Redundancies in Nambu–Goldstone Bosons
Haruki Watanabe1, ∗ and Hitoshi Murayama1, 2, 3, †2

1
Department of Physics, University of California, Berkeley, California 94720, USA
Theoretical Physics Group, Lawrence Berkeley National Laboratory, Berkeley, California 94720, USA
3
Kavli Institute for the Physics and Mathematics of the Universe (WPI),
Todai Institutes for Advanced Study, University of Tokyo, Kashiwa 277-8583, Japan

We propose a simple criterion to identify when Nambu–Goldstone bosons (NGBs) for different symmetries
are redundant. It solves an old mystery why crystals have phonons for spontaneously broken translations but
no gapless excitations for equally spontaneously broken rotations. Similarly for a superfluid, the NGB for
spontaneously broken Galilean symmetry is redundant with phonons. The most nontrivial example is Tkachenko
mode for a vortex lattice in a superfluid, where phonons are redundant to the Tkachenko mode which is identified
as the Boboliubov mode.
PACS numbers: 11.30.Qc, 14.80.Va, 03.75.Kk

Introduction. — In many areas of physics, it is important
to study consequences of microscopic physics on macroscopic
behaviors, sometimes called emergent phenomena. One of
the best examples in this category is the existence of gaplessexcitations, called Nambu–Goldstone bosons (NGBs), when
global continuous symmetries are spontaneously broken [1].
For spontaneously broken internal symmetries in Lorentzinvariant systems, the symmetries dictate the number (nNGB ),dispersion relation, and interactions of NGBs completely. The
present authors have generalized this well-known results tosystems without Lorentz invariance, and proved a general formula [2]1nNGB = nBG − rankρ,2ρab = lim

1

V →∞ Vh0|[Qa , Qb ]|0i.

(1)Here, nBG ≡ dim(G/H) is the number of broken generators. Note that the symmetry breaking pattern itself is not
sufficient to fix the number of NGBs and the additional information on the ground state, ρ, is required. Here and hereafter,
whenever we refer to broken generators
Qa , we mean suitRable large-volume limits limV →∞ V dd x ja0 (x), where j 0 (x)is the Noether charge density.
In the case of spacetime symmetries, however, the counting of NGBs is more subtle. Even for relativistic systems,
some examples elude the above rule for internal symmetries,e.g., spontaneously broken conformal and scale invariance.
There is an empirical prescription called inverse Higgs mechanism that allows one to identify possible constraints that canbe imposed among NGBs [3], while it does not dictate if
they should be imposed. Little is known for theories without
Lorentz invariance.
In this Letter, we propose a simple criterion to determine
what redundancies exist among NGBs in a given system. Redundancies can arise for two separate reasons: (1) special
property of the ground state annihilated by a linear combination of symmetry generators, and (2) identities among Noether
charge densities. It is complementary to the inverse Higgs
mechanism because our criterion requires redundancies.
This result was inspired by the work by Low andManohar [4], which pointed out that a local transformationof different symmetries may lead to the same field configurations. But they did not clearly distinguish the classicalfield configurations and quantum states and operators, and restricted themselves to Lorentz-invariant systems. We need togeneralize their intuition and formulate it more concretely.
Noether constraints. — A symmetry is spontaneously
broken if its generator Qa has an order parameterh0|[Qa , Φ(y)]|0i 6= 0. By inserting a complete set of states,6
one finds the existence of a gapless state hπa (~pa )|ja0 (x)|0i =0 where limp~a →0 Eπa (~pa ) = 0.
We first point out that the above general theorem immediately tells us the NGBs are redundant if a linear combinationof Noether currents annihilate the ground state for non-zerocoefficients ca ,ZXdd x
ca (x)ja0 (x)|0i = 0.
(2)a

We call them Noether constraints. In general, the coefficientsca (x) are spacetime dependent. Since for each spontaneously
broken symmetry
there must be a gapless NGB state |πa i, letP
us multiply a |πa ihπa | on the above equation. Then we findthat the would-be NGB states satisfy
ZX
dd x
ca (x)|πa ihπa |ja0 (x)|0i = 0.
(3)aSince hπa |ja0 (x)|0i 6= 0 by definition, we find |πa i states arelinearly dependent. Therefore, the would-be NGB states haveredundancies by the number of Noether constraints Eq. (2).
The rest of the discussion is how such Noether constraints
arise in two general categories.
Internal Symmetries. — Let us first look at the wellknown example: Heisenberg ferromagnet. When all spins are
lined up along the positive z direction, the Noether charge
density of spin rotations around the x and y axes, sx,y (x),cannot raise the spins anymore at any point in space. Therefore,Z
dd x(sx + isy )(x)|0i = 0.
(4)

2
The states created by two broken charges, sx and sy are hence
not independent. Indeed, it is known that there is only one
magnon (quantized spin wave) state, consistent with Eq. (1).
On the other hand, Eq. (4) is not satisfied in the case of antiferromagnets, and hence Sx and Sy excite independent NGBs.
In general, we consider Noether constraints ca Qa |0i = 0.
If ca is a real vector, it simply means this linear combination
is an unbroken generator. On the other hand, if ca cannot bemade real, it is straightforward to prove that the constraint canbe cast to the form (Qk + iQl )|0i = 0 after suitable changeof basis of generators. Then2

0 = |(Qk + iQl )|0i| = h0|(Q2k + Q2l )|0i + h0|i[Qk , Ql ]|0i.
(5)
For broken generators Qk,l , the first term is positive definite (and is proportional to the spatial volume), and hence
the latter commutator must have a non-vanishing expectation value. We have developed an effective Lagrangian [2]that describes the number, dispersion relation, and interactions in the most general case of internal symmetry breaking. If ρab ∝ h0|[Qa , Qb ]|0i 6= 0, some of broken generatorsform canonically conjugate pairs we call Type-B, and hence
the number of NGBs is reduced as in Eq. (1).
Note, however, that not all cases of ρab 6= 0 can be brought
to the form of a Noether constraint. The constraint requires
the redundancy, while the redundancy occurs with a constraint
for each canonically conjugate pair in ρab .
Spacetime Symmetries. — Another reason for redundancies is when thePNoether charge densities are linearly dependent. Namely a ca (x)ja0 (x) = 0 as an operator identity,
and the redundancy is obviously independent of the propertyof the ground state.
To illustrate the point, let us consider a simple crystal. The
Lagrangian or Hamiltonian is both translationally and rotationally invariant, with six generators in three spatial dimensions. A crystal spontaneously breaks all six symmetries.
However, it is well-known that there are three gapless phonon
excitations (two transverse and one longitudinal), but no more.
We are not aware of satisfactory explanation for the lack of
NGBs for rotational symmetries in the literature.
The crucial observation is that the Noether charge densities
for translation T 0i and rotation R0i are related by
R0i = iji xj T 0k .

(6)

Therefore, what could have been NGBs for spontaneously
broken rotational symmetries are redundant with those for
spontaneously broken translational symmetries, hence only
three NGBs. Note that xi are parameters and not operatorsin quantum field theories.
A more nontrivial example is a superfluid. The matterfield changes its phase under the particle-number symmetry
U(1) as ψ(~x, t) → eiθ ψ(~x, t), while changes both its argument and the phase under the Galilean boost by velocity ~v ,2
1
ψ(~x, t) → ei(m~v·~x− 2 m~v ) ψ(~x − ~v t, t) (we set ~ = 1). The
order parameter h0|ψ(~x, t)|0i = ψ0 hence breaks one phasesymmetry and three boost symmetries. However, there is only

one gapless excitation, namely the Bogoliubov mode. Recall
that consideration of the spontaneously broken Galilean invariance is crucial to the Landau’s criterion for superfluidity.
The lack of independent NGBs for Galilean symmetry
again can be seen in the operator identity that the Noether current for the Galilean boost B iµ is related to the U(1) currentas
B iµ = tT iµ − mxi j µ .

(7)Here and hereafter, the Greek index µ refers to the spacetime index, 0 = t, 1 = x, 2 = y, 3 = z. It is straightforward to derive this identity from the Lagrangian density1∇ψ † ∇ψ − V (ψ † ψ). Since the translational
L = iψ † ψ̇ − 2m
invariance is not broken in the superfluid, T i0 does not createa gapless excitation, while those created by B i0 and j 0 are
linearly dependent, hence redundant.
Vortex lattice. — Perhaps the most nontrivial exampleof the redundancy among NGBs is the Tkachenko modein a vortex lattice in rotating BEC. Rotating superfluids
and atomic BEC form a triangular lattice of quantized vortices [5], spontaneously breaking the translational symmetry. It is known that the vortex lattice system supports a softcollective oscillation with a quadratic dispersion, so-called
Tkachenko mode [6–9]. Since the Tkachenko mode is often
associated with an elliptically-polarized lattice vibration, one
may naively expect the existence of the usual (Bogoliubov)
phonon, which corresponds to the fluctuation of the superfluid
phase. Until today, all prior works on the collective modesin the system have been based on the hydrodynamic theory.
Although they seem to imply the absence of such a gaplessmode, the reason for the missing has been left unclear.
To clarify the low-energy structure of the system, here weconstruct an effective Lagrangian. In order to discuss collective modes from the symmetry-breaking point of view, we donot take into account the inhomogeneity due to trapping potential or the centrifugal potential. In other words, we focus on
the region where the trapping potential almost cancels the centrifugal potential but still retains a finite particle density. Our
system thus can be rephrased as bosons which couple to aneffective uniform magnetic field Beff = 2mΩ/eeff as if they
have a charge eeff . The effective Lagrangian for vortices insuperfluids has been discussed in several papers [10], but they
did not discuss the vortex lattice configuration. They also introduced several fields in addition to NG degrees of freedom,which is not suitable for our purpose.
Let us start with the standard Lagrangian [11],
L=

1
i †
(ψ ψ̇ − ψ̇ † ψ) −
|∇ψ|2
2
2m
1
−Vtrap (~x)ψ † ψ − g(ψ † ψ)2 .
2(8)

We restrict ourselves to 1+2D and the zero temperature. To go~ = Ωẑ,to the corotating frame with the angular frequency Ω~
one makes the substitution ∂t → ∂t − Ω × ~x · ∇.
√ Assuming a
Bose–Einstein condensate, we substitute ψ = ne−iθtot into

3
By equating these two expressions for the topological current, we have d(dθsing ) = −mΩab dX a ∧ dX b , which gives

the Lagrangian and obtaini †1~ × ~x)ψ|2(ψ ψ̇ − ψ̇ † ψ) −|(∇ − imΩ2
2m
1
−Veff (~x)ψ † ψ − g(ψ † ψ)2
2
(∇n)2
1
= nµ −− Veff (~x)n − gn28mn
21[µ − Veff (~x)]2 ,
'
2g

L=

where Veff (~x) ≡ Vtrap (~x) −µ ≡ θ̇tot −dθsing = −mΩab X a dX b + dχ.

(9)

m 2 2
Ω x and2

1 2µ − Eel (∂~u),g~ · ~u × ~u˙µ = θ̇reg − mΩ

~x0 = ~x + ~a,0ψ (~x , t) = ψ(~x, t)e

(10)
~ a
im~
x·Ω×~

.

(14)

1
~ × ~u − mΩkl uk ∇ul )2 .(15)(∇θreg + 2mΩ
2m
The ground state of H − µ0 N (N is the total number of particles) is characterized as θreg = µ0 t and ~u = 0. Leff describes~
the dynamics of fluctuation ϕ ≡ µ0 t − θreg and ~u ≡ ~x − X.
As a nontrivial test, let us derive hydrodynamic equations
as the Euler-Lagrange equations of the effective Lagrangian.
Variation w.r.t θreg gives the continuity equation ∂µ j µ =∂t n + ∇ · (n~v ) = 0, where n ≡ µg and−In the third line (9), we integrated n out, keeping only theleading term in the derivative expansion [12].
If we neglect the effective potential Veff (~x), as we do so forthe rest of the paper, the Lagrangian possesses the magnetictranslational symmetry,0

A smooth function χ corresponds to the ambiguity mentioned
above. We choose χ = mΩjk xj X k so that the explicit coordinate dependence drops from the Lagrangian. Assuming thetriangular lattice and adding the corresponding elastic energy
Eel (∂~u) ≡ (2C1 + C2 )(∇ · ~u)2 + C2 (∇ × ~u)2 (in the notationof Ref. [8]), we arrive at our effective Lagrangian,
Leff =

1~ × ~x)2 .
(∇θtot + mΩ
2m

(13)

(11)

Because of the lack of Galilean invariance, the energy momentum tensor no longer satisfies T 0i = mj i . Instead,
T 0i = mj i − 2mΩij xj j 0 .
(12)Rvortex lattice system, both P i ≡ dd x T 0i and N ≡RIn thedd x j 0 are spontaneously broken. However, according toour general criterion, the operator identity Eq. (12) suggeststhat T 0i and j 0 do not produce independent NGBs. We will
explicitly verify this claim in the following.
In the presence of vortices, the phase θtot contains singularities. We decompose θtot into the regular part θreg and thevortex part θsing ; i.e., θtot = θreg + θsing . Since θsing is only
defined up to a smooth function, this decomposition is notunique; we will fix the ambiguity later. Due to the singularity,θsing does no longer satisfy d2 θsing = 0. In fact, ∗d(dθsing ) (∗is the Hodge dual of 1 + 2D Minkowski space) can be idenµtified as the vortex current jvortex (jvortex= µνλ ∂ν ∂λ θsing )that automatically satisfies the topological conservation lawµ
d ∗ jvortex = ∂µ jvortex= 0.
Now let us introduce a continuum description of the vortex dynamics. Because the crystalline order breaks the magnetic translation, we introduce fields X a that specify the
position of the vortices. Here, we follow the notation in
Ref. [13]: X a is the Lagrangian coordinate frozen on thelattice, while xi is the Eulerian coordinate. We fix the relation between X a and xi in such a way that ~u(~x, t) ≡~ x, t) represents the displacement from the equilibrium~x − X(~
position ~x. The vortex current in the continuum descriptionµcan be expressed as jvortex = ∗ 12 m0 ab dX a ∧ dX b (jvortex
=
m0
1
mΩ
µνλ
a
b
m


∂
X
∂
X
)
[13]
where
=
−
[11]
is
the
0
ab
ν
λ
2
2π
π
number density of the vortices in the equilibrium.

1(∇θreg + 2mΩ × ~u − mΩkl uk ∇ul ). (16)m
Since we implicitly assumed that vortices are massless andhence ~u does not have the kinetic term ∝ ~u˙ 2 , the Equation of Motion (EOM) of the displacement vector requires
the balance between the Magnus force and the elastic force
δEel
F~Magnus + F~el = 0, where F~el ≡and F~Magnus =δ~u~ × [~v − (∂t +~v · ∇)~u]. These equations agree with those2mnΩ
discussed in Refs. [7, 8] based on the linearized hydrodynamic
theory, which in turn verifies our effective Lagrangian. Notethat our expressions are fully non-linear, e.g., the third term in
Eq. (16), beyond the linearized expressions in their papers.
Let us analyze the low-energy collective mode in our effective Lagrangian. If we keep only quadratic terms in thefluctuation ϕ and ~u, the Lagranigan becomesn0  2
Leff 'ϕ̇ − c2s (∂i ϕ + 2mΩij uj )22
2mcs~ · ~u × ~u˙ − Eel (∂~u).
−n0 mΩ(17)~v ≡ −In order to compare our expressions to those in the literature, we have eliminated g and µ0 in terms of the equilibrium density n0 and the superfluid velocity cs by g = nµ00
and µ0 = mc2s . The remarkable feature of the effective Lagrangian is the mass term −2mn0 Ω2 ~u2 . Combined with thesecond term, which makes ux and uy canonically conjugate
to each other, it explains the gapped mode with a gap 2Ω inthe spectrum [7, 8].
Given the gap, one can safely integrate ~u out by using
EOM,
ui =

1 ij
 ∂j ϕ + O(∂0 ∂i , ∂i ∂j ∂k ).
2mΩ

(18)

4
At the leading order in the derivative expansion, the remaining
Lagrangian is
C2 c2s
n0
2
2
2
ϕ̇
−
(∇
ϕ)
,
(19)
Leff '
2mc2s
2mn0 Ω2
which describesqthe Tkachenko mode with the dispersion re-C2 cs 2
lation E(~p) = 2mn
p + O(p4 ) [7, 8]. The Tkachenko
0 Ω
mode thus can be understood as the phase oscillation, and thevortex lattice simply follows transverse to the motion of thesuperfluid through Eq. (18).
After all, there is only one gapless mode in the vortex lattice, as expected from our general criterion. In the derivation,
we introduced the redundant fields in our effective Lagrangianand observed a mass term ∝ ~u2 for them. (In principle, weshould be able to write down the effective Lagrangian based
purely on the symmetry without introducing redundant fields.)
An effective Lagrangian of crystal phonons does not usuallycontain ~u without any derivatives, because the invariance under ~u0 (x0 ) = ~u +~a prohibits it. This is why we usually expect
acoustic (gapless) phonons [14]. However, in the current example, the appearance of the mass term does not contradict
with the symmetry — the original magnetic translation is still
exactly realized in our effective Lagrangian Eq. (14) in a nontrivial manner,~x0 = ~x + ~a,(20)~u (~x , t) = ~u(~x, t) + ~a,(21)~ × [~u(~x, t) − 2~x].
θ (~x , t) = θ(~x, t) − m~a · Ω

(22)

0

0

0

0

One can verify that the associated Noether current T 0i satisfies Eq. (12). This is expected, since the low-energy effective
theory must have the same symmetry structure as the microscopic (high-energy) theory. This symmetry also protects thequadratic dispersion relation of the Tkachenko mode; i.e., thelower order term ∝ (∇ϕ)2 cannot be generated by renormalization process in Eq. (19).
It is instructive to compare the vortex lattice with a supersolid [13]. A supersolid exhibits a similar symmetry-breaking
pattern; namely, it breaks both (usual) translation and U(1)phase rotation. In contrast to the vortex lattice case, each ofd momentum operators P i and the number operator N independently produces a NGB, giving rise to d + 1 NGBs in total
in d-space dimensions. This is consistent with our criterion,
since in the case of supersolid, the Galilean invariance [13](more precisely, the non-relativistic general-coordinate invariance [15]) leads to T 0i = mj i . Therefore phonons originated from the translational symmetry breaking and Bogoliubov mode are not redundant.
We thank T. Brauner for fruitful discussions. H.W. is also
grateful to M. Nitta, M. Kobayashi and S. Furukawa. H.W.

would like to appreciate the support from Honjo international scholarship foundation. The work of HM was supported in part by the U.S. DOE under Contract DE-AC0376SF00098, by the NSF under grant PHY-1002399, the JSPS
grant (C) 23540289, and by the FIRST program Subaru Measurements of Images and Redshifts (SuMIRe), CSTP, and by
WPI, MEXT, Japan.

∗

hwatanabe@berkeley.edu
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[12] If interested in the Bogoliubov spectrum ω 2 = c2s k2 (1 +k2), one should not integrate n out and leave it as a con4m2 c2sjugate field to the phase. In this case, the quantity µ in the text
loses the meaning of the chemical potential. Instead, we define
the chemical potential as µ(n) ≡ ∂ε(n), using the internal en∂n
1 2ergy ε(n) = 2g
n . The EOM of n describes the time-evolution√
u×~
u˙ + m ~v 2 − 1√ ∇2 n.
of the phase, θ̇reg = µ(n)+ mΩ·~
†

2

2m

n

Other EOMs remain unchanged.
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