S 1 ,S 2 ,...,S N

13 Task Theoretical Physics VI - Statistics

13.1 (Exact solution of the Ising model in

d = 1

⁾

a)

Werstwantto show,thatthepartitionfunction canbewritten as:

Z _N (T, B 0 ) = X

S 1 ,S 2 ,...,S N

e ^{β P i (gµ B B 0 S i +JS i S i+1 )}

= X

S 1 ,S 2 ,...,S N

τ (S 1 , S 2 ) τ (S 2 , S 3 ) . . . τ (S N , S 1 )

theHamiltonianisgivenwith

H ˆ = −J

N− 1

X

i=1

S ˆ i S ˆ i+1 − gµ B B 0 N

X

i=1

S ˆ i

Usingthedenition ofpartitionfunction ():

Z _N = T r(e ^−βH )

= X

S 1 ,S 2 ,...,S N

e ^{β P i (gµ B B 0 S i +JS i S i+1 )}

= X

S 1 ,S 2 ,...,S N

e ^{β(gµ B B 0 S 1 +JS 1 S 2 )+β(gµ B B 0 S 2 +JS 2 S 3 )+...+(gµ B B 0 S N +JS N S N+1 )}

= X

S 1 ,S 2 ,...,S N

e ^β ( ¹ 2 gµ B B 0 (S 1 +S 2 )+JS 1 S 2 ) ^+β ( ¹ 2 gµ B B 0 (S 2 +S 3 )+JS 2 S 3 ) ^+...+ ( ¹ 2 gµ B B 0 (S N +S 1 )+JS N S N+1 )

Thisrearrangementwillprovetobeuseful. Nowwewanttousetheperiodic

boundarycondition

S ˆ N +1 = ˆ S 1

^{and wefurther denethefunction:}

τ (S i , S i+1 ) = exp

β

J S _i S i+1 + 1

2 gµ _B B 0 (S i + S i+1 )

whichleadsusdirectlyto:

Z _N = X

S 1 ,S 2 ,...,S N

e ^{β P i (gµ B B 0 S i +JS i S i+1 )}

= X

S 1 ,S 2 ,...,S N

e ^β ( ¹ 2 gµ B B 0 (S 1 +S 2 )+JS 1 S 2 ) ^+β ( ¹ 2 gµ B B 0 (S 2 +S 3 )+JS 2 S 3 ) ^+...+ ( ¹ 2 gµ B B 0 (S N +S 1 )+JS N S 1 )

= X

S 1 ,S 2 ,...,S N

τ (S 1 , S 2 ) τ (S 2 , S 3 ) . . . τ (S N , S 1 )

TheTransfermatixisdenedwith:

T =

τ ¹ ₂ , ¹ ₂

τ − ¹ ₂ , ¹ ₂ τ ¹ ₂ , − ¹ ₂

τ − ¹ ₂ , − ¹ ₂

thisfollowsfromthe4possiblespincombinations,whichevery

S i , S i+1

^can

have.

Wearemeantto show,that

X

S i ,S j ,S k

τ (S i , S _j ) τ (S j , S _k ) = X

S i ,S j

T ²

S i ,S k

with

T ²

S i ,S k = P

S j τ (S i , S _j ) τ (S j , S _k )

^{. Wecansimplywrite:}

X

S i ,S j ,S k

τ (S i , S _j ) τ (S j , S _k ) = X

S j

X

S i ,S k

τ (S i , S _j ) τ (S j , S _k )

= X

S i ,S j

T ²

S i ,S k

c)

Wecandenethespinstates

|S i = 1 2 i =

1 0

|S i = − 1 2 i =

0 1

andwiththoseweget:

1 2 |T | − 1

2

= 1 0

τ ¹ ₂ , ¹ ₂

τ ¹ ₂ , − ¹ ₂ τ ¹ ₂ , − ¹ ₂

τ − ¹ ₂ , − ¹ ₂ 0 1

= τ 1

2 , − 1 2

orgenerally

hS i |T |S i+1 i = τ (S, S i+1 )

Thiscanbeinsertedinthepartitionfunction:

Z N = X

S 1 ,S 2 ,...,S N

τ (S 1 , S 2 ) τ (S 2 , S 3 ) . . . τ (S N , S 1 )

= X

hS 1 |T |S 2 i hS 2 |T |S 3 i . . . hS N |T |S 1 i

Z N = X

S 1

S 1 |T ^N |S 1

= T r T ^N

the trace denition. While the trace is the sum of the eigenvalues of a

matrixandthisoneis2x2wegettwodierenteigenvalues

λ 1,2

^for

T

^,usingthe

multiplicitythisleadsto

Z N = T r T ^N

= λ ^N ₁ + λ ^N ₂

Togettheeigenvaluesof

T

^,werstwrite

T

^{in fullform:}

T =

exp

β ^J ₄ + ¹ ₂ gµ B B 0

exp

β − ^J ₄ exp

β − ^J ₄

exp

β ^J ₄ − ¹ ₂ gµ _B B 0

Calculatingtheeigenvalues

λ 1,2

^{ofthetransfermatrix}

det |T − λE| = 0 =

exp

β ^J ₄ + ¹ ₂ gµ B B 0

− λ exp

β − ^J ₄ exp

β − ^J ₄

exp

β ^J ₄ − ¹ ₂ gµ _B B 0

− λ

leadingto

0 =

exp

β J

4 − 1 2 gµ B B 0

− λ exp

β J

4 + 1 2 gµ B B 0

− λ

− exp

2β

− J 4

0 = exp βJ

2

− λ exp

β J

4 − 1 2 gµ _B B 0

− λ exp

β J

4 + 1 2 gµ _B B 0

+ λ ² − exp

− βJ 2

0 = λ ² − λ exp βJ

4 exp

β 2 gµ _B B 0

+ exp

− β 2 gµ _B B 0

+ exp

βJ 2

− exp

− βJ 2

0 = λ ² − 2 exp βJ

4

cosh β

2 gµ _B B 0

λ + 2 sinh βJ

2

meaningtheeigenvaluesare:

λ 1,2 = exp βJ

4

cosh β

2 gµ B B 0

± s

exp βJ

2

cosh ² β

2 gµ B B 0

− 2 sinh βJ

2

λ 1,2 = exp βJ

4

cosh β

2 gµ B B 0

± s

cosh ² β

2 gµ B B 0

− 2 exp

− βJ 2

sinh

βJ 2

!

d)

We are considering the limit

λ 2

λ 1 1

^{which means}

λ 1 λ 2

^{, meaningwecan}

neglect

λ 2

^{andonlyneedtoconsider}

λ 1

^{. Usingfromc)}

Z _N (T, B 0 ) = λ ^N ₁ + λ ^N ₂

≈ λ ^N ₁

=

"

exp βJ

4

cosh β

2 gµ B B 0

+

s cosh ²

β 2 gµ B B 0

− 2 exp

− βJ 2

sinh

βJ 2

!# ^N

Denitionoffreeenergy:

F (T, B 0 ) = −k B T ln Z N (T, B 0 )

= −N J

4 ln cosh β

2 gµ _B B 0

+

s cosh ²

β 2 gµ _B B 0

− 2 exp

− βJ 2

sinh

βJ 2

!

Wecan look at specialcases

β → ∞

^{, meaning}

T → 0

^{thereforethe expo-}

nentialtermwill vanishandweget:

F (T, B 0 ) = −N J 4 ln

cosh

β 2 gµ _B B 0

+ cosh

β 2 gµ _B B 0

= −N J 4 ln

2 cosh

β 2 gµ _B B 0

= −N J 4

ln 2 + ln

cosh β

2 gµ _B B 0

Ortheotherspecialcase

B 0 → 0

^{,wethengetusing onlyrstorder taylor-}

expansion(

cosh (x) = 1 + ^x ₂ ² + . . .

^):

Z _N (T, 0) ≈

"

exp βJ

4

1 + s

1 − 2 exp

− βJ 2

sinh

βJ 2

!# ^N

=

"

exp βJ

4

1 + s

1 − exp

− βJ 2 exp

βJ 2

− exp

− βJ 2

!# ^N

=

exp βJ

4

1 + p

1 − 1 + exp (−βJ ) N

=

exp βJ

4

+ exp

− βJ 4

N

=

2 cosh βJ

4 N

andafreeenergyof

F (T, 0) = −k B T ln Z _N (T, 0)

Magnetization:

M (T, B 0 ) = 1 β

∂

∂B 0

ln Z N (T, B 0 )

= 1 β

∂

∂B 0

ln λ ^N ₁

= N

β 1 λ 1

∂

∂B 0

λ 1

= N

βλ 1

exp βJ

4 ∂

∂B 0

cosh β

2 gµ _B B 0

+

s cosh ²

β 2 gµ _B B 0

− 2 exp

− βJ 2

sinh

βJ 2

!

CASmeaningmathematica:

M (T, B 0 ) = β

2 gµ B sinh β

2 gµ B B 0



1 + exp

− ^βJ ₄ cosh

β 2 gµ _B B 0

ζ cosh

β 2 gµ _B B 0

+ ζ





with

r

exp (−βJ ) + cosh ²

β 2 gµ B B 0

− 1 = ζ

^{. Whiletheresultof}

M

^seems

tobewrongweskiptheisothermalsusceptibility:

χ T (B 0 ) = ∂M

∂B 0

skipped