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Маслов ИНТРОДУЦТИОН ТО ПХЫСИЦС ОФ СЕЦОНД-ОРДЕР МАГНЕТИЦ 2015

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At H > 0 the ground state (the state with the lowest energy) is with

µz = µ0.

Let us evaluate the value of µz at T 0:

 

 

 

 

µ

z

(T ) ≡ µ

z

 

= µ

P

−µ

P

,

(4.8)

 

 

T

0

0

 

 

where P= C exp(E

 

T ) and

 

P= C exp(E

T )

are the statistical

(thermodynamic) probabilities to find the magnetic moment in the state

 

and

 

respectively (hereafter the Boltzmann constant kB = 1).

 

 

Calculating

the constant C from the

 

normalization

condition

 

P+ P=1, we have

 

 

 

 

 

 

 

 

 

µz (T ) = µ0

 

µ

H

(4.9)

 

 

 

 

tanh

0

 

.

 

 

 

 

 

 

T

 

 

In the limiting case T 0 one has µz → µ0 at H > 0 (as shown above)

and µz µ0 at H < 0, while µz 0 at T → ∞. At H = 0, the magnetic moment is zero regardless of T.

In the case that magnetic field is oriented along the z-axis, the magnetic moment projections on x and y axes are zero at any H and T:

 

 

 

= 0 .

(4.10)

µx

T

= µy

 

 

T

 

This is a consequence of zero values of quantum-mechanical averages

of the spin operators S x

and S y in the states

 

and

 

, see Eq. (1.8).

 

 

5. Interplay of quantum mechanics and statistics

In order to calculate the temperature dependence of a physical quantity A, we first should solve the Schrödinger equation

 

(5.1)

H n = En n ,

is the Hamiltonian of the system under consideration, and where H

find all eigenenergies En and eigenvectors n . Next we should work

out the quantum-mechanical averages of the operator in the states

A

n :

11

 

 

 

An = n

 

 

 

n

(5.2)

 

 

 

 

 

 

 

 

 

A

 

(note that the states

 

n

 

 

 

 

 

 

 

 

 

 

 

 

need not be the eigenstates of A , that is,

A and

need not commute). Finally, the statistical (thermodynamic) average

H

is obtained as

A(T )

 

 

 

,

 

 

(5.3)

A = An Pn (T )

 

 

 

T

n

 

 

 

 

where

 

 

 

 

 

 

 

 

 

 

 

 

Pn (T ) = C exp(En T )

 

 

 

(5.4)

is the probability to find the system in the state

 

n

at temperature T.

 

The constant C is determined from the condition Pn (T ) =1 :

 

 

 

 

n

 

C =

 

1

,

 

 

 

(5.5)

exp(En / T )

 

 

 

so that

n

 

 

 

 

 

 

An exp(En / T )

 

 

 

 

A(T ) =

.

(5.6)

n

exp(En / T )

n

Let us derive a more general expression for A(T). It follows from Eq.

(5.1) that

 

 

n = exp(En / T )

 

n . Therefore, with account

 

 

exp(H / T )

 

for normalization n | n

 

=1, we have

 

 

 

exp(En / T )= n

 

 

 

n

 

 

 

 

 

exp(H / T )

 

Tr exp(H / T ) (5.7)

n

 

 

n

 

 

 

 

 

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

An exp(

En / T )

= n

 

 

n

 

 

 

 

 

Aexp(H / T )

 

 

n

 

 

 

n

 

 

(5.8)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Tr Aexp(H / T )

 

 

 

 

 

 

 

 

 

 

so that

12

 

 

 

 

 

A(T ) =

Tr Aexp(H / T )

.

(5.9)

 

 

/ T )

 

Tr exp(H

 

 

Since the trace (the sum of the diagonal matrix elements) of any operator does not depend on a specific set of states {n} used for its calcula-

tion, those states need not be the eigenstates of the Hamiltonian, they should just form the complete set and satisfy the orthonormality condi-

tion n m = δnm .The use of one or another set of states is a matter of

convenience. For example, in the case of isolated magnetic moment in an external magnetic field (§ 4), it is reasonable to choose

{

 

n }={

 

,

 

}. Then

for

 

 

 

 

 

 

 

 

 

 

we

 

have from

 

 

 

 

 

 

 

 

 

 

A = µz

and H

= −µz H

 

Eq. (5.9):

 

 

 

 

 

 

 

 

 

 

 

H / T )

 

 

 

 

 

 

 

 

 

exp(

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

↑ + ↓

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

µz (T ) =

 

µz exp(µz

 

 

 

 

µz

µz H / T )

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

exp(µz

H / T )

 

↑ + ↓

 

exp(µz H / T )

 

 

 

 

(5.10)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

µ

0

exp(

µ

H / T )−µ

0

exp(−µ

H / T )

 

 

 

µ

H

 

=

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

= µ0

tanh

0

 

 

,

 

 

exp(µ0 H / T )+ exp(−µ0 H / T )

 

 

 

 

 

 

 

 

 

 

 

 

T

 

 

in accordance with Eq. (4.9).

6. A system of noninteracting local magnetic moments in an external magnetic field. Curie law

Having discussed the behavior of a single local magnetic moment in

an external magnetic field H , we now turn to a macroscopic system of N 1 local magnetic moments in a solid. Our purpose is to calculate

the experimentally measured magnetization M as a function of H and T.

Two types of interactions contribute to the total energy E of such a system: 1) the interaction of each magnetic moment with the magnetic field; 2) the interaction of the magnetic moments to one another, so that

N

 

 

(6.1)

E = (−µi H )+Uint ({µi }),

i=1

13

and hence the Hamiltonian is

 

N

 

 

 

 

 

 

 

 

(6.2)

H

= −H µi

+Uint ({µi }).

 

i=1

 

 

 

 

Taking the z-axis along the magnetic field, H = Hez , we have

 

 

N

 

 

}).

 

 

 

(6.3)

H

= −H µiz +Uint ({µi

i=1

In the case that magnetic moments weakly interact with one another (for example, because of their low concentration and/or small absolute values), one can neglect the term Uint in Eq. (6.3). Then the Hamiltonian of the whole system is just the sum of Hamiltonians of individual magnetic

moments, each being in the field H , and for the thermodynamic average of an arbitrary k-th magnetic moment we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Trµk exp(H / T )

 

 

 

 

 

 

 

 

 

 

µk (T ) =

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Tr exp(H / T )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

N

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Tr

 

k µk exp(

µk z H / T )Tr

 

ik exp µi z H / T

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ik

 

 

=

(6.4)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

N

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Tr

 

k

exp(µk z H

/ T )Tr

 

ik exp

µi z

H / T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ik

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

Tr

 

k

µk exp(

µk z H / T )

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Tr

 

k exp(µk z H / T )Tr

 

ik

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where Trk and Trik stand for the trace over the states of, respectively,

k-th and all other magnetic moments.

Let every magnetic moment in the system be associated with spin S =12 . Then the value of µkz(T) is given by Eq. (4.9) for an isolated

magnetic moment, while µkx(T) = µky(T) = 0, see § 4. Since all magnetic moments are identical, the total magnetic moment of the system is N

µk (T ) , and the magnetization equals to

 

 

 

N

 

 

M (T, H ) =

 

µk (T )

= M (T, H )ez ,

(6.5)

V

14

where

M (T, H ) = VN

In weak magnetic fields H << T/µ0 have from Eq. (6.6)

M (T, H )

µ0

 

µ

H

(6.6)

tanh

0

 

.

 

 

T

 

 

(note that T/µ0 ~ 1 T at T = 1 K), we

N

µ2

H

,

(6.7)

V

T

 

0

 

 

that is, M is linear in H, and the magnetic susceptibility χ = M/H equals to

χ(T ) =

N

µ2

1

.

(6.8)

V

 

 

0

T

 

Such temperature dependence of χ is referred to as Curie law. The fact that χ > 0 implies that materials containing noninteracting or weakly interacting magnetic moments are paramagnetic, see § 3.

7. Effective Weiss field

At H = 0 , magnetization M of a system of noninteracting local magnetic moments is zero at any temperature, see Eq. (6.6). Meanwhile,

there exist materials (ferromagnets) for which M 0 at H = 0 provid-

ed the temperature is below some characteristic value θ called the Curie temperature. It seems reasonable to assume that the failure to explane the occurrence of such materials within the model of noninteracting moments is a consequence of ignoring the interaction of magnetic moments with each other.

Weiss was the first who appreciated this. He noted that once the magnetic field H is applied to the sample, each moment µi interacts

not only with this field but with the field created by all other moments µk i as well. Since, first, the magnetic field induced in the sample by an

external field H is

 

4π N

 

 

4πM =

 

 

(7.1)

V

µk

 

k =1

 

T

15

and, second, the field generated by the i-th magnetic moment has no effect on this moment itself (there is no self-action), the net field acting on i-th moment is

 

H / = H + λM ,

(7.2)

 

 

 

 

 

 

 

 

 

does not de-

where λ < 4π. Taking into account that the value of µk

pend on k, we have

 

 

 

T

 

 

 

 

/

 

N

 

 

H

= H + λ

 

 

 

 

µi T .

(7.3)

V

The second term in Eq. (7.3) is called the molecular field or the Weiss field, and the net field H / is named the mean field or the effective field.

Each magnetic moment experiences the field H / , so that the Hamiltonian is

 

 

 

 

 

N

 

 

/

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.4)

 

 

 

H

= (−µi H ).

 

 

 

 

 

 

 

i=1

 

 

 

 

 

/

 

 

 

 

 

 

 

 

/

Let H = Hez and

 

 

 

 

 

ez . Then

H = H ez , where

µi

T

= µiz

T

 

 

 

 

 

 

 

 

 

 

 

 

 

H

/

= H + λ

N

 

T ,

 

(7.5)

 

 

 

 

µiz

 

and the Hamiltonian is

 

 

V

 

 

 

 

 

 

 

 

N

 

 

 

 

 

 

 

 

 

 

 

/

 

 

 

(7.6)

 

 

 

 

H

= −H

µiz .

 

 

i=1

This is the Hamiltonian of a system of noninteracting magnetic moments in an effective magnetic field. By analogy with Eq. (4.9) we have

 

 

 

 

 

 

 

 

 

 

 

N

 

 

 

 

 

 

 

µ0 H

/

 

 

 

 

µ0

H + λ

 

 

µiz

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V

 

T

 

. (7.7)

µiz

= µ0

tanh

T

 

 

= µ0

tanh

 

 

T

 

 

 

 

T

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Contrary to Eq. (4.9), this is not expression but equation for µiz T .

Let us examine whether Eq. (7.7) has nonzero solutions at H = 0. Denoting

16

 

 

µiz

T

 

x

 

(7.8)

and

µ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

N

 

 

 

 

 

 

 

 

 

2

≡ θ,

(7.9)

µiz T

= λ

 

 

µ0

V

we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

θ

 

 

 

x = tanh

 

 

 

x .

(7.10)

 

 

 

 

 

 

 

T

 

 

 

Taking into account that tanh(z) is a monotonically increasing function of z, being linear in z at z << 1 and approaching ±1 when z → ±∞, we conclude that Eq. (7.10) has just one trivial solution x = 0 at T < θ, while at T > θ there are three solutions: one x = 0, one x < 0, and one x > 0. It

can be shown that solutions with x 0 correspond to the lower free energy of the system, that is, they are thermodynamically more stable.

So, at T < θ the average values of all magnetic moments and hence the magnetization M are nonzero even in the absence of the external

field H . This is just what is referred to as ferromagnetism. Let us esti-

mate the value of the Curie temperature θ, see Eq. (7.9). Taking λ ~ 10, N/V ~ 1023 cm-3, and µ0 ~ 10-20 erg/G, we have θ ~ 1 K. Meanwhile fer-

romagnetic materials with θ ~ 1000 K do exist. To explain such high Curie temperatures within the Weiss approach one should assume unre-

alistically large values of λ, N/V, and µ0 in Eq. (7.9). Hence, the Weiss idea appears to be at variance with experimental data. Moreover, numerous experiments revealed that the “molecular field” generated by the local moments is, if any, orders of magnitude weaker than that required to account for the phenomenon of ferromagnetism. Surprisingly, based on the mistaken assumption, Weiss arrived at qualitatively (although not quantitatively) correct results. The main merit of Weiss is that he has pointed to the necessity of account for interactions between the local magnetic moments. Below we shall see how such interactions naturally arise when the problem is treated quantum-mechanically.

17

8. Exchange interaction

For simplicity sake assume that the local magnetic moment of every atom in the solid-state system results from a single valence electron occupying the energy level with the s-wave wave function ϕ(r ) and thus

having no orbital magnetic moment. In this case the atomic magnetic moment is entirely due to the electron spin S = 1/2, and the magnetic moment operator is

 

 

 

 

 

(8.1)

µ = −2µB S ,

see § 1, so that µz = ± µB.

 

 

Now consider two atoms, i and j, with coordinates Ri

and Rj , hav-

ing magnetic moments µi and µj

respectively. Our objective is to de-

termine the interaction energy Uij

of those moments as a function of

their mutual orientation and the distance

 

Ri Rj

 

 

between them. In fact,

 

 

 

 

 

 

 

 

 

 

 

we should express the operator U ij

in terms of operators µi

and µj :

 

 

 

 

 

,

(8.2)

µi

= −2µB Si , µj = −2µB S j

where Si = Sj = 1/2. The (normalized to unity) wave functions ϕi (r ) and jj (r ) of an electron in i-th and j-th atoms can be found from the corre-

sponding Schrödinger equation. They are identical in form but localized in different regions of space:

ϕ (r ) = ϕ(r R ) , j

(r ) =j(r R

) ,

(8.3)

i

i

j

j

 

 

having their maxima at r = Ri and r = Rj respectively.

Let us now consider the two-electron wave function for a pair of electrons in atoms (i, j):

Ψ(1,2) = Ψ(r1,σ1;r2 ,σ2 ) ,

(8.4)

where "1" = (r1,σ1 ) and "2" = (r2 ,σ2 ) are the sets of spatial and spin

coordinates of the first and second electron respectively, σ1 = 2Sz1 = ±1, σ2 = 2Sz2 = ±1. Since every electron has a half-integer spin, the electrons

18

obey the Fermi-Dirac statistics, and the many-electron wave function changes sign upon permutation of any two electrons. In our case

Ψ(1,2) = −Ψ(2,1) ,

(8.5)

that is,

 

Ψ(r1,σ1;r2 ,σ2 ) = −Ψ(r2 ,σ2 ;r1,σ1 ) .

(8.6)

In the absence of spin-orbit interaction, the wave function can be represented as a product of coordinate and spin wave functions:

Ψ(r1,σ1;r2 ,σ2 ) = Φ(r1,r2 )Χ(σ1,σ2 )

(8.7)

From Eqs. (8.6) and (8.7) one has

 

Φ(r1,r2 )Χ(σ1,σ2 ) = −Φ(r2 ,r1 )Χ(σ2 ,σ1 ) ,

(8.8)

that is, either

 

Φ(r1,r2 ) = −Φ(r2 ,r1 ) and Χ(σ1,σ2 ) = Χ(σ2 ,σ1 )

(8.9)

or

 

Φ(r1,r2 ) = Φ(r2 ,r1 ) and Χ(σ1,σ2 ) = −Χ(σ2 ,σ1 ) .

(8.10)

It is known that symmetric and antisymmetric functions Χ(σ1,σ2 )

correspond to the total spin of two electrons S = 1 and 0 respectively. Hence, as follows from Eqs. (8.9) and (8.10), the function Φ(r1,r2 ) is

antisymmetric if S = 1 and symmetric if S = 0, that is, the symmetry of the coordinate wave function depends on the spin state. Since from the viewpoint of classical physics the values of S = 1 and 0 correspond, respectively, to parallel and antiparallel spins of two electrons, the sym-

metry of Φ(r1,r2 ) may be thought of as being dependent on mutual spin

orientation.

Let us construct the function Φ(r1,r2 ) from one-electron wave functions ϕi (r ) and jj (r ) . Taking into account that at i j these functions overlap weakly, that is,

we have

 

ji (rd)jj (rd)drd ≈ dij ,

 

 

 

(8.11)

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

Φ(r1

,r2 ) =

 

 

 

ji (r1 )jj (r2 ) + ηjj (r1 )ji (r2 )

,

(8.12)

 

 

 

2

 

 

 

 

 

 

 

 

 

 

19

where η = 1 at S = 0 and η = –1 at S = 1. In order to express η in terms

 

 

of electron spin at

atoms

i and j, note that

of operators Si

and S j

 

 

 

 

 

 

(Si S j )= = 1/4 and –3/4 at S = 1 and 0, respectively. Hence,

 

 

η = −1 1

 

 

 

 

 

+ 4(Si S j )

,

(8.13)

 

 

2

 

 

 

 

 

 

 

so that the function Φ(r ,r ) depends explicitly on Si

and S j .

 

1

2

 

 

 

Since electrons at atoms i and j repulse each other, from Coulomb’s law we have for their interaction energy:

*

d

d

e2

 

 

d

d d d

Uij = Φ

(r1

,r2 )

 

d

 

d

 

Φ(r1

,r2 )dr1dr2 .

 

 

 

 

 

 

 

r1

r2

 

 

 

Making use of Eqs. (8.11) and (8.12), we arrive at

Uij = Aij + ηJij ,

where

Aij =

 

ji (rd1 )

 

2

 

jj (rd2 )

 

2

de2 d

 

drd1drd2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r1 r2

 

 

is the energy of Coulomb repulsion, and

*

d * d

d

d

e2

 

d d

Jij = ji

(r1 )jj (r2 )jj (r1 )ji (r2 )

d

 

d

dr1dr2

 

 

 

 

r1

r2

 

(8.14)

(8.15)

(8.16)

(8.17)

is the so called exchange energy (or exchange integral, or the energy of exchange interaction). The latter is of purely quantum-mechanical origin and arises because of change of sign in the electron wave function upon permutation (i.e., exchange) of two electrons, see Eq. (8.5). Since both

functions ϕi (r )

and

jj (r )

are exponentially localized at a distance

a 1 Å from the corresponding atoms, it can be shown that

 

 

 

 

 

e

2

 

 

 

 

 

 

e

2

 

 

 

 

R R

j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A

 

 

 

 

and

J

 

 

 

 

 

exp

 

 

i

 

 

(8.18)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ij

 

 

 

 

 

 

 

 

 

 

ij

 

 

 

 

 

 

 

 

 

 

a

 

 

 

 

 

 

Ri Rj

 

 

 

 

Ri Rj

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

at Ri Rj a .

20

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