superposition of crystal eigenstates which has a well-defined centre-of-mass
postion. In terms of the collective Hamiltonian, this amounts to adding a
perturbation which tends to localise the crystal:

$\displaystyle{H}^{\prime}_{\text{CoM}}$
$\displaystyle=\frac{{\mathbf{P}}^{2}_{\text{tot}}}{2mN}+\mu{\mathbf{X}}^{2}_{\text{CoM}}.$
(2.11)

Here, ${\bf X}_{\text{CoM}}=1/N\sum_{\bf x}{\bf X}({\bf x})$ is the centre of
mass position of the crystal, while $\mu$ is the strength of a potential that
tends to localise the crystal at the origin of our coordinate system. We will
consider $\mu$ to be a very small perturbation, and take the limit $\mu\to 0$
at the end of our calculation.

######  Exercise 2.2 (Commutator of ${\bf P}_{\text{tot}}$ and ${\bf
X}_{\text{CoM}}$).

Show that ${\bf P}_{\text{tot}}$ and ${\bf X}_{\text{CoM}}$ obey canonical
commutation relations.

Since $\mathbf{P}_{\text{tot}}$ and $\mathbf{X}_{\text{CoM}}$ obey canonical
commutation relations, the perturbed Hamiltonian ${H}^{\prime}_{\text{CoM}}$
is that of a harmonic oscillator, with the well-known energies and ground
state:

$\displaystyle E_{n}$ $\displaystyle=\hbar\omega\left({n}+1/2\right)$
$\displaystyle\psi_{0}({\bf x})$
$\displaystyle=\langle\mathbf{x}|n=0\rangle=\left(\frac{2mN\mu}{\pi^{2}\hbar^{2}}\right)^{3/8}\mathrm{e}^{-\sqrt{\frac{mN\mu}{2\hbar^{2}}}|{\bf
x}|^{2}},$ (2.12)

with $\omega=\sqrt{\frac{2\mu}{mN}}$. The ground state of the harmonic crystal
in the presence of a perturbation is a Gaussian wave packet. Its width in
position space, $\sigma^{2}=\hbar/\sqrt{2mN\mu}$, decreases as the number of
particles in the crystal grows. The occurrence of spontaneous symmetry
breaking is again signalled by two non-commuting limits:

$\displaystyle\lim_{N\to\infty}\lim_{\mu\to 0}|\psi_{0}({\bf x})|^{2}$
$\displaystyle={\text{constant}},$ $\displaystyle\lim_{\mu\to
0}\lim_{N\to\infty}|\psi_{0}({\bf x})|^{2}$ $\displaystyle=\delta({\bf x}).$
(2.13)

As before, this is an indication that the thermodynamic limit $N\to\infty$ is
singular. For a system of any finite size, no matter how large, the
perturbation can be made small enough for the ground state wave function to be
essentially spread out over the entire universe. But for an infinitely large
system, _any_ perturbation, no matter how weak, is enough to completely
localise the wave function in a single position. In the thermodynamic limit,
the localisation happens even in the presence of only an infinitesimal
potential, which in effect means that the wave function localises
spontaneously. Notice that the energy of the state does not depend on the
order of limits. In both cases $E_{0}=\hbar\omega/2=0$. This implies that
indeed in the thermodynamic limit the symmetry-broken, localised state of the
crystal has the same energy as the exact, plane-wave ground state.

Real materials are not infinitely large, and thus neither of the limits in
equation (2.3) strictly speaking applies to real pieces of matter. This is not
just a practical consideration, but a point of view that is increasingly
advocated even in formal mathematical approaches to spontaneous symmetry
breaking [23, 24]. The importance of the non-commuting limits, is to signal
that even if $N$ is not yet truly infinite, the approach towards the
thermodynamic limit is singular. This implies in particular that as you
consider larger and larger pieces of matter, a weaker and weaker perturbation
suffices to make its ground state a localised wave packet. Imagine for example
a typical ‘finite size’ iron crystal, with a volume of one cubic centimeter.
Iron has an atomic mass of $55.8u=9.27\times 10^{-26}$ kg, and a lattice
constant of $a=2.856\times 10^{-10}$ m. This means the iron crystal contains
approximately $N=4\times 10^{22}$ atoms. So how strong does the symmetry
breaking field need to be in order to reasonably localise such a crystal? The
units of $\mu$ are kg s-2 or N m-1 (Newton per meter). The weakest possible
forces that can currently be measured are of the order of zeptonewtons
($10^{-21}$ N). The weakest possible symmetry breaking field for our piece of
iron that would still be measurable, is thus something like a zeptonewton per
centimeter, or $\mu\sim 10^{-19}$ N m-1. Even with such a weak perturbation,
the width of the crystal’s ground state wave function is constrained to be
about $2\times 10^{-12}$ m. That is, two orders of magnitude smaller than the
unit cell of the iron crystal itself. Clearly, it is practically impossible to
find any sort of everyday-sized object in a momentum eigenstate.

######  Exercise 2.3 (Elitzur’s theorem for a free gas).

We stated in Section 1.4.3 that, in contrast to global symmetries, local
symmetries cannot be spontaneously broken. To understand why this is the case,
consider an ideal gas of $N$ non-interacting particles with positions
$\mathbf{X}_{j}$ and momenta $\mathbf{P}_{j}$, described by:

$\displaystyle{H}_{0}=\sum_{j}\frac{{\bf P}_{j}^{2}}{2m}.$ (2.14)

a. The ideal gas has a local translation symmetry. Write down the operator
${U}$ which describes local translations, and show that the Hamiltonian is
invariant under it.

We can try and break the local symmetry by adding a perturbation which acts on
one particle only:

$\displaystyle{H^{\prime}}=\sum_{j}\left(\frac{{\bf
P}_{j}^{2}}{2m}\right)+\frac{1}{2}\kappa{\bf X}_{j=2}^{2}.$ (2.15)

b. What is the ground state $\mathop{\lvert\psi\rangle}$ of the perturbed
Hamiltonian?
c. Calculate the expectation value in the symmetry-broken state of both the
unperturbed energy and the uncertainty in position of the perturbed particle:

$\displaystyle\bar{E}$
$\displaystyle=\mathop{\langle\psi\rvert}{H}_{0}\mathop{\lvert\psi\rangle},$
$\displaystyle\Delta{\bf X}_{j=2}^{2}$
$\displaystyle=\mathop{\langle\psi\rvert}{\bf
X}_{j=2}^{2}\mathop{\lvert\psi\rangle}-\left(\mathop{\langle\psi\rvert}{\bf
X}_{j=2}\mathop{\lvert\psi\rangle}\right)^{2}.$ (2.16)

d. What are the values of $\bar{E}$ and $\Delta{\bf X}_{j=2}^{2}$ in the limit
$\kappa\to 0$?
e. Is there any limit that does not commute with the limit $\kappa\to 0$? In
particular, does $\kappa\rightarrow 0$ commute with the thermodynamic limit
$N\rightarrow\infty$? What do your answers imply for the local symmetry of
${H}_{0}$?

The exercise above is suggestive of the general fact that continuous local
symmetries cannot be spontaneously broken. This result is known as Elitzur’s
theorem [25], ††margin:  Elitzur’s theorem and it has the status of a
mathematical theorem. Its only assumption is that the physics of the system is
described by a minimum action principle, which is the case for all presently
known theories of physics. The central ingredient needed to prove the theorem
is the realisation that for locally symmetric systems, there cannot exist a
singular limit of the kind we encountered for global symmetries. Systems with
a local symmetry therefore lack the instability that comes with global
symmetry and they are always robust against small perturbations. This is not a
result that can be negotiated with by “smart engineering” of a model or
system. Spontaneously breaking a continuous local symmetry is just impossible.

###  2.4 The Heisenberg antiferromagnet

A second instructive example of spontaneous symmetry breaking in quantum
physics, is that of the Heisenberg antiferromagnet. Consider a cubic lattice
with a spin-half degree of freedom on every site, and isotropic interactions
between neighbouring spins: ††margin:  Heisenberg model

$\displaystyle{H}$ $\displaystyle=J\sum_{\langle
i,j\rangle}{\mathbf{S}}_{i}\cdot{\mathbf{S}}_{j}.$ (2.17)

Here ${\mathbf{S}}_{i}=\hbar\bm{\sigma}_{i}$ is the spin operator on site $i$,
with $\bm{\sigma}_{i}$ the Pauli matrices as introduced in Exercise 1.8, and
the usual commutation relations
$[S^{a}_{i},S^{b}_{j}]=\mathrm{i}\hbar\delta_{ij}\epsilon_{abc}S^{c}_{i}$. The
indices $i$ and $j$ label all sites of the cubic lattice, and $\langle
i,j\rangle$ indicates that we sum over nearest-neighbour pairs of spins only.
Expanding the product of spin vectors, the Hamiltonian becomes:

$\displaystyle{H}$ $\displaystyle=J\sum_{\langle
i,j\rangle}{S}^{z}_{i}{S}^{z}_{j}+{S}^{y}_{i}{S}^{y}_{j}+{S}^{x}_{i}{S}^{x}_{j}$
$\displaystyle=J\sum_{\langle
i,j\rangle}{S}^{z}_{i}{S}^{z}_{j}+\frac{1}{2}\left({S}^{+}_{i}{S}^{-}_{j}+{S}^{-}_{i}{S}^{+}_{j}\right).$
(2.18)

The operators ${S}^{\pm}_{i}={S}^{x}_{i}\pm\mathrm{i}{S}^{y}_{i}$ are the spin
raising and lowering operators. The strength of the interaction between
neighbouring spins is denoted by the coupling constant $J$. If it is negative,
the energy is minimised whenever neighbouring spins point in the same
direction. The ground state is then a ferromagnet, with all spins in the
entire material pointing in the same direction, which is spontaneously chosen.
We will come back to the very special case of spontaneous symmetry breaking in
ferromagnets in Exercise 2.8.

For positive $J$, the energy is lowered by neighbouring spins being anti-
parallel. If we divide the cubic lattice into two sublattices, as shown in the
Fig. 2.4, then all nearest-neighbour pairs of spins can be made anti-parallel
by choosing all spins on the $A$-sublattice to point up, and all spins on the
$B$-sublattice to point down. Such a perfect antiferromagnetic ††margin:
antiferromagnet arrangement of spins is known as a Néel state. Again the axis
along which the spins point either up or down is spontaneously chosen. Global
rotations of the spins around this axis are still symmetry transformations in
the Néel state (they are “unbroken”), while spin-rotations around all other
axes are spontaneously broken.

######  Exercise 2.4 (Noether current of the Heisenberg magnet).

The Heisenberg Hamiltonian of Eq. (2.17) is not in canonical form. That is, it
is not expressed in term of operators and their conjugate momenta. We then
cannot perform a Legendre transformation to obtain a Lagrangian and action,
and we cannot use the method of Section 1.2.3 to obtain the Noether current.
However, we already have the symmetry generators $Q^{a}=S^{a}_{\mathrm{tot}}$,
expressed as volume ‘integrals’ over what must be the Noether charge density
$j^{t}_{a}(i)=S^{a}_{i}$. We can then obtain the conservation law by
calculating the time derivative explicitly.

a. Calculate the Heisenberg equation of motion
$\partial_{t}S^{a}_{i}=\frac{\mathrm{i}}{\hbar}[H,S^{a}_{i}]$, using the spin
commutation relations.

The vector connecting a point $i$ on a lattice with a neighbouring point
$i+\delta$ may be written as $v_{i,i+\delta}$. In this notation, the (lattice)
divergence of a vector at $i$ is given by $\sum_{\delta}v_{i,i+\delta}$.

b. Show that the right-hand side of the equation of motion has the form of the
lattice divergence of a vector. That vector corresponds to the spatial Noether
current $j^{m}_{a}$, and the equation of motion is the lattice version of
$\partial_{t}j^{t}_{a}+\partial_{m}j^{m}_{a}=0$.

Figure 2.4:  A square lattice can be divided into two inequivalent sublattices
$A$ and $B$. In the case of a perfect Néel antiferromagnet, the spins on the
two sublattices point in opposite direction.

If the spins were classical bar magnets, the Néel state would have been the
ground state of the antiferromagnet. The quantum mechanical Hamiltonian
however, contains terms like ${S}^{+}_{i}{S}^{-}_{j}$, of which the Néel state
is not an eigenstate. The Néel state can therefore not be the ground state of
the Heisenberg Hamiltonian. In fact, for a three dimensional lattice, it is
still an open question what the precise ground state of the antiferromagnetic
Heisenberg Hamiltonian is.

The Hamiltonian of equation (2.17) is invariant under global, simultaneous
rotations of all spins around any axis. You can check that the Hamiltonian
indeed commutes with the generators
${S}^{a}={S}^{a}_{\mathrm{tot}}=\sum_{i}{S}^{a}_{i}$ of spin-rotations around
all axes $a\in x,y,z$. It follows that the Heisenberg Hamiltonian also
commutes with the operator for total spin
${S}^{2}={\mathbf{S}}\cdot{\mathbf{S}}$:

$\displaystyle\left[{H},{S}^{2}\right]=0.$ (2.19)

Since the Hamiltonian is invariant under global rotations of all spins, you
might expect that like in the case of a crystal, it will be sufficient to
consider the collective part of the Hamiltonian in order to see if its
symmetry can be spontaneously broken. This is true, but in the case of the
antiferromagnet we do need to be careful about what this collective part is
precisely. Because the $A$ and $B$ sublattice are different, there is a
collective mode which rotates all spins on the $A$ sublattice, while the ones
on the $B$ sublattice are left invariant, and the other way around. These two
modes correspond to two components in the Fourier transfrom of the spin
rotation operators, rather than only the one component that defined the
collective motion of the crystal. Together, the two Fourier components make up
the collective part of the Heisenberg Hamiltonian in equation (2.17):

$\displaystyle{H}$ $\displaystyle={H}_{\text{coll}}+\sum_{\mathbf{k}\neq
0,\pi/a}{H}(\mathbf{k}),$ $\displaystyle{H}_{\text{coll}}$
$\displaystyle=\frac{J}{N}{\mathbf{S}}_{A}\cdot{\mathbf{S}}_{B}.$ (2.20)

This collective Hamiltonian is known as the Lieb-Mattis model. ††margin:
Lieb-Mattis model Here, the collective operator for the total spin on the $A$
sublattice is defined as ${\mathbf{S}}_{A}=\sum_{i\in A}{\mathbf{S}}_{i}$, and
similarly for ${\mathbf{S}}_{B}$. The non-collective, internal modes which
make up ${H}(\mathbf{k})$, are the spin-wave analogue to the phonons in a
crystal, and are known as ††margin:  magnon magnons.

The eigenstates of the collective Hamiltonian can be easily found by re-
writing it in terms of the total spin operator
${\mathbf{S}}={\mathbf{S}}_{A}+{\mathbf{S}}_{B}$:

$\displaystyle{H}_{\text{coll}}$
$\displaystyle=\frac{J}{2N}\left({S}^{2}-{S}_{A}^{2}-{S}_{B}^{2}\right).$
(2.21)

Because all operators in this expression commute, we can immediately write the
eigenstates and the corresponding energies in terms of the quantum numbers for
total spin and total spin on the sublattices:

$\displaystyle{H}_{\text{coll}}\mathop{\lvert S_{A},S_{B},S,S^{z}\rangle}$
$\displaystyle=E(S,S_{A},S_{B})\mathop{\lvert S_{A},S_{B},S,S^{z}\rangle}$
$\displaystyle E(S,S_{A},S_{B})$
$\displaystyle=\frac{J\hbar^{2}}{2N}\left(S(S+1)-S_{A}(S_{A}+1)-S_{B}(S_{B}+1)\right).$
(2.22)

The spectrum of low-energy eigenstates is shown schematically in Figure 2.5
below. The ground state is the state with $S_{A}$ and $S_{B}$ equal to their
maximal value of $N/4$, and the total spin $S$ equal to zero. Because $S=0$
implies that $\langle S^{x}\rangle=\langle S^{y}\rangle=\langle
S^{z}\rangle=0$, the ground state does not break spin-rotation symmetry. It is
a unique, non-degenerate state in which the spins on each sublattice are all
exactly aligned with each other, and anti-aligned with neighbours on a
different sublattice, but in which no direction in space is different from any
other. You can think of the total spin singlet ground state as a superposition
of infinitely many Néel states, pointing in all possible directions. Clearly,
this is a highly non-local state.

Figure 2.5: The states with lowest energy in the Lieb-Mattis Hamiltonian.

Flipping one of the individual spins on for example the $A$ sublattice would
decrease $S_{A}$ by one. Such an excitation costs an energy of the order of
$J$, and if the temperature is low enough, these excitations will not be
present. It is also possible, however, to make collective excitations which
increase the value of the total spin $S$ by one. Such excitations cost an
energy that is only proportional to $J/N$, and for a large enough system these
excitations can never be neglected. In fact, in the limit of an infinitely
large system, with $N\to\infty$, there is a again a tower of states labeled by
different values of $S$ which all become degenerate with the ground state777In
this context it is also referred to as the Anderson tower of states after P.W.
Anderson, who argued its existence already in 1952 [26]. These entire lecture
notes in fact are very much in the spirit of the first chapter of Anderson’s
seminal textbook [11].. This suggests that again, in that limit even an
infinitesimally small perturbation might have a qualitative effect and
spontaneously break the spin-rotational symmetry.

The perturbation or symmetry breaking field to consider in this case, is a
field that prefers the spins to be arranged into a Néel state. As before, the
reasoning is that in principle we should consider all possible perturbations
and find that none of them have any effect in the limit where they are
infinitesimally small, except for the one that happens to stabilise the
classically expected ground state. Since we already know what a classical
antiferromagnet looks like, we will straight away consider the only relevant
perturbation:

$\displaystyle{H}^{\prime}_{\text{coll}}=\frac{J}{N}{\mathbf{S}}_{A}\cdot{\mathbf{S}}_{B}-\mu\left({S}_{A}^{z}-{S}_{B}^{z}\right).$
(2.23)

Since the perturbation breaks spin-rotational symmetry, it does not commute
with the original Hamiltonian. The states $\mathop{\lvert
S,S_{A},S_{B},S^{z}\rangle}$ are therefore no longer eigenstates of the
perturbed Hamiltonian. In order to find the new eigenstates, we need to
calculate the matrix elements of the perturbed Hamiltonian in the basis of the
old eigenstates. Because the excitations that change $S_{A}$ and $S_{B}$ cost
far more energy than an infinitesimal perturbation is expected to provide, we
can safely assume that $S_{A}=S_{B}=N/4$ throughout. And because the quantum
number $S^{z}=S_{A}^{z}+S_{B}^{z}$ is a good quantum number even for the
perturbed Hamiltonian, we can additionally fix $S^{z}=0$. The only remaining
relevant quantum number is then the total spin $S$, and we can abbreviate the
eigenstates of the original Hamiltonian by writing $\mathop{\lvert
S,S_{A}=S_{B}=N/4,S^{z}=0\rangle}\equiv\mathop{\lvert S\rangle}$. Calculating
the matrix elements of the symmetry-breaking field in this basis involves some
rather tedious exercises in the addition of angular momentum, but the result
is known to be:

$\displaystyle\mathop{\langle
S^{\prime}\rvert}{S}_{A}^{z}-{S}_{B}^{z}\mathop{\lvert S\rangle}$
$\displaystyle=\delta_{S^{\prime},S-1}~{}S\sqrt{\frac{(N/2+1)^{2}-S^{2}}{4S^{2}-1}}+\delta_{S^{\prime},S+1}~{}S^{\prime}\sqrt{\frac{(N/2+1)^{2}-S^{\prime
2}}{4S^{\prime 2}-1}}$
$\displaystyle\approx\left(\delta_{S^{\prime},S-1}+\delta_{S^{\prime},S+1}\right)\frac{N}{4}.$
(2.24)

In the final line we simplified the matrix element by using the fact that only
the states with $1\ll S\ll N$ will contribute to the perturbed ground state.
That this is indeed the case will become clear shortly.

The Hamiltonian in the $\mathop{\lvert S\rangle}$ basis can now be explicitly
written as:

$\displaystyle{H}^{\prime}_{\text{coll}}\approx\sum_{S}\frac{J\hbar^{2}}{2N}S^{2}\mathop{\lvert
S\rangle}\mathop{\langle S\rvert}-\mu\frac{N}{4}\mathop{\lvert
S\rangle}\mathop{\langle S+1\rvert}-\mu\frac{N}{4}\mathop{\lvert
S+1\rangle}\mathop{\langle S\rvert},$ (2.25)

where again we used the assumption $1\ll S\ll N$ to simplify the diagonal
term, and we left out the constant contribution to the energy from the $S_{A}$
and $S_{B}$ quantum numbers. We can simplify the expression further by writing
the eigenstates of the perturbed Hamiltonian as $\mathop{\lvert
n\rangle}=\sum_{S}\psi_{n}(S)\mathop{\lvert S\rangle}$, and by approximating
$S$ to be a continuous variable and taking the continuum limit in the
Schrödinger equation:

$\displaystyle\mathop{\langle S\rvert}{H}\mathop{\lvert n\rangle}$
$\displaystyle=\mathop{\langle S\rvert}E_{n}\mathop{\lvert n\rangle}$
$\displaystyle\frac{J\hbar^{2}}{2N}S^{2}\psi_{n}(S)-\frac{\mu
N}{4}\left(\psi_{n}(S+1)+\psi_{n}(S-1)\right)$
$\displaystyle=E_{n}\psi_{n}(S)$
$\displaystyle\Rightarrow~{}~{}~{}\left(\frac{J\hbar^{2}}{2N}S^{2}-\frac{\mu
N}{2}\right)\psi_{n}(S)-\frac{\mu N}{4}\frac{\partial^{2}\psi_{n}(S)}{\partial
S^{2}}$ $\displaystyle=E_{n}\psi_{n}(S)$
$\displaystyle-\frac{1}{2}\frac{\partial^{2}\psi_{n}(S)}{\partial
S^{2}}+\frac{1}{2}\omega^{2}S^{2}\psi_{n}(S)$
$\displaystyle=\epsilon_{n}\psi_{n}(S).$ (2.26)

In the third line we used the discrete version of the second derivative to
take the continuum limit, while in the fourth line we defined the variables
$\omega^{2}=2J\hbar^{2}/(\mu N^{2})$ and $\epsilon_{n}=(2E_{n}+\mu N)/(\mu
N)$.

The final expression is just the Schrödinger equation for a harmonic
oscillator with $m=\hbar=1$, whose eigenstates and energies are well known. We
should keep in mind the subtlety that in the present problem, the quantum
number $S$ can only be positive. We should therefore only accept solutions of
equation (2.4) which have a node at the origin. That is, the eigenstates of
the perturbed antiferromagnet are the (positive $S$ part of the) harmonic
oscillator eigenstates with odd values for the quantum number $n$, and the
eigenvalue corresponding to the lowest allowed state is
$\epsilon_{1}=(1+1/2)\omega$. The ground state of the perturbed
antiferromagnet is therefore the state $\mathop{\lvert n=1\rangle}$, with
ground state energy:

$\displaystyle E_{1}$ $\displaystyle=\frac{\mu
N}{2}\left(\frac{3}{2}\omega-1\right)=\frac{3}{4}\sqrt{2\mu
J\hbar^{2}}-\frac{\mu N}{2}.$ (2.27)

From the ground state energy we can again understand what happens in the
thermodynamic limit. If we first take the limit $\mu\to 0$, then the energy of
the perturbed system is just equal to the energy of the unperturbed system,
and the symmetric ground state is unaffected. However, if we first take the
limit $N\to\infty$ while keeping $\mu$ non-zero (but infinitesimally small),
then the first term in the energy can be neglected, and we find $E_{1}=-\mu
N/2$, with $N\to\infty$. But this is precisely the maximum amount of energy
that can only possibly by gained by having all spins on the $A$ sublattice
point upwards, and all spins on the $B$ sublattice point downwards. The ground
state in this limit thus must be the Néel state, and the spin-rotational
symmetry is spontaneously broken.

An alternative way of seeing the broken symmetry, is to calculate the
expectation value of the difference in magnetisation between the two
sublattices in the thermodynamic limit:

$\displaystyle\lim_{N\to\infty}\lim_{\mu\to 0}\mathop{\langle
n=1\rvert}{S}_{A}^{z}-{S}_{B}^{z}\mathop{\lvert n=1\rangle}$ $\displaystyle=0$
$\displaystyle\lim_{\mu\to 0}\lim_{N\to\infty}\mathop{\langle
n=1\rvert}{S}_{A}^{z}-{S}_{B}^{z}\mathop{\lvert n=1\rangle}$
$\displaystyle=\frac{N}{2}$ (2.28)

Again, the final line indicates that for a large enough piece of
antiferromagnetic material, even an infinitesimally small symmetry breaking
field suffices to completely align all of the spins. The spin-rotational
symmetry of the Heisenberg Hamiltonian is thus spontaneously broken in the
thermodynamic limit.

###  2.5 Symmetry breaking in the thermodynamic limit

The spontaneous breakdown of symmetry in both the harmonic crystal and the
antiferromagnet is signalled by the fact that the thermodynamic limit in those
systems is singular. As emphasised repeatedly however, the practical
implication of this singularity is that symmetric states are unstable and
symmetry-breaking states may be stable already in large, but finite, systems.
The thermodynamic limit thus foreshadows the behaviour of finite-sized
objects. Because the stable and unstable states become degenerate in the
thermodynamic limit, many aspects of symmetry breaking are easier to describe
there, even though they actually apply more generally. We will make use of
this fact here, and follow the standard approach of classifying broken-
symmetry states entirely within the thermodynamic limit.

####  2.5.1 Classification of broken-symmetry states

For mathematical convenience, we will from here on consider only Hamiltonians
and broken-symmetry states that have some degree of translational invariance.
It is sufficient for the invariance to be discrete. That is, we consider
states with a periodic arrangement of unit cells on a regular lattice. The
examples of the crystal and the antiferromagnetic Néel state clearly fall in
this class, but it also applies more generally to almost any symmetry-breaking
system of interest, at least on a course-grained level. Translational
invariance allows for a Fourier transformation with well-defined wave numbers,
which is a prerequisite for the introduction of Nambu–Goldstone modes in
Chapter 3. It also ascertains that we can write the broken-symmetry state as a
product state of the form
$\mathop{\lvert\psi\rangle}=\otimes_{i}\mathop{\lvert\psi_{i}\rangle}$, with
$\mathop{\lvert\psi_{i}\rangle}$ the same local wave function for every unit
cell $i$. If the translational symmetry happens to be continuous, rather than
discrete, the index $i$ may be replaced by a continuous parameter ${\bf x}$.
In the following, we will use $i$ and ${\bf x}$ interchangeably.

Just the assumption of translational invariance is sufficient to guarantee
that distinct symmetry-breaking states are orthogonal in the thermodynamic
limit. To see this, consider the overlap between two normalised broken-
symmetry states $\mathop{\lvert\psi\rangle}$ and
$\mathop{\lvert\psi^{\prime}\rangle}$:

$\displaystyle{\mathop{\langle\psi^{\prime}|\psi\rangle}}$
$\displaystyle=\left(\otimes_{{\bf
x}^{\prime}}\mathop{\langle\psi^{\prime}({\bf
x}^{\prime})\rvert}\right)\left(\vphantom{\psi^{\prime}}\otimes_{\bf
x}\mathop{\lvert\psi({\bf x})\rangle}\right)$ $\displaystyle=\prod_{\bf
x}{\mathop{\langle\psi^{\prime}({\bf x})|\psi({\bf
x})\rangle}}={\mathop{\langle\psi^{\prime}({\bf x})|\psi({\bf
x})\rangle}}^{N}.$ (2.29)

Here we used the vanishing overlap between local states
$\mathop{\lvert\psi({\bf x})\rangle}$ and $\mathop{\lvert\psi({\bf
x}^{\prime})\rangle}$ for ${\bf x}\neq{\bf x}^{\prime}$ to write the overlap
of product states as a product of local overlaps. In the final step, we also
used translational invariance by taking $\mathop{\lvert\psi({\bf x})\rangle}$
to be independent of ${\bf x}$. The number of unit cells is denoted $N$. If
the two symmetry-breaking states $\mathop{\lvert\psi\rangle}$ and
$\mathop{\lvert\psi^{\prime}\rangle}$ are distinct, then the local overlap
${\mathop{\langle\psi^{\prime}({\bf x})|\psi({\bf x})\rangle}}$ must be
smaller than one, and the full inner product
${\mathop{\langle\psi^{\prime}|\psi\rangle}}$ vanishes as $N$ is taken to
infinity. On the other hand, if the states are not distinct, and differ only
by a phase $\mathop{\lvert\psi^{\prime}({\bf
x})\rangle}=\mathrm{e}^{\mathrm{i}\varphi}\mathop{\lvert\psi({\bf
x})\rangle}$, the inner product is one, up to a phase factor. We thus find
that any two symmetry-breaking states in the thermodynamic limit must be
either equivalent or orthogonal.

At this point, we can apply the group theory of Section 1.6.1 to begin the
classification of all possible distinct symmetry-breaking states. Recall that
symmetry transformations make up a group $G$, with elements $g$. As we saw
before, a specific symmetry transformation $g$ is represented in quantum
physics by a unitary operator acting on Hilbert space. With a slight abuse of
notation, and to avoid introducing too many symbols, we will write $g$ both
for the element in the group of symmetry transformations, and for the operator
it represents. We can then consider a state $\mathop{\lvert\psi\rangle}$ that
breaks some of the symmetries in the group $G$, but is invariant under others.
For symmetry transformations $g$ that are broken, $\mathop{\lvert\psi\rangle}$
and $g\mathop{\lvert\psi\rangle}$ are distinct, inequivalent states. Assuming
some degree of translational invariance, they must then be orthogonal in the
thermodynamic limit. For unbroken symmetry transformations $g$, on the other
hand, we know from the definition of a symmetric state in equation (1.1), that
$g\mathop{\lvert\psi\rangle}=\mathrm{e}^{\mathrm{i}\varphi}\mathop{\lvert\psi\rangle}$.
The set of all such unbroken transformations together forms a subgroup
$H\subset G$ called the residual symmetry group. ††margin:  residual symmetry
group

######  Exercise 2.5 (Subgroup of unbroken transformations).

Recall the definition of a subgroup from Section 1.6.1 and show that the set
of transformations that leave $\mathop{\lvert\psi\rangle}$ invariant indeed
form a subgroup.

Now consider two transformations, $g_{1}$ and $g_{2}$, which happen to satisfy
the relation $g_{1}=g_{2}h$ for some element $h$ of the residual symmetry
group. We then see that
$g_{1}\mathop{\lvert\psi\rangle}=\mathrm{e}^{\mathrm{i}\varphi}g_{2}\mathop{\lvert\psi\rangle}$,
which implies that $g_{1}\mathop{\lvert\psi\rangle}$ is equivalent to
$g_{2}\mathop{\lvert\psi\rangle}$. Conversely, if $g_{1}$ and $g_{2}$ do not
satisfy $g_{1}=g_{2}h$ for any $h\in H$, it follows that
$g_{1}\mathop{\lvert\psi\rangle}$ and $g_{2}\mathop{\lvert\psi\rangle}$ are
distinct and orthogonal broken-symmetry states. Since the operators
corresponding to a broken symmetry transform a given symmetry-breaking state
into an inequivalent one, they are often said to correspond to abstract
‘rotations’ within the space of broken-symmetry states. Starting from the
initial state $\mathop{\lvert\psi\rangle}$, we can label each symmetry-
breaking state $\mathop{\lvert\psi^{\prime}\rangle}$ that is distinct from it
by some $g$ for which
$\mathop{\lvert\psi^{\prime}\rangle}=g\mathop{\lvert\psi\rangle}$. The
resulting list of labels for inequivalent broken-symmetry states consists of a
subset of the elements $g$ of the symmetry group $G$, in which no two elements
$g$ and $g^{\prime}$ satisfy the relation $g=g^{\prime}h$ for any $h\in H$.
Looking back at Section 1.6.1, this coincides precisely with the definition of
the quotient set $G/H$.

Inequivalent symmetry-broken states are classified by the cosets $gH$ as
elements of the quotient set $G/H$, where $G$ is the group of all symmetry
transformations and $H\subset G$ is the subgroup of unbroken transformations.

Notice that this classification of symmetry-breaking states in terms of cosets
only applies to the thermodynamic limit. In a system of finite size, distinct
broken-symmetry states need not be precisely orthogonal. This is easy to see
in the case of a finite object breaking for example continuous rotational
symmetry. The Hilbert space in that case is finite-dimensional, but there are
infinitely many symmetry-broken states labelled by all possible directions in
space. These correspond to the infinitely many cosets of the rotational
symmetry group.

To make the classification more concrete, consider for example a Hamiltonian
with a global phase-rotational symmetry, described by the continuous group of
symmetry transformations $G=U(1)$, see Eq. 1.19. For a stable state that
breaks the phase rotations, so that there is no symmetry left, the residual
symmetry group is just the trivial group $H=e$, containing only the identity
operator. The quotient set is then $G/H=U(1)$ and distinct or inequivalent
broken-symmetry states may be labeled by their phase factors
$\mathrm{e}^{\mathrm{i}\varphi}$. This global phase rotation actually is the
broken symmetry characterising a superfluid, and superfluids with different
phase values may be distinguished by observing a Josephson current between
them, as we will see in Section 2.5.5. More generally, whenever all symmetry
transformations in a group are simultaneously broken, the symmetry-breaking
states are simply labelled by the elements of the full symmetry group.

As a second example, suppose that the spin-rotation symmetry in the Heisenberg
antiferromagnet of Section 2.4 is broken down to only rotations around a
single axis. This is the case for example in the classical Néel state. Then
the group describing the spin-rotational symmetry of the Hamiltonian is
$G=SU(2)$, while the residual symmetry group describing the leftover rotations
around a single axis is $H=U(1)$. It can be shown that the quotient set $G/H$
equals $SU(2)/U(1)\simeq S^{2}$, which corresponds to the set of points on the
surface of a sphere. These points indicate the possible directions of the
residual rotation axis, or equivalently, the direction of the sublattice
magnetisation. Note that while $S^{2}$ classifies all possible symmetry-broken
states, it does not have a group structure itself.

For continuous groups, the classification of broken symmetry states can also
be expressed in terms of the generators of the continuous symmetry
transformations, defined in Eq. (1.5). In this context, generators $Q$ of
which the broken-symmetry state under consideration is an eigenstate, are
called unbroken generators or unbroken Noether charges, and any finite
transformation generated by $Q$ is also unbroken. Conversely, generators that
do not leave the state invariant are called broken. The continuous symmetry
group may be broken down to either a continuous or a discrete subgroup, and
even to the trivial group. The dimension of the quotient set $G/H$ for
continuous groups $G$ is said to equal the number of broken generators. The
algebraic relations between broken and unbroken generators will play an
important role in the classification of Nambu–Goldstone modes in Chapter 3.

####  2.5.2 The order parameter

Having classified all possible inequivalent symmetry-breaking states, it would
be useful to have an operator whose expectation value can be used to
distinguish between them. Ideally, such an operator would have expectation
value zero in any symmetric state, and a unique non-zero expectation value for
each of the sets of equivalent broken-symmetry states. As it turns out, such
an operator can be defined, and is called the order parameter operator.
††margin:  order parameter operator Actually the word order parameter is often
used to denote any quantity whose expectation value is non-zero in the broken-
symmetry phase, and zero in the symmetric state, without having the additional
benefit of distinguishing between inequivalent symmetry-breaking
configurations. A well-known example can be found in the theory of
superconductivity, where the amplitude of the gap function (see Section 7.2)
is often called the superconducting order parameter. In these lecture notes,
we will stick to the more narrow definition, which has the added advantage
that it will be instrumental in deriving the Goldstone theorem in Chapter 3.

To define a fool-proof recipe for identifying an order parameter operator, we
first need to slightly update our definition of what constitutes a symmetric
state. ††margin:  symmetry – of states

######  Definition 2.6.

Let $U=\mathrm{e}^{\mathrm{i}\alpha{Q}}$ be a symmetry transformation that
commutes with the Hamiltonian, and which is parameterised by a discrete or
continuous variable $\alpha$. A state $\mathop{\lvert\psi\rangle}$ breaks this
symmetry if there exists any operator $\Phi$ such that:

$\mathop{\langle\psi\rvert}[{Q},\Phi]\mathop{\lvert\psi\rangle}\neq 0.$ (2.30)

If no such operator $\Phi$ exists, the state is symmetric under the
transformation $U$.

This definition is consistent with our earlier intuitive definition of
symmetric states being eigenstates of the symmetry transformation, because if
$U\mathop{\lvert\psi\rangle}=\mathrm{e}^{\mathrm{i}\varphi}\mathop{\lvert\psi\rangle}$,
the left-hand side of Eq. (2.30) vanishes for any operator $\Phi$, and the
state is said to be symmetric.

††margin:  interpolating field

The operator $\Phi$ appearing in Eq. (2.30) will be a field $\Phi({x})$ acting
locally in space in all cases we shall encounter in these lecture notes. It is
called the interpolating field. With it, we can define the order parameter
operator ${\mathcal{O}}({x})$ related to a broken symmetry ${Q}$, and its
expectation value ${O}({x})$, which is known as the (local) order parameter:
††margin:  order parameter

$\displaystyle{\mathcal{O}}({x})$ $\displaystyle=[{Q},\Phi({x})]$
$\displaystyle{O}({x})$
$\displaystyle=\mathop{\langle\psi\rvert}{\mathcal{O}}({x})\mathop{\lvert\psi\rangle}.$
(2.31)

Notice that if $\mathop{\lvert\psi\rangle}$ is translationally invariant, then
so is ${O}({x})$. Also, $\Phi({x})$ and $\mathcal{O}({x})$ are not necessarily
Hermitian, but they can always be used to construct an observable quantity,
such as $\mathcal{O}+\mathcal{O}^{\dagger}$ or
$\mathcal{O}\mathcal{O}^{\dagger}$. The order parameter ${O}({x})$ is just the
left-hand side of Eq. (2.30). Therefore, the order parameter is automatically
zero if $\mathop{\lvert\psi\rangle}$ is symmetric and non-zero if
$\mathop{\lvert\psi\rangle}$ breaks a symmetry under the new definition of the
symmetry of states. It thus clearly satisfies the requirement of
distinguishing symmetric from symmetry-breaking states.

To make sure that the order parameter also distinguishes beween inequivalent
broken-symmetry states, we can require that any good order parameter operator
$\mathcal{O}$ has eigenvalues that map in a one-to-one fashion onto the
quotient space $G/H$ introduced in Section 2.5.1. This way, the order
parameter $O({x})$ will not only be different for distinct broken-symmetry
states and equal for states related by residual symmetry transformations, but
it will also inherit the structure of the quotient space. In particular, this
means that states which are close to each other, according to the topological
structure of the Lie group $G$, will have only a small difference in their
corresponding order parameter values.

As it turns out, it is always possible to find an order parameter operator
that satisfies these constraints, because Eq. (2.30) does not uniquely
determine the order parameter and interpolating field. For example,
multiplying $\Phi$ by a constant, or taking its Hermitian conjugate, yield
alternative definitions of an interpolating field that still obey Eq. (2.30).
In almost all cases, a convenient choice for the order parameter operator,
which maps onto the quotient space $G/H$, is suggested by the physics of the
symmetry-breaking system itself.

To make the formal definitions of the order parameter and interpolating field
more concrete, consider the example of the Heisenberg antiferromagnet, which
we discussed in Section 2.5.1. The Hamiltonian has $SU(2)$ spin-rotational
symmetry, which is broken down in the antiferromagnetic state to just $U(1)$
rotations around the axis of sublattice magnetisation. Inequivalent broken-
symmetry states correspond to antiferromagnetic configurations with the
sublattice magnetisation pointing in different directions. All possible
directions constitute the set of points on the surface of a sphere, $S^{2}$,
which indeed coincides with the quotient $SU(2)/U(1)\simeq S^{2}$. For a
specific broken-symmetry state with the sublattice magnetisation along the
$z$-axis, the symmetry generators $S^{x}$ and $S^{y}$ are spontaneously
broken, while $S^{z}$, which generates rotations around the $z$-axis, is
unbroken. We can then define the staggered magnetisation ††margin:  staggered
magnetisation as $N^{a}_{i}=(\pm 1)^{i}S^{a}_{i}$, where $i$ is a position
index that is even on the $A$-sublattice and odd on the $B$-sublattice, and
$a$ denotes a direction in spin space. To describe the breaking of rotations
generated by $S^{x}$, we can choose $N^{y}$ to be the interpolating field,
which yields
$\sum_{ij}[S^{x}_{i},N^{y}_{j}]=\mathrm{i}\sum_{ij}\delta_{ij}N^{z}_{i}=\mathrm{i}\sum_{i}N^{z}_{i}$
as the order parameter operator. Similarly, for the breakdown of rotations
around the $y$-axis, we can choose $N^{x}$ as the interpolating field, which
also corresponds to $N^{z}$ as the order parameter operator. The staggered
magnetisation identified as the order parameter in this way, is also the
natural choice for an antiferromagnet, because the Néel state is precisely an
eigenstate of the staggered magnetisation operator.

As a second example, recall the Schrödinger field theory of Section 1.3.1.
This model has a global $U(1)$ symmetry describing uniform rotations of the
phase of a complex scalar field $\psi(x)$. To find out what it means for this
symmetry to be broken, we can for example consider the field $\psi(x)$ itself
to be an interpolating field. The associated order parameter is then:

$[Q,\psi(x)]=\int\mathrm{d}^{d}x^{\prime}\;\hbar[\psi^{*}(x^{\prime})\psi(x^{\prime}),\psi(x)]=-\hbar\psi(x).$
(2.32)

Here we used the commutation relation
$[\psi^{*}(x^{\prime}),\psi(x)]=-\delta(x-x^{\prime})$. The order parameter
operator thus turns out to be given by the field $\psi(x)$ itself. Because an
eigenstate of the order parameter is a symmetry-breaking state, the states
that break the global phase rotation symmetry must be eigenstates of the
operator $\psi(x)$. But that operator is just the annihilation operator for
quanta of the $\psi$-field. Within Fock space, it is possible to construct
eigenstates of the annihilation operator, and these are called coherent
states. ††margin:  coherent state You can check that the state
$\mathrm{e}^{\int\mathrm{d}^{D}x\,\phi(x)\psi^{*}(x)}\mathop{\lvert\text{vac}\rangle}$
is an eigenstate of the annihilation operator $\psi(x)$, with complex
eigenvalue $\phi(x)$. Because this coherent state is an eigenstate of the
field operator, $|\phi(x)|^{2}$ is the expectation value of the number of
field quanta, or particles, at position ${\bf x}$. Notice that if we assume
translational invariance as before, $\phi$ is independent of ${\bf x}$.
Expanding the exponential in the definition of the coherent state, it is seen
to be a superposition of infinitely many states with different numbers of
excited field quanta. Conversely, an eigenstate of the number or density
operator $\psi^{*}(x)\psi(x)$ can be written as a superposition of infinitely
many coherent states which all have the same absolute value of $\phi(x)$, but
which differ in phase. The phase and modulus of $\phi$ can in fact be shown to
be conjugate variables, with canonical commutation relations.

The symmetry-breaking state of the Schrödinger field is a coherent state,
which has an indefinite number of particles or quanta. This is in direct
accordance with the fact that the symmetry generated by
$Q\propto\int\psi^{*}\psi$ is associated with the conservation of the number
of field quanta. Distinct symmetry-breaking states are characterised by the
phase of $\phi$. Performing rotations with $\exp(\mathrm{i}\alpha Q)$ will
lead to other order parameter operators $\mathrm{e}^{\mathrm{i}\alpha}\psi(x)$
which indeed correspond to broken-symmetry states with different phase values
that can be labelled by elements of the coset $U(1)/1\simeq U(1)$. The
formation of a state with indeterminate particle number and precise phase-
value is a good interpretation of what happens in Bose–Einstein condensates
such as superfluids and superconductors. In these symmetry-broken states of
matter, it costs zero energy to add or remove a particle in the condensate.

####  2.5.3 The classical state

Given the definition of the order parameter operator ${\mathcal{O}}$ in Eq.
(2.31), you might be tempted to believe that broken-symmetry states are simply
the eigenstates of ${\mathcal{O}}$, with eigenvalues ${O}$. This would
certainly justify the definition of the order parameter as the expectation
value of the order parameter operator, and in most cases it also agrees very
well with our expectation for a perfectly ordered state. This is because in
translationally invariant systems, the eigenstates of the local operator
${\mathcal{O}}(x)$ are tensor products of local eigenstates
$\mathop{\lvert\psi\rangle}=\otimes_{\mathbf{x}}\mathop{\lvert\psi(\mathbf{x})\rangle}$,
and they correspond directly to the states of a classical Hamiltonian in which
all operators are replaced by their expectation values. ††margin:  classical
state We will call the eigenstates of the order parameter operator classical
states.

The symmetry-breaking states encountered in real quantum systems are typically
not classical states. This is easy to understand, because although a symmetry-
breaking perturbation may dominate the shape of the ground state for
sufficiently large systems, it is not the only contribution to the
Hamiltonian. The remaining, symmetric part of the Hamiltonian contributes to
the ground state and takes it away from the classical ideal. We have already
seen this for the antiferromagnet in Section 2.4. The true, quantum, broken-
symmetry states typically have order parameter expectation values close to
those of the classical state. The quantum states can therefore be thought of
as arising in a perturbation theory around the classical states. The
differences between the classical state and the true quantum broken-symmetry
state, are then known as ††margin:  quantum corrections quantum corrections to
the classical state. These corrections consist of a part at zero wave number,
related to the tower of states that will be discussed in more detail in
Section 2.6, and a part at nonzero momentum, which is the topic of Chapter 4.

####  2.5.4 Long-range order

For any given uniform ground state, the order parameter identifies whether it
has broken or unbroken symmetries, and distinguishes between inequivalent
symmetry-breaking states. In practical calculations however, one often does
not know the exact ground state of a given Hamiltonian, and many systems exist
in conditions that are not perfectly uniform. In those cases, a particularly
useful alternative way of quantifying the occurrence of spontaneous symmetry
breaking, is through the behaviour of the so-called two-point correlation
function:

$C(\mathbf{x},\mathbf{x}^{\prime})=\mathop{\langle\psi\rvert}{\mathcal{O}}^{\dagger}(\mathbf{x}){\mathcal{O}}(\mathbf{x}^{\prime})\mathop{\lvert\psi\rangle}.$
(2.33)

Here ${\mathcal{O}}$ is the local order parameter operator. Clearly, if
$\mathop{\lvert\psi\rangle}$ is uniform in space and an eigenstate of the
order parameter, the correlation function is the same for any choice of the
coordinates $\mathbf{x}$ and $\mathbf{x}^{\prime}$, and equal to the square of
the uniform order parameter. The advantage of the two-point function, however,
is that it can be used also in less clear-cut cases. The behaviour of the two-
point function can be divided roughly into two distinct classes, depending on
its functional form as the distance
$\lvert\mathbf{x}-\mathbf{x}^{\prime}\rvert$ is taken to infinity:

$\displaystyle
C(\mathbf{x},\mathbf{x}^{\prime})\propto\begin{cases}\@add@raggedright\marginpar{\footnotesize\raggedright
long-rangeorder}\text{constant}&\text{\em{long-range ordered}}\\\
\mathrm{e}^{-\lvert\mathbf{x}-\mathbf{x}^{\prime}\rvert/l}&\text{\em{disordered}}\end{cases}\qquad\lvert\mathbf{x}-\mathbf{x}^{\prime}\rvert\to\infty.$
(2.34)

Here $l$ is a length scale called the ††margin:  correlation length
correlation length. For long-range ordered systems, the spatial average of the
local order parameter will be non-zero, and the correlation length $l$
diverges. The presence of long-range order is therefore associated with the
breaking of a symmetry. In fact, the two-point correlation function signals
the propensity to break symmetry already for finite-size systems that really
have a symmetric ground state. Even though the order parameter expectation
value is exactly zero, the two-point function shows correlations for long
separations. This can be easily understood by considering the extreme example
of the two-spin singlet state
$\lvert\uparrow\downarrow\rangle-\lvert\downarrow\uparrow\rangle$, which is
the ground state of the two-site Heisenberg antiferromagnet. Clearly, the
singlet state has no preferred direction of staggered magnetisation, but the
two spins are definitely anti-parallel. Especially in numerical
investigations, the two-point function is typically easy to ‘measure’ even
with only limited information about the spectrum, and is widely used in
establishing the presence of order and spontaneous symmetry breaking. We will
come back to this in more detail when we discuss stability in Section 2.7.

Notice that the separation of states into long-range ordered and disordered is
not exhaustive. A special case may occur in low dimensions when the two-point
function is proportional $\lvert\mathbf{x}-\mathbf{x}^{\prime}\rvert^{c}$, for
some exponent $c$. This is called algebraic long-range order ††margin:
algebraic long-range order or quasi long-range order. We will discuss this
type of order in Section 6.3.

You may have also encountered the term “off-diagonal long-range order” or
“ODLRO” ††margin:  long-range order – off-diagonal in the literature. This
term was introduced by Oliver Penrose and Lars Onsager in the 1950s [27, 28],
in the context of superfluidity in helium, as a way of contrasting the
superfluid order with ordering in solids. The concept is largely historical,
and the distinction between ODLRO and other types of order has become obsolete
with the modern definition of the order parameter in Eq. (2.31). To see this,
we will first consider the usual definition of ODLRO in terms of the
$N$-particle wave function $\Psi(\mathbf{x}_{1},\ldots,\mathbf{x}_{N})$. We
can then define:

$\rho(\mathbf{x}_{1},\ldots,\mathbf{x}_{N},\mathbf{y}_{1},\ldots,\mathbf{y}_{N})=\Psi^{*}(\mathbf{x}_{1},\ldots,\mathbf{x}_{N})\Psi(\mathbf{y}_{1},\ldots,\mathbf{y}_{N}),$
(2.35)

for two sets of coordinates $\mathbf{x}_{i}$ and $\mathbf{y}_{i}$. For
coinciding coordinates $\mathbf{y}_{i}=\mathbf{x}_{i}$, the matrix
$\rho(\mathbf{x}_{i})\equiv\rho(\mathbf{x}_{i},\mathbf{x}_{i})$ is just the
usual ††margin:  density matrix density matrix, giving the probability for
finding the $N$ particles at positions $\mathbf{x}_{i}$. The two-particle
reduced density matrix can be found by integrating over all but two of the
coordinates:

$\rho_{\mathrm{D},2}(\mathbf{x}_{1},\mathbf{x}_{2})=\int\mathrm{d}\mathbf{x}_{3}\cdots\mathrm{d}\mathbf{x}_{N}\;\Psi^{*}(\mathbf{x}_{1},\mathbf{x}_{2},\mathbf{x}_{3},\ldots,\mathbf{x}_{N})\Psi(\mathbf{x}_{1},\mathbf{x}_{2},\mathbf{x}_{3},\ldots,\mathbf{x}_{N}).$
(2.36)

If space is uniform, the reduced density matrix is invariant under global
translations, and can only depend on the difference of the two coordinates, so
that
$\rho_{\mathrm{D},2}(\mathbf{x}_{1},\mathbf{x}_{2})=\rho_{\mathrm{D},2}(\mathbf{x}_{1}-\mathbf{x}_{2})$.
††margin:  long-range order – diagonal Diagonal long-range order is said to
occur if $\rho_{2}(\mathbf{x}_{1}-\mathbf{x}_{2})$ is periodic in
$\mathbf{x}_{1}-\mathbf{x}_{2}$. This is the usual ordering we find for solids
such as the harmonic crystal introduced in Section 2.3. It is called diagonal
because we only need to consider diagonal elements of
$\rho(\mathbf{x}_{i},\mathbf{y}_{i})$, with $\mathbf{y}_{i}=\mathbf{x}_{i}$.

Alternatively, we can consider another type of two-point function:

$\rho_{\mathrm{O},2}(\mathbf{x},\mathbf{y})=\int\mathrm{d}\mathbf{x}_{2}\cdots\mathrm{d}\mathbf{x}_{N}\;\Psi^{*}(\mathbf{x},\mathbf{x}_{2},\ldots,\mathbf{x}_{N})\Psi(\mathbf{y},\mathbf{x}_{2},\ldots,\mathbf{x}_{N})$
(2.37)

That is, we choose all wave function coordinates except the first to coincide,
and integrate over them. Note that this entails one more integration than the
definition of the reduced density matrix in Eq. (2.36). In fact,
$\rho_{\mathrm{O},2}$ and $\rho_{\mathrm{D},2}$ represent two very different
physical quantities that have little to do with one another. For superfluids
in particular, $\rho_{\mathrm{O},2}$ is almost identical to the two-point
correlation function of Eq. (2.33), if we choose the order parameter operator
${\mathcal{O}}(x)$ to be the field operator $\Psi(x)$. ††margin:  long-range
order – off-diagonal Off-diagonal long-range order (ODLRO) is said to occur if
$\rho_{\mathrm{O},2}(\mathbf{x},\mathbf{y})$ does not vanish as
$\mathbf{x}-\mathbf{y}$ is taken to infinity. It was called off-diagonal
because it involves off-diagonal elements of
$\rho(\mathbf{x}_{i},\mathbf{y}_{i})$.

Although we need to be careful when applying Eq. (2.34) to crystals, because
they retain discrete translational symmetry and
$C(\mathbf{x},\mathbf{x}^{\prime})$ approaches a periodic function rather than
a true constant at large separations, the two-point function does capture the
breakdown of symmetry in both crystals and superfluids in essentially the same
way. In fact, both diagonal and off-diagonal long-range order are part of a
much larger family of possible types of ordered states that are all classified
by the behaviour of the two-point correlation function at large separation.
There is nothing special about either the order occurring in crystals (DLRO),
or that in superfluids (ODLRO). When off-diagonal long-range order was
originally introduced, the concept of symmetry breaking for internal degrees
of freedom and its embedding within the larger theory of symmetry breaking in
general were not yet developed, and ODLRO was a way to capture the long-range
ordering of the internal $U(1)$-phase degree of freedom contained in the
$N$-body wave function.

####  2.5.5 The Josephson effect

When we first discussed symmetric states in Section 1.1, we pointed out that
symmetry must always be defined with respect to some reference. In the example
of a crystal breaking translational symmetry, the broken translations are
really defined with respect to an outside observer, who can for example
measure the distance between the crystal and herself. More generally, in order
to observe the breakdown of a symmetry in a given state, an observer needs
some reference frame with respect to which the broken symmetry can be
measured. For the reference frame to be able to distinguish between
inequivalent symmetry-breaking states, it must itself be in broken-symmetry
state. That is, you cannot measure the position of a crystal with respect to a
uniform fluid permeating all of space, and it is not possible to measure a
direction of magnetisation with a piece of plastic that cannot itself be
magnetised. Although this might seem obvious for crystals and magnets, one
could wonder what it implies for materials hosting less intuitive forms of
broken symmetry, like the $U(1)$ phase-rotation symmetry associated with
conserved particle number, which we argued in Section 2.5.2 to be broken in
superfluids? This question found a literal manifestation in the spontaneous
tunnelling current that was predicted by Josephson in 1962 to occur between
two separated pieces of superconducting material [29]. ††margin:  Josephson
effect

Since the origin of the Josephson effect lies in the broken $U(1)$ symmetry,
we will discuss it here for neutral superfluids rather than superconductors.
As we saw before, a superfluid is a state with broken $U(1)$ phase-rotation
symmetry and a complex order parameter
$\psi=\lvert\psi\rvert\mathrm{e}^{\mathrm{i}\varphi}$. In terms of observable
properties, the superfluid is characterised by its ability to host
supercurrents that flow without viscosity. One way to understand this, is by
noticing that the supercurrent is identical to the conserved Noether current
associated with the broken $U(1)$ symmetry:

$j^{n}=\mathrm{i}\left((\partial_{n}\psi^{*})\psi-\psi^{*}(\partial_{n}\psi)\right).$
(2.38)

The order parameter field $\psi$ is the expectation value of the field
operator, but is often referred to in the more popular literature as a
“macroscopic wave function”. Although somewhat misleading, this terminology
does emphasise the fact that like a quantum wave function, the field $\psi(x)$
satisfies equations of motion with spatial derivatives, which force it to be
continuous. As a consequence, the field does not abruptly vanish at the
boundary of a sample, but rather falls off exponentially into the vacuum. For
two samples of superfluid separated by a small gap, the order parameter fields
extending into the gap from both sides can overlap. Just like for quantum
mechanical wave functions, this implies the possibility for field quanta to
tunnel from one sample to the other, which is the essence of the Josephson
effect.

For a junction of width $w$ between superfluids with constant order parameters
$\psi_{1}$ and $\psi_{2}$, the order parameter field inside the junction can
be written as:

$\psi(x)=A\mathrm{e}^{-x/\xi}+B\mathrm{e}^{x/\xi}.$ (2.39)

Here, $A$ and $B$ are complex constants, and the decay length of the order
parameter field in the vacuum is $\xi$. The samples are assumed to extend
indefinitely in the $y$ and $z$ direction while being semi-infinite in the $x$
direction, with one sample having an edge at $x=-w/2$ and the other at
$x=w/2$. You can think of the field in the junction as a superposition of the
decaying order parameter fields from either side. The boundary conditions are
given by the field values in the samples, $\psi(-w/2)=\psi_{1}$ and
$\psi(w/2)=\psi_{2}$, so that we find:

$\displaystyle A$
$\displaystyle=\frac{\mathrm{e}^{w/2\xi}\psi_{1}-\mathrm{e}^{-w/2\xi}\psi_{2}}{2\sinh(w/\xi)},$
$\displaystyle B$
$\displaystyle=\frac{\mathrm{e}^{w/2\xi}\psi_{2}-\mathrm{e}^{-w/2\xi}\psi_{1}}{2\sinh(w/\xi)}.$
(2.40)

The current density in the junction is given by Eq. (2.38), and equals
$j^{x}=2\mathrm{i}(B^{*}A-A^{*}B)/\xi$, independent of the position $x$.
Substituting the values of $A$ and $B$ yields the expression:

$j^{x}=\mathrm{i}\frac{\psi_{2}^{*}\psi_{1}-\psi_{1}^{*}\psi_{2}}{\xi\sinh(w/\xi)}=\frac{2\lvert\psi_{1}\rvert\lvert\psi_{2}\rvert}{\xi\sinh(w/\xi)}\sin(\varphi_{2}-\varphi_{1}).$
(2.41)

We thus find a current per unit area $j_{x}$ flowing through the junction,
which is proportional to the sine of the phase difference between the two
superfluid order parameters.

The flow of supercurrent without a chemical potential difference between the
superfluid samples is an interesting physical observation in and of itself. In
the context of symmetry breaking however, it also gains a more fundamental
interpretation. The phase of the order parameter for one superconducting
sample can be determined with respect to the phase of a second sample by
measuring the Josephson current flowing between them. This is precisely
analogous to the way that the position of a crystal can be determined only
with respect to the position of some other object with broken translational
symmetry. Historically, the discovery of the Josephson effect was therefore
the deciding factor in settling the debate of whether or not a symmetry was
spontaneously broken in superconductors (see for instance Ref. [30]). More
generally, the calculation of the Josephson effect should actually be
applicable in some form to two pieces of material with any type of
spontaneously broken symmetry[31]. There are very few known examples of
generalised Josephson effects outside of superconductivity and superfluidity,
but at least conceptually, the Josephson effect offers an unambiguous general
way of measuring the order parameter of any sample with respect to a reference
broken-symmetry state.

######  Exercise 2.7 (Josephson effect).

The Josephson effect equations for describing the generalised Josephson
current between any two symmetry-breaking objects can be derived from a very
simple model due to Feynman [32]. Here you write the global order parameter
operators of the two systems at the left ($\mathrm{L}$) and right
($\mathrm{R}$) as $\Psi_{\mathrm{L}}(t)$ and $\Psi_{\mathrm{R}}(t)$ (no
$\mathbf{x}$-dependence). The Hamiltonian is taken to be of the form:

$H=H_{\mathrm{L}}+H_{\mathrm{R}}+H_{K},$ (2.42)

where the left and right systems are described by local Hamiltonians
$H_{\mathrm{L}}$ and $H_{\mathrm{R}}$. The coupling between the order
parameters across the junction is described by $H_{K}$.

a. For superconductors, the order parameter operators correspond to field
operators $\psi_{i}$, with commutation relation
$[\psi_{i},\psi^{*}_{i^{\prime}}]=\delta_{i,i^{\prime}}$. The coupling is
given by:

$H_{K}=K(\psi_{\mathrm{R}}^{*}\psi_{\mathrm{L}}+\psi^{*}_{\mathrm{L}}\psi_{\mathrm{R}}),$
(2.43)

where $K$ is a coupling constant with units of energy. Using the Heisenberg
equations of motions $-\mathrm{i}\hbar\partial_{t}A=[H,A]$, derive an
expression for the Josephson current
$I_{\mathrm{J}}=\partial_{t}(\psi^{*}_{\mathrm{L}}\psi_{\mathrm{L}})$. Compare
your result with Eq. (2.41).

b. Now consider two ferromagnets, with magnetisation vectors $\mathbf{M}_{i}$,
and commutation relations
$[M^{a}_{i},M^{b}_{i^{\prime}}]=\mathrm{i}\epsilon_{abc}\delta_{i,i^{\prime}}M^{c}_{i}$.
They are coupled to each other via the interaction described by:

$H_{K}=K\mathbf{M}_{\mathrm{L}}\cdot\mathbf{M}_{\mathrm{R}}.$ (2.44)

Derive an expression for the “spin Josephson current”
$\partial_{t}\mathbf{M}_{\mathrm{L}}$. Your result will agree with the much
more sophisticated calculation based on a microscopic description of the
tunnelling of electrons between two ferromagnets [33].

###  2.6 The tower of states

As emphasised in several places already, the true ground state of a finite-
sized quantum system without any symmetry-breaking field present, is typically
symmetric. Except for symmetry-breaking states associated with conserved order
parameters that will be introduced in Exercise 2.8 and discussed in more
detail in Section 3.4, the true ground state of a symmetric Hamiltonian is
unique, and is an eigenstate of the symmetry generators. Since the symmetry
transformations are global, the symmetric ground state also has a global
structure. In particular, it typically contains ††margin:  entanglement long-
range entanglement between distant parts of the system, which therefore all
strongly depend on each other (as discussed in more detail in Section A.2).
Furthermore, as we will see in the next section, the symmetric state is
unstable. In terms of the spectrum of eigenstates of the symmetric
Hamiltonian, however, the long-range entangled nature of the ground state is
in no way exceptional. There is a whole set of eigenstates that can be seen as
low-energy, global excitations on top of the ground state, which all share the
same feature.

Crudely speaking, if the order parameter can be defined in terms of some
canonical observable, the symmetric Hamiltonian must contain a kinetic energy
proportional to the total canonical momentum squared. This is easy to see,
because the eigenstates of momentum are symmetric combinations of all possible
canonical positions. Within a finite volume $V$, the Fourier transform of the
canonical momentum operator is given by
${\mathbf{p}}(\mathbf{x})=\sum_{\mathbf{k}}\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot\mathbf{x}}\mathbf{p}_{\mathbf{k}}$.
The total momentum is proportional to the $\mathbf{k}=0$ component of this
decomposition:

$\displaystyle\mathbf{p}_{\mathrm{tot}}$
$\displaystyle=\int\mathrm{d}\mathbf{x}\;\mathbf{p}(\mathbf{x})=\sum_{\mathbf{k}}\int\mathrm{d}\mathbf{x}\,\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot\mathbf{x}}\mathbf{p}_{\mathbf{k}}$
$\displaystyle=V\sum_{\mathbf{k}}\delta_{\mathbf{k}}\mathbf{p}_{\mathbf{k}}=V\mathbf{p}_{\mathbf{k}=0}.$
(2.45)

Here we used the representation of the Kronecker delta function given by
$\frac{1}{V}\int\mathrm{d}\mathbf{x}\;\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot\mathbf{x}}=\delta_{\mathbf{k}}$888Note
that
$\frac{1}{V}\sum_{\mathbf{k}}e^{-i\mathbf{k}\cdot\mathbf{x}}=\delta(\mathbf{x})$
yields the Dirac delta function.. The term in the Hamiltonian proportional to
the total momentum comes from the $\mathbf{k}=0$ part of the usual kinetic
energy operator:

$\displaystyle H_{\mathrm{kin}}$
$\displaystyle\propto\int\mathrm{d}\mathbf{x}\;\mathbf{p}^{2}(\mathbf{x})=V\sum_{\mathbf{k}}\mathbf{p}_{\mathbf{k}}\cdot\mathbf{p}_{-\mathbf{k}}$
$\displaystyle=\frac{1}{V}\mathbf{p}_{\mathrm{tot}}^{2}+V\sum_{\mathbf{k}\neq
0}\mathbf{p}_{\mathbf{k}}\cdot\mathbf{p}_{-\mathbf{k}}$ (2.46)

The second term in the final line combines with the potential energy to
describe internal excitations like phonons, magnons, supercurrents, and so on.
The first term on the other hand, is just the (canonical) kinetic energy of
the object as a whole.

Since $V\propto N$, the modes corresponding to the total momentum will be
quantised in any confining potential with spacing $1/N$, as in Figs. 2.3 and
2.5. This set of global eigenstates of canonical total momentum is referred to
as the tower of states, or Anderson tower of states, or occasionally as the
thin spectrum. In the thermodynamic limit, all states in the tower become
exactly degenerate.

As we will see in Section 3, systems with a spontaneously broken symmetry have
gapless, propagating excitations, called Nambu–Goldstone modes. These modes
exist at non-zero wave number and for a finite system of linear size $L$, the
lowest possible energy they can take is proportional to $1/L$. In spatial
dimension $D>1$, the states in the $\mathbf{k}=0$ tower of states have
energies proportional to $1/V\propto 1/L^{D}$ and these are therefore much
lower in energy than even the $\mathbf{k}>0$ states with the lowest possible
energies. Since all states in the tower are eigenstates of the total canonical
momentum at zero wave number, they all have a global structure, and they are
all symmetric. The classical symmetry-broken states are superpositions of the
states in the tower, with the special property that they can be written as
local product states of the form
$\otimes_{\mathbf{x}}\mathop{\lvert\psi(\mathbf{x})\rangle}$. They are not
energy eigenstates, but because the energy spacing between the states in the
tower is so small, the classical states are very narrow wavepackets in energy
space, which take on a single, well-defined energy expectation value in the
thermodynamic limit.

Because the existence of a tower of global excitations is so intrinsically
linked to spontaneous symmetry breaking, it is reasonable to ask whether these
states influence any other measurable properties of a symmetry-broken object.
To see this, consider the free energy of a quantum system with symmetric
Hamiltonian ${H}$, which at temperature $T$ can be calculated using:

$\displaystyle F$ $\displaystyle=-k_{\mathrm{B}}T\ln Z$ $\displaystyle Z$
$\displaystyle=\sum_{\mathop{\lvert\psi\rangle}}\mathop{\langle\psi\rvert}e^{-\frac{{H}}{k_{\mathrm{B}}T}}\mathop{\lvert\psi\rangle}.$
(2.47)

The sum in the partition function $Z$ runs over all energy eigenstates
$\mathop{\lvert\psi\rangle}$ of the Hamiltonian. The free energy associated
with the collective part of the Hamiltonian for an object consisting of $N$
interacting particles, is generically proportional to $\ln N$. Again crudely,
this can be seen by defining the collective part of the Hamiltonian to be the
kinetic energy associated with some total canonical momentum, so that:

$\displaystyle Z_{\text{coll}}$
$\displaystyle=\sum_{\mathbf{p}_{\mathrm{tot}}}\mathop{\langle\mathbf{p}_{\mathrm{tot}}\rvert}\mathrm{e}^{-\frac{{H}_{\text{coll}}}{k_{\mathrm{B}}T}}\mathop{\lvert\mathbf{p}_{\mathrm{tot}}\rangle}\sim\int\mathrm{d}\mathbf{p}_{\mathrm{tot}}\,\mathrm{e}^{-\frac{\mathbf{p}_{\mathrm{tot}}^{2}}{k_{\mathrm{B}}TV}}$
$\displaystyle\propto\sqrt{k_{\mathrm{B}}TN}.$ (2.48)

Since $Z_{\text{coll}}$ is proportional to $\sqrt{N}$, the corresponding
contribution of the tower of collective states to the free energy
$F_{\text{coll}}$ is proportional to $\ln N$. The free energy $F$ associated
with the full Hamiltonian must always be proportional to $N$, because it is an
extensive quantity. The relative contribution of the collective states to the
total free energy, $F_{\text{coll}}/F$, is then proportional to $\ln N/N$, and
disappears in the limit of large system size. In other words, even though they
are the only states with energies as low as $1/N$, there are so few states in
the tower that they do not contribute to the free energy at any non-zero
temperature, no matter how low. This part of the spectrum in fact is so ‘thin’
that it cannot be observed in any thermodynamic properties of the material,
such as specific heat or conductivity, which are all determined by the free
energy.

This observation is again fully general for collective states governing the
spontaneous breakdown of any continuous symmetry. They always form an
exceedingly thin part of the spectrum that is practically undetectable for any
realistically sized system in our everyday world. Paradoxically, one of the
few ways in which the presence of this part of the spectrum does have an
influence on measurable quantities, is due precisely to its undetectable
nature. If a material with a broken continuous symmetry is used to store
quantum information, for example using the presence or absence of a magnon in
an antiferromagnet as the zero and one states of a hypothetical qubit, then
the presence of many states beyond any experimental control acts as a sort of
environment to the qubit. Even if one could entirely isolate such a system
from any external influences, the qubit will decohere, because the information
about the magnon state becomes entangled with the unmeasurable, thin part of
the spectrum [34].

###  2.7 Stability of states

In spontaneous symmetry breaking, the exact ground state of a system is
infinitely sensitive to perturbations, which therefore always yield a broken-
symmetry state in the thermodynamic limit. However, such symmetric states are
also generically unstable all by themselves. To illustrate this, consider a
magnetic system with some local magnetisation $\sigma^{z}({\bf x})$ defined at
each position. ††margin:  instability – against local measurement If a
measurement of the magnetisation at ${\bf x}$ can influence a subsequent
measurement at a far-away positions ${\bf y}$, the system is unstable against
local measurements. In other words, stability requires the expectation value
$\langle\sigma^{z}({\bf y})\rangle$ at position ${\bf y}$ to be independent of
the measurement of magnetisation $\sigma^{z}({\bf x})$ at a position ${\bf x}$
far away from ${\bf y}$.

††margin:  Ising model

A simple example of an unstable state is the ground state of the transverse
field Ising model,

$H=-J\sum_{\langle
ij\rangle}\sigma^{z}_{i}\sigma^{z}_{j}-\mu\sum_{i}\sigma^{x}_{i}.$ (2.49)

This Hamiltonian is defined on any lattice of spin-$\tfrac{1}{2}$ states, with
$\langle ij\rangle$ denoting nearest neighbours, and $\sigma^{a}_{i}$ Pauli
matrices on site $i$. The coupling $J$ is positive and the transverse field is
represented by $\mu$. This model has a discrete, global $\mathbb{Z}_{2}$
symmetry of simultaneously flipping all spins in the $z$-direction, so that
$\sigma^{z}_{i}\rightarrow-\sigma^{z}_{i}$. If the transverse field is small,
$0<\mu\ll J$, the ground state of this model is approximately a superposition
of all spins up and all spins down,

$|\psi_{0}\rangle\approx|\uparrow\uparrow\uparrow\cdots\rangle-|\downarrow\downarrow\downarrow\cdots\rangle.$
(2.50)

Adopting the continuum limit, in which $\sigma^{z}_{i}$ becomes the function
$\sigma^{z}({\bf x})$ of the continuous variable ${\bf x}$, notice that the
expectation value of $\sigma^{z}({\bf x})$ at any position equals zero,
$\langle\sigma^{z}({\bf x})\rangle=0$, as expected for a system with spin-flip
symmetry. Measuring the $z$-component of the spin at position ${\bf x}$ will
collapse the superposed ground state onto the component corresponding to the
observed value of $\sigma^{z}({\bf x})$. For example, if we happen to measure
an up spin at ${\bf x}$, the entire state after the measurement has collapsed
to $|\uparrow\uparrow\uparrow\cdots\rangle$, and subsequently measuring the
$z$-component of spin at any position ${\bf y}$ will always yield up. The
expectation value of $\sigma^{z}({\bf y})$ has thus qualitatively changed
because $\sigma^{z}({\bf x})$ was measured, and the state of Eq. (2.50) is
concluded to be unstable against local measurements.

Conversely, the broken-symmetry state $|\uparrow\uparrow\uparrow\cdots\rangle$
itself is stable against local measurements, since no measurement of
$\sigma^{z}({\bf x})$ for any ${\bf x}$ will influence the result of
subsequent measurements at any other positions. In fact, this pattern is
general, and the stability of the symmetry-breaking state is a direct
consequence of its long-range order, discussed in Section 2.5.4. Local
measurements will generically rapidly collapse an unstable symmetric state
onto one of the possible broken-symmetry states. The definition of stability
with respect to local measurements is especially relevant when considering the
embedding of any given system in its local environment. Even the weakest
interactions with an environment can easily amount to an effective measurement
of local observables like the magnetisation $\sigma^{z}({\bf x})$, and thus
prevent symmetric states from being observed in any realistic setting.

The central ingredient in the definition of stability against local
measurements, is the requirement for an unstable state, a single measurement
influences the outcome of many subsequent measurements. To quantify the
meaning of ‘many’, the concept of cluster decomposition ††margin:  cluster
decomposition can be used. A state is said to satisfy the cluster
decomposition property if and only if for all local observables $a({\bf x})$
and $b({\bf y})$ we have

$C_{ab}({\bf x},{\bf y})=\langle a({\bf x})b({\bf y})\rangle-\langle a({\bf
x})\rangle\langle b({\bf y})\rangle\rightarrow 0\;\;\mathrm{when}\;\;|{\bf
x}-{\bf y}|\rightarrow\infty.$ (2.51)

This means that measurements of any $a({\bf x})$ and $b({\bf y})$, provided
${\bf x}$ and ${\bf y}$ are far apart, will be independent. Cluster
decomposition can therefore be considered a requirement for macroscopic
stability.

It is easy to check that the ground state of the transverse field Ising model
in Eq. (2.50) does not satisfy the cluster decomposition property. States like
these are sometimes called cat states, ††margin:  cat states in reference to
Schrödinger’s cat. The exact ground states of Hamiltonians susceptible to
spontaneous symmetry breaking are almost always cat states. Conversely, the
broken-symmetry states that may be stabilised in the thermodynamic limit
always do satisfy the cluster decomposition property.

The concept of cluster decomposition is itself closely related to a
thermodynamic restriction on fluctuations of extensive observables. In
thermodynamics, the extensive observables of two subsystems can be added to
find the corresponding extensive observable associated with the system as a
whole. In other words, extensive observables can be written as a sum of local
observables, $A=\sum_{\bf x}a({\bf x})$. The expectation value of $A$ must
therefore scale with the volume of the system, $\langle
A\rangle\sim\mathcal{O}(V)$. In general, the variance ††margin:  variance of
an observable scales as $\mathrm{Var}(A)=\langle A^{2}\rangle-\langle
A\rangle^{2}\propto V^{\alpha}$. If the exponent $\alpha$ is two or greater,
$\sqrt{\mathrm{Var}(A)}/\langle A\rangle$ does not vanish in the thermodynamic
limit, and the fluctuations in $A$ are as large as its expectation value. The
state is then said to be ††margin:  instability – thermodynamic
thermodynamically unstable. On the other hand, for states with $\alpha=1$, the
fluctuations vanish in comparison to the expectation value, and these states
are thermodynamically stable. Product states are always of this type. Some
special states may have $1<\alpha<2$. These are thermodynamically stable, but
they may be fragile in other senses. For instance, the critical systems that
we will discuss in Section 5.5 fall in this class.

Unsurprisingly, a system that violates the cluster decomposition property is a
superposition of macroscopically distinct states, and thus possesses
macroscopic fluctuations of an extensive variable. This can be easily seen by
writing the variance in terms of the two-point correlation function
$C(\mathbf{x},\mathbf{y})$,

$\displaystyle\mathrm{Var}(A)$ $\displaystyle=$ $\displaystyle\langle
A^{2}\rangle-\langle A\rangle^{2}$ (2.52) $\displaystyle=$
$\displaystyle\sum_{{\bf x},{\bf y}}\langle a({\bf x})a({\bf
y})\rangle-\langle a({\bf x})\rangle\langle a({\bf y})\rangle$ (2.53)
$\displaystyle=$ $\displaystyle\sum_{{\bf x},{\bf y}}C_{aa}({\bf x},{\bf y}).$
(2.54)

In a product state, such as the ordered, symmetry-breaking states, the two-
point function becomes equal to the product of local expectation values for
${\bf x}$ and ${\bf y}$ far apart. The variance is then dominated by
contributions with ${\bf x}\sim{\bf y}$, and therefore scales as
$\mathrm{Var}(A)\sim\mathcal{O}(V)$. On the other hand, if $C({\bf x},{\bf
y})$ does not equal the uncorrelated product of expectation values for large
${\bf x}-{\bf y}$, the variance has contributions from all terms in the double
sum and scales as $\mathrm{Var}(A)\sim\mathcal{O}(V^{2})$. That is,
fluctuations in a measurement of $A$ are as large as the observed average $A$
itself, indicating a highly unstable situation.

We can thus define stability in three equivalent ways: using the stability
against local measurements, examining the cluster decomposition property, and
considering the variance of extensive variables. The symmetric ground state of
models that exhibit SSB are generically unstable, while classical broken-
symmetry states are always stable under any of these three definitions of
stability. The applicability of the rule that symmetric ground states are
inherently unstable is further reaching than many other results presented in
these lecture notes. For example, Noether’s theorem only applies to continuous
symmetries, and the tower of states is only relevant to systems in which the
order parameter does not commute with the Hamiltonian (see Exercise 2.8).
However, the instability of symmetric states is a property of all systems that
exhibit spontaneous symmetry breaking.

######  Exercise 2.8 (Heisenberg Ferromagnet).

††margin:  ferromagnet

We end this chapter by highlighting a special case within the realm of
symmetry breaking. In daily parlance, the ferromagnet is often used as the
simplest example of symmetry breaking. Unfortunately, as you will see, the
properties of a ferromagnet make it exceptional, and unlike most other forms
of symmetry broken states, such as antiferromagnets, superfluids, or even the
symmetry breaking in the Standard Model of particle physics.

Consider the Heisenberg Hamiltonian Eq. (2.17), but now with negative coupling
$J<0$, so that pairs of spins prefer to be aligned, rather than anti-aligned.
This can easily be accommodated by having all spins aligned, say in the
$z$-direction.

a. Show that $S^{x}$ and $S^{y}$ are spontaneously broken according to Eq.
(2.30), by finding appropriate interpolating fields.

The order parameter operator for the state with all spins aligned in the
$z$-direction is $S^{z}$, which is itself one of the symmetry generators of
the symmetric Heisenberg Hamiltonian. The order parameter operator thus
commutes with the Hamiltonian. The ferromagnet is truly exceptional, however,
due to the following property:

b. Show that the state with all spins aligned (in the $z$-direction) is an
eigenstate of the Hamiltonian.
Hint: use the second line of Eq. (2.18).

This result implies that the fully magnetised, classical state in the sense of
Section 2.5.3 is an exact ground state of the symmetric quantum mechanical
Hamiltonian, for any system size. In fact, any fully magnetised state, with
all spins simultaneously pointing in any direction, is a ground state. The
ground state is thus far from unique, even for finite-sized ferromagnets.
There is no tower of states and there are no quantum corrections. The system
merely chooses a state that is stable against local perturbations from the
degenerate set of ground states.

We will examine the ferromagnet in more detail in Section 3.4. For now, the
moral of this exercise is that you should mistrust any text that uses the
ferromagnet as an archetype of spontaneous symmetry breaking. It truly is an
exceptional case.

##  3 Nambu–Goldstone modes

Every symmetry of the Hamiltonian or Lagrangian corresponds to a conserved
quantity, regardless of whether or not the state of the system respects the
symmetry. In homogeneous space for example, both a classical ball with
spontaneously broken translational symmetry, and an electron in a symmetric,
plane-wave state, will have a conserved total momentum. The intimate relation
between the conserved global quantity and the possibility of spontaneously
breaking a symmetry, was elucidated in Section 2.6, where we discussed the
tower of states. This collective, $k=0$, part of the spectrum consists of
eigenstates of the conserved global quantity, which in the thermodynamic limit
can be combined into a coherent-state superposition. Both the individual
eigenstates and the symmetry-breaking superposition conserve the (expectation
value of the) global quantity.

As we saw in Section 1.2.2, however, Noether’s theorem has implications far
beyond the global aspects of the system. For every continuous global symmetry,
it guarantees the existence of a locally conserved current, obeying a local
continuity equation. Again, this form of Noether’s theorem holds regardless of
whether or not the state of the system respects the symmetry. Moreover, the
local conservation law is intimately tied to a generic property of the
spectrum of systems with a spontaneously broken continuous symmetry. Rather
than affecting the collective states, however, the local continuity equation
impacts the excitations at non-zero wave number, and guarantees the appearance
of gapless modes known as Nambu–Goldstone (NG) modes. ††margin:
Nambu–Goldstone mode In particle physics and relativistic quantum field
theory, these modes are referred to as (Nambu–)Goldstone bosons, and they are
said to be massless instead of gapless. The difference is purely a matter of
nomenclature.

To understand the nature of the NG modes, consider the temporal component of
the Noether current operator $j^{t}(x)$ related to the symmetry generator
$Q=\int\mathrm{d}^{D}x\,j^{t}$ that is spontaneously broken. The NG mode
$\lvert\pi(\mathbf{k})\rangle$ can then be viewed as a plane-wave
superposition of local excitations created by acting with the Noether current
operator on the broken-symmetry state $\mathop{\lvert\psi\rangle}$:

$\mathop{\lvert\pi(\mathbf{k},t)\rangle}\propto\int\mathrm{d}^{D}x\;\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot\mathbf{x}}j^{t}(\mathbf{x},t)\mathop{\lvert\psi\rangle}.$
(3.1)

Goldstone’s theorem, which we will introduce below, shows these states to be
gapless. That is, their energy goes to zero as $\mathbf{k}\to 0$. Because low-
energy excitations of a symmetry-broken state necessarily correspond to
creating local Noether charge density, Noether’s continuity equation
guarantees that they will be dispersed over time. In other words, low-energy
disturbances in the order parameter will be carried away like waves in a
puddle carry away the local excitation of a raindrop, and systems with a
spontaneously broken symmetry are thus endowed with a form of rigidity [11].
††margin:  rigidity

###  3.1 Goldstone’s theorem

Before delving into the proof for Goldstone’s theorem and discussing some of
its implications and more modern aspects, let us simply state the theorem and
define its realm of applicability:

######  Theorem 3.1.

(Goldstone’s theorem). If a global, continuous symmetry is spontaneously
broken in the absence of long-ranged interactions, and leaving some (discrete)
translational symmetry intact, then there exists a mode in the spectrum whose
energy vanishes as its wave number approaches zero.

The theorem includes many assumptions, and in cases where these do not hold,
the NG mode either ceases to exist or to be gapless. If a symmetry is
explicitly broken by an external field $\mu$, for example, the NG mode will
exist, but with a gap of size $\mu$ at $k\to 0$. If the broken symmetry is
discrete, rather than continuous, there is no NG mode at all. And if the
symmetry appears in conjunction with a gauge freedom encoding a long-ranged
interaction, the NG mode may couple to the gauge field and develop a gap (this
is called the Anderson–Higgs mechanism and will be addressed in Section 7.3).
The original theorem also required Lorentz invariance, but non-relativistic
versions have been derived later, which we shall address in Section 3.2.

The requirement that some translational invariance remains in the broken-
symmetry state is the same as the one we needed to prove Noether’s theorem in
Section 1.2.3. We again need translational invariance only on a coarse-grained
level, so that momentum is a good quantum number, and modes will have a
definite value of momentum. We can then define a complete set of eigenstates
of the Hamiltonian, $\mathop{\lvert n,\mathbf{k}\rangle}$, labelled by their
momentum $\mathbf{k}$ and energy $E_{n}(\mathbf{k})$, with $n$ encoding all
relevant quantum numbers other than momentum. These states are orthogonal,
$\braket{n^{\prime},\mathbf{k}^{\prime}}{n,\mathbf{k}}=(2\pi)^{D}\delta_{nn^{\prime}}\delta(\mathbf{k}-\mathbf{k}^{\prime})$,
and can be used to write a resolution of the identity:

$\mathbb{I}=\sum_{n}\int\frac{\mathrm{d}^{D}k}{(2\pi)^{D}}\;\mathop{\lvert
n,\mathbf{k}\rangle}\mathop{\langle n,\mathbf{k}\rvert}.$ (3.2)

We can insert this into the definition of a broken-symmetry state of Eq.
(2.30), in terms of the interpolating field:

$\displaystyle\mathop{\langle\psi\rvert}[Q,\Phi]\mathop{\lvert\psi\rangle}$
$\displaystyle=\sum_{n}\int\frac{\mathrm{d}^{D}k}{(2\pi)^{D}}\Big{(}\mathop{\langle\psi\rvert}Q(t)\mathop{\lvert
n,\mathbf{k}\rangle}\mathop{\langle
n,\mathbf{k}\rvert}\Phi\mathop{\lvert\psi\rangle}-\text{ c.c.}\Big{)}$
$\displaystyle=\int_{\Omega}\mathrm{d}^{D}x\;\sum_{n}\int\frac{\mathrm{d}^{D}k}{(2\pi)^{D}}\Big{(}\mathop{\langle\psi\rvert}j^{t}(x,t)\mathop{\lvert
n,\mathbf{k}\rangle}\mathop{\langle
n,\mathbf{k}\rvert}\Phi\mathop{\lvert\psi\rangle}-\text{ c.c.}\Big{)}\neq 0.$
(3.3)

Here, c.c. indicates the complex conjugate, and in the second line the global
conserved charge $Q=\int\mathrm{d}x\,j^{t}$ is written as an integral over the
Noether charge density. Goldstone’s theorem addresses the modes as
$\mathbf{k}$ approaches zero, but it is not concerned with the tower of states
at precisely $\mathbf{k}=0$. It is thus related to the behaviour of the
Noether charge density integrated over a large, but finite part of
space999Formally, we should consider both the limit of the volume $V$ of our
system tending to infinity, and that of the integration volume $\Omega$
tending to $V$. Taking $V\to\infty$ before taking $\Omega\to V$ then
guarantees that the point $k=0$ is excluded from any momentum integrals
appearing in this section. This singular limit is discussed in detail in Ref.
[35]. The primary assumption in the derivation of Goldstone’s theorem then, is
that we can take the integration volume $\Omega$ in the expression above to be
large but finite. Because the interpolating field ${\Phi}$ is local, any
contributions to the expectation value from outside the volume $\Omega$ are
guaranteed to vanish in relativistic theories by causality. In non-
relativistic, or effective, theories, it vanishes as long as the theory does
not contain any long-ranged interactions. That is, all interactions should
decay sufficiently quickly with distance.

With this caveat in mind, we can again use translational invariance, and
write:

$\displaystyle\langle[Q,\Phi]\rangle$
$\displaystyle=\int_{\Omega}\mathrm{d}^{D}x\;\sum_{n}\int\frac{\mathrm{d}^{D}k}{(2\pi)^{D}}\left(\mathop{\langle\psi\rvert}\mathrm{e}^{-\frac{\mathrm{i}}{\hbar}(Ht-\mathbf{P}\cdot\mathbf{x})}j^{t}(0,0)\mathrm{e}^{\frac{\mathrm{i}}{\hbar}(Ht-\mathbf{P}\cdot\mathbf{x})}\mathop{\lvert
n,\mathbf{k}\rangle}\mathop{\langle
n,\mathbf{k}\rvert}\Phi\mathop{\lvert\psi\rangle}-\text{c.c.}\right)$
$\displaystyle=\int_{\Omega}\mathrm{d}^{D}x\;\sum_{n}\int\frac{\mathrm{d}^{D}k}{(2\pi)^{D}}\;\left(\mathrm{e}^{\frac{\mathrm{i}}{\hbar}(E_{n}t-\mathbf{k}\cdot\mathbf{x})}\mathop{\langle\psi\rvert}j^{t}(0,0)\mathop{\lvert
n,\mathbf{k}\rangle}\mathop{\langle
n,\mathbf{k}\rvert}\Phi\mathop{\lvert\psi\rangle}-\text{c.c}\right)$
$\displaystyle=\sum_{n}\int\mathrm{d}^{D}k\;\delta_{\Omega}(\mathbf{k})\left(\vphantom{\mathrm{e}^{\frac{\mathrm{i}}{\hbar}E_{n}t-\frac{\mathrm{i}}{\hbar}\mathbf{k}\cdot\mathbf{x}}}\mathrm{e}^{\frac{\mathrm{i}}{\hbar}E_{n}t}\mathop{\langle\psi\rvert}j^{t}(0,0)\mathop{\lvert
n,\mathbf{k}\rangle}\mathop{\langle
n,\mathbf{k}\rvert}\Phi\mathop{\lvert\psi\rangle}-\text{c.c.}\right)\neq 0.$
(3.4)

In the first line, the local Noether charge $j^{t}(\mathbf{x},t)$ was
translated in time and space using the shift operators
$\mathrm{e}^{-\mathrm{i}{H}t/\hbar}$ and
$\mathrm{e}^{-\mathrm{i}\mathbf{P}\cdot\mathbf{x}/\hbar}$. In the second line
we set $E_{n}$ to be the energy of the state $\mathop{\lvert
n,\mathbf{k}\rangle}$ relative to that of the state
$\mathop{\lvert\psi\rangle}$, and we invoked translational invariance to see
that $\mathop{\lvert\psi\rangle}$ is a zero-momentum state. In the final line
we defined
$(2\pi)^{D}\delta_{\Omega}(\mathbf{k})\equiv\int_{\Omega}\mathrm{d}^{D}x\exp(\mathrm{i}\mathbf{k}\cdot\mathbf{x})$
to be a strongly peaked function tending towards a Dirac delta function in the
limit of large integration volume. Because $\mathop{\lvert\psi\rangle}$ is
assumed to be a broken-symmetry state, the order parameter cannot be zero.
This implies there should be at least one state $\lvert n,\mathbf{k}\rangle$
such that the integrand in the final line also does not vanish, even for large
$\Omega$, when only contributions with momentum $\mathbf{k}$ tending to zero
can contribute. This is the first part of the theorem: there must exist some
state near zero momentum that is excited from the broken-symmetry state by
both the local Noether charge $j^{t}(0,0)$ and the interpolating field $\Phi$.

Noether’s theorem guarantees the global charge $Q$ to be time-independent. If
$\Phi$ also does not depend on time, then in the thermodynamic limit where the
broken state $\mathop{\lvert\psi\rangle}$ is an energy eigenstate, the entire
order parameter $\langle[Q,\Phi]\rangle$ is time-independent. For the right-
hand side of Eq. (3.4) we then find:

$\displaystyle\partial_{t}\langle[Q,\Phi]\rangle$
$\displaystyle=\partial_{t}\sum_{n}\int\frac{\mathrm{d}^{D}k}{(2\pi)^{D}}\;\delta_{\Omega}(\mathbf{k})\left(\mathrm{e}^{\mathrm{i}E_{n}t}\mathop{\langle\psi\rvert}j^{t}(0,0)\mathop{\lvert
n,\mathbf{k}\rangle}\mathop{\langle
n,\mathbf{k}\rvert}\Phi\mathop{\lvert\psi\rangle}-\text{c.c.}\right)$
$\displaystyle=\sum_{n}\int\frac{\mathrm{d}^{D}k}{(2\pi)^{D}}\;\delta_{\Omega}(\mathbf{k})\mathrm{i}E_{n}\left(\mathrm{e}^{\mathrm{i}E_{n}t}\mathop{\langle\psi\rvert}j^{t}(0,0)\mathop{\lvert
n,\mathbf{k}\rangle}\mathop{\langle
n,\mathbf{k}\rvert}\Phi\mathop{\lvert\psi\rangle}-\text{c.c.}\right)=0.$ (3.5)

We already found that there must be at least one state, the NG mode, for which
the term between brackets does not vanish. The final line then implies that
the NG mode must have vanishing energy, $E_{n}(\mathbf{k})\to 0$ as
$\mathbf{k}\to 0$. This completes the proof of Goldstone’s theorem: a system
with a spontaneously broken symmetry has at least one excitation whose energy
vanishes as its wave number approaches zero.

Notice that Goldstone’s theorem is constructive, in the sense that it not only
tells us that gapless modes exist whenever a symmetry is spontaneously broken,
but also indicates how to find these modes. They can be excited from the
symmetry-broken state by acting on it with either the local Noether charge
operator, or the interpolating field.

###  3.2 Counting of NG modes

The derivation of Goldstone’s theorem may at first sight seem to suggest that
there is always one NG mode for each broken symmetry generator. This cannot be
the case, however, since the Heisenberg ferromagnet is known to have only a
single NG mode, while two spin-rotation symmetries are broken. Similarly, one
may be tempted the assume that the energy of NG modes always vanishes linearly
in momentum, $E_{n}\propto k$. For relativistic systems, this certainly is the
case, since Lorentz symmetry dictates that time and space derivatives appear
on equal footing in the action. However, in non-relativistic systems the
Heisenberg ferromagnet again provides a counterexample to the general rule.
Its single NG mode is quadratic in momentum, rather than linear.

Goldstone’s theorem as we derived it above, actually just states there is at
least one NG mode whenever any symmetry is broken, and it does not specify its
dispersion relation other than that it is gapless, so there is no real
contradiction with the observed properties of ferromagnets. How many NG modes
should really be expected in any given system, and what replaces the seemingly
intuitive rule of one mode per broken symmetry, was cleared up only recently.
It cannot yet be found in any of the standard text books, but is readily
accessible through either the original literature in Refs [36, 37, 38, 39, 40,
41, 42, 43], or in the short review of Ref. [44].

In the derivation of Goldstone’s theorem, we found that NG modes can be
excited from the broken-symmetry state by either the generator of a broken
symmetry, or the interpolating field. A special case then arises if the
interpolating field $\Phi$ is itself also a generator of a broken symmetry. A
clear example is again the Heisenberg ferromagnet, in which one of the spin-
rotation operators, say $S^{z}$, obtains a non-zero expectation value. The
commutator of the broken generators $S^{x}$ and $S^{y}$ is proportional to
$S^{z}$, and can thus be used as an order parameter. The broken generators in
this case act as interpolating fields for each other, and Eq. (3.4) shows that
they must excite the same NG mode [38].

More generally, take any two symmetry generators $Q_{a,b}=\int_{\bf
x}j^{t}_{a,b}({\bf x})$, and consider the commutator expectation value

$\displaystyle\langle[Q_{a},j^{t}_{b}({\bf x})]\rangle$
$\displaystyle=\int\mathrm{d}^{D}y\;\langle[j^{t}_{a}({\bf y}),j^{t}_{b}({\bf
x})]\rangle=\int\mathrm{d}^{D}y\;\delta({\bf x}-{\bf
y})\sum_{c}\mathrm{i}f_{abc}\langle j^{t}_{c}({\bf y})\rangle$
$\displaystyle=\sum_{c}\mathrm{i}f_{abc}\langle j^{t}_{c}({\bf
x})\rangle=\langle[j^{t}_{a}({\bf x}),Q_{b}]\rangle.$ (3.6)

If this commutator has non-zero expectation value in the broken-symmetry
state, they are again seen to excite the same NG mode. After Watanabe and
Murayama we call such NG modes type-B, ††margin:  Nambu–Goldstone mode –
type-B while ‘ordinary’ NG modes are said to be type-A [41].

From Eq. (3.2), it is clear that type-B modes cannot arise for Abelian
symmetry groups, in which all generators commute with one another. To
systematically count the number of NG modes of either type, we should
construct the Watanabe–Brauner matrix [40] ††margin:  Watanabe–Brauner matrix

$M_{ab}=-\mathrm{i}\langle\psi\rvert[Q_{a},j^{t}_{b}({\bf
x})]\lvert\psi\rangle.$ (3.7)

Here $a$ and $b$ label all broken symmetry generators, and $\lvert\psi\rangle$
is the SSB state. The total number of broken generators is equal to the
dimension of the quotient space $G/H$. Notice that since the broken-symmetry
state is assumed to be translationally invariant, the matrix elements do not
depend on the position $\mathbf{x}$. The numbers $n_{\mathrm{A}}$ and
$n_{\mathrm{B}}$ of type-A and type-B Nambu–Goldstone modes are now given by:

$\displaystyle n_{\mathrm{A}}$
$\displaystyle=\mathrm{dim}\,G/H-\mathrm{rank}\,M,$ $\displaystyle
n_{\mathrm{B}}$ $\displaystyle=\frac{1}{2}\mathrm{rank}\,M.$ (3.8)

The two independent proofs [41, 42] also show that (in almost all cases)
type-A NG modes have linear dispersion while type-B modes have quadratic
dispersion. This can be understood using the low-energy effective Lagrangian
method [45, 39, 41, 43]. ††margin:  Lagrangian – effective A detailed
derivation is beyond the scope of these notes, but in short, one writes down
the most general Lagrangian allowed by the symmetry of the problem, in terms
of fields $\pi_{a}(x)$ which take values in the space of broken symmetry
generators, the quotient space $G/H$. The number of these fields then equals
the number of broken symmetry generators, but the fields are not necessarily
independent. The gapless modes in the spectrum of this low-energy effective
Lagrangian will correspond to the NG modes. The lowest order terms are

$\mathcal{L}_{\mathrm{eff}}=m_{ab}(\pi_{a}\partial_{t}\pi_{b}-\pi_{b}\partial_{t}\pi_{a})+\bar{g}_{ab}\partial_{t}\pi_{a}\partial_{t}\pi_{b}-g_{ab}\nabla\pi_{a}\cdot\nabla\pi_{b}.$
(3.9)

Here $m_{ab}$, $\bar{g}_{ab}$ and $g_{ab}$ are coefficients, some of which are
constrained by symmetry. For example, there are no terms linear in gradients,
since we assume space to be isotropic. For the same reason, it is clear that
the first term in $\mathcal{L}_{\mathrm{eff}}$ breaks Lorentz invariance, and
can only be non-zero in non-relativistic systems. Watanabe and Murayama have
shown that the coefficients $m_{ab}$ are given precisely by the corresponding
elements of $M_{ab}$ in Eq. (3.7) above [41].

######  Exercise 3.2 (Number of type-B NG modes).

Part of the proof of Eq. (3.8) in Ref. [41] is the following. The matrix
$M_{ab}$ is real and antisymmetric. Then there exists an orthogonal
transformation $O$ such that $\tilde{M}=OMO^{\mathrm{T}}$ takes the form

$\displaystyle\tilde{M}$ $\displaystyle=\begin{pmatrix}M_{1}&&&&&\\\
&\ddots&&&&\\\ &&M_{m}&&&\\\ &&&0&&\\\ &&&&\ddots&\\\ &&&&&0\\\
\end{pmatrix},$ $\displaystyle M_{i}$
$\displaystyle=\begin{pmatrix}0&\lambda_{i}\\\ -\lambda_{i}&0\end{pmatrix},$
(3.10)

where $m=\tfrac{1}{2}\mathrm{rank}\;M$ and the
$\lambda_{1},\ldots,\lambda_{m}$ are real and non-zero. To prove this:

a. Show that the eigenvalues of $M$ are purely imaginary. This means that the
matrix $E$ with the imaginary eigenvalues on the diagonal can be obtained by
some unitary transformation $E=UMU^{\dagger}$.

b. Show that the non-zero eigenvalues come in conjugate pairs
$\mathrm{i}\lambda_{i},-\mathrm{i}\lambda_{i}$. Since there are
$\mathrm{rank}\;M$ non-zero eigenvalues, this implies $\mathrm{rank}\;M$ is
even, so $n_{\mathrm{B}}$ in Eq. (3.8) is integer.

c. For each $2\times 2$-submatrix $e_{i}$ with a conjugate pair on the
diagonal elements, find a unitary matrix $w_{i}$ such that
$w_{i}e_{i}w_{i}^{\dagger}=M_{i}$.

You have found that $M$ is unitarily equivalent to $\tilde{M}$ by the unitary
transformation $WU$ where $W$ has the submatrices $w_{i}$ on the top-left
diagonal and other entries 0. Since $M$ and $\tilde{M}$ are both real they are
then also orthogonally equivalent, and the orthogonal matrix $O$ can be
constructed from $WU$. See Problem 160 in Ref. [46].

From the effective Lagrangian, we can find the equations of motion for the
fields and their dispersion relations. In systems where $m_{ab}$ is zero,
including all relativistic systems, the effective Lagrangian describes modes
with linear dispersions. More precisely, Fourier transforming the Lagrangian
will introduce a frequency $\omega$ for every time derivative and a momentum
$k$ for every gradient, so the dispersion will obey $\omega^{2}\propto k^{2}$.
If the coefficients $m_{ab}$ are not zero, their contribution to the
Lagrangian will always dominate the second order derivatives at sufficiently
low energies, and the dispersion will be quadratic, $\omega\propto k^{2}$.
Notice that terms like $\pi_{a}\partial_{t}\pi_{a}$ (no summation over $a$)
are total derivatives in the Lagrangian, and vanish in the action. Therefore,
$m_{ab}$ must be antisymmetric, and the first term in Eq. (3.9) can only be
non-zero in systems where two fields are coupled. The reduction in the number
of gapless NG modes because two generators excite the same mode, and the fact
that type-B modes have quadratic dispersion, are thus seen to go hand-in-hand.

Finally, notice that in the effective Lagrangian, it is possible for the
coefficients $g_{ab}$ to be zero. In that case, higher-order terms must be
taken into account, and it is therefore possible that type-A modes have
quadratic dispersion, or rather $\omega^{2}\propto k^{4}$. This is the case,
for example, for so-called Tkachenko modes in vortex lattices in rotating
superfluids [43].

###  3.3 Examples of NG modes

To see how the formal considerations of Sections 3.1 and 3.2 impact the
observable properties of real materials, we will present a short selection of
practical examples. This list is far from exhaustive, but should give you a
feeling for the extent to which the Goldstone’s theorem shapes the physics of
all systems subject to spontaneous symmetry breaking.

##### Superfluid

The superfluid was argued in Section 2.5.2 to be described by a complex scalar
field theory, in which the field operator itself acts as the order parameter.
The action is invariant under rotations of the phase of the field, and
Noether’s theorem shows this symmetry to be associated with the conservation
of particle number. In the superfluid phase, the $U(1)$ phase-symmetry is
spontaneously broken and the number of particles in the superfluid condensate
is indeterminate. There is one broken symmetry generator and one NG mode,
which may be excited by finite-wave-number rotations of the phase variable in
the complex scalar field. The NG mode is type-A and its dispersion is linear
in momentum. The supercurrent (a particle current that flows without
viscosity) is a direct manifestation of this NG mode.

##### Crystal

Crystals in $D$ spatial dimensions break the symmetries of space, $D$
translations and $\frac{1}{2}D(D-1)$ rotations. The translation group is
Abelian, so the associated NG modes are all type-A, with linear dispersions.
They are called phonons, or sound waves, and there is one for each direction
of space. The rigidity due to breaking of translational symmetry is shear
rigidity, whose non-zero value is the traditionally used quantity for
distinguishing solids from liquids.

The broken rotational symmetries in the crystal do not lead to any additional
NG modes. As was shown only recently [47, 48, 49], rotations and translations
are not independent symmetry operations. ††margin:  Nambu–Goldstone mode –
redundant Consequently, the NG fields excited by broken translations and
broken rotations are also not independent, and contain redundant degrees of
freedom. The broken rotations do therefore not lead to independent NG modes.
Intuitively, this means that if you try to excite a rotational NG mode by
applying torque stress to a crystal, you instead end up exciting transverse
sound modes. In fact, Lorentz boosts are also spontaneously broken in the
crystal (or any other medium), but like rotations, they do not lead to
independent NG modes [50].

##### Antiferromagnet

The Heisenberg antiferromagnet of Section 2.4 breaks two out of three spin-
rotational symmetries, say $S^{x}$ and $S^{y}$. The commutator in the off-
diagonal elements of the Watanabe-Brauner matrix is the magnetisation $S^{z}$,
which vanishes. The NG modes excited by the broken symmetry generator are thus
independent, so there are two type-A NG modes with linear dispersions, called
spin waves. These can be viewed as plane waves of precessions for the spins on
each sublattice.

##### Ferromagnet

The Heisenberg ferromagnet of Exercise 2.8 breaks the same spin-rotation
symmetries as the antiferromagnet. This time, however, the magnetisation
$S^{z}$ is an order parameter whose expectation value does not vanish in the
broken-symmetry state. The Watanabe-Brauner matrix is therefore non-zero and
the modes excited by the two broken symmetry generators are not independent.
There is then one type-B NG mode with a quadratic dispersion.

##### Canted antiferromagnet

Adding a term that favours orthogonal alignment of neighbouring spins to the
antiferromagnetic Heisenberg Hamiltonian will lead to a symmetry-broken state
in which all spins uniformly cant away from the preferred direction in the
Néel state. The result is a state with a total uniform magnetisation as well
as a staggered or sublattice magnetisation in a perpendicular direction. This
state breaks all three spin-rotation symmetries. In this case, the broken
generator of rotations around the axis of uniform magnetisation will excite
one type-A NG mode, while the remaining two broken generators excite one
type-B NG mode.

######  Exercise 3.3 (Chiral symmetry breaking).

Consider a complex scalar doublet
$\Phi=\begin{pmatrix}\phi_{1}&\phi_{2}\end{pmatrix}^{\mathrm{T}}$ where
$\phi_{1}$ and $\phi_{2}$ are complex scalar fields, with Lagrangian:

$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\Phi^{\dagger})(\partial^{\mu}\Phi)-\frac{1}{2}r\Phi^{\dagger}\Phi-\frac{1}{4}u(\Phi^{\dagger}\Phi)^{2}.$
(3.11)

a. Show that this Lagrangian is invariant under $\Phi\to L\Phi$ where $L\in
SU(2)$ is a unitary $2\times 2$ matrix with determinant 1.

b. Show that this Lagrangian is furthermore invariant under

$\displaystyle\phi_{1}$ $\displaystyle\to r_{2}\phi_{1}+r_{1}\phi_{2}^{*},$
$\displaystyle\phi_{2}$ $\displaystyle\to-r_{1}\phi_{1}^{*}+r_{2}\phi^{2},$
(3.12)

with $r_{1}^{*}r_{1}+r_{2}^{*}r_{2}=1$.

If we write $\Phi$ as a $U(2)$ matrix $\breve{\Phi}$ and collect $r_{1}$,
$r_{2}$ in an $SU(2)$-matrix $R$:

$\displaystyle\breve{\Phi}$
$\displaystyle=\begin{pmatrix}\phi_{2}^{*}&\phi_{1}\\\
-\phi_{1}^{*}&\phi_{2}\end{pmatrix},$ $\displaystyle R$
$\displaystyle=\begin{pmatrix}r_{2}&-r_{1}\\\
r_{1}^{*}&r_{2}^{*}\end{pmatrix},$ (3.13)

then
$\Phi^{\dagger}\Phi=\frac{1}{2}\mathrm{Tr}\;\breve{\Phi}^{\dagger}\breve{\Phi}$,
and the Lagrangian is invariant under:

$\breve{\Phi}\to L\breve{\Phi}R^{\dagger}.$ (3.14) ††margin:  chiral symmetry

This is called chiral symmetry. The Lagrangian is also invariant under global
U(1) phase rotations $\Phi\to\mathrm{e}^{\mathrm{i}\alpha}\Phi$, and the full
symmetry group is $SU(2)_{\mathrm{L}}\times SU(2)_{\mathrm{R}}\times U(1)$,
but we will disregard the $U(1)$ symmetry here.

The groups $SU(2)_{\mathrm{L}}$ and $SU(2)_{\mathrm{R}}$ are generated by
$Q_{a}^{\mathrm{L}}$ and $Q_{a}^{\mathrm{R}}$, which satisfy the
$SU(2)$-relations
$[Q_{a}^{E},Q_{b}^{F}]=\mathrm{i}\epsilon_{abc}\delta_{EF}Q_{c}^{E}$,
$E,F=\mathrm{L},\mathrm{R}$ (see Exercise 1.8). Define the vector and axial
charges $Q_{a}^{\mathrm{V}}=Q_{a}^{\mathrm{L}}+Q_{a}^{\mathrm{R}}$ and
$Q_{a}^{\mathrm{A}}=Q_{a}^{\mathrm{L}}-Q_{a}^{\mathrm{R}}$.

c. Show that these satisfy the algebra relations:

$\displaystyle[Q^{\mathrm{V}}_{a},Q_{b}^{\mathrm{V}}]$
$\displaystyle=\mathrm{i}\epsilon_{abc}Q_{c}^{\mathrm{V}},$ (3.15)
$\displaystyle[Q^{\mathrm{A}}_{a},Q_{b}^{\mathrm{A}}]$
$\displaystyle=\mathrm{i}\epsilon_{abc}Q_{c}^{\mathrm{V}},$ (3.16)
$\displaystyle[Q^{\mathrm{V}}_{a},Q_{b}^{\mathrm{A}}]$
$\displaystyle=\mathrm{i}\epsilon_{abc}Q_{c}^{\mathrm{A}}.$ (3.17)

This implies the $Q_{a}^{\mathrm{V}}$ generate a subgroup but the
$Q_{a}^{\mathrm{A}}$ do not.

If $r<0$, for $u>0$, the potential has a minimum at
$\langle\Phi\rangle=\begin{pmatrix}0&v\end{pmatrix}^{\mathrm{T}}$ with
$v\in\mathbb{R}$, or $\langle\breve{\Phi}\rangle=\mathrm{diag}(v,v)$ (see
Section 5.2).

d. Show that the vector transformations ($L=R$ in Eq. (3.14)) leave the
expectation value $\langle\breve{\Phi}\rangle$ invariant, while the axial
transformations ($L=R^{\dagger}$ in Eq. (3.14)) do not.

The symmetry is spontaneously broken by $\langle\breve{\Phi}\rangle$ from
$SU(2)_{\mathrm{L}}\times SU(2)_{\mathrm{R}}$ to the diagonal subgroup
$SU(2)_{\mathrm{L}+\mathrm{R}}$ generated by the vector charges
$Q_{a}^{\mathrm{V}}$. Since there are three broken generators
$Q_{a}^{\mathrm{A}}$ we expect three type-A NG modes. The Lagrangian expressed
in terms of $\breve{\Phi}$, with the symmetry of Eq. (3.14) describes the
Higgs field in the Standard Model of elementary particles, before coupling to
other fields. The reason that these NG bosons are not massless particles in
the Standard Model, will be explained in Section 7.3.

$SU(3)\times SU(3)\to SU(3)$ chiral symmetry breaking occurs in quantum
chromodynamics (QCD), in the limit where quark masses go to zero.

####  3.3.1 NG-like excitations

There are several systems which harbour excitations that are clearly related
to the physics of spontaneous symmetry breaking and NG modes, but that do not
satisfy all of the assumptions underlying Goldstone’s theorem. Again, we give
a short selection of examples to give you a feeling for how NG-like
excitations extend into systems that strictly speaking fall just outside the
realm of spontaneous symmetry breaking.

##### Gapped NG mode

If a system with a spontaneously broken symmetry is exposed to an external
field that explicitly breaks the same symmetry, there is an NG-like excitation
with an energy gap proportional to the external field. This has been dubbed a
gapped or massive NG mode [51, 52]. The typical example is that of spin waves
in a ferromagnet exposed to a magnetic field parallel to the magnetisation.
This situation may be interpreted as a model without explicit symmetry
breaking, at the cost of having modified, time-dependent symmetry generators
[53].

##### Pseudo NG mode

If a symmetry is broken explicitly, due to a weak coupling to other fields (or
other degrees of freedom) rather than by an external field, there is a bosonic
particle with an energy gap, which otherwise has all the characteristics of a
NG mode. This is now called a pseudo NG mode or pseudo NG boson. The most
famous example are the lightest eight pseudoscalar mesons in the Standard
Model, pseudoscalar particles whose mass is small because the approximate
$SU(3)\times SU(3)$ chiral symmetry is broken spontaneously (see Exercise
3.3).

##### Quasi NG mode

In some cases, the ground state of a system may have a larger symmetry group
than the Hamiltonian itself, and that symmetry may be broken spontaneously. In
that case, an NG-like excitation emerges, which is now called a quasi NG mode
[54], although confusingly these same modes used to be called a pseudo NG
boson [55]. This occurs in particle physics, where the charged pions obtain
part of their mass in this way, as well as in certain technicolour and
supersymmetry models. In condensed matter physics quasi NG modes are found in
helium-3 superfluids and spinor Bose-Einstein condensates.

##### Goldstino

The ††margin:  Goldstino broken generators $Q_{a}$ are creation operators for
NG modes when acting on the broken-symmetry state. Since they generally obey
some set of commutation relations, the NG modes are bosons. Usually they are
scalar particles, although sometimes it makes sense to assign a vector or
tensor structure to several NG modes. However, if the symmetry generators
happen to satisfy anti-commutation relations, the NG modes are fermions. This
is the case for ††margin:  supersymmetry supersymmetry, which is a possible
extension of the Poincaré algebra of spacetime symmetries with fermionic
supersymmetry generators $Q_{a}$. If supersymmetry is spontaneously broken,
the associated fermionic NG modes are called Goldstinos.

###  3.4 Gapped partner modes

Type-B Nambu–Goldstone modes may be excited from the broken-symmetry states by
two distinct broken symmetry generators, whose commutator has a non-zero
expectation value. In the effective Lagrangian description of Eq. (3.9), this
was signalled by two fields $\pi_{1}$ and $\pi_{2}$ not being independent.
Even though the modes are not independent, they do originate from two distinct
symmetry transformations, and one might therefore wonder whether there should
not be two degrees of freedom or modes associated with the two fields in the
Lagrangian. In fact, there generally are two modes, but the second mode is
gapped [56, 57, 58, 59]. ††margin:  gapped partner mode To see this, consider
the effective Lagrangian for a system with two coupled broken symmetry
generators, and nothing else [60]:

$\mathcal{L}_{\mathrm{eff}}=2M(\pi_{1}\partial_{t}\pi_{2}-\pi_{2}\partial_{t}\pi_{1})+\frac{1}{c^{2}}(\partial_{t}\pi_{1})^{2}+\frac{1}{c^{2}}(\partial_{t}\pi_{2})^{2}-\frac{1}{2}(\nabla\pi_{1})^{2}-\frac{1}{2}(\nabla\pi_{2})^{2}.$
(3.18)

The dispersion relations for the two modes described by this Lagrangian are:

$\omega_{\pm}=\sqrt{c^{2}k^{2}+M^{2}c^{4}}\pm Mc^{2}.$ (3.19)

That is, there is one gapless NG mode with dispersion
$\omega_{-}=\frac{k^{2}}{2M}+\ldots$, and one gapped partner mode with
dispersion $\omega_{+}=2Mc^{2}+\frac{k^{2}}{2M}+\ldots$. The coefficients $2M$
and $c$ can be interpreted as an effective mass and velocity.

In the limit $M\to 0$, the two modes decouple and we obtain a degenerate,
linear dispersion $\omega_{\pm}=ck$. This corresponds to having two type-A NG
modes, which is indeed consistent with $M$ being the expectation value of the
off-diagonal element in the Watanabe-Brauner matrix and going to zero in this
limit. In the opposite limit of $c\to\infty$, the gap goes to infinity and we
are effectively left with only a single, gapless mode. This corresponds to
having only terms with single time derivatives in Eq. (3.18), which is the
case for example for the Heisenberg ferromagnet, and indeed there is no gapped
mode in the spectrum of the ferromagnet. Physically, the fact that the second
mode disappears altogether for the ferromagnet can be understood by realising
that its NG modes are always excited by a lowering of the maximally polarised
spins. That is, in both $S^{x}=(S^{+}+S^{-})/2$ and
$S^{y}=(S^{+}-S^{-})/2\mathrm{i}$ only the $S^{-}$ part actually excites a
mode. Acting with $S^{+}$ on a maximally polarised ferromagnet annihilates the
state, and does not correspond to a physical excitation, recall Exercise 2.8.
In this limit, the action of the two broken symmetry generators on the
symmetry-breaking state are thus entirely equivalent, and there really is only
a single excitation associated with them. More generally, however, non-zero
values for both $M$ and $c$ always indicate the presence of both a type-B NG
mode and an accompanying gapped partner mode.

Notice that at first sight, the existence of a gapped mode excited by a broken
symmetry generator seems to invalidate the proof of Goldstone’s theorem in
Section 3.1, in which we argued that all modes excited by broken generator
must be gapless to ensure that the order parameter would be time-independent.
Although there is no general proof of which we are aware, it has been checked
in several cases that this paradox is resolved by observing that the terms in
Eq. (3.5) always contain a product of two matrix elements, of the form:

$\displaystyle E_{n}\mathop{\langle\psi\rvert}j^{t}(0,0)\mathop{\lvert
n,\mathbf{k}\rangle}\mathop{\langle
n,\mathbf{k}\rvert}\Phi\mathop{\lvert\psi\rangle}.$ (3.20)

In all verified cases where $j^{t}(0,0)$ excites a gapped partner with non-
zero energy $E_{n}$, it turns out that
$\mathop{\langle\psi\rvert}\Phi\mathop{\lvert n,\mathbf{k}\rangle}$ is
proportional to the energy of the accompanying gapless mode, which vanishes
for $\mathbf{k}\to 0$. The existence of the gapped partner mode therefore does
not contradict the time-independence of the order parameter101010We thank
Tomas Brauner for discussions concerning this point..

######  Exercise 3.4 (Heisenberg Ferrimagnet).

A ferrimagnet ††margin:  ferrimagnet can be described by the Heisenberg
Hamiltonian $H=J\sum_{\langle i,j\rangle}\mathbf{S}_{i}\cdot\mathbf{S}_{j}$ on
a square lattice with positive coupling $J$, but for a system in which the
spins on the A- and B-sublattices have different sizes.

a. Let the spins on the $A$-sites have spin-$S_{A}$ and those on the $B$-sites
have spin-$S_{B}$. Calculate the average magnetisation per unit cell $\langle
S^{z}_{\mathrm{tot}}\rangle/(N/2)$ for the Néel-like state in which all spins
are in eigenstates of their $S^{z}_{j}$ operators with eigenvalues
$m_{A}=S_{A}$ and $m_{B}=-S_{B}$.
b. This state breaks the spin-rotation symmetries generated by $S^{x}$ and
$S^{y}$. Calculate the matrix elements of the Watanabe-Brauner matrix defined
in Eq. (3.7) (take the average per unit cell).
c. The local magnetisation $S^{z}_{i}$ is an order parameter operator for this
state, but it turns out that there is also a second order parameter: calculate
the expectation value per unit cell for the staggered magnetisation operator
$N^{z}_{i}=(-1)^{i}S^{z}_{i}$.

We thus find that the breaking of spin-rotational symmetry in the ferrimagnet
may be described by (at least) two distinct order parameters, one of which
commutes with the Hamiltonian. Clearly then, it is not sufficient to find an
order parameter operator that does not commute with the Hamiltonian to claim
that any symmetry-breaking system is of type-A. One should calculate all the
elements of the Watanabe-Brauner matrix, as you did in this exercise.

With a bit more effort, you can see that the ferrimagnet has one quadratically
dispersing NG mode and one gapped partner mode, whose gap scales with the
difference in spin sizes, $\Delta\propto\lvert S_{A}-S_{B}\rvert$. If the
spins on the two sublattices have the same size, the system is an ordinary
antiferromagnet with two gapless, linearly dispersing modes (as described by
Eq. (3.19) in the limit $M\to 0$). In the opposite limit, as $\Delta$ grows,
exciting the gapped partner mode costs ever more energy, and the spectrum
looks more and more like that of a ferromagnet with only a single gapless,
quadratically dispersing, NG mode.

####  3.4.1 The tower of states for systems with type-B NG modes

NG modes may be seen as the $k>0$ cousins of the collective excitations with
zero wave number that make up the tower of states in systems with a
spontaneously broken symmetry. The close relation between the internal and
collective modes persists in the distinction between systems with type-A and
those with type-B modes. We have seen that the Heisenberg antiferromagnet and
the harmonic crystal, for example, have unique ground states and a tower of
low-energy states, all of which are unstable. This is general for systems with
type-A NG modes. We have also seen in Exercise 2.8 that the ferromagnet has a
macroscopically degenerate ground state and no tower of states. The number of
exact ground states is infinite in the thermodynamic limit, but of order $N$
for finite systems, with $N$ the number of particles.

As it turns out, in systems with a type-B NG mode and a gapped partner mode
(which have both $M$ and $c$ non-zero in Eq. (3.18)) there is a ground state
degeneracy, and still no tower of states. While we are not aware of a general
proof, the relation to the collective modes can be seen in the explicit
example of an antiferromagnet on the Lieb lattice111111We thank Hal Tasaki for
pointing out this example, which is discussed also in his textbook [61]. The
Hamiltonian is the same as that of the usual Heisenberg antiferromagnet, Eq.
(2.17), but on the Lieb lattice there are twice as many A sites as there are B
sites, as shown in Fig. 3.1. The classical ground state is a Néel-type state
with spins pointing antiparallel on the two sublattices. It has non-zero
magnetisation $\langle S^{z}\rangle=S(N_{A}-N_{B})=\frac{1}{3}SN$, with
$N_{A,B}$ denoting the number of sites on the $A$ and $B$ sublattices, and $N$
the combined total number of sites. Because the magnetisation is finite, the
NG modes of the Lieb-lattice antiferromagnet will be type-B. Just like the
usual square-lattice antiferromagnet discussed in Section 2.4, any exact
ground states of the full model have finite overlap with those of a
corresponding Lieb–Mattis model (the $k=0$-part of the Hamiltonian, with
infinte-range interactions). Using that fact, the ground states can be shown
to have total spin $S_{\mathrm{tot}}=S(N_{A}-N_{B})$, and degeneracy
$2S_{\mathrm{tot}}+1$, which is of order $N$. The energy required to excite
any of the remaining $k=0$ states is of order $\mathcal{O}(J)$, in contrast to
the usual antiferromagnet with type-A NG modes, where $k=0$ excitations have
energy vanishing as $\mathcal{O}(J/N)$. In other words, there is no tower of
states. The relation between types of NG modes and the spectrum of collective
excitations is summarised in Table 3.1.

Figure 3.1:  The Lieb lattice. Each A site has only B neighbours and vice versa, but there are twice as many A sites as there are B sites. | NG mode | ground state | tower of
---|---|---|---
| dispersion | degeneracy | states
type-A | $\propto k$ | no | yes
type-B ferro | $\propto k^{2}$ |  yes, $\mathcal{O}(N)$ | no
type-B ferri |  $\propto k^{2}$, $M+k^{2}$ |  yes, $\mathcal{O}(N)$ | no
Table 3.1:  Properties associated with different types of Nambu–Goldstone
modes. The accurate forms of the dispersion relations can be found in Eqs.
3.19.

##  4 Quantum corrections and thermal fluctuations

The eigenstates of the local order parameter operator, which are the
classically expected symmetry-broken states, are not in general eigenstates of
their corresponding symmetric Hamiltonian. Except for systems with conserved
order parameters and type-B NG modes, such as ferromagnets and ferrimagnets,
symmetric Hamiltonians have unique, symmetric ground states. That the
‘classical states’ nevertheless typically bear a large resemblance to the
states actually realised in nature, is owing to their stability, in the sense
of Section 2.7.

Despite their particular stability, however, the classical states are not
exactly the states found in actual quantum materials, due to the somewhat
subtle effect of excitations with finite wave number reducing the perfect
local order preferred by the $k=0$, collective part of the Hamiltonian in the
thermodynamic limit. The differences between the classically expected and
actually encountered state at zero temperature are known as ††margin:  quantum
corrections quantum corrections. At non-zero temperature, these may be
supplemented by thermal fluctuations, which further suppress the local order
parameter.

Notice the distinction made between thermal fluctuations and quantum
corrections. The latter are often referred to as “quantum fluctuations” in the
literature. The reasoning being that the effect of quantum corrections in
reducing the order parameter amplitude is quite similar to that of the thermal
fluctuations. Indeed, they may even lead to a quantum phase transition at zero
temperature which is in some ways analogous to the usual, thermally induced,
phase transition (see also the discussion in Section A.4). We believe this
terminology is misleading. Thermal fluctuations describe actual small, random
fluctuations within a thermal state due to, for example, Brownian motion. At
zero temperature, on the other hand, nothing ever fluctuates. The ground state
may be unique all the way from a maximally ordered phase to just before a
quantum phase transition, despite the expectation value of the local order
parameter strongly decreasing. For any given value of system parameters,
nothing about the ground state evolves or fluctuates in time. We therefore
prefer the term quantum corrections to denote the difference between the
actual quantum ground state and the perfectly ordered, classical state.

Rather than describing the generic properties of quantum corrections to a
general ordered state, we will study the specific example the Heisenberg
antiferromagnet in Section 4.1. This allows us both to introduce some of the
commonly encountered mathematical techniques in studying the actual ground
states of ordered systems, and to show in detail a realistic description of a
system in which quantum corrections are of significant magnitude. Indeed, in
most ordered states that we encounter in daily life, quantum corrections to
the classical state are “utterly negligible” [11]. For example, a chair or
table, even at low temperatures, is extremely well-described by its classical
state with infinitely well-defined position.

In stark contrast to these everyday objects, stand systems in low dimensions,
for which quantum corrections and thermal fluctuations are generically so
large that they altogether prevent the formation of any ordered state and the
spontaneous breaking of continuous symmetries. A heuristic derivation of the
Mermin–Wagner–Hohenberg–Coleman theorem which explains both the effect of
dimensionality, and the link between quantum corrections and thermal
fluctuations, is given in Section 4.2.

###  4.1 Linear spin-wave theory

There is no known exact expression for the broken-symmetry state realised by
the Heisenberg antiferromagnet in the thermodynamic limit, in dimensions
higher than one. One thing we can do to describe it, however, is to start from
the classical state (the eigenstate of the Néel order parameter operator), and
use the variational principle to look for deviations that lower the energy
expectation value. This is done in a systematic wave in spin-wave theory,
which is based on a reformulation of the Heisenberg Hamiltonian in terms of
boson operators acting on the classical state, so that each bosonic excitation
lowers the order parameter expectation value. ††margin:  linear spin-wave
theory In linear spin-wave theory, the bosonic Hamiltonian is additionally
linearised, allowing for a ground state to be identified by direct
diagonalisation. We can thus find the exact ground state of an approximate
Hamiltonian, and trust that it can serve as an approximate ground state to the
exact Hamiltonian. As the name suggests, the linear spin-wave theory considers
only the first non-trivial order in a systematic expansion of the order
parameter. For Heisenberg spin-$S$ antiferromagnets on bipartite lattices with
$z$ nearest neighbours, it turns out that the results obtained this way are
good beyond expectation, even for $S=\frac{1}{2}$ and $z=4$, where neither
$1/S$ nor $1/z$ are expected to be very good expansion parameters.

The Hamiltonian for the spin-$S$ Heisenberg antiferromagnet on a bipartite
lattice in $d$ dimensions, is given by Eq. 2.17:

$\displaystyle H$ $\displaystyle=J\sum_{\langle
i,j\rangle}\left(S^{x}_{i}S^{x}_{j}+S^{y}_{i}S^{y}_{j}+S^{z}_{i}S^{z}_{j}\right)=\frac{J}{2}\sum_{j\delta}\left(S^{x}_{j}S^{x}_{j+\delta}+S^{y}_{j}S^{y}_{j+\delta}+S^{z}_{j}S^{z}_{j+\delta}\right).$
(4.1)

Here, the coupling constant $J$ is positive, so that neighbouring spins prefer
to anti-align. The lattice vectors $\delta$ run over the $z$ connections of
any site to all of its nearest neighbours, and the factor $\frac{1}{2}$ is
included to avoid double counting. Because the lattice is bipartite, it can be
divided into $A$\- and $B$-sublattices. For the square lattice, we saw this
before in Figure 2.4. Anticipating antiferromagnetic Néel order, we expect the
spins on the $A$-sublattice to have a positive magnetisation in the broken-
symmetry state, and spins on the $B$-sublattice to have negative
magnetisation. To avoid the inconvenience of having to keep track of the local
magnetisation direction on each sublattice, we introduce rotated spin
operators $\mathbf{N}_{j}$, defined as:

$\displaystyle N^{a}_{j\in A}$ $\displaystyle=S^{a}_{j};$ $\displaystyle
N^{x}_{j\in B}$ $\displaystyle=S^{x}_{j},$ $\displaystyle N^{y}_{j\in B}$
$\displaystyle=-S^{y}_{j},$ $\displaystyle N^{z}_{j\in B}$
$\displaystyle=-S^{z}_{j}.$ (4.2)

That is, the coordinate system for spins on the $B$ sublattice is rotated over
an angle of $\pi$ around the $x$-axis with respect to the coordinate system
used for spins on the $A$-sublattice. The new spin operators $\mathbf{N}_{j}$
are still proper spin-$S$ operators, obeying the commutation relations
associated with the $SU(2)$ algebra,
$[N^{a}_{i},N^{b}_{j}]=\delta_{ij}\mathrm{i}\epsilon_{abc}N^{c}_{j}$.121212Note
that in Section 2.5.2 and Exercise 3.4, we introduced the order parameter
operators $N^{a}_{i}=(-1)^{i}S^{a}_{i}$. While that former definition is