not energy eigenstates.

However, the broken states are all related by (broken) symmetry transformations and therefore have the same energy expectation value. Moreover, this expectation value is very close to the groundstate energy. In other words, these broken-symmetry states will be superpositions of energy eigenstates which are almost degenerate (they become exactly degenerate in the thermodynamic limit). Therefore the fluctuations in energy are extremely small.

On the other hand, the exact ground state of finte-size SSB systems has extremely large fluctuations of its order parameter (in almost all cases). For the case of the crystal, the energy eigenstates are also momentum eigenstates, and the ground state has total momentum 0. This means that its center of mass position is completely undefined! So this would be a superposition of crystals with all possible centers of mass.

So the symmetry must be broken, because the system must be in a stable state. The question of how the symmetry is broken is not answered. As mentioned by others, some external disturbance must be taken into account, which will fix the center-of-mass position. But once it has been fixed, it is stable against fluctuations by the above reasoning.

Summarizing: symmetry is broken because the system must be in a stable state. Energy fluctuations are nevertheless small, because the broken symmetry leads to a large set of almost-degenerate energy eigenstates.


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