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Маслов ИНТРОДУЦТИОН ТО ПХЫСИЦС ОФ СЕЦОНД-ОРДЕР МАГНЕТИЦ 2015

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T < θ the

solution x = 0 is

maximum.

Actually,

in

this case

2 F (x,T )

 

= N [θ +T ]< 0 . For

x 0 let us

consider

two

cases for

x2

 

 

 

 

 

which we know the approximate analytical expressions for order parameter: 1) T < θ and T → θ; 2) T θ.

1. We earlier obtained (see § 14) that at T < θ and T → θthe order

parameter was x

 

 

T

. Therefore

 

 

 

 

 

 

3 1

θ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 F (x,T )

 

 

 

 

 

 

 

T

 

 

 

 

 

T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

N −θ+

 

 

 

 

 

 

 

= N −θ+

 

 

 

 

=

 

x2

 

 

 

 

 

 

 

T

 

 

 

 

 

 

1

 

T

 

3

2

 

 

 

 

 

 

 

 

3 1

 

θ

 

 

 

 

 

 

 

 

 

 

 

 

θ

 

 

 

 

(16.13)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=N 3T + 2θ+T 2N (θ−T )> 0.3Tθ 2

So, this is the minimum.

2. We earlier obtained (see § 14) that at T θ the order parameter

was

 

 

 

12e2

θ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x

 

 

. Therefore

 

 

 

 

 

 

 

 

 

 

 

 

 

 

T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

F (x,T )

 

 

 

T

 

 

 

 

 

T

 

 

 

 

 

 

 

 

 

 

= N −θ+

 

 

 

 

 

 

= N −θ+

 

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

θ

 

 

 

θ

 

 

 

 

 

 

 

 

x2

 

1

 

2

 

 

4e

2

 

 

 

 

 

 

 

 

 

 

 

 

(16.14)

 

 

 

 

 

 

 

 

 

 

 

14e

 

T

 

 

T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

T 2 θ

=N −θ+ 4 e T > 0.

So, this is the minimum as well.

It can be shown that at any given temperature 0 T < θ solution of the Curie-Weiss equation with x 0 corresponds to local minimum of

F = F (x,T ), and solution with x = 0 corresponds to local maximum.

41

Finally, we obtain that at T > θ the paramagnetic state is realized, and at T < θ the ferromagnetic state occurs.

17. Free energy of a ferromagnet near the critical temperature

We found the Helmholtz free energy

F = F (x,T, H ) at arbitrary T

and H :

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

F = F (T, H )+ N

 

 

x2

 

θ− xh +T 1+ x ln 1+ x

+T 1x ln 1

x .

 

 

 

0

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

2

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(17.1)

The order parameter

 

x = 0 at T > θ and

 

 

 

x

 

1 at T < θ and T → θ.

 

 

 

Therefore we can approximately expand

 

 

F = F (x,T, H )

into Taylor

series at T ≈ θ up to O(x4 )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1+ x

 

 

 

 

 

+ x

 

 

 

 

 

 

 

 

1+ x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+ x

 

 

 

 

 

 

 

 

1

 

 

=

 

 

ln

(1+ x)

1

ln 2

 

 

 

 

 

2

 

ln

 

2

 

 

 

 

 

2

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 (1

+ x) x

 

x2

 

+

x3

 

 

x4

 

1 (1+ x)ln 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

3

 

 

 

 

 

 

 

4

 

 

 

 

2

 

 

 

 

 

(17.2)

1

 

 

 

x

2

 

 

 

 

x

3

 

 

 

 

x

4

 

 

 

 

 

 

 

 

 

 

 

x

3

 

 

 

 

 

 

x

4

 

 

 

1

(1+ x)ln 2 =

 

 

 

 

 

 

 

 

 

 

 

+ x2

 

 

 

 

 

 

 

 

 

x

 

 

+

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

2

 

 

2

 

3

 

 

 

4

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

x

2

 

 

 

 

 

x

3

 

 

 

x

4

 

 

 

 

 

 

 

1

(1+ x)ln 2.

 

 

 

 

 

=

x +

 

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

2

2

 

 

6

 

 

 

 

 

 

 

2

 

 

Analogously,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

12

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1x

 

1x

 

 

 

 

1

 

x +

 

x2

 

 

 

 

x3

 

 

 

 

 

x4

 

1

(1x)ln 2.

(17.3)

 

 

ln

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

+

 

 

 

 

 

 

2

 

2

 

 

2

2

 

 

 

 

6

 

 

12

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Combine

42

 

 

 

1+ x

 

 

1+ x

 

 

 

1x

 

1x

 

 

 

 

 

 

 

2

 

ln

2

 

+

2

 

ln

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

x2

 

x3

 

 

x4

 

1

(1+ x)ln 2 +

(17.4)

 

 

 

x +

 

 

 

 

+

 

 

 

 

 

 

2

2

6

12

2

 

1

x +

x2

x3

x4

 

1

(1x)ln 2 =

x2

x4

 

+

2

 

+

 

 

+

 

 

 

 

+

 

ln 2.

 

2

 

6

12

2

2

12

 

For the Helmholtz free energy we obtain

F = F

(T, H )+ N

 

x2

θ− xh +T

x2

+T

x4

 

,

 

 

 

 

0

 

2

2

12

 

 

 

 

 

 

 

where F0 (T, H )= F0 (T, H )T ln 2 . And finally,

F (x,T, H )

=

F0 (T, H )

+

1

(T −θ)x

2

+

 

T

x

4

xh.

N

N

2

 

12

 

 

 

 

 

 

 

 

 

In order not to increase accuracy, let us assume that T = θ

T

x4

 

 

 

 

 

 

 

 

 

 

 

 

12

 

 

 

 

 

 

 

 

 

 

 

 

 

F (x,T, H )

 

F0 (T, H )

 

 

 

 

 

 

 

 

 

 

 

 

=

+

1

(T −θ)x

2

+

 

θ

x

4

xh.

 

 

N

N

2

 

12

 

 

 

 

 

 

 

 

 

 

 

(17.5)

(17.6) in the term

(17.7)

The Eq. (17.7) is correct at T > θ (x = 0) and at T → θ. Let us analyze graphically the dependence F (x) at various T and H .

1. At T > θ and H = 0 , F (x) has the minimum at x = 0 (see Fig. 17.1), i.e., the atoms do not have the mean magnetic moments.

2. At T > θ and H > 0 , F (x)

has the minimum at x > 0 due to the

term xh (see Fig. 17.2).

H < 0 , F (x) has the minimum at

3. Analogously, at T > θ and

x < 0 (see Fig. 17.3).

 

4. At T < θ and H = 0 , F (x)

has two minima (see Fig. 17.4) divid-

ed by barrier. Thus, the ground state is doubly degenerated: x > 0 or x < 0 .

43

Fig. 17.1. Free energy dependence on x at T > θ and H = 0

Fig. 17.3. Free energy dependence on x at T > θ and H < 0

Fig. 17.2. Free energy dependence on x at T > θ and H > 0

Fig. 17.4. Free energy dependence on x at T <θ and H = 0

5. At T < θ and H 0 one of the minima becomes deeper (see Fig. 17.5), i.e., the magnetic field removes the degeneracy. Now, if the magnetic field tends to zero H 0 , the macroscopic system remains in the state corresponding to this deeper minimum, since the probability of tunneling through the barrier is very low.

44

Fig. 17.5. Free energy dependence on x at T <θ and H > 0

18. Spontaneous symmetry breaking

at the paramagnetic-ferromagnetic transition

Earlier we showed (see § 13) that at T < θ and H = 0 the order parameter x 0 . This corresponds to two physically unequivalent states

with magnetization M = NVµB x > 0 and M < 0 for the magnetic mo-

ments alignment along and against the selected axis z, respectively. In other words, at T < θ and H = 0 the system state is doubly degenerated, i.e., two different states have the same values of energy. The reason

is the parity of Helmholtz free energy function

F (x,T )= F (x,T ) at

H = 0. This is apparent from the expression for F (x,T )

near the Curie

temperature

 

 

 

 

 

 

 

 

F(x,T ) = F0 (T )+ N T −θ x2

+

 

 

 

θ

x4 .

 

(18.1)

12

 

2

 

 

 

 

Note that at H 0 the expression for F (x,T ) takes the form

 

F(x,T ) = F0 (T )+ N T −θ x2 +

θ

 

 

 

x4 xh

,

(18.2)

12

 

 

2

 

 

 

 

 

 

45

 

 

 

 

 

 

 

 

and one can see that F( x ,T ) < F(x ,T ), thus the state with x > 0 , i.e., with the magnetic moments aligned along magnetic field, is more favorable than the state with x < 0 , i.e., with the magnetic moments

aligned against magnetic field. So, the state with x = − x < 0 is the local

minimum. Therefore the magnetic field removes the degeneracy. And what happens in the real systems at H = 0 ? In the real systems the only one of the two different macroscopic states ( M > 0 or M < 0 ) is realized. What is it? It turns out that under cooling the paramagnet to T = θ

the “random” selection of the state with x = x or with x = − x occurs.

And this choice depending on many random factors is made by Nature. In terms of the degree of degeneracy at T > θ we have one

(nondegenerate) state ( x = 0 ), and at T < θ we have two states ( x > 0 and x < 0 ) with the same energies, i.e., doubly degenerated state (see Fig. 18.1).

Fig. 18.1. The variant

of “two-dimensional bottle”: the ball falls to one of the wells

In terms of symmetry, what do we mean by symmetric and asymmetric states when we talk about the system of magnetic moments. At

T > θ, M = 0 , thus the magnetization does not change the sign at replacing the selected axis z → −z . This is the symmetric state. At T < θ, magnetization M 0 , thus the magnetization changes the sign at replacing the selected axis z → −z . This is the asymmetric state. In this case the preferred direction occurs. Therefore at the paramagneticferromagnetic transition M = 0 M 0 the spontaneous symmetry

46

breaking takes place. Note that in real solids the preferred orientations, so called easy magnetization axes, are always presented. On these axes the magnetization M is oriented at T < θ.

Let us look at this issue from the other side. Let us calculate σkz ,

where k is the number of arbitrary magnetic moment, without the meanfield approximation in the Ising model at H = 0

 

 

 

 

 

 

 

ˆ

)

 

 

 

 

 

 

 

 

H T

 

 

 

 

 

 

 

Tr (σkze

 

 

 

.

 

 

 

σkz

=

 

 

 

 

 

 

(18.3)

 

 

ˆ

 

 

 

 

 

 

 

 

 

Tr (eH T )

 

 

 

 

Let us consider the numerator

 

 

 

 

 

 

 

 

 

 

 

 

 

)= σkze

1

/ Jijσizσjz

 

 

 

 

ˆ

2T

 

 

 

H T

 

i , j

 

.

(18.4)

Tr (σkze

 

 

 

 

 

 

 

 

 

 

 

{σiz }

 

 

 

 

 

 

 

This is the sum of 2N configurations, i.e., the sum of 2N

different sets

of N numbers σ1z ,σ2z ,...,σNz = ±1. Divide this sum on pairs of summands which differ in their signs. For example, one of these configura-

tions and the paired configuration will be

+1

–1

–1

+1

–1

+1

+1

+1

–1

σ1z

 

 

 

 

 

 

σkz

 

 

 

σNz

–1

+1

+1

–1

+1

–1

–1

–1

+1

σ1z

 

 

 

 

 

 

σkz

 

 

 

σNz

Every pair in the sum

results zero, i.e., sum

/ Jij σiz σjz

does

 

 

 

 

{σiz }

 

 

 

i, j

 

 

 

 

 

 

 

 

 

 

 

 

= 0 at any

not change, while σkz has the different sign. Therefore σkz

T. This is in contrast with our previous result (which is obtained in the mean-field approximation) and with the experiment. The reason is the availability of two degenerated states with the different orientation of

magnetic moments. In the limiting case

T = 0 we have either state

 

1 =

↑↑↑ ↑ ↑↑ or state

 

2 =

↓↓↓

↓ ↓↓ . Energies of these

 

 

 

 

 

 

 

 

 

 

 

" "

 

 

 

" "

47

states are equal, so the probabilities of their realization are P1 = P2 = 12 .

Therefore

 

= (σkz ) P1 +(σkz )

 

P2 = +1

1

+(1)

1

= 0.

Here we

σkz

 

2

2

 

 

1

2

 

 

 

 

 

 

 

 

 

 

 

 

do not consider that at phase transition system “selects” one of the two states rather than it is located at both of them simultaneously. We can draw an analogy with the “two-dimensional bottle” (see Fig. 18.2).

Fig. 18.2. “Two-dimensional bottle” again

The probability that the ball will fall to the left or to the right minimum is equal to 12 . But the “mean coordinate” is x = −a 12 + a 12 = 0

which can not be.

So, calculating the σkz we took the mean values for all configura-

tions, half of which corresponds to σkz = +1 and the other one corresponds to σkz = −1, and the energies of these configurations are the

same at H = 0 . Thus, σkz h=0 = 0 . On the other hand, if H 0 , then

the degeneracy is removed, and the configuration with the magnetic moments aligned along the z-axis becomes more favorable. While calcu-

lating

 

this configuration makes the main contribution, so,

σkz

 

 

0 . Even the magnetic field is removed, the system will re-

σkz

h0

main

in

this configuration, i.e.,

lim

 

0 . Why do we obtain

σkz

 

 

 

h0

 

h0

48

 

0

in

the mean-field approximation

at

h = 0 ? Recall that

σkz

 

 

 

. Therefore even at h = 0 the

heff

is nonzero, since we

heff = h + θ σkz

supposed that σkz 0 . This is the analogy to the magnetic field re-

moving the degeneracy.

Finally, at T < θ the order parameter x 0 (i.e., the magnetization is nonzero as well). But we can not determine the sign of the order parameter (or magnetization) beforehand, it is fully random.

19. Phenomenological Landau theory of second-order phase transitions

Let us discuss this theory briefly. In the previous paragraphs we developed the microscopic theory of phase transitions into ferromagnetic state, i.e., we took into account the real interactions between the magnetic moments (exchange interaction) and obtained the equation for the order parameter x , which was valid in the whole range 0 x 1 (in

other words at temperatures 0 T ≤ θ and T > θ). We made this in the mean-field approximation. Also we found the Helmholtz free energy

function F (x,T ) in the whole range 0 x 1 and derived F (x,T ) at T near θ, i.e., x 1 . If we do not know the general form of F (x,T ) from the microscopic theory, we can not expand into series F (x,T ) at

x 1 , i.e., at T = θ.

Lev Landau did not know the microscopic theory of any secondorder phase transition known in the middle of XX century. He suggested that for all second-order phase transitions the Helmholtz free energy dependence on the corresponding order parameter (different for various second-order phase transitions) possessed the same form. But this

universal form function F (,T ) possessed only at small , i.e., near the critical temperature TC (for ferromagnetic transition we denoted it

θ). So, ∆ = 0 at T >TC , and 1 at T <TC , T TC, and further increasing as T decreases. At H = 0 this universal form is the following

49

F (,T )= F0 (T )+ A(T )2 + B(T )4 ,

(19.1)

where F0 (T ) is the component of free energy that is not related to the

effects of ordering. Actually this is the series expansion in powers of .

But the powers of and 3 are absent in this expansion, due to the fact that for all second-order phase transitions states with and −∆ are degenerated. And what are the parameters A and B? The experimental

facts tell that phase transition occurs at T =TC . So,

∆ = 0 at T >TC and

∆ ≠ 0 at critical temperature TC . In other words,

F (,T )

must have

minima at ∆ = 0

and T >TC , and at ∆ ≠ 0 and T <TC . It will be if:

a) A(T )> 0 and

B(T )> 0 at

T >TC ; b) A(T )< 0 and B(T )> 0 at

T <TC . The simplest choice is

 

 

 

 

A(T )= a(T T ), where a > 0;

 

 

 

C

 

(19.2)

 

B(T )= b = const (T )> 0.

 

 

 

 

 

 

Actually it is the first terms in the expansion of A(T ) and

B(T ) into

series in powers of (T TC ) considering only the first nonzero terms.

Therefore the free energy can be described as

 

F (,T )= F0 (T )+ a(T TC )2 +b(T )4 .

(19.3)

In the case of ferromagnetic transition we earlier obtained a =12 and b = 12θ . For the other second-order phase transitions these parameters

will be different. If the microscopic theory is absent then these coefficients are unknown. Nevertheless, we can obtain or estimate them from numerical calculations or from the experimental data.

From Eq. (19.4) we can obtain

(T ) by minimizing the function

F (,T )

 

 

 

 

F (,T )

= 0 2a(T TC )∆ + 4b3 = 0;

 

 

∂∆

(19.4)

 

 

 

2a(T TC )+ 2b2 = 0.

50

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