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Маслов ИНТРОДУЦТИОН ТО ПХЫСИЦС ОФ СЕЦОНД-ОРДЕР МАГНЕТИЦ 2015

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well. But the differential magnetic susceptibility

 

M

χ =

 

 

equal to zero. Let us show this fact.

 

 

 

 

 

H H 0

 

 

 

 

 

 

 

Differential magnetic susceptibility is equal to

 

 

 

 

M

 

M

=

µ2 N

x

x

 

χ =

 

= µB

 

B

 

 

A +

B

.

 

H H 0

 

h h0

 

2V

 

h

h

h0

Considering Eq. (28.12) and taking into account that TN = z J tain

x

A

= tanh

 

h xBTN

 

 

;

 

 

 

 

T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h xATN

.

x

B

= tanh

 

 

 

 

T

 

 

 

 

 

 

 

 

 

 

Partial derivatives of xhA and xhB are

may not

(29.1)

,we ob-

(29.2)

xAh

xBh

Hence

=

1tanh2

h xBTN

 

 

 

T

 

 

 

=1tanh2 h xATNT

 

1

 

T

x

 

;

 

 

 

N

B

 

 

 

T

 

T

h

 

 

 

1

 

T

x

 

 

 

 

 

N

A

.

 

 

T

 

T

h

 

 

 

xA

 

2

 

 

1

 

 

TN

xB

;

 

 

 

 

 

h

= 1

xA

 

 

T

 

 

 

 

 

T

 

h

 

 

xB

 

 

 

 

 

1

 

 

TN

xA

 

 

 

2

 

 

 

 

xB

 

 

 

 

 

h

= 1

 

 

 

T

.

 

 

 

 

T

 

h

 

Sum up the latter equations considering that xA = x and xB = −x

x

x

 

2

 

T

x

x

A +

B = (1

x2 )

 

N

 

A +

B .

 

 

h

h

T

 

T

h

h

(29.3)

(29.4)

(29.5)

81

Expressing the sum of partial derivatives, we have

xA

 

xB

 

2 1x2

)

 

 

(

.

h

+

h

=

 

T +(1x2 )TN

Finally, we obtain the differential magnetic susceptibility

χ =

µB2 N

 

2(1x2 )

 

.

2V

 

1

x2

)

T

 

T +

 

 

 

 

 

(

 

 

N h0

(29.6)

(29.7)

Let us analyze the Eq. (29.7) in more detail. In the absence of magnetic

field the expression for the parameter order takes form x = tanh x TTN . In the case of T >TN the order parameter x = 0 . Therefore

 

2

 

1

 

 

χ =

µB N

 

.

(29.8)

 

 

V

T +TN

 

In the case of

TN T

1, the order parameter squared takes the follow-

T

ing form

N

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x

2

 

T

 

 

 

 

3 1

 

 

(29.9)

 

 

 

 

 

 

 

 

 

TN

 

Therefore differential magnetic susceptibility is

 

 

 

 

 

2

 

3T 2TN

 

 

 

 

 

χ =

µB N

 

.

(29.10)

 

 

 

 

2V

 

2T TN

 

 

 

So if T T

, differential magnetic susceptibility tends to

χ →

µ2 N

.

B

 

N

 

 

 

 

 

2V

 

 

 

 

 

 

 

 

 

Let us show that if the temperature decreases below TN , differential magnetic susceptibility decreases linearly

 

 

2

 

 

3T 2TN

 

 

 

2

 

 

2T TN +T

TN

 

 

 

χ = µB N

 

 

=

µB N

 

 

=

 

 

2T TN

 

 

 

2V

 

 

2T TN

 

 

 

2V

 

 

 

 

 

(29.11)

 

µB2 N 1

 

 

T TN

 

 

 

µB2 N

1

 

T TN

= µB2 N

 

 

=

+

+

 

 

T

.

 

 

 

 

 

 

 

2V

 

 

 

 

 

 

 

2V

 

 

 

 

 

2V TN

 

 

 

 

2T TN

 

 

 

 

 

 

TN

82

In the case of T TN , the order parameter squared takes the following form

 

 

x2 14e

2TN

 

 

 

 

 

 

 

T

,

 

 

 

(29.12)

 

 

 

 

 

 

 

and differential magnetic susceptibility takes the following form

 

 

µ2 N

4e

2TN

 

 

 

 

 

4µ2 N

 

 

2TN

 

 

T

 

 

 

 

e

 

 

 

 

 

 

 

 

χ =

B

 

 

 

 

 

 

 

B

 

T .

(29.13)

T + 4T e

2T

 

 

 

 

V

T

 

 

 

VT

 

 

 

 

 

 

 

 

 

 

N

 

 

 

 

 

 

N

Fig. 29.1. Resulting temperature dependence of differential magnetic susceptibility for antiferromagnets

The resulting differential magnetic susceptibility characteristic dependence on temperature for antiferromagnets is shown on Fig. 29.1.

30. Spin waves in the Heisenberg model

Spin waves are propagating disturbances in the ordering of magnetic materials. These low-lying collective excitations occur in magnetic lattices with continuous symmetry. From the equivalent quasiparticle point of view, spin waves are known as magnons. In this paragraph, the spin wave dispersion for a ferromagnet is derived based on the Heisenberg approach. The Heisenberg Hamiltonian for the energy of interaction of atoms i and j in the absence of magnetic field can be written in the following form (see § 10)

 

 

 

 

 

 

 

 

 

 

(30.1)

H

= −Jij (Si S j )= −2Jij (Si S j ),

 

ij

 

i< j

 

 

83

where Jij represents the exchange integral (see § 8), and Si and S j are

the spin operators for the i-th and j-th atoms. The spin wave is a particular excited state solution of the Hamiltonian (30.1). Here we assume that

Jij > 0 for any pair of atoms (ferromagnetic ordering), and Jij = J for the nearest neighbors and Jij = 0 for other atomic pairs. So, we account

only for the exchange interaction between the nearest neighbors.

Let us consider the simplest model of ferromagnet: the onedimensional chain. Interaction energy for the system consisting of N spins can be expressed as

 

N

 

 

j )

 

U = −2J

(

i

S

(30.2)

 

S

.

i, j=1

Let us choose an arbitrary atom at site p. This atom has the exchange interaction with atoms at sites (p 1) and (p +1). So, for the interaction energy in accordance with the Eq. (30.2) we have

U = −2J (Sp1Sp )2J (Sp Sp+1 )= −2JSp (Sp1 + Sp+1 ). (30.3) Recall that µp = −gµBSp , therefore the Eq. (30.3) takes the following form

 

 

 

2J

 

 

 

 

U = −µp

 

 

(Sp1

+ Sp+1 ).

(30.4)

 

 

 

 

gµB

 

 

 

The Eq. (30.4) is similar to the expression for the energy of isolated

magnetic

moment

in the external field U = −µpbp . Thus the value

bp = −

2J

 

(Sp1 + Sp+1 )

can be considered as the some effective field

 

 

 

gµB

 

 

 

acting on the magnetic moments µp . The magnetic exchange interac-

tion exerts a torque

µ

b

on the spins that causes the precession of

 

 

 

 

p

 

p

the spins. The spin precesses around the constant magnetic field according to the torque expression Td = dSdt . The torque acting on a p-th spin is

84

 

dSp

 

d d

 

 

d d

 

 

d d

 

(30.5)

dt

= −gµB Spb

 

= 2J ( Sp Sp1

 

+ Sp Sp+1

).

Let us choose the coordinate system as follows: the z-axis coincides with the direction of magnetic moments ordering, the x-axis is directed along the chain of magnetic moments, and the y-axis is perpendicular to

the z and x-axes. In this case the components of vectors Sp , Sp+1 are expressed as follows

dSpx

 

= 2J

Spy (Spz 1 + Spz +1 )Spz (Spy1

+ Spy+1 )

;

 

dt

 

 

 

 

 

 

 

 

 

 

 

dSpy

 

= −

2J

Spx

(Spz 1 + Spz +1 )Spz

(Spx

1

+ Spx+1 )

;

dt

 

 

 

 

 

 

 

 

 

 

 

 

 

dSpz

 

= −

2J

Spx

(Spy1 + Spy+1 )Spy

(Spx

1

+ Spx+1 ) .

dt

 

 

 

 

 

 

 

 

 

 

 

Sp1 and

(30.6)

The magnetization is assumed to be aligned along the z-direction and the deviation of magnetic moments from this direction is taken to be

small, hence

Spx , Spy Spz . Neglecting quadratic terms in Spx and Spy

 

dS z

 

one obtains

p

= 0. Therefore the z-components of the spin are time

dt

 

 

independent and because of the assumption of identical atoms, Spz

S

.

Thus the Eqs. (30.6) take the following form

 

 

 

 

 

 

 

dSpx

 

 

 

2J

 

S

 

 

 

 

y

 

y

y

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dt

=

 

 

 

 

 

 

 

 

2Sp

Sp1 Sp+1

;

 

 

 

 

 

 

 

 

 

 

 

 

 

d

 

 

 

 

 

 

 

 

 

 

 

dSpy

 

 

 

 

2J

S

 

 

x

x

x

 

 

(30.7)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dt

 

= −

 

 

 

 

 

2Sp

Sp1

Sp

+1

;

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dS z

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

p

=

0.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dt

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let us try to find the solution of this system of equations in the form of travelling waves

85

Spx = u exp{i(pka −ωt)};

(30.8)

Spy = vexp{i(pka −ωt)},

 

where u and v are constants, p is the whole number, k is the wave vector and a is the lattice constant. Upon substitution of Eqs. (30.8) to Eqs. (30.7), we obtain

iωu =

2J

S

(2 eika eika )v =

4J

 

S

 

 

(1cos ka)v;

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2J

 

S

 

 

 

 

4J

 

S

(30.9)

iωv = −

 

 

 

(2 eika eika )u = −

 

 

 

(1cos ka)u.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

These equations have the nontrivial solution if

 

 

 

 

 

 

d

 

 

 

 

 

 

 

 

4J

S

 

 

 

 

 

 

 

iω

(1cos ka)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d

 

 

 

= 0.

(30.10)

det

 

 

 

 

 

 

 

 

4J

S

(1cos ka)

 

 

iω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This determinant defines the spin wave dispersion ω(k ). It has the following form

ω(k )=

4J

S

 

(1cos ka).

(30.11)

 

 

 

 

 

 

 

 

From the solution obtained one can see that v = −iu , i.e., the solution describes the spin precession relative to the z-axis.

86

Literature

1.Kittel C. Introduction to Solid States Physics. 8th ed. John Wiley

&Sons, 2004.

2.Kittel C. Quantum Theory of Solids. Second Revised Printing. John Wiley & Sons, 1987.

3.Landau L.D., Lifshitz L.M. Quantum Mechanics: NonRelativistic Theory (Volume 3). 3rd ed. Butterworth-Heinemann, 1981.

4.Landau L.D., Lifshitz L.M. Statistical Physics (Volume 5). 3rd ed. Butterworth-Heinemann, 1980.

5.Aharoni A. Introduction to the Theory of Ferromagnetism. 2nd ed. Oxford University Press, 2001.

6.Miyazaki T., Jin H. The Physics of Ferromagnetism. Springer,

2012.

7.Kronmüller H., Fähnle M. Micromagnetism and the Microstructure of Ferromagnetic Solids. 1st ed. Cambridge University Press, 2009.

87

Mikhail M. Maslov

Konstantin P. Katin

Leonid A. Openov

INTRODUCTION TO PHYSICS OF SECOND-ORDER

MAGNETIC PHASE TRANSITIONS

Оригинал-макет подготовлен М.В. Макаровой

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