Материал: Маслов ИНТРОДУЦТИОН ТО ПХЫСИЦС ОФ СЕЦОНД-ОРДЕР МАГНЕТИЦ 2015

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namely

+1,+1,+1

+1,+1 ≡

↑,↑,↑ ↑ ↑,↑ . The

" "

slightest excitement will be

−1

+1,+1

↑,↑,↑ ↓ ↑,↑ ,

+1,+1,+1

≡

" "

+1,	i ≠ k;	Therefore for the energy			E1 we obtain
i.e., σiz =	i = k.	Therefore for the energy			E1 we obtain
−1,	i = k.
	E1 = − 1 ∑			N
	E1 = − 1 ∑		/ Jij σiz σjz	= −1 ∑Jij σiz σjz =
		2 i, j		2 i, j=1
				i≠ j
= −	1 ∑Jij σiz σjz		− 1 ∑J j σ zσjz −		1 ∑Ji	σizσ z =
	2 i≠ j		2 j≠		2 i≠
	ij≠≠	+1		−1		−1
	ij≠≠
= −1	∑Jij	+ 1 ∑J j + 1 ∑Ji = − 1			∑Jij −	1 ∑J j −
2 i≠ j		2 j≠	2 i≠	2 i≠ j		2 j≠
	ij≠≠				ij≠≠
− 1 ∑Ji + ∑J j + ∑Ji = − 1 ∑/ Jij + 2∑J j .
2 i≠		j≠	i≠	2 i, j		j≠

Finally, for the energy of the first excited state we find

E1 = E0 + 2θ.

(15.9)

(15.10)

Thus, the energy spectrum has the gap, i.e., the first excited level of energy is separated from the ground level by 2θ.

16. Free energy of a ferromagnet in the Ising model

Helmholtz free energy is defined as F = E −TS , where E is the internal energy of the system and S is the entropy. Note that there is heat exchange with the environment (T = const ), the free energy must be

minimal. The internal energy can be described as E = Hˆ . Taking into

account

that

Hamiltonian

the

Ising

model

∑

, for internal energy we obtain

H = −

Jij σiz σjz − h∑σiz

2 i, j

E = −

∑

Jij

− h∑

(16.1)

σiz σjz

σiz .

2 i, j

Considering the mean-field approximation (see § 11) we can rewrite the mean of the product as the product of the means σiz σjz ≈ σiz σjz .

Taking into account that σiz = x is the order parameter, we obtain for the internal energy

E = −	1 ∑/ Jij x2 − h∑x = −1 x2				∑∑Jij − xh∑1 =
	2 i, j		i	2	i j≠i	i	(16.2)
			x2				(16.2)
		= −	x2	Nθ− xhN.
		= −	2	Nθ− xhN.
			2
Let us find the entropy		S = ln Γ , where			Γ is the number of possible

states of the system at a given temperature (here and after we omit the Boltzmann constant). Denote N↑ and N↓ as the number of magnetic

moments aligned along or against the selected axis z, respectively.

Therefore the total number	of magnetic moments is equal to
N = N↑ + N↓ . The difference	N↑ − N↓ defines the total magnetic mo-

ment of the system, i.e., magnetization:

µB N↑ +(−µB )N↓ = µB (N↑ − N↓ ).

On the other hand, the total magnetic moment of the system can be de-

Actually, if N↑ = N↓ , then x = 0 , therefore the order is missing; on the

	N	↑	= N, N	↓	= 0 x =1;
other hand, if							then the “apple-pie order”
			= N, N		= N	x = −1,
	N	↓		↑

is presented. Since x = x(T ), the difference N↑ − N↓ depends on tem-

perature as well. This fact imposes a limitation on Γ, i.e., the number of possible states of the system at a given temperature implies the number of possible states of the system at a given difference N↑ − N↓ . Let us

find the Γ

Γ = CNN↑ =		N !	=	N !	= CNN↓ .	(16.4)
	N↑	!(N − N↑ )!		N↑ !N↓ !

Taking into account the Stirling’s formula ln(n!)≈ nln n −n we obtain for the entropy

S = ln Γ = ln (N !)−ln (N↑ !)−ln (N↓ !)≈

N↑

N↓

≈ N ln

−

N↑ ln

− N↓ ln

= N↑

N↑

+ N↓

N↓

(16.5)

−ln

= N

+ N

= N

N↑

N↓

↑

↓

↑

↓

↑

= −N

↑

↓

Express

N↑

and

N↓

by the order parameter x

N↑ − N↓

N↑

−

N↓

N − N↓

−

N↓

=1− 2

N↓

(16.6)

Therefore

N↓

N↑

− x

;

1+ x

(16.7)

Finally, for the entropy we obtain

S = −N

1+ x

1− x

(16.8)

Note that at x = 0 (no order), S = N ln 2 , so, Γ = 2N – every magnetic moment has an arbitrary direction. Ultimately the Helmholtz free energy is described as

	x2		1+ x		1+ x		1− x		1− x		(16.9)
F = N −		θ− xh +T		ln		+T		ln		.
	2		2		2		2		2

In general, we must include in the Eq. (16.9) the term F0 (T, H ), which

does not depend on the order parameter, i.e., it is not associated with magnetic subsystem, but includes phonon contribution, etc.

So, we obtained F = F (x,T, H ). Let us find F = F (T, H ). To do this, we should find x = x(T, H ) from the conditions of F = F (x,T, H )

minimum at T = const and H = const . So, let us find

∂F (x,T, H )

∂F0

(T, H )

∂x

considering that

= 0

∂x

∂F (x,T, H )

+ x

− x

= N −xθ− h +

−

∂x

(16.10)

+ x

= N −xθ− h +

− x

From the condition on the extremum ∂F (x,T, H ) = 0 , we obtain

∂x

T				1+ x	= xθ+ h;
2	ln
				1− x
	e	2	(xθ+h) =		1+ x	;
		T
					1− x

		(1− x)e			2	(xθ+h) =1+ x;
		(1− x)e			T	(xθ+h) =1+ x;
	2	(xθ+h)					2	(xθ+h)		(16.11)
eT		(xθ+h)	−1					(xθ+h)	;	(16.11)
eT			−1	= x 1+ eT					;

2(xθ+h)

x = eT2 (xθ+h) −1;

eT +1

x = tanh xθT+ h .

Finally, we obtain the Curie-Weiss equation for the order parameter. We already know that this equation for H = 0 and T > θ has only one solu-

tion	x = 0 , and for H = 0 and T < θ there are three solutions x = 0 ,
x > 0	and x < 0 , i.e.,	∂F (x,T, H )	= 0	at three different		x . To obtain

		∂x			∂2 F (x,T )
the minimum, let us find the second derivative						(next, we as-
					∂x2

sume that the magnetic field is zero). Recall that the condition on the

minimum is	∂2 F (x,T )	> 0
minimum is	∂x2	> 0
	∂x2
	∂2 F (x,T )			T 1	− x (1			− x)+(1+ x)
		= N	−θ+					(1− x)2	=
	∂x2	= N	−θ+	2 1	+ x				=
	∂x2			2 1	+ x				(16.12)
				T					(16.12)
				T
		= N	−θ+				.
		= N	−θ+	1− x		2	.
				1− x

Let us check the results obtained. At T > θ we have only one solution

x = 0 , thus	∂2 F (x,T )	= N [−θ+T ]> 0 , therefore this is minimum. At
	∂x2

Источник: https://studfile.net/preview/16708691/

scribed as

N = xµB N . Thus, we can rewrite the order

N = µB σiz

parameter in the following form

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