Solution 2

October 14, 2015

1 Antiferromagnetic Spin Waves

1.1

The Heisenberg Hamiltonian is

H = J

N/2∑n=−N/2+1

Sk(n) · Sk(n+ 1)

=J

2

∑j

(S+(n) · S− + S−(n) · S+(n+ 1)

)+ JS3(n) · S3(n+ 1) (1)

where S− = S1 − iS2 and S+ = S1 + iS2.To calculate the quantum mechanical equations of motion obeyed by

S±(j) and S3(j), we first need to calculate the commutators

[S+(j), H] =J

2S3(j)(S+(j − 1) + S+(j + 1))− J

2S+(j)(S3(j − 1) + S3(j + 1))

(2)

where we used

[S3, S±] = ±S±[S+, S−] = 2S3 (3)

Similarly, we can calculate [S−(j), H] and [S3(j), H].

1

The Heisenberg equations of motion are

∂0S+(j) = −iJ

2

[S3(j)(S+(j − 1) + S+(j + 1))− S+(j)(S3(j − 1) + S3(j + 1))

]∂0S

−(j) = −iJ2

[−S3(j)(S−(j − 1) + S−(j + 1))− S−(j)(S3(j − 1) + S3(j + 1))

]∂0S3(j) = −iJ

4

[S+(j)(S−(j − 1) + S−(j + 1) + S−(j)(S+(j − 1) + S+(j + 1))

](4)

They are non-linear equations. This suggests that the Heisenberg model ingeneral is not a free bosonic system.

1.2

For the even sites, by using Eq.(8), we have

S+|n〉 =√

2S

[1− n(j)

2S

] 12

a(j)|n〉 =

[2S

(1− n− 1

2S

)n

] 12

|n− 1〉

S−|n〉 =[2S(n+ 1)(1− n

2S)] 1

2 |n+ 1〉 (5)

From the above equation, we have (for the even sites)

n(j)|S, S〉 = (S − S3(j))|S, S〉 = 0

n(j)|S,−S〉 = 2S|S,−S〉 (6)

Hence Eq.(7) and Eq.(8) are consistent with Eq.(4).

1.3

The Heisenberg Hamiltonian is

H = Heven +Hodd

=∑

j∈evenJS[a†(j)a(j) + b†(j + 1)b(j + 1)

]+ JS

[(1− n(j)

2S)1/2(1− n(j + 1)

2S)1/2a(j)b(j + 1) + a†(j)b†(j + 1)(1− n(j)

2S)1/2(1− n(j + 1)

2S)1/2

]+ Ja†(j)a(j)b†(j + 1)b(j + 1)− S2 +

∑j∈odd

(a↔ b) (7)

2

1.4

In the semiclassical limit S → ∞, for the above Hamiltonian, if we onlyinclude terms which are of order 1/S, the Hamiltonian is

Heven = JS∑

j∈even

[a(j)b(j + 1) + a†(j)b†(j + 1) + a†(j)a(j) + b†(j + 1)b(j + 1)− S

]Hodd = JS

∑j∈odd

[b(j)a(j + 1) + b†(j)a†(j + 1) + b†(j)b(j) + a†(j + 1)a(j + 1)− S

](8)

The above Hamiltonian takes a quadratic form.

1.5

In the semiclassical limit, the equations of motion becomes

∂0a(j) = −iJS2

[b†(j − 1) + b†(j + 1)− 2a(j)

]∂0a

†(j) = −iJS2

[b(j − 1) + b(j + 1)− 2a†(j)

](9)

The above result is for the even j. For the odd j, we only need to switcha↔ b to get the similar result. The equations of motion are now linear. Theterm n(j)n(j + 1) is neglected.

1.6

By using the Fourier transformation

a(q) =

√2

N

∑j∈even

eiqj a(j)

b(q) =

√2

N

∑j∈odd

e−iqj b(j) (10)

The Hamiltonian can be written as

H = 2SJ∑q

[a(q)b(q) cos(q) + a†(q)b†(q) cos(q) + a†a(q) + b†(q)b(q)− N

2S

](11)

3

By performing the canonical transformation

c(q) = cosh(θ)a(q) + sinh(θ)b†(q)

d(q) = cosh(θ)b(q) + sinh(θ)a†(q) (12)

The Hamiltonian has off-diagonal term cos(q)(cosh2(θ) + sinh2(θ))c(q)d(q)−2 cosh(θ) sinh(θ)d(q)c(q). To diagonalize the Hamiltonian, this term is equalto zero and this requires that

cos(q) = tanh(2θ) (13)

The Hamiltonian after diagonalization takes the following form

HSW = E0 +

∫ π2

−π2

dq

2πω(q) (nc(q) + nd(q)) (14)

where ω(q) = | sin(q)|.

1.7

The ground state satisfies

c(q)|GS〉 = d(q)|GS〉 = 0 (15)

1.8

The single particle eigenstate can be generated by c†(q) or d†(q).

c†(q)|GS〉 = |q, 1〉,d†(q)|GS〉 = |q, 2〉 (16)

The dispersion relation is

E(q) = 2NJS| sin(q)| (17)

It equals to zero when q = nπ. Since q ∈ [−π2, π

2], when q is around zero, the

energy of excited state goes to zero. Around q = 0,

E(q) ≈ 2NJS|q| (18)

The energy vanishes linearly as the momentum q approaches to zero.

4

1.9

In the spin wave approximation,

S+(j) =∑q

e−iqj√

2S(cosh θc†(q)− sinh θd(q))

S−(j) =∑q

eiqj√

2S(cosh θc(q)− sinh θd†(q)) (19)

When both n and n′ are even or odd numbers,

D+−(jt, j′t′) = −i〈GS|T S+(j, t)S−(j′, t′)|GS〉

= −4Si∑q

e−iEq(t−t′)θ(t− t′)〈GS| cosh2 θc†(q)c(q) + sinh2 θd(q)d†(q)|GS〉

− 4Si∑q

eiEq(t−t′)θ(t′ − t)〈GS| cosh2 θc†(q)c(q) + sinh2 θd(q)d†(q)|GS〉

(20)

The propagator in the momentum space is

D+−(q) = 4Si

∫dt[θ(∆t)e−i(Eq+ω)∆t cosh2(θ) + θ(−∆t)ei(Eq−ω)∆t sinh2(θ)

]=

4S

| sin(q)|

[1− | sin(q)|ω − Eq + iε

+1 + | sin(q)|ω + Eq − iε

](21)

When one is odd and one is even, the calculation is similar,

D+−(jt, j′t′) = −i〈GS|T S+(j, t)S−(j′, t′)|GS〉

= −4Si∑q

e−iEq(t−t′)θ(t− t′)〈GS| cosh θ sinh θc†(q)c(q) + cosh θ sinh θθd(q)d†(q)|GS〉

− 4Si∑q

eiEq(t−t′)θ(t′ − t)〈GS| cosh θ sinh θθc†(q)c(q) + cosh θ sinh θθd(q)d†(q)|GS〉

(22)

The propagator in the momentum space is

D+−(q) =4S cos(q)

| sin(q)|

[1

ω − Eq − iε+

1

ω + Eq + iε

](23)

5

For D33,

D33(jt, j′t′) = −i〈GS|T (S − n(j, t))(S − n(j′, t′))|GS〉 (24)

It includes term 〈GS|a†(j)a(j)|GS〉 and term 〈GS|a†(j)a(j)a†(j′)a(j′)|GS〉.The propagator in the momentum space is

D33 = NS2δq,πδq′,π

[− 1

ω − iδ+

1

ω + iδ

]+δq+q′,0N

∑k

=(cosh θk−q/2 sinh θk+q/2 − cosh θk+q/2 sinh θk−q/2

)×(

1

ω − Ek−q/2 − Ek+q/2 + iε− 1

ω − Ek−q/2 − Ek+q/2 − iε

)(25)

1.10

The propogator is already calculated in the above problem.

1.11

From the result on D+−(p, ω), there is a pole when cos2(q/2) → 0 . Thisrequires that q = π.

2 The electron gas

2.1

The Feynman propagator for the non-interacting system

Gσ,σ′

0 = −i0〈G|Tψσ(x)ψ†σ′(x′)|G〉0

= −iσσ,σ′

∑α

ϕ∗α(r)ϕα(r′)×[θ(α−G)ei(Eα−EG)(t−t′))θ(t− t′)− θ(G− α)ei(Eα−EG)(t−t′))θ(t′ − t)

]= sigmaσ,σ′

∑α

ϕ∗α(r)ϕα(r′)

[θ(α−G)

ω − (Eα − EG) + iδ+

θ(G− α)

ω − (Eα − EG − iδ)

](26)

6

For free fermion, we have

Gσ,σ′

0 = δσ,σ′

∫d3peip(r−r

′)

[θ(|p| − pF )

ω − p2

2m+ iδ

+θ(pF − |p|)ω − p2

2m− iδ

](27)

Hence after Fourier transformation, we have

GF (p, ω) =θ(|p| − pF )

ω − p2

2m+ iδ

+θ(pF − |p|)ω − p2

2m− iδ

=1

ω − E(p) + iε(|p| − pF )=

1

ω − E(p) + iεsign ω(28)

2.2

(a) The density operator is

〈ρ(x, y)〉 = −i limt′→t

GF (x, t, σ, y, t′, σ′) (29)

(b) Using the result in Eq.(28), we have

ρ(x, y) ∼∫ π

0

dθ sin θ

∫dp p2

(2π)3ei cos θp|x−y|θ(pF − |p|)

∼ 1

(2π)3

∫ pF

0

dp psin(p|x− y|)|x− y|

∼ 1

r2(−pF cos(pF r) +

1

rsin(pF r)) (30)

where r = |x− y|.In the short distance limit when r << 1/pF , we have

ρ(r) ≈ p3F

3π2(31)

In the long distance limit when r >> 1/pF , the second term is negligible,we have

ρ(r) ≈ −pF cos(rpF )

π2r2(32)

It decays as 1/r2 and is also oscillatory.

7

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2.3

(a) For any operator O, the expectation value is

〈O(t)〉 = 〈Ψ(t)|O|Ψ(t)〉 (33)

Applying some perturbation term to the Hamiltonian, we have

δO(t) = 〈G|OH′ |G〉 − 〈G|OH |G〉

= i

∫dt′〈G

[Hext(t′), O(t)

]||G〉 (34)

For an electron system interacts with an external potential, the changeof the local density δρ(x) caused by the external potential is

δρ(p, ω) = Π(p, ω)V (p, ω) = Π0(p, ω)Veff (p, ω) (35)

δρ(x) can be obtained by performing Fourier transformation on δρ(p, ω).(b) δρ(x) is

δρ(x) =

∫d3pΠ(p, ω = 0)V (p)e−i~p·~x (36)

In our model, V (x) is a point charge of strength Q at the origin, hence

V (p) =4πQ

p2(37)

Inserting the potential in δρ(x), we have

δρ(x) =

∫d3pΠ(p, ω = 0)V (p)e−i~p·~x

=

∫d3pΠ0(p, ω = 0)Veff (p, ω = 0)e−i~p·~x

=

∫d3p

V (p, ω = 0)Π0(p, ω = 0)

1− V (p, ω = 0)Π0(p, ω = 0)e−i~p·~x

(38)

where the density propagator Π0 equals to

Π0(p, ω) = 2

∫d3k

(2π)3θ(|k + q| − pF )θ(pF − |k|)×[

1

ω + (E(k)− E(k + q)) + iε− 1

ω − (E(k)− E(k + q)− iδ)

](39)

8

Veff equals to

Veff (p, ω) =V (p)

1− V (p)Π0(p, ω)=

V (p)

ε(p, ω)(40)

In the static limit,

Π0(q, ω = 0) = −8

∫d3k

(2π)3

θ(|k + q| − pF )θ(pF − |k|)q2 + 2k · q

(41)

According to Eq.(4.88) in the lecture notes, the static dielectric functionε(q, ω = 0) is

ε(q, 0) = 1 +

(4

9π

)1/3

rsu(x)

x2(42)

where x = |q|2pF

and rs satisfies

pFa0rs =

(9π

4

)1/3

(43)

The dimensionless function u(x) is

u(x) =1

2

(1 +

1

2x(1− x2) log

∣∣∣∣1 + x

1− x

∣∣∣∣) (44)

when x→ 0,

u(x) =1

2+

1

4x(1− x2) log

∣∣∣∣1 + x

1− x

∣∣∣∣ =1

2+

1

2= 1 (45)

Therefore we have(4

9π

)1/3

rsu(x)

x2≈(

4

9π

)1/3

rs1

x2=λ2

p2(46)

where λ is proportional to the Thomas-Fermi screening length ξTF definedin the lecture notes.

Combing Eq.(39), Eq.(40) and Eq.(46) together, we have

δρ(x) = −∫

d3p

(2π)3

λ2

p2 + λ2ei~p·~x = −λ

2

4π

1

|x|e−|x|/εTF (47)

9

where we used Eq.(4.95) in the lecture notes. Actually there is some issuewith the above calculation. There is a weak singularity as x → 1. If theFermi surface is sharp, there will be some extra contribution to the dielectricfunction in the denominator of Veff .

Below we do the calculation more carefully,

δρ(x) =

∫d3p

(2π)3δρ(p)e−i~p·~x =

∫ ∞0

dp

(2π)2p2

∫ 1

−1

d cos(θ)δρ(p)e−ip|x| cos θ

=

∫ ∞0

dp

2π2|x|1− ε(p)ε(p)

sin(p|x|) (48)

where ε(p) = ε(p, 0).Define

g(p) =1

2π2p

1− ε(p)ε(p)

(49)

We have

δρ(x) =1

|x|

∫ ∞0

dpg(p) sin(p|x|) = − 1

|x|3

∫ ∞0

dpg′′(p) sin(p|x|) (50)

where the second term is obtained by performing integration by parts twice.Since there is a singularity in ε(p) when x→ 1, when we expand around

x = 1, the most singular part in g′′ is

g′′(p) ∼ A

p− 2pF(51)

By using sin(px) = sin[(p− 2pF )x] cos(2pFx) + cos[(p− 2pF )x] sin(2pFx),we can write Eq.(50) as

δρ(x) =A

|x|3

∫ 2p+Λ

2pF−Λ

sin[(p− 2pF )|x|] cos(2pF |x|) + cos[(p− 2pF )|x|] sin(2pF |x|)p− 2pF

dp

(52)

When pF |x| >> 1, we have

δρ(x) ∼ A cos(2pF |x|)|x|3

(53)

It decays as 1/|x|3. The oscillation behavior of the charge density is due thesharp Fermi surface and this behavior is called Friedel oscillation.

10