Маслов ИНТРОДУЦТИОН ТО ПХЫСИЦС ОФ СЕЦОНД-ОРДЕР МАГНЕТИЦ 2015

∂2 F (∆,T )

= 2a(T −T

)+12b∆2

.

(19.5)

∂∆2

C

А.

∆ = 0

∂2 F (∆,T )

= 2a(T −T

)=

> 0 at T

>TC ;

So,

∆ = 0 is

< 0 at T

<T .

∂∆2

C

C

the solution at

T >TC (minimum), and it is not the solution at T <TC

(maximum).

∂2 F (∆,T )

B. ∆ = ±

a

(T

−T )

= 4a(T −T )> 0

at T <T . So,

2b

∂∆2

C

C

C

T <TC is doubly degenerated. Therefore

(

T

)

C

∆

=

0 at T >T ;

(19.6)

∆

(T )−

at T <TC , T →TC−.

T −TC

By definition

dE (T )

C (T )=

.

(20.1)

At H = 0

dT

1

ˆ

(T )−

∑

/

Jij

≈

E (T )= H

= E0

σiz σjz

T

2 i, j

(20.2)

C (T )=

∂E

0 (T )

−

1

kBθN

∂(x2 (T ))

= C0

(T )+CS (T ),

(20.3)

∂T

2

∂T

x

= 3

θ−T

, thus x

2

T

. In this case

θ

= 3 1−

θ

CS

(T )= −

1

−

3

3

kB N.

(20.4)

2

kBθN

θ

=

2

Abrupt jump of heat capacity at T =θ

is concerned with the appearance

new degrees of freedom in the system. So, at T > θ we have

µˆ iz = 0

ciated with it, on an average, energy of

1 kBT . At T →0 (T θ) the

2

=1− 2e−2

θ

≈1− 4e−2

θ

order parameter is

x

T

, thus x2

T

. In this case

k

θN

−2

θ

1

θ2

−2

θ

CS (T )= −

B

−4e

T (−2θ)

−

= 4kB N

e

T . (20.5)

T

2

T

2

2

Therefore CS (T ) exponentially tends to zero at T →0 .

Generally speaking,

C0 (T )≠ 0

thus

total

heat

capacity

C (T )= C0 (T )+CS (T )

takes the form as it is shown on Fig. 20.1, i.e. at

T = θ the abrupt jump takes place ∆C =

3 k

B

N.

2

Recall

the definition of the differential magnetic

susceptibility

χ = (∂M ∂H )H →0 . In

paramagnetic

state (T > θ) the

magnetization

M = 0 at

H = 0 and

M ≠ 0 only at

H ≠ 0 . In this case “conventional”

Earlier we showed that at

T θ magnetic susceptibility could be de-

scribed as χ =

NµB2

1

(Curie-Weiss law). Let us find the expression

V

T −θ

regard to magnetic field

x = tanh

h + xθ

. Since in this equation we

T

have h rather than H , let us make the following conversion

∂M

Nµ2

∂x

χ =

=

B

.

(21.1)

∂H

V

∂h

h→0

Considering that we do not know the dependence x(h)

let us differenti-

ate the both sides of the Curie-Weiss equation

∂x

2 h + xθ

1

θ ∂x

1

− x2

θ

(1− x

2

)

∂x

= 1− tanh

+

=

+

;

∂h

T

T ∂h

T

T

∂h

T

∂x

1

−

θ

(1− x2 ) = 1− x2

;

(21.2)

∂h

T

T

∂x

=

1

− x2

.

∂h

(

)

T −θ 1− x

2

∂tanh z

=

∂ sinh z

=

cosh2 z −sinh2 z

=1− tanh

2

z.

∂z

cosh

2

z

∂z cosh z

Therefore the expression for χ(T ) takes the following form

χ =

NµB2

1− x2

.

(21.3)

V

T −θ 1− x2

)

(

xθ

. We already know this solution and in limiting cases at

x = tanh

T

A. At T > θ the order parameter is x = 0 χ =

NµB2

1

. Thus we

V

T −θ

T

B. At T → θ−

the order parameter is

x

= 3

θ−T

x

2

−

= 3 1

.

In this case

T

θ

θ

χ =

NµB2

3 θ − 2

≈

NµB2

1

=

NµB2

1

.

V

T

T

− 2

V T

−

3T

+ θ

V 2

(

θ−

T )

−θ 3

θ

2

So, at T → θ−

magnetic susceptibility is also diverged, and the charac-

χ(T )−

1

at T → θ.

(21.4)

T −θ

≈1− 2e−2

θ

x2 ≈1− 4e−2

θ

x

.

In this case

T

T

NµB2

4e−2

θ

NµB2 4e−2

θ

χ =

T

≈

T

.

T

V

T −θ

4e

−2

θ

V

T

pacity, in the mean-field approximation it has the abrupt

jump (see

§ 20). Critical exponents α , β and γ are denoted as follows

C (T )−

T −θ

−α ;

∆(T )−(T −θ)β ;

(22.1)

χ(T )−

T −θ

−γ .

Critical exponent

Mean-field

Numerical

Experiment

approximation

calculations

α

0

0,12

≈ 0,1

β

0,5

0,31

0,3 ÷0,4

γ

1

1,25

1,2 ÷1,4

α + 2β + γ

2

1,99

1,9 ÷2,3

α + 2β+ γ = 2 .

(22.2)

Recall that the

Curie temperature, by definition, is

θ = ∑J (Rl )= zJ, where

z is the number of nearest neighbors of each

Rl ≠0

= −

1

∑

/

(23.1)

H

Jij σiz σjz

− h∑σiz ,

2 i, j

i

µiz

,

h = µB H , and

σiz

= ±1 are the eigenvalues of opera-

where σiz =

µB

(23.2)

H

= −J σ1z σ2z − J σ2z σ3z −... − J σN −1,z σNz − h(σ1z +... + σNz ).

Since

N 1 the boundary conditions are not essential. Let us make

the following term

−J

σNz σ1z . Next, let us calculate the partition func-

ˆ

En

−H

∑

−

=

e T

, where n are the numbers of the Hamiltoni-

tion Q =Tr e

T

n

Every state “n” is defined by the set of N numbers σiz

each is equal

+1 or –1, and ∑ is the sum over all number sets {σiz }

n

ˆ

H

Q = ∑

{σiz }

e− T

{σiz } .

(23.3)

{σiz }

i =1,..., N ,

The Hamiltonian

is

the function

of

operators

,

σiz

ˆ

). Since

{σiz }

= σkz

{σiz }

k , then

H

= H (σ1z ,σ2z ,...,σNz

σkz

ˆ

{σiz }

= H (σ1z ,σ2z ,...,σNz ){σiz } .

H

Therefore

Nz )

ˆ

H

σ

,σ

2 z

,...,σ

H

( 1z

e− T

{σiz } = e−

{σiz } .

T

Considering these expressions, for the partition function we obtain

Q = ∑

{σiz }

e−

H (σ1z ,...,σNz )

{σiz }

= ∑e−

H (σ1z ,...,σNz )

=

T

T

{σiz }

{σiz }

(23.4)

H (σ1z ,...,σNz )

= ∑ ∑

... ∑ e

−

,

T

J

J

J

Q = ∑

... ∑ exp

σ1zσ2z

+

σ2zσ3z +... +

σN −1,zσNz +

T

T

σ1z =±1 σNz =±1

T

J

h

h

σ2z +... +

h

+

σNzσ1z +

σ1z +

σNz

=

T

T

T

T

J

h

J

= ∑ ... ∑ exp

σ1z σ2z

+

(σ1z + σ2z )+

σ2z σ3z

+

2T

T

σ1z =±1 σNz =±1

T

+

h

(σ2z

+ σ3z )+... +

J

σN −1,z σNz +

2T

T

+

h

(σN −1,z + σNz )+

J

σNz σ1z +

h

(σNz + σ1z ) .

T

2T

2T

Denote the 2 ×2 matrix A

A =

A

A

(23.5)

1,1

1,−1

,

A

A

−1,1

−1,−1

where the matrix elements are

Aσ

,σ

= exp

J

σ1z σ2z +

h

(σ1z + σ2z ) .

2 z

1z

2T

Thus the matrix takes the following form

T

J

+

h

−

J

eT

T

e

T

(23.6)

A =

J

J

h

.

e−

e

−

T

T

T

Denote the other matrix elements Aσ

2 z

,σ

,..., Aσ

Nz

,σ

in the same way.

For Q we obtain

3 z

1z

Q = ∑ ... ∑ Aσ1z ,σ2 z

Aσ2 z ,σ3 z ,..., AσN −1,z ,σNz

AσNz ,σ1z .

(23.7)

σ1z =±1 σNz =±1

Recall the rule of matrix multiplication (A2 )kn = ∑Akm Amn . Therefore

∑ Aσ1z ,σ2 z

= (A2 )σ

m

Aσ2 z ,σ3 z

,σ

3 z

,

σ2 z =±1

1z

∑ (A2 )σ

,σ

3 z

Aσ3 z ,σ4 z

= (A3 )σ

,σ

, …,

σ3 z =±1

1z

1z

4 z

∑ (AN −1 )σ

,σ

AσNz ,σ1z = (AN )σ

,σ .

σN ,z =±1

1z

Nz

1z

1z

Finally, for the partition function we obtain

Q = ∑ (AN )σ

,

σ .

(23.8)

σ1z =±1

1z

1z