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Homework 6 - Solutions - Physics Department, UCSB

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<strong>Homework</strong> 6 - <strong>Solutions</strong><br />

Problem 1 - Non-Linear M(H)<br />

The Hamiltonian of a Heisenberg ferrromagnet in a magnetic field H is given by<br />

H = −J ∑ ∑<br />

S i · S J − gµ B H · S i , J > 0 (1)<br />

〈ij〉<br />

i<br />

It was seen in class that in mean field theory we can study the system<br />

H = −J ∑ [<br />

]<br />

∑<br />

〈S i 〉 · S j + S i · 〈S j 〉 − 〈S i 〉 · 〈S j 〉 − gµ B H · S i (2)<br />

〈ij〉<br />

i<br />

= −zJ ∑ (<br />

)<br />

∑<br />

m · S i + const. − gµ B H · S i (3)<br />

i<br />

i<br />

= − ∑ i<br />

(zJm + gµ B H) · S i (4)<br />

where m = 〈S i 〉 and z equals the number of nearest neighbour lattice sites.<br />

We saw in class that the magnetization for a spin-1/2 particle interacting with an effective field h is given<br />

by the solution to the equation<br />

m = 1 [ ] h<br />

2 tanh (5)<br />

2kT<br />

We know that the spins will align in the same direction as the magnetic field so that m and H point in<br />

the same direction.<br />

Then, in our system, the effective field is given by h = zJm + ζgµ B |H|, where ζ = sign(H). Therefore,<br />

the magnetization is given by the solution to the equation<br />

m = 1 [ ]<br />

zJm +<br />

2 tanh ζgµB |H|<br />

2kT<br />

(6)<br />

We also saw in class that the critical point for the Heisinberg ferromagnet in mean field theory is at kT c =<br />

zJ/4. Subbing this in for our temperature gives<br />

m = 1 [ ] 2(zJm +<br />

2 tanh ζgµB |H|)<br />

= 1 [<br />

zJ<br />

2 tanh 2m + 2ζgµ ]<br />

B|H|<br />

(7)<br />

zJ<br />

When H is small, we know that the magnetization m is also small near the critical point T c , since<br />

the magnetization is zero above the critical temperature and the magnetization is a continuous function of<br />

temperature for this model. Therefore we can Taylor expand our tanh(x) ≈ x − x 3 /3 function<br />

1


m = 1 [<br />

2m + 2ζgµ B|H|<br />

+ 1 (<br />

2m + 2ζgµ ) 3 ]<br />

B|H|<br />

(8)<br />

2 zJ 3<br />

zJ<br />

0 = ζgµ B|H|<br />

+ 1 (<br />

2m + 2ζgµ ) 3<br />

B|H|<br />

(9)<br />

zJ 6<br />

zJ<br />

( ) 1/3 6ζgµB |H|<br />

⇒ 2m ≈<br />

+ O(H) (10)<br />

zJ<br />

( ) 1/3 6gµB<br />

|H| 1/3 (11)<br />

m = sign(H) 1 2<br />

zJ<br />

Where above I expanded the equation above to lowest nonvanishing order in both m and H.<br />

Therefore<br />

M = gµ B m = sign(H) gµ ( ) 1/3<br />

B 6gµB<br />

|H| 1/3 (12)<br />

2 zJ<br />

2


Problem 2 - Quantum transverse-field Ising model<br />

Part (a)<br />

We start with the spin 1/2 Hamiltonian, and apply the mean field approximation:<br />

H = −J ∑ ∑<br />

Si z Sj<br />

z − h ⊥ Si x (13)<br />

〈ij〉<br />

i<br />

≈ −J ∑ ∑<br />

Si z 〈Sj z 〉 + 〈Si z 〉Sj z − 〈Si z 〉〈Sj z 〉 − h ⊥ Si x (14)<br />

〈ij〉<br />

i<br />

= ∑ (−zJm)Si z − h ⊥ Si x + const (15)<br />

i<br />

= − ∑ i<br />

h · S with h = (h ⊥ , 0, zJm) (16)<br />

where m = 〈Si z 〉. Therefore, our mean field Hamiltonian can be written as a set of spins in an effect field<br />

h eff = (h ⊥ , 0, zJm)<br />

Part (b)<br />

Now, we assume that we are at zero temperature and that each spin is aligned fully with the effect field<br />

h eff .<br />

Therefore, we now that S points in the direction (h ⊥ , 0, Jzm). We also know that |S| = S = 1/2. Then,<br />

we can write<br />

S i = 1 Jzmẑ + h<br />

√ ⊥ˆx<br />

(17)<br />

2 J 2<br />

z 2 m 2 + h 2 ⊥<br />

Since this is the proper normalized spin-1/2 vector which points in the h eff direction.<br />

Since all spins point in the same direction, the magnetization is just the z component of this spin vector<br />

m = 〈Si z 〉 = 1 Jzm<br />

√ (18)<br />

2 J 2<br />

z 2 m 2 + h 2 ⊥<br />

⇒ m 2 J 2 z 2 m 2<br />

=<br />

4(J 2 z 2 m 2 + h 2 ⊥ ) (19)<br />

0 = 4J 2 z 2 m 4 + (4h 2 ⊥ − J 2 z 2 )m 2 (20)<br />

Part (c)<br />

Therefore, for a given transverse field h ⊥ , the magnetization is given by the solution to the equation<br />

Solving this gives<br />

We see that for<br />

occurs when<br />

Therefore<br />

h 2 ⊥<br />

J 2 z 2<br />

4J 2 z 2 m 4 + 42h 2 ⊥ − J 2 z 2 )m 2 = 0 (21)<br />

m 2 (4J 2 z 2 m 2 + (4h 2 ⊥ − J 2 z 2 )) = 0 (22)<br />

⇒ m = 0 or m = ±√<br />

1<br />

4 − h2 ⊥<br />

J 2 z 2 (23)<br />

> 1 4 , the only solution to this equation is at m = 0. Therefore the critical field h c<br />

h2 ⊥<br />

J 2 z<br />

= 1 2 4 (i.e when h ⊥ = Jz/2). When h ⊥ > h c , we have that m(h ⊥ ) = 0<br />

1<br />

m(h ⊥ ) = ±√<br />

4 − h2 ⊥<br />

J 2 z 2 , for h ⊥ < h c where h c = Jz<br />

2<br />

(24)<br />

3


Problem 3 - Low Temperature Magnetization of the Heisenberg Ferromagn<br />

∫ ∞<br />

〈STOT〉/N z = +S − n B (ε)dε (25)<br />

ε=0<br />

(26)<br />

The number of magnons at a given temperature T is given by<br />

∑<br />

∫<br />

n k = dεD(ε)〈n(ε)〉 (27)<br />

k<br />

The factor of D(ε) is the number of frequency modes k which exist at a given energy ω. For a given<br />

wavevector k, we have the number of states within a sphere of radius k is N(k) = (2π) −3 ( 4 3 πk3 ). Then the<br />

number of states within a shell of thickness dε is:<br />

Therefore<br />

D(ε)dε = dN(k) dk<br />

dk dε dε = 1 dk<br />

4πk2 dε (28)<br />

(2π)<br />

3<br />

dε<br />

(29)<br />

〈S z TOT〉/N = S − 4π<br />

(2π) 3 ∫ kF<br />

k=0<br />

k 2<br />

dk (30)<br />

e βck2 − 1<br />

≈ S − 1 ∫ ∞<br />

k 2<br />

2π 2 dk<br />

0 e βck2 − 1<br />

(31)<br />

(32)<br />

where we make the approximation that the upper bound on the integral goes to infinity since n B (k) → 0<br />

exponentially fast as k becomes large.<br />

Now, let x = βck 2 So that dx = 2βck dk and we can write the integral above as<br />

〈S z TOT〉/N = S − 1<br />

2π 2 ∫ ∞<br />

0<br />

dx x 1<br />

√<br />

2βc βc e x − 1<br />

(33)<br />

= S − 1 ∫ ∞<br />

4π 2 c (kT x<br />

)3/2<br />

e x dx (34)<br />

− 1<br />

= S − 1 (kT ) 3/2<br />

12 c<br />

where in the last line I used the fact that the integral over x is equal to π 2 /6<br />

0<br />

(35)<br />

4

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