Маслов ИНТРОДУЦТИОН ТО ПХЫСИЦС ОФ СЕЦОНД-ОРДЕР МАГНЕТИЦ 2015

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scribed as

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

N = µB σiz

parameter in the following form

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we obtain M ≠ 0 (ferromagnetic phase). The value θ is called the Curie temperature or transition temperature into ferromagnetic state.

The physical basis of ferromagnetic transition is the exchange interaction, i.e., it is energetically favorable for magnetic moments to align along one direction and to decrease the energy of the system as well as Helmholtz free energy. Magnetic moments alignment leads to the order appearance in the system, i.e., the entropy decreases. If the temperature is high, then though the intrinsic energy of ordered (ferromagnetic) phase is lower than the intrinsic energy of disordered (paramagnetic) phase, the entropy of disordered phase is higher than the entropy of ordered phase. So, the Helmholtz free energy of disordered phase at high temperatures is lower than the corresponding value of the ordered phase.

Denote the order parameter x as the mean magnetic moment of at-

		µiz
om x = σiz	=		. In the case of the order absence, x = 0 due to the
		µB

magnetic moments have the chaotic orientation, while the mean magnetic moment is the total moment divided by the number of moments. If the magnetic moments have the preferred orientation, then x ≠ 0 and the order is presented.

14. Temperature dependence of the ferromagnetic order parameter in the Ising model

The Curie-Weiss equation for order parameter at zero magnetic field takes the following form

				θ
			x = tanh x		,			(14.1)
			x = tanh x		,			(14.1)
				T

where θ = ∑J (Rl )	is the Curie temperature, and							µiz	for
						x = σiz	=
						x = σiz	=	µB
l≠0								µB
any i =1,..., N . Since			= −µB ÷µB , therefore −1≤ x ≤1				. Recall that
any i =1,..., N . Since		µiz	= −µB ÷µB , therefore −1≤ x ≤1				. Recall that
x = 0 at T > θ, and	x ≠ 0 at T < θ. Magnetization is described as

	N
M =		µiz	=	NµB	x = M0 x ,	(14.2)
		V
				V

where M0 = NVµB is the maximum possible value of magnetization

(i.e., all magnetic moments have the same direction). It is clear that magnetization is equal to zero at T > θ. Let us analyze the temperature dependence of magnetization at temperatures below the Curie tempera-

ture. At T θ, i.e., T → 0 we can expand		θ	in a series for
	tanh x

		T

large values of argument z = x

(we assume

x > 0 for clarity)

tanh z =

sinh z

−e−z

− 2e−z

≈

cosh z

+ e−z

+ 2e−z

(14.3)

≈ 1− 2e−z

1−

2e−z

)

≈1−

2e−2z .

(

)(

Therefore for the order parameter we obtain

x ≈1− 2e−2x

, x > 0.

(14.4)

Let us try to solve this equation iteratively

x(0)

=1;

(14.5)

=1− 2e−2x

(0)

x(1)

=1− 2e−2

From the Curie-Weiss equation one can see that if there is a solution at x > 0 , then there is a solution at x < 0 . Therefore

−2 θ

x =1− 2e T .

Thus, the magnetization is
	M	= M0		−2	θ
				−2
			1− 2e		T .

One can see that if T →0 , then M → M0 .

(14.6)

(14.7)

In the case of T < θ and T → θ−

we can expand

tanh x

in a

series for small values of argument z

= x

tanh z = sinh z

. (14.8)

≈ z 1+

1−

≈ z

1−

cosh z

Therefore for the order parameter we obtain

θ2

x = x

1−

(14.9)

Finally, the order parameter can be described as

= 3

θ−T

(14.10)

and magnetization can be described as

= M0

θ−T

(14.11)

Resulting characteristic temperature dependencies for order parameter and magnetization are shown on Fig. 14.1.

Let us analyze the results obtained. The order parameter and magnetization identically equal zero at T ≥ θ. At temperatures T < θ both x and M increase monotonically with temperature decreasing to values 1 and M0 , respectively. In other words, in the system of magnetic mo-

ments the order occurs under cooling below T = θ, and this order increases as cooling increases:

In other words, the temperature decreasing leads to orientating the magnetic moments along one direction. So, order parameter and magnetization are the quantitative measures of ordering.

Fig. 14.1. Order parameter (left) and magnetization (right) dependencies on temperature for ferromagnets at H = 0

15. Ground and excited states of a ferromagnet in the Ising model

Recall the Hamiltonian in the Ising model in the absence of magnetic field

∑

(15.1)

H = −

Jij σiz

σjz .

2 i, j

At T = 0 the energy of the system is minimal E = E0 , where

E0 is the

energy of the ground state. If

Ψ0

is the ground state, then the follow-

ing expression is right

Ψ0

= E0

Ψ0 .

(15.2)

It is common knowledge that any state of system of magnetic moments is defined by set of numbers σ1z ,σ2z ,...,σNz which are equal to ±1. The

total amount of these numbers {σnz } is equal to

N . What should be

these numbers for the ground state? Recall that

{σnz } = σiz

Ψ .

(15.3)

σiz

Ψ = σiz

Therefore

Ψ = σiz σjz

(15.4)

σiz σjz

Ψ = σizσjz

thus,

∑

Jij

Ψ =

Ψ = −

σiz σjz

2 i, j

(15.5)

Ψ = E

Ψ .

−1 ∑/ Jij σizσjz

2 i, j

nz }

{

is described as

So, energy of the state with arbitrary set of σ

E = − 1 ∑/ Jij σiz σjz .

(15.6)

2 i, j

(i, j). Thus

Since both σiz and σjz

are equal to ±1, then σizσjz = ±1

the energy will be minimal if

σiz = +1 and

σjz

= +1, or

σiz = −1 and

σjz = −1 (i, j). As this takes place, the energy of the ground state is

described as		1 ∑/ Jij .
E0	= −	1 ∑/ Jij .	(15.7)
		2 i, j

In the “language of arrows” σiz = +1 is equivalent to ↑, and σiz = −1 is equivalent to ↓. Hence the ground state of the system is doubly degen-

erated: a)	↑↑↑...↑↑	or b)	↓↓↓...↓↓	. In other words, all magnetic

	+1 +1 +1 +1 +1		−1 −1 −1 −1 −1

moments have the same direction and aligned along or against the se-

lected axis. Since ∑/

Jij = ∑∑J (Rl )= Nθ, the energy of the ground

i, j

i=1 R

≠0

((

state is

E = −1

Nθ.

(15.8)

Ψ0

+1,+1,+1... +1,+1

Ground state is

equal

Ψ0 =

−1,−1,−1... −1,−1 .

Ψ1

. Let us find its energy

E1 .

Now, consider the first excited state

For the sake of definiteness, we shall choose one of the ground states,

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

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