Spontaneous Symmetry Breaking

Abstract

The roots of the modern understanding of symmetries in physics can be traced to the work of Sophus Lie on transformations of differential equations in the nineteenth century. In this chapter, the history of the subject is used to emphasize the distinction between the symmetry of the dynamics of a physical system and the symmetry of its specific state. The central concept of spontaneous symmetry breaking is then introduced as the discrepancy between the two notions of symmetry. It is shown how the presence of spontaneous symmetry breaking can be detected through the sensitivity of measurements on the physical state to external perturbations. This naturally leads to the concepts of the order parameter and the vacuum manifold. The rest of the chapter is devoted to highlighting some subtle features of spontaneous symmetry breaking. These include the properties of the ground state and the tower of low-lying excited states in a finite volume, and the absence of well-defined unitary operators representing broken symmetry in the thermodynamic limit.

In this chapter, I will finally spell out a formal definition of spontaneous symmetry breaking (SSB). While the previous chapter was largely phrased in terms of classical field theory, I will now switch entirely to the quantum-theoretic language. This is more than justified by Sect. 5.3, which gives the reader a flavor of the many subtleties, associated with SSB in quantum systems. Some further details on the physical aspects of SSB can be found for instance in the recent lecture notes [1]. A mathematically oriented reader will find the classic book [2] a unique source of additional information.

To motivate what comes below, let me briefly return to the roots of Lie group theory in the study of symmetries of differential equations. Suppose that we have a set of (partial) differential equations invariant under some group G of transformations. A given solution to the equations can fall into one of two classes:

• (i) The solution is not invariant under G. This is the generic case, in which one can use the symmetry to obtain new solutions by the action of transformations from G on the solution already known.

• (ii) The solution is invariant under G. Such solutions are special, and are usually easier to find than a general solution to the same equations. The reason for this is that assuming a priori G-invariance of the solution reduces the number of independent variables of the differential equations.

The situation in quantum physics maps closely to what one does in the context of the theory of differential equations. The major difference is that one tends to focus on the eigenstates of the quantum Hamiltonian as the primary tool to investigate physical observables. For this reason, it is common to start with the ground state, or at least a metastable equilibrium state, and study the excitations above it. This is the quantum counterpart of the theory of small oscillations in classical mechanics [3]. We will see that quantum physics offers natural analogs to both of the above-mentioned cases (i) and (ii) one encounters in differential equations. In both cases, symmetry has profound consequences for the structure of the excitation spectrum. For most of this chapter, I will focus on the symmetry properties of the equilibrium state. I will return to the excitation spectrum in Sect. 5.3, and then again in detail in Chap. 6.

1 Physical State and Its Symmetry

A proper discussion of SSB requires a number of new concepts. In order to make the narrative as natural as possible, I will augment it with some simple examples from condensed-matter and statistical physics. This will highlight the close connection between SSB and phase transitions. To parallel the language of statistical mechanics, I will represent the state of a quantum system with a density operator (or density matrix) $\varrho$. Readers who need to refresh their memory of the density operator formalism will find an undergraduate-level introduction for instance in [4,5,6].

We already know that symmetries of quantum systems are represented, if only formally, by unitary operators. An operator U is said to constitute a symmetry of the (pure or mixed) state $\varrho$ if the density operators $\varrho$ and $U\varrho {U}^{\dagger }$ are indistinguishable by any measurement. The latter means equal probabilities for any specific outcome of any measurement, and by extension equal averages of all observables. Insofar as our description of the state of the system is free of redundancies, this implies that the density operators must be equal,

$$\displaystyle \begin{aligned} U\varrho {U}^{\dagger}=\varrho \qquad \text{(symmetry of a state)}\;. {} \end{aligned} $$

(5.1)

Suppose that $\varrho$ is a pure state, that is $\varrho =\left \lvert {\psi }\right \rangle \left \langle {\psi }\right \rvert$ for some normalized ket-vector $\left \lvert {\psi }\right \rangle$ in the Hilbert space. Then a unitary operator U represents a symmetry of $\varrho$ if and only if $\left \lvert {\psi }\right \rangle$ is an eigenstate of U. One side of the equivalence is obvious: if $\left \lvert {\psi }\right \rangle$ is an eigenstate of U, then . To prove the opposite implication, write $\left \lvert {\left \langle {\psi }\right \rvert U\left \lvert {\psi }\right \rangle }\right \rvert ^2$ as . Thus, the states $\left \lvert {\psi }\right \rangle$ and $U\left \lvert {\psi }\right \rangle$ are both normalized to unity and have a unit overlap, which is only possible if they are equal up to a phase. It follows that $\left \lvert {\psi }\right \rangle$ is an eigenstate of U.

In the following, I will frequently use a shorthand notation for the average of an observable A in the state $\varrho$,

$$\displaystyle \begin{aligned} \langle{A}\rangle _\varrho \equiv\operatorname{\mathrm{tr}}(\varrho A)\;. \end{aligned} $$

(5.2)

It is a simple corollary of (5.1) that a symmetry transformation of observables, $A\to {U}^{\dagger }AU$, does not affect their average in symmetric states. To see this, write as , which is just $\langle {A}\rangle _\varrho$. This conclusion holds for discrete and continuous symmetries alike.

1.1 Broken and Unbroken Symmetry

Suppose now that the Hamiltonian of a system is invariant under a set of transformations that span a group G. These are represented, even if just formally, by a set of unitary operators $U(g)$, $g\in G$. A randomly chosen state $\varrho$ will not be symmetric under all of these. Those transformations that are symmetries of $\varrho$ will span a subgroup of G called the unbroken subgroup of $\varrho$, denoted by

$$\displaystyle \begin{aligned} H_\varrho \equiv\{g\in G\,\vert\,U(g)\varrho {U(g)}^{\dagger}=\varrho \}\;. \end{aligned} $$

(5.3)

All other elements of G are said to be spontaneously broken in the state $\varrho$. This is our formal definition of SSB.

Example 5.1

Pure spin states of a spin-$1/2$ particle comprise a two-dimensional Hilbert space isomorphic to $\mathbb {C}^2$. This Hilbert space carries a representation of the spin group, $G\simeq \mathrm {SU}(2)$, generated by the spin operator $\boldsymbol S$. Consider the projection of spin to a direction, defined by the unit vector $\boldsymbol n=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$,

$$\displaystyle \begin{aligned} \boldsymbol{n}\cdot\boldsymbol{S}=\frac\hbar2\boldsymbol{n}\cdot\boldsymbol{\tau} =\frac\hbar2 \begin{pmatrix} \cos\theta & \mathrm{e}^{-\mathrm{i}\varphi}\sin\theta\\ \mathrm{e}^{\mathrm{i}\varphi}\sin\varphi & -\cos\theta \end{pmatrix}\;. \end{aligned} $$

(5.4)

Here $\theta ,\varphi$ are the standard spherical angles and $\boldsymbol \tau$ is the vector of Pauli matrices. The projected spin operator has an eigenvector $\left \lvert {\boldsymbol n,+}\right \rangle$ with eigenvalue $+\hbar /2$,

$$\displaystyle \begin{aligned} \boldsymbol{n}\cdot\boldsymbol{S}\left\lvert{\boldsymbol n,+}\right\rangle =\frac\hbar2\left\lvert{\boldsymbol n,+}\right\rangle \;,\qquad \left\lvert{\boldsymbol n,+}\right\rangle =\begin{pmatrix} \cos\theta/2\\ \mathrm{e}^{\mathrm{i}\varphi}\sin\theta/2 \end{pmatrix}\;. \end{aligned} $$

(5.5)

Any nonzero vector in $\mathbb {C}^2$ can, up to an overall factor, be cast in this form with a suitable choice of $\theta ,\varphi$. Hence, any nonzero vector in this Hilbert space is an eigenstate of a projection of the spin operator to some direction. We conclude that for any pure state of a spin-$1/2$ particle, the unbroken subgroup is isomorphic to $\mathrm {U}(1)$, and consists of spin rotations generated by $\boldsymbol {n}\cdot \boldsymbol {S}$.

How do we check whether a given symmetry is spontaneously broken in a given state $\varrho$? For the density operators $\varrho$ and $U\varrho {U}^{\dagger }$ to be different, there must be at least one observable A whose average in the two states differs. This observable must itself not be invariant under the symmetry, as ${U}^{\dagger }AU=A$ would imply

$$\displaystyle \begin{aligned} \langle{A}\rangle _{U\varrho {U}^{\dagger}}=\operatorname{\mathrm{tr}}(U\varrho {U}^{\dagger}A)=\operatorname{\mathrm{tr}}(\varrho {U}^{\dagger}AU)=\operatorname{\mathrm{tr}}(\varrho A)=\langle{A}\rangle _\varrho \;. \end{aligned} $$

(5.6)

In order to detect SSB, one must therefore use an observable that itself breaks the symmetry. This fairly simple observation forms a conceptual foundation of the approach to SSB developed in Sect. 5.2.

One observable that is by definition invariant under all the operators $U(g)$ from the symmetry group is the Hamiltonian itself. Hence, for those $g\in G$ that are spontaneously broken, the state $U(g)\varrho {U(g)}^{\dagger }$ differs from $\varrho$ but has the same average energy. This presents a serious conundrum. Suppose that $\varrho$ is a purported ground state of a quantum system. From quantum mechanics, we are used to thinking about the ground state as a unique vector in the Hilbert space. But now $U(g)\varrho {U(g)}^{\dagger }$ for any $g\notin H_\varrho$ is a different state of the same energy. How do we know which of the different states having the same energy to choose as the ground state?

We noticed already in Chap. 2 that SSB implies the existence of degenerate ground states. This property is so universal that it earned the privilege of being my first “moral lesson” in Sect. 2.4. But now our problem is even worse. In the classical analysis of Chap. 2, the various candidate ground states could be uniquely labeled by the expectation value of a scalar field. In the present fully quantum setting, we have not only the set of density operators $U(g)\varrho {U(g)}^{\dagger }$ with $g\in G$. We can even make linear combinations of these states. In the desire to identify a unique ground state, we might wish to take a “democratic average” of all the would-be ground states, connected by transformations from G. This is an idea that certainly deserves at least consideration, even if it will eventually turn out to be physically incorrect. However, the related mathematical procedure of averaging over different elements of the symmetry group will actually prove quite useful.

1.2 Symmetrization by Group Averaging

When performing an average over the symmetry group, I will repeatedly make use of a simple statement that I will formulate as a standalone lemma. Let G be a finite group or a compact Lie group, and $U(g)$ be its unitary representation on a finite-dimensional vector space. Then

$$\displaystyle \begin{aligned} \mathcal{P}\equiv\frac 1{|G|}\sum_{g\in G}U(g) {} \end{aligned} $$

(5.7)

is a projector to the subspace of vectors invariant under G. In particular, if the representation of G by $U(g)$ does not contain in its decomposition any singlet (that is, a trivial one-dimensional representation of G), then the action of $\mathcal {P}$ on any vector gives zero. The proof of the lemma is very simple and I will therefore leave it up to the reader. Let me just add a comment on the sum over group elements. Equation (5.7) as it stands is only valid for finite groups; $|G|$ then denotes the number of group elements. For compact Lie groups, the sum has to be replaced by group integration with an invariant measure on G; $|G|$ then stands for the volume of G with respect to this measure. An interested reader will find more details in Chap. 3 of [7]. In the following, I will pragmatically use a sum over $g\in G$ as in (5.7) without repeating this disclaimer.

Let us now get back to density operators. The action of the group G on these is defined by $g:\varrho \to U(g)\varrho {U(g)}^{\dagger }$. This suggests that we consider the following statistical average,

$$\displaystyle \begin{aligned} \bar\varrho \equiv\frac 1{|G|}\sum_{g\in G}U(g)\varrho {U(g)}^{\dagger}\;. {} \end{aligned} $$

(5.8)

This average still satisfies the requirements on a density operator: it is Hermitian, positive-semidefinite and has a unit trace. Moreover, it is manifestly invariant under all transformations from G. It therefore represents a (mixed) physical state of the system which is perfectly symmetric under the whole group G, yet has the same average energy as $\varrho$.

In case the original state is pure, $\varrho =\left \lvert {\psi }\right \rangle \left \langle {\psi }\right \rvert$, we have the alternative choice to symmetrize the ket-vector $\left \lvert {\psi }\right \rangle$ rather than the density operator,

$$\displaystyle \begin{aligned} \left\lvert{\psi}\right\rangle \to\mathcal{P}\left\lvert{\psi}\right\rangle =\frac 1{|G|}\sum_{g\in G}U(g)\left\lvert{\psi}\right\rangle \;. {} \end{aligned} $$

(5.9)

This averaged ket-vector, if well-defined, represents a pure state of the same energy as $\left \lvert {\psi }\right \rangle$ that is symmetric under the whole group G. The catch is that the ket-vector may be ill-defined. By our symmetrization lemma, this happens whenever $\left \lvert {\psi }\right \rangle$ belongs to a representation of G that does not contain any singlet. Recall, for instance, Example 5.1. There, the whole Hilbert space $\mathbb {C}^2$ constitutes an irreducible representation of the spin group $G\simeq \mathrm {SU}(2)$. The naive average (5.9) therefore gives zero no matter which pure state we start from. On the other hand, the average density operator (5.8) must describe some physical state of the spin-$1/2$ particle. By Schur’s lemma, $\bar \varrho$ must be proportional to the identity operator on $\mathbb {C}^2$. The requirement that $\operatorname {\mathrm {tr}}\bar \varrho =1$ then fixes the normalization so that $\bar \varrho =(1/2) \operatorname {\mathrm {id}}$, for any choice of the initial state $\varrho$. This mixed state describes an unpolarized distribution on the Hilbert space.

It seems that we managed to have our cake and eat it. By the simple averaging procedure, we can make any state G-invariant without changing its energy. Have we done away with the whole notion of SSB? To see that the group averaging is not so innocuous, we need to work out the consequences. Consider a set of observables $A^i$ such that under the action of the operators $U(g)$, they transform by some (real and orthogonal) matrix representation $\mathcal {R}$ of G,

(5.10)

Suppose that all these observables and their nontrivial linear combinations break (part of) the symmetry under G. In other words, the representation $\mathcal {R}$ does not contain any singlet of G in its decomposition. It then follows as a simple consequence of the definition of the density operator (5.8) and the symmetrization lemma that

(5.11)

We could have anticipated this: in the G-invariant state $\bar \varrho$, the average of any observable breaking the symmetry vanishes.

Why should this be bad? Any quantum system possesses a distinguished state invariant under its symmetry group G: the thermodynamic equilibrium. This is represented by the canonical density operator

$$\displaystyle \begin{aligned} \varrho _\beta\equiv\frac 1Z\mathrm{e}^{-\beta H}\;, \end{aligned} $$

(5.12)

where $\beta$ is the inverse temperature and $Z\equiv \operatorname {\mathrm {tr}}\exp (-\beta H)$ the canonical partition function. The claim therefore is that in thermodynamic equilibrium, any observable breaking the symmetry of the system must average to zero. There cannot be any SSB in thermodynamic equilibrium. This would be very pretty, were it not in a blatant contradiction with empirical evidence.

Example 5.2

Solid ferromagnets can be visualized in terms of a lattice of mutually aligned spins (see the left panel of Fig. 5.1). The ferromagnetic state obviously breaks rotational symmetry. Moreover, the presence of the crystal lattice also breaks spatial translations. Both rotations and translations are fundamental symmetries that are expected to be preserved by the microscopic dynamics of any material. Hence ferromagnetism certainly is a macroscopic manifestation of SSB.

Fig. 5.1

Comparison of the symmetry-breaking states of ferromagnets and nematic liquid crystals. The symmetry of the former corresponds to that of a regular lattice of oriented, mutually aligned spins. Nematics, on the other hand, can be represented by mutually aligned undirected rods with random positions

In nematic liquid crystals (right panel of Fig. 5.1), long organic molecules are aligned along the same direction. However, the positions of the molecules are random as in ordinary liquids. As far as symmetry is concerned, the molecules can be represented by (undirected) rigid rods or ellipsoids with randomized positions. Hence the nematic state breaks rotational symmetry but not spatial translations. See Sect. 2.7 of [8] for basic phenomenology of liquid crystals.

One of the greatest accomplishments of physics is the discovery that the fundamental laws of nature possess a high degree of symmetry. How is it then possible at all that we observe so many manifestly asymmetric macroscopic states of matter? There seem to be two logical possibilities to avoid a contradiction:

• The observed asymmetric states are not in thermodynamic equilibrium, and thus may be at most metastable.

• The asymmetry of the observed states is a result of perturbations breaking the symmetry of the microscopic interactions.

Both of these possibilities are relevant. In order to understand their role, we have to dive deeper into the thermodynamics of physical systems with SSB. In the next section, I will focus solely on the canonical density operator $\varrho _\beta$ as the state of greatest importance for macroscopic observations.

2 Effect of External Perturbations

In Sect. 5.1.2, I introduced the set of operators $A^i$ as mere observables. Let us now see what happens if we allow them to actually affect the dynamics of the system. To that end, I couple each $A^i$ to an external field $\lambda _i$ and perturb the Hamiltonian of the system by a term linear in both,

$$\displaystyle \begin{aligned} H(\lambda)\equiv H-\lambda_iA^i\;. \end{aligned} $$

(5.13)

In the case of a spin-$1/2$ particle discussed in Example 5.1, $A^i$ can be chosen for instance as the components of the spin operator. The perturbations $\lambda _i$ can then be interpreted as an external magnetic field. In general, one can think of $\lambda _i$ as a set of “chemical potentials” associated with the observables $A^i$. In presence of the perturbations, the canonical density operator is deformed to

$$\displaystyle \begin{aligned} \varrho _{\beta,\lambda}=\frac 1{Z(\lambda)}\mathrm{e}^{-\beta H(\lambda)}\;, \end{aligned} $$

(5.14)

and the modified partition function is . The symmetry of the Hamiltonian H under the group G is reflected in the invariance of the partition function under the transformation of its variables. This follows from a simple manipulation,

(5.15)

It is convenient to trade the partition function for the free energy $F(\lambda )$ by . The free energy inherits the symmetry of the partition function under G-transformations of $\lambda _i$. Moreover, it makes it easy to evaluate the statistical average of $A^i$ in presence of the perturbations,

$$\displaystyle \begin{aligned} \langle{A^i}\rangle _{\varrho _{\beta,\lambda}}=\frac 1{Z(\lambda)}\operatorname{\mathrm{tr}}\big\{A^i\exp\big[-\beta(H-\lambda_jA^j)\big]\big\}=-\frac{\partial{F(\lambda)}}{\partial{\lambda_i}}\;. {} \end{aligned} $$

(5.16)

One of the distinguishing features of ferromagnets is that once magnetized by an external magnetic field, the macroscopic magnetization persists even after the external field is turned off. With this in mind, we will be particularly interested in the behavior of $\langle {A^i}\rangle _{\varrho _{\beta ,\lambda }}$ when the volume V  of the system is large and simultaneously $\lambda _i$ is small. In other words, we want to take the double limit $V\to \infty$ and $\lambda _i\to 0$. This is where interesting things start to happen.

2.1 Taking the Thermodynamic Limit

In a finite volume and for any nonzero temperature, the partition function, and thus the free energy, is analytic in $\lambda _i$. When all the $\lambda _i$ are small, one can therefore perform a Taylor expansion,

$$\displaystyle \begin{aligned} F(\lambda)=F(0)+\left.{\frac{\partial{F(\lambda)}}{\partial{\lambda_i}}}\right\rvert_{\lambda=0}\lambda_i+\mathcal{O}(\lambda^2)\;. \end{aligned} $$

(5.17)

But the linear term in the expansion is forbidden by the G-invariance of the free energy combined with the symmetrization lemma. Thus, in a finite volume and at nonzero temperature, one has a well-defined limit

$$\displaystyle \begin{aligned} \lim_{\lambda\to0}\langle{A^i}\rangle _{\varrho _{\beta,\lambda}}=0\;. {} \end{aligned} $$

(5.18)

This should not be particularly surprising.

How does the behavior of free energy as a function of $\lambda _i$ change when we take the limit of infinite volume or zero temperature? Some common mathematical features of these two limits may be discussed jointly. It is however the large-volume (thermodynamic) limit that is responsible for the existence of macroscopically distinct phases of matter. Only in special cases may the zero-temperature limit lead to nontrivial physics even in a finite volume; I will give an explicit example below. The general discussion of the effects of external perturbations will however be phrased in terms of the limit $V\to \infty$. The temperature will be assumed to be fixed and nonzero unless explicitly stated otherwise.

It is convenient to trade the free energy for its density, $\mathcal {F}(\lambda )\equiv {F(\lambda )}/V$, which is expected to be well-defined for any volume, including the limit $V\to \infty$. It is sensible to assume that even in this limit, the free energy density remains continuous as a function of $\lambda _i$. However, the analyticity may be lost. With regard to the behavior of the average of $A^i$ around the point $\lambda _i=0$, there are three possible scenarios:

• The free energy density remains analytic in the limit $V\to \infty$. This automatically guarantees the validity of (5.18). There is no SSB once the external field $\lambda _i$ is turned off.

• The free energy density becomes nonanalytic in the limit $V\to \infty$, yet (5.18) remains valid. This is an interesting borderline case that is relevant for the physics of low-dimensional systems.

• The free energy density becomes nonanalytic in the limit $V\to \infty$. The average $\langle {A^i}\rangle _{\varrho _{\beta ,\lambda }}$ remains nonzero even if the limit $\lambda _i\to 0$ is subsequently taken. This is the paradigm of SSB, illustrated by the right panel of Fig. 5.2.

  Fig. 5.2

  

  Schematic illustration of scenarios in which the free energy density $\mathcal {F}$ is (left panel) and is not (right panel) analytic at $\lambda _i=0$. In the latter case, the observables $A^i$ may retain a nonzero average even in the limit $\lambda \to 0$. Its value may however depend on the way the limit is taken

An alert reader might wonder why I focused on possible nonanalytic behavior of $\mathcal {F}$ at $\lambda _i=0$, yet tacitly assumed that analyticity survives the thermodynamic limit for nonzero $\lambda _i$. This is intimately related to the degeneracy of the ground (equilibrium) state. What I really assumed was that adding the external fields $\lambda _i$ breaks the symmetry of the system sufficiently so as to make the ground state unique. I will further elaborate on this assumption in Sect. 5.2.2. The point is that the free energy density typically becomes nonanalytic as a consequence of a competition of degenerate ground states with different values of $\langle {A^i}\rangle _{\varrho _{\beta ,\lambda }}$. This is something one expects only at the G-invariant point $\lambda _i=0$.

It is instructive to work out at least one concrete example to verify our prediction how SSB may emerge when a particular limit of the equilibrium state is taken. Ideally, this example should be exactly solvable in order to leave no doubt about the validity of the conclusions. It is therefore time to meet the Ising model, the most well-known and best-understood model of ferromagnetism in statistical physics. The price for all its benefits is that the limit in which the free energy density becomes nonanalytic is realized by taking the temperature to zero; the $V\to \infty$ limit itself is nearly trivial. I will largely follow the treatment in Sect. 2.2 of [9].

Example 5.3

Consider a one-dimensional chain of N spin variables $\sigma _i$, allowed to take values $\pm 1$. The Hamiltonian is given by

$$\displaystyle \begin{aligned} H=-J\sum_{i=1}^N(\sigma_i\sigma_{i+1}-1)-B\sum_{i=1}^N\sigma_i\;, {} \end{aligned} $$

(5.19)

where $\sigma _{N+1}$ is identified with $\sigma _1$ to ensure a periodic boundary condition. The (positive) coupling J represents interaction of nearest-neighbor spins, and B an external magnetic field. In the absence of the external field, the Hamiltonian (5.19) has a $G\simeq \mathbb {Z}_2$ symmetry under the spin flip $\sigma _i\to -\sigma _i$.

The partition function of the model is easily calculable thanks to the fact that the exponential of a sum equals a product of exponentials. Thus, the contribution of a spin configuration $\{\sigma _i\}_{i=1}^N$ to the partition function equals $\prod _{i=1}^NV(\sigma _i,\sigma _{i+1})$, where

$$\displaystyle \begin{aligned} V(\sigma,\sigma')\equiv\exp\left[\beta J(\sigma\sigma'-1)+\frac{\beta B}2(\sigma+\sigma')\right]\;. \end{aligned} $$

(5.20)

The latter can be thought of as a matrix element of the so-called transfer matrix,

$$\displaystyle \begin{aligned} V=\begin{pmatrix} \mathrm{e}^{\beta B} & \mathrm{e}^{-2\beta J}\\ \mathrm{e}^{-2\beta J} & \mathrm{e}^{-\beta B} \end{pmatrix}\;. \end{aligned} $$

(5.21)

With the periodic boundary condition on the spins, the partition function of the Ising model is now $Z= \operatorname {\mathrm {tr}} V^N=\lambda _+^N+\lambda _-^N$, where are the eigenvalues of the transfer matrix. The average spin at each site of the chain is obtained by applying (5.16) to $\mathcal {F}(B)\equiv F(B)/N$,

(5.22)

The total number of spins N plays the role of volume, and sending it to infinity amounts to the thermodynamic limit.

It is obvious from (5.22) that the Ising model does not feature SSB at any nonzero temperature. Indeed, whether N is finite or not, the average spin necessarily vanishes in the limit $B\to 0$ whenever the temperature is nonzero. On the other hand, taking the temperature to zero gives

$$\displaystyle \begin{aligned} \lim_{\beta\to\infty}\langle{\sigma_i}\rangle _{\varrho _{\beta,B}}=\operatorname{\mathrm{sgn}} B\;, {} \end{aligned} $$

(5.23)

again regardless of whether N is finite or not. Taking subsequently a (one-sided) limit of vanishing magnetic field will render the average spin nonzero, with a sign aligned with that of the magnetic field. This can be traced to the properties of the canonical density operator, which in the zero-temperature limit becomes a projector to the subspace of states with lowest energy. For the Ising Hamiltonian (5.19),

(5.24)

where $\left \lvert {++\dotsb }\right \rangle$ and $\left \lvert {--\dotsb }\right \rangle$ denote respectively the normalized states with all spins being positive and negative. Note that the $\mathbb {Z}_2$-symmetric state on the second line of (5.24) is unstable: an infinitesimally weak external field will project it onto one of the two fully polarized states.

The example demonstrates that taking an appropriate (infinite-volume or zero-temperature) limit in the presence of a perturbation may lead to an equilibrium state violating the symmetry of the system. The system then remains in an asymmetric state even if the perturbation is subsequently removed. It is very important that the limits of infinite volume and vanishing external field are performed in this order. Indeed, taking $\lambda _i\to 0$ first would delete the average of any symmetry-breaking observable in line with (5.18). Colloquially, one often speaks of the “noncommutativity of limits,”

$$\displaystyle \begin{aligned} \lim_{\lambda\to0}\lim_{V\to\infty}\langle{A^i}\rangle _{\varrho _{\beta,\lambda}}\neq\lim_{V\to\infty}\lim_{\lambda\to0}\langle{A^i}\rangle _{\varrho _{\beta,\lambda}}\;, \end{aligned} $$

(5.25)

as a smoking gun of SSB.

This concludes our excursion to thermodynamics. We have seen why asymmetric macroscopic states of matter can—and why symmetric macroscopic states may not—be stable. Identifying a (meta)stable equilibrium is a necessary first step if we ever want to understand the spectrum of a quantum system. Our next goal is therefore to establish a general operational procedure for finding a stable ground state.

2.2 Order Parameter and the Vacuum Manifold

In the Ising model, it was enough to distinguish the two possible asymmetric ground states by the sign of the external field. This is eventually because the model possesses a discrete $\mathbb {Z}_2$ symmetry. In systems with continuous symmetry we expect a continuum of degenerate equilibrium states. An already familiar example is the isotropic ferromagnet, where the spontaneous magnetization can take an arbitrary direction in space. Thus, we must be careful when taking the limit of vanishing external fields.

Suppose that we are given a specific set of fields, $\hat \lambda _i$. I will define the limit of vanishing perturbations by changing the magnitude but keeping the “direction” of these fields. This can be implemented precisely by using a positive scaling parameter $\epsilon$. We can thus construct a macroscopic thermal state by first taking the thermodynamic limit and then removing the external fields via

$$\displaystyle \begin{aligned} \varrho ^\infty_{\beta,\hat\lambda}\equiv\lim_{\epsilon\to0^+}\lim_{V\to\infty}\varrho _{\beta,\epsilon\hat\lambda}\;. {} \end{aligned} $$

(5.26)

In presence of SSB, the average

$$\displaystyle \begin{aligned} \langle{A^i}\rangle _{\varrho ^\infty_{\beta,\hat\lambda}}=\lim_{\epsilon\to0^+}\lim_{V\to\infty}\langle{A^i}\rangle _{\varrho _{\beta,\epsilon\hat\lambda}} {} \end{aligned} $$

(5.27)

is expected to be nonzero. This is an important proxy of SSB, known as the order parameter. It does exactly what its name suggests: parameterize the appearance of order in the thermodynamic state of matter described by the density operator .

Is the state (5.26) stable? This question is closely related to the choice of observables $A^i$. Namely, the set $A^i$ should be “complete” in the sense that the values of the order parameters determine the thermal state uniquely. In the limit of zero temperature, this means that should be a pure state, analogously to the $B\neq 0$ states in (5.24) for the Ising model.

Example 5.4

In the Ising model, the role of the order parameter is played by the average spin $\langle {\sigma _i}\rangle$. According to (5.23), in the zero-temperature limit in the presence of a nonzero magnetic field, $\langle {\sigma _i}\rangle$ acquires one of two possible values. These uniquely specify the corresponding symmetry-breaking pure ground states via (5.24).

In an isotropic ferromagnet, the order parameter can be taken as the average of the operator of total spin (magnetization) $\boldsymbol S$. Specifying the vector $\langle {\boldsymbol S}\rangle$ is then sufficient to determine the stable macroscopic equilibrium uniquely. This is however not necessarily the only possible choice of order parameter. Instead of the three components of the vector $S_i$, we might try to consider for instance the traceless symmetric tensor $T_{ij}\equiv S_iS_j+S_jS_i-(2/3)\delta _{ij}\boldsymbol S^2$. This set of operators transforms under a nontrivial irreducible representation of the spin group $G\simeq \mathrm {SU}(2)$, hence it satisfies the requirements we have imposed on the operators $A^i$. The average $\langle {T_{ij}}\rangle$, however, does not specify a unique equilibrium state. Namely, this order parameter cannot distinguish between the two states that only differ by the overall sign of $\langle {\boldsymbol S}\rangle$.

The order parameter is usually much easier to determine, or at least estimate, than the actual equilibrium state, . Its real value lies in the fact that it carries the same information about the symmetry of the equilibrium state as the density operator itself. To see this, just note that

(5.28)

This shows that for any $g\in G$ that is a symmetry of , the matrix $\mathcal {R}(g)$ leaves the order parameter unchanged. But the opposite is true as well. For any $g\in G$ such that , the assumed completeness of the set of operators $A^i$ guarantees that the states and are the same. The pattern of SSB in a macroscopic equilibrium state is therefore completely determined by the associated order parameter(s).

Finally, let me briefly return to the correspondence between the order parameter and the pure ground state (vacuum) obtained by taking the zero-temperature limit of (5.26). It is in principle possible that the physical system possesses two completely unrelated ground states of equal energy. (This happens for instance at a first-order phase transition.) Barring such accidental degeneracy, however, one expects the possible candidate vacua to be mutually related by (broken) symmetry transformations. It is therefore sufficient to know just one of the vacuum states; it should be possible to reconstruct all the others by the action of the operators $U(g)$, $g\in G$. We will see in Sect. 5.3 that this procedure is subtle in the limit of infinite volume. What one can however do safely is to trade the chosen vacuum state $\varrho$ for the corresponding order parameter $\langle {A^i}\rangle _\varrho$. All other possible values of the order parameter can then be generated by acting on $\langle {A^i}\rangle _\varrho$ with the matrices $\mathcal {R}(g)$.

In case G is a Lie group, the possible vacuum values of the order parameter span a manifold of dimension $\dim G-\dim H_\varrho$; see Appendix A.1 for the mathematical background. This is known as the vacuum manifold. The central tenet of this book is that the physics of systems with spontaneously broken continuous symmetry can be captured by a low-energy effective field theory (EFT). The form of this EFT is largely fixed by the geometry of the vacuum manifold. The latter can in turn be understood in terms of G and its unbroken subgroup $H_\varrho$. To develop the EFT framework based on the pattern of SSB is the goal of Parts III and IV of the book.

Example 5.5

Quantum chromodynamics (QCD) is a gauge theory of the strong nuclear interaction. Strongly-interacting matter is represented by the quark field $\Psi ^\alpha _i$. This is a Dirac spinor where $\alpha$ is an index of the fundamental representation of the $\mathrm {SU}(3)$ gauge group of QCD. The index i represents quark flavor and belongs to the fundamental representation of $\mathrm {SU}(n_{\mathrm {f}})$. In applications, the number of relevant (light) quark flavors $n_{\mathrm {f}}$ is usually two or three.

In the limit of vanishing quark masses, QCD possesses a $G\simeq \mathrm {SU}(n_{\mathrm {f}})_{\mathrm {L}}\times \mathrm {SU}(n_{\mathrm {f}})_{\mathrm {R}}$ symmetry, consisting of independent flavor transformations of the left- and right-handed components of $\Psi ^\alpha _i$. In the ground state of QCD, this symmetry is spontaneously broken by the complex order parameter . Note that is the usual mass term for a Dirac spinor, here with equal masses for all quark flavors. We can therefore think of “switching on” small equal quark masses as an analog of the external field $\lambda _i$ that selects a particular vacuum state. In this vacuum state, $\langle {\Sigma _{ij}}\rangle =\sigma \delta _{ij}$, where the constant $\sigma$ is an intrinsic scale of QCD. The unbroken, “vector” subgroup of G, $H\simeq \mathrm {SU}(n_{\mathrm {f}})_{\mathrm {V}}$, consists of identical unitary transformations of left- and right-handed quarks. Broken symmetry transformations convert this vacuum to other vacuum states, in which $\langle {\Sigma _{ij}}\rangle /\sigma$ is unitary. The vacuum manifold is therefore diffeomorphic to the Lie group $\mathrm {SU}(n_{\mathrm {f}})$.

2.3 Intermediate Summary

We have taken quite a conceptual tour to arrive at the notions of order parameter and vacuum manifold, essential for understanding the physics of SSB. Let us therefore make a little break and review what we have done in this chapter so far.

I started Sect. 5.1 with a basic definition of a state of a quantum system in terms of a density operator, and of its symmetry. Section 5.1.1 makes the distinction between broken and unbroken symmetries, and highlights energetic degeneracy of different states as a universal feature of SSB. A naive way out of the puzzle this poses might be to consider only states respecting the full symmetry G of the system. In Sect. 5.1.2, I motivated the consideration of such G-symmetric states by recalling the thermal (canonical) density operator. The latter however renders the average of any observable breaking the symmetry zero. This is in a striking contradiction with the existence of asymmetric, macroscopically stable states of matter in nature. A careful consideration of the effects of small external perturbations in Sect. 5.2.1 shows that these may drive the system towards a symmetry-breaking state. In fact, in the large-volume limit, even an infinitesimally small perturbation may be sufficient to destabilize the G-invariant thermal state.

For most applications, it is mandatory to build a physical description of a given system upon a stable ground state. Such vacuum states are constructed in Sect. 5.2.2 by taking the thermodynamic limit in presence of a perturbation, before the latter may be switched off. The different, degenerate vacuum states can now be labeled uniquely by different values of an order parameter, corresponding to the average of a suitably chosen set of observables. The choice of the order parameter itself is not unique. Depending on this choice, the same vacuum state may be represented by a point in different order parameter spaces. However, the set of all degenerate vacuum states spans a vacuum manifold, whose geometric structure is determined solely by the symmetry group G and its unbroken subgroup.

This completes the background needed to understand the following chapters. When talking about a vacuum or ground state, I will always implicitly have in mind a pure state, obtained by the limiting procedure in (5.26). Also, I will from now on always assume vanishing thermodynamic temperature. Nonzero temperature was only used in this chapter as a tool to avoid singularities in the partition function.

The story of SSB, however, does not end here. A closer look reveals a number of intriguing aspects that I have skipped so far in the desire to provide a minimal self-contained introduction to SSB. While not of direct relevance for the rest of the book, it would be a pity to omit these aspects altogether. I therefore mention some of them at least briefly in the next section, in the form of a case study.

3 Some Subtle Features of Spontaneous Symmetry Breaking

In the previous section, the thermodynamic limit played a crucial role in selecting a unique symmetry-breaking ground state. This begs the question of what one can expect in volumes that are large yet finite. After all, real macroscopic systems in nature certainly are finite, albeit possibly large from the point of view of microscopic physics. To what extent are then symmetry-breaking states stabilized by an external perturbation? Besides, how does the finite-volume quantum ground state look in the absence of perturbations?

These are some of the questions I will address here. I am not aware of any formal framework that would allow one to tackle these questions in full generality. In fact, the answer to some of them may depend on the specific physical system. Instead of trying to be general, I will therefore work out in detail one concrete example, just to illustrate what is at stake. Some rigorous results on the spectrum of quantum systems with SSB in a finite volume can be found in [10] or in Part I of [11].

3.1 Free Schrödinger Field in Finite Volume

Following Sect. 3 of [12], consider the theory of a free Schrödinger field $\psi$, defined by the Lagrangian density

$$\displaystyle \begin{aligned} \mathcal{L}=\mathrm{i}{\psi}^{\dagger}\partial_0\psi-\frac 1{2m}\boldsymbol\nabla{\psi}^{\dagger}\cdot\boldsymbol\nabla\psi\;, {} \end{aligned} $$

(5.29)

where m is a mass parameter. I will quantize this theory in a d-dimensional box of size L, with volume $V=L^d$, endowed with a periodic boundary condition. The boundary condition admits only discrete values of momentum $\boldsymbol p$, given component-wise by

$$\displaystyle \begin{aligned} p_r=\frac{2\pi n_r}L\;,\qquad n_r\in\mathbb{Z}\;. {} \end{aligned} $$

(5.30)

The field operator can be Fourier-expanded in terms of plane-wave solutions to the Schrödinger equation. In the Schrödinger picture,

$$\displaystyle \begin{aligned} \psi(\boldsymbol x)=\frac 1{\sqrt{V}}\sum_{\boldsymbol p}a_{\boldsymbol p}\mathrm{e}^{\mathrm{i}\boldsymbol{p}\cdot\boldsymbol{x}}\;, {} \end{aligned} $$

(5.31)

where $a_{\boldsymbol p}$ is the annihilation operator for a single-particle state of momentum $\boldsymbol p$. This is normalized so that together with the creation operator , it satisfies $[a_{\boldsymbol p},{a}^{\dagger }_{\boldsymbol q}]=\delta _{\boldsymbol p\boldsymbol q}$. The normalization of the plane-wave expansion by the factor $1/\sqrt {V}$ then guarantees that the canonical coordinate $\psi (\boldsymbol x)$ and its conjugate momentum obey the canonical commutation relation $[\psi (\boldsymbol x),\mathrm{i} {\psi }^{\dagger }(\boldsymbol y)]=\mathrm{i} \delta ^d(\boldsymbol x-\boldsymbol y)$.

Within second quantization, all these operators act on a Hilbert space, built using the Fock construction. The starting point is the Fock vacuum $\left \lvert {0}\right \rangle$, defined by the condition $a_{\boldsymbol p}\left \lvert {0}\right \rangle =0$ for all allowed values of $\boldsymbol p$. An orthogonal basis of the space is then obtained by acting on $\left \lvert {0}\right \rangle$ with a finite number of creation operators in all possible ways. The Schrödinger Hamiltonian has a simple expression in terms of the annihilation and creation operators,

$$\displaystyle \begin{aligned} H=\frac 1{2m}\int\mathrm{d}^d\!\boldsymbol x\,\boldsymbol\nabla{\psi}^{\dagger}(\boldsymbol x)\cdot\boldsymbol\nabla\psi(\boldsymbol x)=\sum_{\boldsymbol p}\frac{\boldsymbol p^2}{2m}{a}^{\dagger}_{\boldsymbol p}a_{\boldsymbol p}\;. {} \end{aligned} $$

(5.32)

This is a positive-semidefinite operator. The lowest (zero) energy is reached by any state $\left \lvert {\Omega }\right \rangle$ in the Hilbert space that satisfies

$$\displaystyle \begin{aligned} a_{\boldsymbol p}\left\lvert{\Omega}\right\rangle =0\quad \text{for all }\boldsymbol p\neq\mathbf0\qquad \text{(ground state)}\;. {} \end{aligned} $$

(5.33)

There are infinitely many such states in the Hilbert space, in particular any state obtained from $\left \lvert {0}\right \rangle$ using the zero-momentum creation operator ${a}^{\dagger }_{\mathbf 0}$. Hence, the subspace of states of zero energy is identical to the Fock space of a linear harmonic oscillator.

The infinite degeneracy of the ground state hints at SSB. To see what symmetry might be spontaneously broken here, let us think what conserved charges there are in the first place. There turn out to be infinitely many of them. One can take for instance any operator of the form

$$\displaystyle \begin{aligned} Q_f\equiv\sum_{\boldsymbol p}f(\boldsymbol p){a}^{\dagger}_{\boldsymbol p}a_{\boldsymbol p}=\int\mathrm{d}^d\!\boldsymbol x\,{\psi}^{\dagger}(\boldsymbol x)f(-\mathrm{i}\boldsymbol\nabla)\psi(\boldsymbol x)\;, {} \end{aligned} $$

(5.34)

where f is assumed to be a real analytic function. This class of operators includes the Hamiltonian itself, with $f(\boldsymbol p)=\boldsymbol p^2/(2m)$, the momentum operator $\boldsymbol P=\sum _{\boldsymbol p}\boldsymbol p{a}^{\dagger }_{\boldsymbol p}a_{\boldsymbol p}$, and the operator of particle number . From (5.34) it is obvious that all the operators $Q_f$ mutually commute with each other, hence all are conserved. By (4.34), they induce a local transformation of $\psi (\boldsymbol x)$ with parameter $\epsilon$, $\updelta \psi (\boldsymbol x)=-\mathrm{i} \epsilon f(-\mathrm{i} \boldsymbol \nabla )\psi (\boldsymbol x)$. In addition to the class (5.34), any Hermitian operator constructed out of $a_{\mathbf 0}$ and ${a}^{\dagger }_{\mathbf 0}$ commutes with the Hamiltonian, and thus represents a conserved charge as well. Not all of these, however, descend from a local Noether current. Two notable exceptions are the two possible Hermitian operators linear in $a_{\mathbf 0}$ and . I will list them alongside the symmetry transformations they generate:

$$\displaystyle \begin{aligned} \begin{alignedat}{2} Q_{\mathrm{R}}&\equiv\mathrm{i}\int\mathrm{d}^d\!\boldsymbol x\,[\psi(\boldsymbol x)-{\psi}^{\dagger}(\boldsymbol x)]=\mathrm{i}\sqrt{V}(a_{\mathbf0}-{a}^{\dagger}_{\mathbf0})\;,\qquad &&\psi\to\psi-\epsilon\;,\\ Q_{\mathrm{I}}&\equiv\int\mathrm{d}^d\!\boldsymbol x\,[\psi(\boldsymbol x)+{\psi}^{\dagger}(\boldsymbol x)]=\sqrt{V}(a_{\mathbf0}+{a}^{\dagger}_{\mathbf0})\;,\qquad &&\psi\to\psi-\mathrm{i}\epsilon\;. \end{alignedat} {} \end{aligned} $$

(5.35)

With all the symmetry transformations at hand, the associated Noether currents can be extracted from the Lagrangian (5.29) using the technique outlined in Sect. 4.2.1.

The Lie algebra of our conserved charges is defined by the commutators

$$\displaystyle \begin{aligned} [Q_f,Q_{\mathrm{R}}]=-\mathrm{i} f(\mathbf0)Q_{\mathrm{I}}\;,\qquad [Q_f,Q_{\mathrm{I}}]=+\mathrm{i} f(\mathbf0)Q_{\mathrm{R}}\;,\qquad [Q_{\mathrm{R}},Q_{\mathrm{I}}]=2\mathrm{i} V\;. {} \end{aligned} $$

(5.36)

These make it natural to split the space of charges $Q_f$ into the one-dimensional subspace of charges proportional to $Q_1$, and a subspace of charges with $f(\mathbf 0)=0$. The latter commute with $Q_{\mathrm {R}},Q_{\mathrm {I}}$ and moreover annihilate any ground state $\left \lvert {\Omega }\right \rangle$ as defined by (5.33). We can therefore drop them and focus on the set $Q_1,Q_{\mathrm {R}},Q_{\mathrm {I}}$. By (5.36), this set furnishes a central extension of the Lie algebra $\mathfrak {iso}(2)$, corresponding to Euclidean transformations in the complex plane of $\psi$. The generator $Q_1$ induces (phase) rotations, whereas $Q_{\mathrm {R}},Q_{\mathrm {I}}$ generate respectively translations of the real and imaginary parts of $\psi$.

The Fock vacuum $\left \lvert {0}\right \rangle$ is annihilated by $Q_1$, but not by $Q_{\mathrm {R}},Q_{\mathrm {I}}$. This suggests that alternative vacuum states may be obtained from $\left \lvert {0}\right \rangle$ by transformations generated by $Q_{\mathrm {R}},Q_{\mathrm {I}}$. We can pick any $z\equiv z_1+\mathrm{i} z_2\in \mathbb {C}$ and define a new vacuum state by

$$\displaystyle \begin{aligned} \left\lvert{z}\right\rangle \equiv\exp\big[\mathrm{i}(z_1Q_{\mathrm{R}}+z_2Q_{\mathrm{I}})\big]\left\lvert{0}\right\rangle =\exp\big[\sqrt{V}(z{a}^{\dagger}_{\mathbf0}-z^*a_{\mathbf0})\big]\left\lvert{0}\right\rangle \;. {} \end{aligned} $$

(5.37)

This is a coherent state in the Fock space of zero-energy states. Indeed, it is easily seen to be an eigenstate of the annihilation operator, $a_{\mathbf 0}\left \lvert {z}\right \rangle =\sqrt {V}z\left \lvert {z}\right \rangle$. In fact, $\left \lvert {z}\right \rangle$ is simultaneously an eigenstate of the field operator at any point in space, $\psi (\boldsymbol x)\left \lvert {z}\right \rangle =z\left \lvert {z}\right \rangle$. We thus end up with a continuum of vacuum states, labeled by a complex parameter z. The order parameter distinguishing these states from each other is provided by the expectation value of $\psi (\boldsymbol x)$. The vacuum manifold is the complex plane $\mathbb {C}$. For any choice of z, the state $\left \lvert {z}\right \rangle$ possesses an unbroken $\mathrm {SO}(2)$ symmetry, corresponding to rotations of the complex plane around the point z. Its generator is ${ }^{z}{}Q_1\equiv Q_1+z_2Q_{\mathrm {R}}-z_1Q_{\mathrm {I}}$. The state $\left \lvert {z}\right \rangle$ is an eigenstate of this operator, ${ }^{z}{}Q_1\left \lvert {z}\right \rangle =-V\left \lvert {z}\right \rvert ^2\left \lvert {z}\right \rangle$.

Interestingly, it is not possible to select the state $\left \lvert {z}\right \rangle$ by adding to the Hamiltonian a perturbation linear in $\psi$ as I did in Sect. 5.2. That would make the Hamiltonian unbounded from below. This is the price to pay for the vacuum manifold not being compact. An alternative is to add to the Hamiltonian a chemical potential $\mu$ for the conserved charge ${ }^{z}{}Q_1$, so that it becomes $H_\mu \equiv H-\mu { }^{z}{}Q_1$. The expectation value of such a perturbed Hamiltonian in the state $\left \lvert {z'}\right \rangle$ with any $z'\in \mathbb {C}$ is

$$\displaystyle \begin{aligned} \left\langle{z'}\right\rvert H_\mu\left\lvert{z'}\right\rangle =-V\mu\big[\left\lvert{z-z'}\right\rvert ^2-\left\lvert{z}\right\rvert ^2\big]\;. \end{aligned} $$

(5.38)

An infinitesimally small negative$\mu$ will then select $\left \lvert {z}\right \rangle$ as the unique ground state with energy $V\mu \left \lvert {z}\right \rvert ^2$.

3.2 Pathologies of the Infinite-Volume Limit

The analysis of the free Schrödinger theory has been perfectly clean so far. Most of the properties of the quantized theory match their classical counterparts, in particular the existence of infinitely many conservation laws, and infinitely many degenerate ground states. What happens when we try to take the limit of infinite volume? Let us evaluate the overlap of two different coherent states of the type (5.37),

$$\displaystyle \begin{aligned} \begin{array}{rl} \left\lvert{\left\langle{z'}\middle\vert{z}\right\rangle }\right\rvert ^2&=\left\lvert{\left\langle{z'}\middle\vert{U(z)}\middle\vert{0}\right\rangle }\right\rvert ^2=\exp\big(-V\left\lvert{z-z'}\right\rvert ^2\big)\;,\\ U(z)&\equiv\exp\big[\mathrm{i}(z_1Q_{\mathrm{R}}+z_2Q_{\mathrm{I}})\big]\;. \end{array} {} \end{aligned} $$

(5.39)

We therefore find that

$$\displaystyle \begin{aligned} \lim_{V\to\infty}\left\lvert{\left\langle{z'}\middle\vert{z}\right\rangle }\right\rvert ^2=\lim_{V\to\infty}\left\lvert{\left\langle{z'}\middle\vert{U(z)}\middle\vert{0}\right\rangle }\right\rvert ^2=\delta_{zz'}\;. \end{aligned} $$

(5.40)

In the limit $V\to \infty$, any two states $\left \lvert {z}\right \rangle$ and $\left \lvert {z'}\right \rangle$ with $z\neq z'$ become orthogonal. It is easy to generalize the above calculation to show that, in fact, any vector in the Fock space built above $\left \lvert {z}\right \rangle$ must be orthogonal to $\left \lvert {z'}\right \rangle$ in this limit.¹ In other words, the ket-vector $\left \lvert {z'}\right \rangle$ does not lie in the Fock space built above $\left \lvert {z}\right \rangle$ at all. Moreover, the matrix element of any observable constructed using a finite number of annihilation and creation operators between the $\left \lvert {z}\right \rangle$ and $\left \lvert {z'}\right \rangle$ states vanishes. Altogether, the states $\left \lvert {z}\right \rangle$ with $z\in \mathbb {C}$ therefore give rise to a completely disconnected Fock space each. Transitions between Fock spaces with different vacuum states are forbidden. This is certainly not what we expected!

This singular behavior of the $V\to \infty$ limit is easy to understand. Just remember that $\left \lvert {z}\right \rangle$ is an eigenstate of $a_{\mathbf 0}$ with the eigenvalue $\sqrt {V}z$. Thus, if we insist on keeping z fixed, then scaling V  towards infinity makes the eigenvalue diverge as well. This is ultimately the reason why $\left \lvert {z'}\right \rangle$ does not lie in the Fock space built above $\left \lvert {z}\right \rangle$. We might, on the other hand, try to keep the eigenvalue of $a_{\mathbf 0}$ fixed, which would require scaling z as $1/\sqrt {V}$ as the volume grows. Roughly speaking, this means that in a large but finite volume V , transitions between Fock spaces with different z are allowed, but only within a range $\left \lvert {z-z'}\right \rvert \lesssim 1/\sqrt {V}$.

There is another way of looking at this, emphasizing the properties of the operator $U(z)$ defined by (5.39). Since the states $\left \lvert {0}\right \rangle$ and $\left \lvert {z'}\right \rangle$ do not lie in the same Fock space, the operator $U(z)$ becomes ill-defined in the limit $V\to \infty$. Spontaneously broken symmetry may not be realized by unitary operators on the Hilbert space of the physical system! That however does not mean that in presence of SSB, it makes no sense to speak of symmetry transformations that are spontaneously broken. The transformation of the field operator by $U(z)$,

$$\displaystyle \begin{aligned} {U(z)}^{\dagger}\psi(\boldsymbol x)U(z)=\psi(\boldsymbol x)+z\;, \end{aligned} $$

(5.41)

is perfectly well-defined even in the limit $V\to \infty$. The same applies to any local operator built out of a finite number of $\psi$ and ${\psi }^{\dagger }$. In each of the Fock spaces built above any $\left \lvert {z}\right \rangle$, we thus have a well-defined notion of broken symmetry transformations in terms of averages of observables. We just have to be careful not to think of broken symmetry as an operator connecting different states in the same space.

The distinction between the realizations of broken symmetries by operators on states and by transformations of observables is not a mere mathematical curiosity one can safely ignore. It has observable consequences. One of the earliest applications of group theory to physics was to classify energy levels of a quantum system in terms of representations of its symmetry group. In quantum mechanics, the so-called Wigner theorem guarantees that any symmetry is represented by a unitary or antiunitary operator on the Hilbert space (see Chap. 2 of [13]). Energy levels are therefore classified by irreducible representations of the symmetry group G. In quantum field theory, on the other hand, only the unbroken subgroup H is as a rule realized by unitary operators on the Hilbert space. Thus, energy levels form irreducible multiplets of H. SSB implies degeneracy of the ground state, accompanied by a reduced degeneracy of the excitation spectrum.

For example, the spectrum of hadrons is organized into (approximately degenerate) multiplets of the isospin group $\mathrm {SU}(2)$. We know from Example 5.5 that this is the unbroken subgroup of QCD with two light quark flavors. There is an even higher degeneracy, corresponding to multiplets of $\mathrm {SU}(3)$. This is however less accurate due to the large splitting between the masses of the up, down and strange quarks. There are no multiplets of the $\mathrm {SU}(n_{\mathrm {f}})_{\mathrm {L}}\times \mathrm {SU}(n_{\mathrm {f}})_{\mathrm {R}}$ symmetry of QCD. Such multiplets would be easily noticeable. For instance, the pseudoscalar mesons such as pions or kaons would have to appear in the spectrum together with particles of opposite parity. The lack of such particles is a directly observable manifestation of the nonexistence of unitary operators representing spontaneously broken symmetry.

3.3 Uniqueness of the Finite-Volume Ground State

The above excursion into the subtleties of SSB was purposefully limited to a single, exactly solvable example. The conclusion that in infinite volume, spontaneously broken symmetry is not realized by unitary operators on the Hilbert space is, however, general [14]. Likewise, it is generally true that different symmetry-breaking vacuum states belong to disconnected Hilbert spaces.

One property of the free Schrödinger theory (5.29) that is not general is the existence of exactly degenerate ground states already in a finite volume. This is typical for systems where the order parameter is constructed using an operator that commutes with the Hamiltonian. In our case, this operator can be taken as the spatial average of the Schrödinger field,

$$\displaystyle \begin{aligned} \bar\psi\equiv\frac 1V\int\mathrm{d}^d\!\boldsymbol x\,\psi(\boldsymbol x)=\frac 1{\sqrt{V}}a_{\mathbf0}\;. \end{aligned} $$

(5.42)

Another system of this type is the ferromagnet, where the order parameter is supplied by a selected component of total spin. Historically, the very special properties of such systems led to a controversy (involving some famous individuals [15]) as to whether they should count as SSB at all or not.

An alternative is that in any finite volume, the system possesses a unique ground state, which is then necessarily G-invariant. Above this ground state, there is a tower of closely spaced energy levels, which all become exactly degenerate in the limit $V\to \infty$. A notable, much studied example of such a system are antiferromagnets. The realization of SSB is then quite nontrivial. Although a detailed discussion would be beyond the scope of this book, let me make at least a few remarks. The reader is encouraged to consult Sect. 19.1 of [16] or Sect. 2 of [1] for further information.

If one tries to take the naive limit $V\to \infty$, the unique, finite-volume G-invariant ground state turns into a likewise G-invariant state in infinite volume. The latter is, however, unstable under infinitesimally small external perturbations. The instability has its precursor already in a finite volume. Namely, the magnitude of a perturbation needed to drive the system away from the G-invariant vacuum scales as the inverse of the system’s volume. Such a perturbation may be supplied in particular by the measurement of any symmetry-breaking observable. This makes the system relax into one of the pure symmetry-breaking states constructed in Sect. 5.2.2.

Another way to understand the distinction between the various would-be ground states is through correlations of spatially separated observables. For any two local observables $A(\boldsymbol x)$ and $B(\boldsymbol x)$, one may construct the spatial correlation function $\langle {A(\boldsymbol x)B(\boldsymbol y)}\rangle$; the angular brackets indicate expectation value in a chosen ground state. The so-called cluster decomposition principle dictates that

$$\displaystyle \begin{aligned} \lim_{\left\lvert{\boldsymbol x-\boldsymbol y}\right\rvert \to\infty}\big[\langle{A(\boldsymbol x)B(\boldsymbol y)}\rangle -\langle{A(\boldsymbol x)}\rangle \langle{B(\boldsymbol y)}\rangle \big]=0\;. {} \end{aligned} $$

(5.43)

It encodes the fundamental requirement that the outcomes of distant measurements should not be correlated with each other. The naive G-invariant ground state obtained by taking the $V\to \infty$ limit violates this principle, and thus displays long-range entanglement. The entanglement is destroyed by any measurement of a local symmetry-breaking observable, which brings the system to a stable, symmetry-breaking ground state satisfying (5.43).

To illustrate some of the above general observations, I will conclude the discussion of SSB by working out one more simple toy model. We return to the free Schrödinger theory, but twist the boundary condition to

$$\displaystyle \begin{aligned} \psi(\boldsymbol x+\boldsymbol e_rL)=\mathrm{e}^{\mathrm{i}\theta_r}\psi(\boldsymbol x)\;. {} \end{aligned} $$

(5.44)

Here $\boldsymbol e_r$ is a unit vector along the r-th Cartesian coordinate axis, and $\theta _r\in (-\pi ,+\pi ]$ is an arbitrarily chosen phase. I assume that at least one of $\theta _r$ is nonzero, otherwise (5.44) would be just the periodic boundary condition we used previously.

The quantization of the theory proceeds as before, except for the allowed values of momenta, where (5.30) is replaced with

$$\displaystyle \begin{aligned} p_r=\frac{\theta_r}L+\frac{2\pi n_r}L\;,\qquad n_r\in\mathbb{Z}\;. {} \end{aligned} $$

(5.45)

The Hamiltonian (5.32) now has a unique ground state, namely the Fock vacuum $\left \lvert {0}\right \rangle$. The lowest-lying excitation of this ground state corresponds to $n_r=0$ for all r, that is, $\boldsymbol p=\boldsymbol \theta /L$. This is a one-particle state $\left \lvert {\boldsymbol \theta /L}\right \rangle \equiv {a}^{\dagger }_{\boldsymbol \theta /L}\left \lvert {0}\right \rangle$ with excitation energy

$$\displaystyle \begin{aligned} E_{\boldsymbol\theta/L}=\frac{\boldsymbol\theta^2}{2mL^2}\;. \end{aligned} $$

(5.46)

As long as all the $\theta _r$ are sufficiently small, the low end of the excitation spectrum will consist entirely of states with $n\in \mathbb {N}$ quanta of momentum $\boldsymbol \theta /L$ and excitation energy $nE_{\boldsymbol \theta /L}$. Only above these will one find states for which some of the $n_r$ are nonzero. See Fig. 5.3 for a visualization of the spectrum of energy levels.

Fig. 5.3

Schematic visualization of the spectrum of the free Schrödinger theory with the twisted boundary condition (5.44), for small (positive) $\theta _r$. Each of the indicated groups of energy levels consists of states with multiple quanta of momentum $\boldsymbol \theta /L$; the number of such quanta is denoted as n. The lowest-lying multiplet includes no other particles. The higher multiplets include one or more particles where some of the $n_r$ in (5.45) are nonzero. For simplicity, all the $\theta _r$ are assumed to be equal; otherwise the spectrum would feature additional structure

How is the Fock vacuum $\left \lvert {0}\right \rangle$ affected by perturbations? Since the $\boldsymbol p=\mathbf 0$ state is excluded by the boundary condition, we can now add a perturbation linear in the field without running into an instability. Let us therefore add to (5.29) the term

$$\displaystyle \begin{aligned} \mathcal{L}_\epsilon[\psi,\boldsymbol x](x)\equiv\frac{\epsilon^*}{\sqrt{V}}\psi(x)\mathrm{e}^{-\mathrm{i}\boldsymbol{\theta}\cdot\boldsymbol{x}/L}+\frac\epsilon{\sqrt{V}}{\psi}^{\dagger}(x)\mathrm{e}^{+\mathrm{i}\boldsymbol{\theta}\cdot\boldsymbol{x}/L}\;, \end{aligned} $$

(5.47)

where $\epsilon$ is a small constant parameter with the dimension of energy. The exponential, coordinate-dependent factors are added in order that $\mathcal {L}_\epsilon$ itself satisfies a periodic boundary condition. The corresponding correction to the Hamiltonian (5.32) is

$$\displaystyle \begin{aligned} H_\epsilon=-\big(\epsilon^*a_{\boldsymbol\theta/L}+\epsilon{a}^{\dagger}_{\boldsymbol\theta/L}\big)\;. \end{aligned} $$

(5.48)

This perturbation will change the energy eigenstates within the multiplets shown in Fig. 5.3. It will however not lead to any mixing between states from different multiplets, that is states with different $n_r$. Let us focus on the lowest-lying multiplet, which is where the ground state resides. Here the Hamiltonian is projected down to

$$\displaystyle \begin{aligned} \begin{array}{rl} H_{\boldsymbol\theta/L}&\equiv E_{\boldsymbol\theta/L}{a}^{\dagger}_{\boldsymbol\theta/L}a_{\boldsymbol\theta/L}-\big(\epsilon^*a_{\boldsymbol\theta/L}+\epsilon{a}^{\dagger}_{\boldsymbol\theta/L}\big)\\ &=E_{\boldsymbol\theta/L}\big[({a}^{\dagger}_{\boldsymbol\theta/L}-z^*_\epsilon)(a_{\boldsymbol\theta/L}-z_\epsilon)-\left\lvert{z_\epsilon}\right\rvert ^2\big]\;, \end{array} \end{aligned} $$

(5.49)

where $z_\epsilon \equiv \epsilon /E_{\boldsymbol \theta /L}$. There is still a unique ground state, but it is now the coherent state $\left \lvert {z_\epsilon }\right \rangle$. The perturbation has lowered the energy compared to the Fock vacuum $\left \lvert {0}\right \rangle$ to $-E_{\boldsymbol \theta /L}\left \lvert {z_\epsilon }\right \rvert ^2=-\left \lvert {\epsilon }\right \rvert ^2/E_{\boldsymbol \theta /L}$.

Let us see what we have found. The original vacuum $\left \lvert {0}\right \rangle$ is nondegenerate. Its separation from the nearest excited states is however only of the order of $E_{\boldsymbol \theta /L}\sim 1/L^2$. In the limit of infinite volume, all the different multiplets of states shown in Fig. 5.3 become degenerate. Accordingly, the Fock vacuum becomes unstable under small perturbations. For small fixed $\theta _r$, one can always find $\epsilon$ that satisfies the bounds

$$\displaystyle \begin{aligned} 1\ll\frac{\left\lvert{\epsilon}\right\rvert }{E_{\boldsymbol\theta/L}}\ll\frac{2\pi}{\left\lvert{\boldsymbol\theta}\right\rvert }\;. {} \end{aligned} $$

(5.50)

The lower bound makes the change in energy caused by the perturbation much larger than the level spacing within a given multiplet. Thus, even a small perturbation can drive the ground state from $\left \lvert {0}\right \rangle$ to a coherent state $\left \lvert {z_\epsilon }\right \rangle$ with large $z_\epsilon$. The upper bound in (5.50) ensures that the shift of energy eigenvalues induced by the perturbation is much smaller than the separation of the different multiplets in the spectrum.

The free Schrödinger theory with the twisted boundary condition (5.44) serves as a good example of the instability of the finite-volume ground state with respect to small perturbations. Likewise, it provides a decent illustration of the structure of the spectrum of systems with SSB in a finite volume. One important feature that it does not illustrate is the G-invariance of the finite-volume ground state. This is because the symmetries (5.35) of the free Schrödinger theory are broken by the boundary condition (5.44) for any nonzero $\theta _r$. In fact, insofar as the expectation value $\langle {\psi (\boldsymbol x)}\rangle$ is well-defined, shift symmetries of the type (5.35) must always be spontaneously broken.