Quantum Matter / Matéria Quântica

Solutions to Problem set 7 — Localized magnetism, spin-wave excitations

Notation recap. We write the Heisenberg Hamiltonian using either of the equivalent notations
introduced in lecture,

1X 1X
Ĥ = − Jij Ŝi · Ŝj = − Jn ŜR · ŜR+n , (1)
2 2
i,j R,n

where i, j (or R and n) runs over all the spins (or lattice sites) and Jii = Jn=0 = 0 because there is no
“self-exchange” interaction. Ŝi represents the dimensionless spin operator at site i (position R) and
Jij has dimensions of energy. Unless stated otherwise, consider a general magnitude S of the spin at
each site (common to all spins), and that the spins sit at the points of a simple Bravais lattice. The
exchange coupling is typically truncated beyond nearest-neighbors (NN), which in practice means
that n (in the second notation above) comprises only the set of nearest neighbors of a given site.
When Jij > 0, the coupling is designated as “ferromagnetic” (FM), and “antiferromagnetic” (AFM)
if Jij < 0. In the AFM case, we frequently write Jij = −|Jij | to emphasize the overall sign.

The Holstein-Primakoff (HP) transformation relative to a spin S fully polarized along +ẑ is

Ŝi+ = (2S − n̂i )1/2 ai , Ŝi− = a†i (2S − n̂i )1/2 , Ŝiz = S − n̂i , (2)

where ai , a†i are boson operators: n̂i ≡ a†i ai ; [ai , a†j ] = δij ; [a†i , a†j ] = [ai , aj ] = 0.

Lattice Fourier transforms and related definitions are needes to make use of translation invariance.
In a Bravais lattice with N unit cells and coordination z (number of nearest neighbors):

1 X −ik·R X 1 X ik·n
aR ≡ √ e ak , Jk ≡ Jn eik·n , γk ≡ e . (3)
N k∈BZ n
z n

In a system with N spins, we designate |S, Mi i(i) the eigenstates of Si2 and Siz for particle i, and
use the notation

|M1 , M2 , . . . , MN i ≡ |S, M1 i(1) ⊗ · · · ⊗ |S, MN i(N ) , Mj ∈ {−S, −S + 1, . . . , +S},

to designate a basis of eigenstates common to Si2 and Siz of all the N spins.

The following angular momentum commutators are frequently needed,

     
Ŝiα , Ŝjβ = δij iεαβµ Ŝiµ , Ŝi+ , Ŝj− = 2δij Ŝiz , Ŝiz , Ŝj± = ±δij Ŝi± , (4)

as well as the basic characteristics of a an eigenstate |S, M i common to Ŝ 2 and Ŝ z :

Ŝ 2 |S, M i = S(S + 1) |S, M i, Ŝ z |S, M i = M |S, M i,

p
Ŝ ± |S, M i = S(S + 1) − M (M ± 1) |S, M i.

1
Problem 1 (FM Bloch waves)

In the presence of a constant external magnetic field B, the field orientation fixes the direction of
spin polarization in the broken-symmetry FM state of the Heisenberg model. Setting B = B ẑ, the
Hamiltonian gains a Zeeman term that captures the coupling to the magnetic field:

1X X
Ĥ = − Jij Ŝi · Ŝj − h Ŝiz , h ≡ gµB B > 0.
2
i,j i

Consider this description for a system of localized spins on a simple Bravais lattice L containing N
unit cells. Consider arbitrary exchange couplings Jn > 0, not truncated beyond nearest neighbors.
The ket representing the fully polarized ground-state is

|Φ0 i ≡ |M1 = S, M2 = S, . . . , MN = Si = |S, S, . . . , Si.

Consider the following state with an elementary spin deviation at site i ∈ L:

1 1
|χi i ≡ √ Ŝi− |Φ0 i ⇔ −
|χR i ≡ √ ŜR |Φ0 i.
2S 2S

Note — In this problem you should work only with spin operators (no HP transformation).

a) Verify that the classical ground-state energy is

N J0 S 2 X
E0 ≡ − − hN S, where J0 ≡ Jk=0 = Jn .
2 n

b) Verify that the kets |χi i are normalized.

c) By computing the action of Ĥ on the state |χi i, show that

 
S 2 N J0 X
Ĥ|χi i = SJ0 − + h(N S − 1) |χi i − S Jin |χn i,
2
n6=i

where and Jk was defined in eq. (3) above. This result shows that Ĥ propagates the spin
deviation at site i to any of its neighbors with Jij 6= 0.
+ − −
Suggestion — Use commutators to reduce products such as Ŝm Ŝn Ŝi , etc. to a single (lowering) operator.

d) The following linear superposition of local spin deviations is known as a Bloch spin wave:

1 X −ik·R 1 X
−
|χk i = √ e |χR i = √ e−ik·R ŜR |Φ0 i.
N R∈L 2SN R ∈ L

Show that |χk i is an exact eigenstate of the Heisenberg Hamiltonian with energy given by

Ek = E0 + εk , εk ≡ S(J0 − Jk ) + h,

where E0 is the FM ground-state energy and J0 ≡ Jk=0 .

2
 Comment — These Bloch spin-wave states are the elementary excitations of the FM Heisenberg model,
each having an excitation energy εk relative to the ground state.

e) Set the external field to zero: h = 0. Given the result above for the energy cost εk to excite
a spin-wave, show that the fully polarized state |Φ0 i is only guaranteed to be the zero-field
ground state if Jij > 0 for all neighbors. Otherwise, |Φ0 i might not be the ground state.

Solution

a) In the classical version, the spin operators Ŝi are conventional vectors in R3 with magnitude S. With
Jij > 0 and h > 0, the lowest energy state is obtained when all spins are parallel to the direction of the
external field: Si = S ẑ. That energy is

1X X S2 X S2 X
E0 = − Jij S 2 − h S=− Jij − hN S = − Jn − hN S
2 i,j i
2 i,j 2
R,n

N S2 X N S 2 J0
=− Jn − hN S = − − hN S.
2 n 2

b) Compute the norm, remembering that (Ŝi− )† = Ŝi+ :

  
1 1 1
hχi |χi i = √ hΦ0 |Ŝi+ −
√ Ŝi |Φ0 i = hΦ0 |Ŝi+ Ŝi− |Φ0 i,
2S 2S 2S

but
(i) p (i)
Ŝi− |Φ0 i = Ŝi− |S, · · · , S , · · · , Si = S(S + 1) − S(S − 1)|S, · · · , S − 1, · · · , Si
√ (i)
= 2S|S, · · · , S − 1, · · · , Si,

so
2S (i) (i)
hχi |χi i = hS, · · · , S − 1, · · · , S|S, · · · , S − 1, · · · , Si = 1.
2S

c) Writing the Hamiltonian in the form

 
1X X 1X 1 + − 1 − + X
Ĥ = − Jij Ŝi · Ŝj − h Ŝiz = − Jij z z
Ŝi Ŝj + Ŝi Ŝj + Ŝi Ŝj − h Ŝiz ,
2 i,j i
2 i,j 2 2 i

we have
 
1 X 1 + − 1 − + h X z −
Ĥ|χi i = − √ Jmn Ŝm Ŝn + Ŝm Ŝn + Ŝm Ŝn Ŝi− |Φ0 i − √
z z
Ŝm Ŝi |Φ0 i.
2 2S m,n 2 2 2S m

Let us compute each of the 4 contributions.

3
 z z
(i) Terms arising from Ŝm Ŝn :
 
z z −
Ŝm z
Ŝn Ŝi |Φ0 i = Ŝm Ŝnz Ŝi− |Φ0 i


y use the commutator Ŝiz , Ŝi± = ±2Ŝi±
 

 
z
= Ŝm −Ŝi− δin + Ŝi− Ŝnz |Φ0 i
z − z − z
= −δin Ŝm Ŝi |Φ0 i + Ŝm Ŝi Ŝn |Φ0 i
   
= −δin −δim Ŝi + Ŝi Ŝm |Φ0 i + −δim Ŝi− + Ŝi− Ŝm
− − z z
Ŝnz |Φ0 i
= δim δin Ŝi− |Φ0 i − δin Ŝi− Ŝm
z
|Φ0 i − δim Ŝi− Ŝnz |Φ0 i + Ŝi− Ŝm
z z
Ŝn |Φ0 i

y since |Φ0 iis the fully polarized state, Ŝjz |Φ0 i = S|Φ0 i



= δim δin − δin S − δim S + S 2 Ŝi− |Φ0 i,

and we obtain
1 X z z − 1 X

− √ Jmn Ŝm Ŝn Ŝi |Φ0 i = − √ Jmn δim δin − δin S − δim S + S 2 Ŝi− |Φ0 i
2 2S m,n 2 2S m,n
1X 2


=− Jmn ✘ ✘δ✘
δim in − δin S − δim S + S |χi i
2 m,n
!
SX S2 X
= Jmn (δin + δim ) − Jmn |χi i
2 m,n 2 m,n


y use the property Jij = Jji
!
X S2 X
= S Jin − Jmn |χi i
n
2 m,n
 
S 2 N J0
= SJ0 − |χi i,
2

where we used the fact that X X X

Jk=0 = Jn ei0·k = Jn = Jij .
n n j

+ − − +
(ii) Terms arising from Ŝm Ŝn + Ŝm Ŝn :
   
Ŝn Ŝi− |Φ0 i = Ŝm
+ −
Ŝm + −
Ŝi Ŝn− |Φ0 i
 
= 2δim Ŝiz + Ŝi− Ŝm +
Ŝn− |Φ0 i
 
= 2δim Ŝiz Ŝn− |Φ0 i + Ŝi− 2δmn Ŝm z
+ Ŝn− Ŝm
+
|Φ0 i
   
= 2δim −δin Ŝn− + Ŝn− Ŝiz |Φ0 i + 2δmn Ŝi− Ŝm z
+ Ŝi− Ŝn− Ŝm
+
|Φ0 i
 ✘ 
−✘
= −2δim δin Ŝn− + 2δim Ŝn− Ŝiz + 2δmn Ŝi− Ŝm z
+✘ Ŝi−✘
Ŝ✘ +
n Ŝm |Φ0 i
 
= −2δim δin Ŝn− + 2Sδim Ŝn− + 2Sδmn Ŝi− |Φ0 i

and  
− + − ✟
Ŝm −
Ŝn Ŝi |Φ0 i = Ŝm Ŝi−✟
2δin Ŝiz + ✟ −
Ŝn+ |Φ0 i = 2Sδin Ŝm |Φ0 i,

4
 so that
1 X  
− √ Jmn Ŝm+ − − +
Ŝn + Ŝm Ŝn Ŝi− |Φ0 i
4 2S m,n
1X  
✘δ✘ − ✟|χi i + 2Sδin |χm i
=− Jmn −2✘ δim δmn
in Ŝn + 2Sδim |χn i + 2S✟
4 m,n
SX S X
=− Jin |χn i − Jmi |χm i
2 2
n6=i m6=i


y use Jij = Jji
X
= −S Jin |χn i.
n6=i

(iii) Terms arising from the Zeeman coupling:

h X z − h X 
√ Ŝm Ŝi |Φ0 i = √ −δim Ŝi− + Ŝi− Ŝm
z
|Φ0 i
2S m 2S m
h  − 
= √ −Ŝi + N S Ŝi− |Φ0 i
2S
= h(N S − 1)|χi i

Collecting the partial results (i) to (iii) we finally obtain

 
S 2 N J0 X
Ĥ|χi i = SJ0 − − h(N S − 1) |χi i − S Jin |χn i.
2
n6=i

d) In the previous question we found that

 
S 2 N J0 X
Ĥ|χR i = SJ0 − − h(N S − 1) |χR i − S Jn |χR+n i,
2 n

which we may use to compute Ĥ|χk i:

1 X −ik·R
Ĥ|χk i = √ e Ĥ|χR i
N R∈L
  !
1 X −ik·R S 2 N J0 X
= √ e SJ0 − − h(N S − 1) |χR i − S Jn |χR+n i
N R∈L 2 n
1 X −ik·R X 1 X −ik·R
= [· · · ] √ e |χR i − S Jn √ e |χR+n i
N R∈L n N R∈L


y change variable R → R − n in the second term
X 1 X −ik·(R−n)
= [· · · ] |χk i − S Jn √ e |χR i
n N R∈L
!
X
= [· · · ] |χk i − S Jn eik·n |χk i
n
| {z }
Jk
 
S 2 N J0
= − − h(N S − 1) + SJ0 − SJk |χk i.
2

5
 This can be written as

S 2 N J0
Ĥ|χk i = Ek |χk i, Ek ≡ − − hN S + S(J0 − Jk ) + h = E0 + εk .
2 | {z }
εk

Comment — We have shown that a Bloch spin-wave state |χk i is an exact eigenstate with energy εk above
the FM ground state energy, E0 . Note that the excitation energy, εk = S(J0 − Jk )+ h, is precisely what we
obtain in the linearized bosonic theory after we perform the Holstein-Primakoff transformation, and take the
formal limit of large S. This is why, even though the large-S limit may seem a questionable approximation
at first, the magnon description that we obtain by taking that limit after the H-P transformation actually
provides the exact energies of the elementary excitations. In other words, the result above explains why we
may safely take the large-S limit in the bosonic/magnon description without compromising the nature and
the energy spectrum of the elementary (low energy) excitations.

e) The previous calculation shows that a Bloch spin-wave state has energy Ek = E0 + S(J0 − Jk ) + h.
Without external field (h = 0), exciting such a collective state costs an energy S(J0 − Jk ) relative to the
ground state. Of course, this is only consistent if S(J0 − Jk ) ≥ 0 for all k. If this is condition is violated
for some k, it means that the fully polarized FM state is not the ground state, since the system could then
lower its energy by exciting spin waves, thereby reaching the actual ground state. The condition can be
written as X X
J k ≤ J0 ⇔ Jn eik·n ≤ Jn .
n n

We now note that, since we are considering the spins on a simple Bravais lattice, for each neighbor vector
n we have another neighbor at −n. So we may symmetrize any summation over neighbors as follows:
X 1X 1X
f (n) = f (n) + f (−n), for any arbitrary summand f (n).
n
2 n 2 n

Applying this to the inequality above, we get

X X
Jk ≤ J0 ⇔ Jn eik·n ≤ Jn
n n
1X 1X X
⇔ Jn eik·n + J−n e−ik·n ≤ Jn
2 n 2 n n


y note that J−n = Jn
X eik·n + e−ik·n X
⇔ Jn ≤ Jn
n
2 n
X X
⇔ Jn cos(k · n) ≤ Jn .
n n

Since | cos(k · n)| ≤ 1, it is clear that if all Jn > 0, then the inequality is automatically satisfied. On the
other hand, if some Jn are negative, it could be violated. Hence, the fully polarized FM state |Φ0 i is the
ground state of the Heisenberg Hamiltonian for arbitrary magnitudes and range of the exchange couplings
Jn , provided all the couplings are non-negative.

6
Problem 2 (Exact Holstein-Primakoff transformation)

Consider the Holstein-Primakoff (HP) transformation defined in eq. (2), which is adequate to de-
scribe the low-energy excitations of the FM Heisenberg model. Do not make any approximations.

a) Imagine we forgot the correct position of the square root, and so we consider both possibilities
in the definition of Ŝi− : Ŝi− = a†i (2S − n̂i )1/2 and Ŝi− = (2S − n̂i )1/2 a†i . Show that, while
the first definition recovers the correct commutator [Ŝ − , Ŝ z ], the second doesn’t.

b) Show that the HP transformation reproduces the angular momentum commutators (4) exactly
(i.e., without any approximation for the square root that appears in the HP transformation).

Solution

a) We wish to verify which definition yields the (exact) commutator

 
Ŝ − , Ŝ z = Ŝ − .

Since the operators at different sites i commute, it is sufficient to consider a given site i; for convenience,
we may thus drop the site index i (a†i → a† , etc.). With the first definition we would have
     
Ŝ − , Ŝ z = a† (2S − n̂)1/2 , S − n̂ = − a† (2S − n̂)1/2 , n̂
   ✘✘✘ ✿ 0
1/2 ✘1/2
= − a† , n̂ (2S − n̂) − a† ✘ (2S
✘− ✘n̂) , n̂
  1/2   1/2
= − a† , n̂ (2S − n̂) = − a† , a† a (2S − n̂)
1/2
= a† (2S − n̂)
= Ŝ − .

Using the second definition:

     
Ŝ − , Ŝ z = (2S − n̂)1/2 a† , S − n̂ = − (2S − n̂)1/2 a† , n̂ =
= (2S − n̂)1/2 a†
= Ŝ − .

This could suggest that either definition works. But see the following question.
 
b) In the previous question we verified the HP transformation reproduces the commutator Ŝ − , Ŝ z = Ŝ −

7
  
exactly. We just need to establish it also gives the exact commutator Ŝ + , Ŝ − = 2Ŝ z :
 + −  1/2 1/2 
Ŝ , Ŝ = (2S − n̂) a, a† (2S − n̂)
1/2 1/2 1/2 1/2
= (2S − n̂) aa† (2S − n̂) − a† (2S − n̂) (2S − n̂) a
1/2
1/2
= (2S − n̂) 1 + a† a (2S − n̂) − a† (2S − n̂) a
= (2S − n̂) + n̂ (2S − n̂) − a† (2S − n̂) a
✘ †
 ✘✘  
= (2S − n̂) + ✘ ✘✘−✘n̂) − a ✘
n̂ (2S a (2S
✘✘− n̂) + (2S − n̂) , a
 
= (2S − n̂) + a† n̂, a
= (2S − n̂) − a† a
= 2 (S − n̂)
= 2Ŝ z .

However, if we used the second definition, we would obtain

 + −  1/2 1/2 
Ŝ , Ŝ = a (2S − n̂) , (2S − n̂) a†
1/2 1/2
= a (2S − n̂) a† − (2S − n̂) a† a (2S − n̂)
  
= a a† (2S − n̂) + (2S − n̂) , a† − n̂ (2S − n̂)
 
= aa† (2S − n̂) − a n̂, a† − n̂ (2S − n̂)
= aa† (2S − n̂) − aa† − n̂ (2S − n̂)
= (1 + n̂) (2S − n̂) − (1 + n̂) − n̂ (2S − n̂)
= (2S − n̂) − 1 − n̂
= (2S − 2n̂) − 1
6= 2Ŝ z .

8
Problem 3 (FM magnons with magnetic anisotropy)

Localized magnetism in anisotropic crystals is frequently characterized by anisotropic exchange

couplings. For example, in a system with axial anisotropy along the direction ẑ, the corresponding
Heisenberg Hamiltonian could be

1 Xh  x x  i X
Ĥ = − J Ŝi Ŝj + Ŝiy Ŝjy + J ′ Ŝiz Ŝjz − h Ŝiz , h ≡ gµB B > 0,
2
hi,ji i

where J 6= J ′ to reflect the exchange anisotropy and hi, ji in the summation means the coupling is
only between nearest neighbors. The last term accommodates the possibility of having an external
magnetic field along the anisotropy axis: B = B ẑ. Consider J and J ′ positive (strictly FM coupling).

a) Use the HP transformation in the large-S approximation to show that, to quadratic order in
the magnon operators, we can write

N S 2 zJ ′ X

Ĥ ≃ −hN S − + εk a†k ak , where εk ≡ Sz J ′ − Jγk + h.
2
k

b) Suppose that J ′ > J and let h → 0:

i) Show that the magnon spectrum, εk , is gapped and obtain the excitation gap, ∆.
Note — The excitation gap (∆) is defined as the lowest value that εk might take, since that is the
lowest energy necessary to excite the system from the ground-state.

ii) Explain why the magnon spectrum we obtain in this case tells us that the ground state
is the fully polarized FM state with all spins pointing along ẑ.

c) Suppose that J ′ < J and let h → 0:

i) Obtain the magnon excitation gap, ∆, in this case.

ii) Explain why, in contrast with the previous question, the current result for ∆ tells us the
ground state is not the fully polarized FM state with all spins pointing along ẑ.
Hint — Note that the HP transformation is done relative to a particular reference state of the spins.

d) Consider now the isotropic case (J ′ = J, h 6= 0) and let us analyze the average spin polarization

1 X z
m(T ) ≡ hŜi iT ,
N
i

as a function of temperature, where h· · · iT is the average at thermal equilibrium at tem-

perature T . For this calculation: recall that magnons are non-conserved bosons, so their
occupations at thermal equilibrium follow a Planck distribution; consider the spins are on a
hypercubic lattice in d dimensions with lattice constant a; assume the thermodynamic limit
(N → ∞); assume low temperatures.

i) Show that, in the absence of the magnetic field (h = 0), m(T ) diverges for d ≤ 2.
Comment on the physical significance of such divergence.

9
 ii) Show that, for d > 2 and zero field (h = 0), it is given approximately by
 d/2 Z
ζ( d2 )
Ld kB T 1 ∞
xs−1
m(T ) ≃ S − d , ζ(s) = dx.
Γ( 2 ) N π d/2 D Γ(s) 0 ex − 1

iii) Verify that a finite external magnetic field (h 6= 0) suppresses the divergence for d = 2.
Show that, in this case, the average spin polarization per particle is given by
   
kB T a2 1 h≪kB T kB T a2 kB T
m(T ) ≃ S − log −−−−−−→ m(T ) ≃ S − log .
4πD 1 − eβh 4πD h

Solution

a) The procedure is just what we explicitly discussed in the lectures:

1 Xh  x x y y
 i X
Ĥ = − Jn ŜR ŜR+n + ŜR ŜR+n + Jn′ ŜRz z
ŜR+n − h z
ŜR
2
R,n R
 
1 X Jn  + − − +
 X
=− ŜR ŜR+n + ŜR ŜR+n + Jn′ ŜRz z
ŜR+n − h z
ŜR
2 2
R,n R

 √ √ z
y in the large-S approximation we replace ŜR ≃ 2S aR , ŜR ≃ 2S a†R , ŜR
+ −
= S − n̂R ,

X Jn S   J′    X 
† † † †
≃− aR aR+n + aR aR+n + n
S − aR aR S − aR+n aR+n − h S − a†R aR
2 2
R,n R


y expand and drop products of more than two boson operators
X  J−n S † Jn S † J′   X †
≃− aR aR−n + aR aR+n + n S 2 − 2Sa†R aR − hN S + h aR aR
2 2 2
R,n R

N S2 X ′ X  J−n S † Jn S †
 X †
= −hN S − Jn − aR aR−n + aR aR+n − Jn′ S a†R aR + h aR aR
2 n 2 2
R,n R


y introduce the lattice Fourier transforms
N S 2 J0′ h XX † ′
= −hN S − + ak ak′ ei(k−k )·R
2 N
k,k′ R
   
XX † J−n S ik·R−ik′ ·(R−n) Jn S ik·R−ik′ ·(R+n) J′ S
− n a†k ak′ ei(k−k )·R
′
− a k a k′ e + e
2N 2N N
k,k′ R,n
   
N S 2 J0′ X † XX † J−n S ik·n Jn S −ik·n ′ †
= −hN S − +h ak ak − ak ak e + e − Jn a k a k
2 n
2 2
k k
N S 2 J0′ X † X
= −hN S − + (h + SJ0′ ) ak ak − S Jk a†k ak .
2
k k

Rearranging, this can be cast as

N S 2 J0′ X
Ĥ ≃ −hN S − + εk a†k ak
2
k

with the magnon dispersion given by

εk ≡ h + S (J0′ − Jk ) .

10
 Since the exchange couples only nearest-neighbors, we replace Jn → J and the summation over n includes
only the nearest-neighbors. Hence,
X X
Jk = Jn eik·n = J eik·n = zJγk , and J0′ = zJ ′ γ0 = zJ ′ .
n n

So, finally, we have

N S 2 zJ ′ X
Ĥ ≃ −hN S − + εk a†k ak , εk ≡ h + Sz (J ′ − Jγk ) .
2
k

2 ′
Remark — You may note that the constant term −hN S − N S2zJ coincides with the ground state energy
of the classical Heisenberg ferromagnet in the presence of the external magnetic field.
P
b) Begin by noting that, from the definition γk ≡ 1z n eik·n , it follows that −1 < γk < 1 and, in addition,
that γk is maximum at k = 0. When J ′ > J > 0, the minimum of εk therefore occurs at k = 0 (εk is
smallest when γk is largest).

i) Consider then the small k behavior of γk :

1 X ik·n 1X 1 X (ik · n)2

γk = e =1+ ik · n + + ...
z n z n z n 2
1 X 1 X
≃ 1 + ik · n− (k · n)2
z n
2z n
 X

y in a Bravais lattice n=0
n
1 X
≃1− (k · n)2 .
2z n

From here, the magnon dispersion near the minimum is (h → 0)

JS X
εk≃0 ≃ Sz (J ′ − J) + (k · n)2 .
2 n

If J ′ > J, then εk=0 = Sz (J ′ − J) > 0. So the magnon excitation gap is ∆ = Sz (J ′ − J) .

ii) We found that ∆ = Sz (J ′ − J) and, according to the text of the problem J ′ > J > 0. The magnon
gap is thus positive, which means it costs energy to excite the system out of the fully polarized spin
state, which is also the vacuum state of the boson operators ak . The ground state is clearly the
magnon vacuum state which, tracing back to the HP transformation, represents the state with all
spins are fully polarized along ẑ. The ground state energy with zero magnetic field is

N S 2 zJ ′
Egs = − .
2

c) The analytical result for the magnon dispersion is the one obtained in (a), again letting h → 0:

εk = Sz (J ′ − Jγk ) .

i) The magnon gap is still given by the minimum value of the dispersion, εk , which we now rewrite as

z =0
}| { 
εk = Sz J − J + J ′ − Jγk = −Sz (J − J ′ ) + SzJ (1 − γk ) .

11
 The second term is positive because 1−γk ≥ 0, and reaches its minimum value at k = 0. In contrast,
the first term is manifestly negative, because we are now considering the situation J > J ′ . So the
excitation gap is negative in this case (!) with value ∆ = −Sz (J − J ′ ) .

ii) Recall that the Hamiltonian is (setting h → 0)

N S 2 zJ ′ X
Ĥ ≃ − + εk a†k ak .
2
k

Consider the state without any magnons (the vacuum of the operators ak ):

N S 2 zJ ′ N S 2 zJ ′
Ĥ|0i = − |0i −→ energy = − .
2 2
And now a state containing n0 magnons with wavevector k = 0 (but no other magnon occupations):

N S 2 zJ ′ X N S 2 zJ ′
Ĥ|n0 i = − |n0 i + εk a†k ak |n0 i = − |n0 i + εk=0 n0 |n0 i.
2 2
k
   
N S 2 zJ ′ N S 2 zJ ′
= − + εk=0 n0 |n0 i = − + n0 ∆ |n0 i,
2 2

which is another energy eigenstate (of course!). However, we established above that ∆ < 0, which
means that
N S 2 zJ ′ N S 2 zJ ′ N S 2 zJ ′
− + n0 ∆ = − − n0 |∆| < − .
2 2 2
In other words, the vacuum state |0i cannot be the ground state of the system when J ′ < J, because:

• There clearly are energy eigenstates with lower energy; specifically states containing a finite
number of magnons; this is true for all magnons such that εk < 0.

• When we translate from magnons back to the spin language, states with non-zero magnon
occupation represent spin states that are not fully polarized!

• Since magnons have bosonic character, and the system can lower its energy by exciting any of
the magnons that have εk < 0, it is clear that the fully polarized state cannot be the ground
state in this case. For example, if we prepared the system in the FM state fully polarized along
ẑ, an arbitrarily large number of magnons would be spontaneously excited in order to lower the
energy of the system. Hence, we may at least conclude that the fully polarized FM state is
unstable.

Comment — Note that, because of their bosonic character, the magnon occupation numbers, nk ,
can be arbitrarily large. Considering only the magnons with k = 0, the above calculation shows that
we may bring the energy of the system to an arbitrarily large negative value relative to the FM state
by spontaneously exciting magnons:

N S 2 zJ ′ n →∞
− − |∆|n0 −−0−−−→ −∞.
2
This divergence is unphysical. The anomaly signals (at least) three important things:

• the FM state is not the ground state, as discussed above;

• the large-S approximation (that we did to obtain the quadratic bosonic Hamiltonian) breaks
down, because the approximation is only valid when the average number of magnons is small
compared to the spin magnitude S;

12
 • the reference FM state, relative to which we did the HP transformation and defined the magnon
quasiparticles, is not a helpful starting point (reference state) to define the elementary excitations
of the system in the case J ′ < J, because the anomaly above suggests that the true ground-state
is far from the spins polarized along ẑ.

d) In the bosonic representation offered by the HP transformation,

1 X z 1 X  1 X 1 X
m(T ) ≡ hŜi iT = S − ha†i ai iT = S − hni iT = S − hnk iT .
N i N i N i N
k∈BZ

When N → ∞,
 d Z  d Z kmax
1 L (d) 1 1
εk is isotropic L k d−1
m(T ) = S − dk −−−−−−−−→ S − Ad−1 dk ,
N 2π BZ eβεk −1 N 2π 0 eβεk − 1

where Ad−1 ≡ 2π d/2 /Γ( d2 ) is the area of the (d − 1)-sphere. Furthermore, since we are considering a
hypercubic lattice with N unit cells, each having lattice constant a, we have Ld = N ad and the general
expression becomes
Z kmax
 a d k d−1
m(T ) = S − Ad−1 dk βε .
2π 0 e k −1

i) The integral can diverge due to the behavior of the integrand when k → 0. For small ka ≪ 1, we
saw in (a) that the magnon dispersion becomes

d
JS X X
εk≃0 ≃ (k · n)2 = JSa2 kj2 = Dk 2 , D ≡ JSa2 .
2 n j=1

where a is the lattice constant of the assumed hypercubic lattice. Hence, at any nonzero temperature,
Z kmax Z ··· Z ··· Z ···
k d−1 k d−1 kB T k d−1
dk ≃ dk = dk ∝ dk k d−3 ,
0 eβεk − 1 0 βεk D 0 k2 0

which, near k = 0, converges only if d − 3 > −1, or d > 2. Conversely, it diverges if d ≤ 2. Such
“infrared” (low-energy) divergence for d ≤ 2 would imply that

m(T ) = S − ∞ (!).

Physically, this means that in lower dimensions thermal fluctuations are always too strong and do not
allow FM long-range order.

The exception happens if T = 0. In that case, we have hnk iT =0 = 0 before taking the limit
N → 0 and, therefore, we obtain m(T ) = S for any dimension d. This result at T = 0 is expected
because the FM state is the ground state of the isotropic Heisenberg Hamiltonian. The computation
above shows that this FM order is lost in 1 and 2 dimensions for arbitrarily small temperature.

ii) If d > 2 we have

 a d Z kmax k d−1
m(T ) = S − Ad−1 dk βε
2π 0 e k −1
2
where kmax ∼ 1/a. For temperatures such that kB T ≪ Dkmax we may replace kmax → ∞ because

13
 the integrand decays exponentially with k. Hence,
 d Z ∞
Ad−1 L k d−1
m(T ) ≃ S − dk βDk2
N 2π 0 e −1

 2 d d
y let u ≡ βDk and replace L = N a
 d/2 Z ∞
Ad−1  a d 1 ud/2−1
=S− du u
2 2π βD e −1
|0 {z }
Γ(d/2) ζ(d/2)
 d/2
ζ( d2 ) ad
kB T
=S− .
Γ( d2 ) π d/2 D

iii) We have found in a previous question that εk ≡ h + Sz (1 − γk ). For ka ≪ 1 this approximates to

εk ≃ h + Dk 2 , D ≡ JSa2 .

Therefore, in the specific case d = 2, we now have

 d Z ∞
Ad−1 L k d−1
m(T ) ≃ S − dk βh βDk2
N 2π 0 e e −1

 2
y let u ≡ βDk and put d = 2
  Z ∞
a2 1 du
=S− e−βh u − e−βh
4π βD 0 e

By changing variable to x ≡ eu , we may evaluate the integral as follows

Z ∞ Z ∞ Z ∞ 
du dx βh dx dx
= =e −
0 eu − e−βh 1 x(x − e−βh ) 1 x − e−βh x
Z xm  
dx dx
= eβh lim −
xm →∞ 1 x − e−βh x
 −βh

xm − e
= eβh lim log − log x m
xm →∞ 1 − e−βh
e−βh 1
= eβh log = eβh log .
e−βh −1 1 − eβh

So we obtain,
   
kB T a2 1 h≪k T
B kB T a2 kB T
m(T ) ≃ S − log −−−−−−→ m(T ) ≃ S − log .
4πD 1 − eβh 4πD h

14
Problem 4 (AFM magnons)

Consider the Heisenberg Hamiltonian describing an antiferromagnet,

JX X
z
X
z
Ĥ = Ŝi · Ŝj − h ŜA,i +h ŜB,i h, J > 0.
2
hi,ji i∈A i∈B

where hi, ji means the coupling is only between nearest neighbors. The spins occupy the sites of
(bipartite) hypercubic lattice in d dimensions with primitive lattice constant a, and each sublattice
comprises NA = NB = N sites. The two terms proportional to h represent a (hypothetical) coupling
to a staggered field, which allows us to set a preferred symmetry-breaking direction along ẑ.

a) Extending the HP transformation defined in (2), express the spin operators Ŝ ± , Ŝ z on each
sublattice (A and B) in terms of bosonic operators (ai and bi ), which should describe the
elementary spin deviations relative to the AFM Néel state.

b) Show that, in the large-S limit and retaining only up to quadratic terms in the boson operators,
the Hamiltonian becomes
X  †  X  
Ĥ ≃ −N zS 2 J −2hN S +(JSz + h) ak ak + b†−k b−k +JSz γk a†k b†−k + γk∗ ak b−k .
k∈BZ k∈BZ

where ak (bk ) represent magnon excitations belonging to sublattice A (B).

c) Use the appropriate Bogoliubov-Valatin transformation to show we can write

X  
Ĥ = −N zJS(S + 1) − hN (2S + 1) + εk α†k αk + βk† βk + 1 ,
k∈BZ

where αk and βk are the Bogoliubov operators for the elementary excitations (AFM magnons)
whose energy is given by
r d
h 2 1X
εk ≡ JSz 1+ − γk2 , and γk = cos(ki a).
JSz d
i=1

d) Show that the ground state energy is lower than that of the classical Néel state.

e) Here and in the subsequent questions, consider h = 0. Verify that, for ka ≪ 1,

d
√ X
εk ≃ JS 2z ka, k ≡ |k| = ki2 .
i=1

f) Consider the staggered spin polarization per particle, defined as

1 X z 1 X z
ms (T ) ≡ hŜA,i iT − hŜB,i iT ,
N N
i∈A i∈B

15
 as a function of temperature, where h· · · iT is the average at thermal equilibrium at tempera-
ture T . Show that it has the form

ms (T ) = 2S − δm(0) (T )
s − δms ,

(0) (T )
where δms captures the effect of quantum fluctuations and δms that of thermal fluctu-
ations in reducing the staggered polarization relative to the Néel state. Obtain the explicit
expressions
1
1 Xh 1
i 2 X (1 − γk2 )− 2
δm(0) ≡ (1 − γk2 )− 2 − 1 , δm(T ) ≡ .
N N eβεk − 1
k k

(0) (T )
g) By studying the convergence of δms and δms for a given dimension d, comment on the
stability of AFM order in d dimensions, as predicted by this linear spin-wave theory.
Note — For this calculation: recall that magnon occupations at thermal equilibrium follow a Planck distri-
bution; consider the spins are on a hypercubic lattice in d dimensions with lattice constant a; assume the
thermodynamic limit (N → ∞); assume low temperatures.

Solution

a) The staggered field represented by the last two terms in Ĥ selects ẑ as the relevant direction, relative to
which we should parameterize the spin deviations. The structure of the staggered coupling favors positive
(negative) spin polarization along ẑ on the A (B) sublattice. The appropriate HP transformation is thus
 1/2  1/2
sublattice A: +
ŜA,i = 2S − a†i ai âi , −
ŜA,i = a†i 2S − a†i ai , z
ŜA,i = S − a†i ai ,
 1/2  1/2
sublattice B: +
ŜB,i = b†i 2S − b†i bi , −
ŜB,i = 2S − b†i bi b̂i , z
ŜB,i = −S + b†i bi .

In the large-S limit, these are approximated as

√ √
sublattice A: +
ŜA,i ≃ 2S ai , −
ŜA,i ≃ 2S a†i , z
ŜA,i = S − a†i ai ,
√ √
sublattice B: +
ŜB,i ≃ 2S b†i , −
ŜB,i ≃ 2S bi , z
ŜB,i = −S + b†i bi .

b) Replace the HP transformation above in the original Heisenberg Hamiltonian. The part containing the

16
 Heisenberg coupling becomes

J X J X z z z z
 J X
+ − − +

Ŝi · Ŝj = ŜA,R ŜB,R+n + ŜB,R ŜA,R+n + ŜA,R ŜB,R+n + ŜA,R ŜB,R+n + H.c.
2 2 4
hi,ji R,n R,n


y use the HP transformation, drop terms beyond quadratic order
J X h     i
= S − a†R aR −S + b†R+n bR+n + −S + b†R bR S − a†R+n aR+n
2
R,n
2SJ X  
+ aR bR+n + a†R b†R+n + H.c.
4
R,n
JS X  † 
2
= −N zS J + aR aR + b†R+n bR+n b†R bR + a†R+n aR+n
2
R,n
2SJ X  
+ aR bR+n + a†R b†R+n + H.c.
4
R,n


y lattice Fourier transforms
X †  SJ X X  
= −N zS 2J + JSz ak ak + b†k bk + e−ik·n ak b−k + eik·n a†k b†−k + H.c.
2 n
k k
X † †
 X 
2
= −N zS J + JSz ak ak + b−k b−k + zSJ γk a†k b†−k + γk∗ ak b−k .
k k

The part describing the staggered field coupling becomes

X X X  X 
−h z
ŜA,i +h z
ŜB,i = −h S − a†i ai + h −S + b†i bi
i∈A i∈B i∈A i∈B
X † 
= −2hN S + h aR aR + b†R+n bR+n
R
X † 
= −2hN S + h ak ak + b†k bk .
k

Combining both, we obtain

X  X 
Ĥ ≃ −N zS 2J − 2hN S + (JSz + h) a†k ak + b†−k b−k + zSJ γk a†k b†−k + γk∗ ak b−k .
k k

Note that in a Bravais lattice, for each lattice vector n there is a corresponding −n (all Bravais lattices
have inversion symmetry). Hence,
" #
1 X ik·n 1 X ik·n X −ik·n 1X
γk = e = e + e = Re eik·n −→ γk ∈ R.
z n 2z n n
z n

c) The appropriate transformation here is to define the following Bogoliubov operators (see previous problem
set for details)
αk ≡ uk ak − vk b†−k , βk† ≡ −vk ak + uk b†k ,

with real coefficients (because γk ∈ R in this problem) given by

r r q
JSz + h 1 JSz + h 1 h
uk = + , vk = − , εk ≡ JSz (1 + h̄)2 − γk2 , h̄ ≡ .
2εk 2 2εk 2 zSJ

17
 The diagonal Hamiltonian is then
X X  
Ĥ = −N zS 2J − 2hN S + [εk − (JSz + h)] + εk α†k αk + βk† βk
k k
X  
2
= −N zS J − 2hN S − N (JSz + h) + εk α†k αk + βk† βk + 1
k
X  
= −N zJS(S + 1) − hN (2S + 1) + εk α†k αk + βk† βk + 1 .
k

Here, q
εk ≡ JSz (1 + h̄)2 − γk2

and, for a hypercubic lattice in d dimensions

d d
1 X ik·n 1X 1X 2X 1X
γk = e = Re eik·n = cos(k · n) = cos(ki a) = cos(ki a),
z n z n z n z i=1 d i=1

where we used the fact that a hypercubic lattice has coordination z = 2d.

d) The energy in the classical Néel state would be

ENéel = −N zJS 2 − 2hN S.

Since Ĥ above is diagonalized in the occupation representation of the Bogoliubov operators, the ground
state is their vacuum. Consequently,
X
Egs = −N zJS(S + 1) − hN (2S + 1) + εk
k
Xq
= ENéel − N (h + zJS) + zSJ (1 + h̄)2 − γk2
k
 q

y note that γk2 ≤ 1 −→ (1 + h̄)2 − γk2 = 1 + h̄ − δ, δ>0

= ENéel − N (h + zJS) + (zSJN + hN − δ)

= ENéel − δ,

where q
δ ≡ 1 + h̄ − (1 + h̄)2 − γk2 > 0.

Consequently,
Egs < ENéel .

p
e) Recalling from above that εk ≡ zSJ 1 − γk2 when h = 0 and expanding for ka ≪ 1,
" #
2
1 X ik·n 1X (k · n) (ka)2
γk = e = 1 + ik · n − + ... ≃ 1 − ,
z n z n 2 z

and s r
q  2 r
(ka)2 2(ka)2 2
1 − γk2 ≃ 1− 1− ≃ = ka.
z z z
We thus obtain
√
εk ≃ JS 2z ka.

18
f) Begin writing Ŝ z in the bosonic representation and express the operators in k space:

1 X z 1 X z
ms (T ) ≡ hŜA,i iT − hŜB,i iT
N i N i
1 X  1 X 
= S − ha†i ai iT − −S + hb†i bi iT
N i N i
1 X  
= 2S − ha†i ai iT + hb†i bi iT
N i
1 X † 
= 2S − hak ak iT + hb†k bk iT
N
k

To compute these thermal averages we must express the operators in terms of αk and βk (because these
are the ones that diagonalize the statistical operator), Inverting the Bogoliubov-Valatin transformation
introduced above,

ak = uk αk + vk βk† , b†−k = vk αk + uk βk† , uk , vk ∈ R.

Hence,
  
ha†k ak iT = h uk α†k + vk βk uk αk + vk βk† iT

= u2k hα†k αk iT + vk2 hβk βk† iT

= u2k hα†k αk iT + vk2 hβk† βk iT + vk2
u2k + vk2
= + vk2 ,
eβεk − 1

and we similarly obtain,

u2k + vk2
hb†k bk iT = + vk2 .
eβεk − 1
The staggered spin polarization becomes

2 X 2 2 X u2k + vk2
ms (T ) = 2S − vk − .
N N eβεk − 1
k k

The second term is T -independent and thus represents quantum fluctuation effects; the third is explicitly
T -dependent and reduces to zero at T = 0. Both terms are manifestly negative, thus reducing the staggered
magnetization relative to the Néel state (for which it would be exactly 2S). In conclusion, we may write

ms (T ) = 2S − δm(0) − δm(T ) ,

where
2 X 2 2 X u2k + vk2
δm(0) ≡ vk , δm(T ) ≡ ,
N N eβεk − 1
k k

respectively quantify the effect of quantum and thermal fluctuations in reducing the staggered magnetization
relative to that of the Néel state. Recalling that the BV transformation coefficients read
q
JSz 1 JSz 1
u2k = + , vk2 = − , εk ≡ zSJ 1 − γk2 ,
2εk 2 2εk 2

we have !
(0) 1 X 1 2 X 1 1
δm = p −1 , δm(T ) = p βεk − 1
.
N 1 − γk2 N 2
1 − γk e
k k

19
g) Let us analyze the convergence of each from the behavior of the respective integrand at small k.

Quantum fluctuations:
!
1 X 1
δm(0) = p −1
N 1 − γk2
k


y limit N → ∞
Z kmax !
Ad−1 Ld 1
= dk k d−1 p −1
N (2π)d 0 1 − γk2

 p
y behavior near k = 0, using (1 − γk2 )1/2 ≃ 2/z ka.
r Z ···
Ad−1 Ld z
∼ k d−2 dk.
N (2π)d 2a2 0

This has a logarithmic divergence in d = 1 and is finite for d ≥ 2.

Thermal fluctuations:
2 X 1 1
δm(T ) = p βεk − 1
N 2
1 − γk e
k


y limit N → ∞
Z kmax
2Ad−1 Ld k d−1 1
= p dk
N (2π)d 0 1 − γk2 eβεk − 1

 p
y behavior near k = 0, using (1 − γk2 )1/2 ≃ 2/z ka.
r Z ···
2Ad−1 Ld z kB T
∼ √ k d−3 dk.
N (2π)d 2a2 JS 2za 0

The integral exhibits a power-law divergence for d = 1 and a logarithmic divergence for d = 2, remaining
finite for d ≥ 3.

We conclude that linear spin-wave theory forecasts the total absence of Néel-type AFM order for d = 1.
In 2D, while the contribution from quantum fluctuations (at T = 0) doesn’t diverge, an arbitrarily small
finite temperature would also destabilize the AFM state due to the logarithmic divergence of the thermal
fluctuations. According to this theory, finite-temperature Néel-type AFM order is only stable in d ≥ 3.

20