Chapter 11. Magnetism Diamagnetism 1. Magnetisation M of diamagnetic material is opposite to the total magnetic field B (and applied field H), hence the magnetic susceptibility χ is negative. Magnetic susceptibility is defined as χ= ∂M/∂B. 2. Diamagnetism is a characteristic of atoms with closed shell. Electrons will response to external field by Faraday’s Law. Currents will arrange themselves to oppose the change of increasing field, and hence M is in the opposite direction of H. 3. Classical theory of diagmagnetism: Consider an electron in a circular orbit of radius r and angular frequency ω.

If a magnetic field δB is turned on, and assume it affect ω only.

2mcBq 2mR B

cR

q

,order termfirst only Taking

)2(mR Bc

)(RqmR

)(mR Bc

)(RqmR

mR Bc

RqmR R

mv BcvqmR

00

20

20

020

20

020

220

22

0

δ=δω⇒δωω=δ

ω∴

δω+δωω+ω=δδω+ω

+ω⇒

δω+ω=δδω+ω

+ω⇒

ω=δω

+ω⇒=δ+ω

R ω

20mR F force lCentripeta ω=

R ω

δB

Classically, the B field will make the charge q revolve faster if q is positive. If the particle is an electron, it will revolve slower. Current formed by the loop = I = Charge × revolution / second

smdiamagnetifor equation Langevin - rmc6

NZe - BM

rmc6

BNZe - N M

sample, theofion magnetizat M and e,unit volumper atoms ofNumber N If

rmc6

BZe - rmc4

BZe -32

r32R

zyx Ifzyxr

shell,electron ldimensiona three theof radius r IfyxR

loops.electron theof radius theis R

Rmc4

BZe -Rmc4Be -

cZ RI

cZ

,B field magnetic a applingBy

RIc

Z

rIc

loop) orbital theof Area(A AIc1 atom theofmoment Magnetic

sm)diamagneti torise givessign negative (the mc4Be -

2mcBe-

2e I atom,an in electron an For

2q I

2q f q I

22

2

22

2

22

22

2

2

22

222

2222

222

22

22

22

2

2ii

Z

1i

ii

Z

1i

2

><=δδ

=χ

><δ

=δµ=δ∴

==

><δ

=⎟⎟⎠

⎞⎜⎜⎝

⎛><

δ⋅=δµ

><>=<∴

>>=<>=<<

><+><+>>=<<

=><+>>=<<

><δ

>=<⎟⎟⎠

⎞⎜⎜⎝

⎛πδπ

=><δπ

=δµ

δ∴

><π

=

π=

==µ=

πδ

=

δ⋅

π=δ∴

δωπ

=δ

πω

=⋅=∴

∑

∑

=

=

Paramagnetism 1. When atoms possess their own magnetic moment, paramagnetism will occur. 2. Intrinsic magnetic moment if related to the total angular momentum (including orbital and spin) of the electrons in an atom.

γ is the gyromagnetic ratio, g is the g-factor, and µB is the Bohr magneton.

3. For the spin of a free electron, g=2. For the orbital momentum of an electron, g=1. For a free atom, g is given by the Lande g-factor:

Jg- J B

vvh

v µ=γ=µ

2mce Bh

=µ

( )

( )

( ) ( )

)1j(j2)1()1s(s)1j(j1

)1j(j2)1()1s(s)1j(j3 g

)1j(gj)1(21)1s(s

21)1j(j

23

JgL21S

21J

23

JJgLSJ212SLJ

21

equation, previous theinto theseSubstitute

LSJ21JS JS2SJ )S-J( L

,Similarly

SLJ21JL JL2LJ )L-J( S

JJgJS2JL

)S S ,L L ( JgS2L

J-gS 2 L

J-g hand,other On the

S 2 L

2g and 1g

Sg- Lg- coupling, LSIn

Jg-

2222

222222

2222222

2222222

ii

ii

Bii

ii

B

B

ii

ii

B

sL

iBsi

iBLi

B

++−+++

+=+

+−+++=⇒

+=+−+++⇒

=−+⇒

⋅=−+⋅+−+∴

−+=⋅⇒⋅−+==

−+=⋅⇒⋅−+==

⋅=⋅+⋅⇒

===+⇒

µ=⎟⎠

⎞⎜⎝

⎛+µ−∴

µ=µ

⎟⎠

⎞⎜⎝

⎛+µ−=µ∴

==

µ+µ=µ

µ=µ

∑∑

∑∑

∑∑

∑∑

llll

ll

vvvv

vvvvvvvv

vvvvvvvvvvvv

vvvvvvvvvvvv

vvvvvv

vvvvvvv

vvv

vv

vvv

vvv

vv

4. Original quantum number: L,Lz, S, Sz. Quantum number after LS coupling: L, S, J, Jz.

Atomic notation: 2S+1Lj L = S, P, D, F, G, H, …. for l = 0,1, 2, 3, 4, …. sespectively. 5. Quantum number j for the total angular momentum is determined by Hund’s rule: Let there be x electrons in the outer shell. Each shell can hold y electrons. s shell can hold y=2 electrons.

p shell can hold y=6 electrons. d shell can hold y=10 electrons.

f shell can hold y=14 electrons. Draw y/2 boxes. Example, for d-shell: (10/2=5 boxes) Under each boxes, label Lz (according to the L of the shell) from maximum to

minimum. Example, L=2 for d shell: Hund’s rule #1 (how to fill up the boxes): Always fill up boxes one by one from left to right. Do not double occupy the

boxes until the shell is half full. Start to double occupy the boxes after the shell is half full, start from left to right again.

Hund’s rule #2 (how to calculate L, S, and J): S=ΣSz, L=ΣLz, J=|L-S| if shell is less than half full or half full, and J=L+S if shell is more than hall full or half full. Esample 1: 7 electrons in d-shell L=ΣLz = +2+1+0-1-2+2+1 =3 (or F) S=ΣSz = 1/2×5-1/2×2=3/2 (2S+1=4) J=L+S =9/2 Ground level of the atom: 4F9/2 6. There are 2j+1 sub-levels with Jz = -J, -J+1,…,-1,0,1,…,J-1, J. If we define the energy for Jz =0 as 0, then the energy of each of this state is given by

Lz: +2 +1 0 -1 -2

Lz: +2 +1 0 -1 -2

BJg B - E BBJzµ=⋅µ=

vv

The relative population in level Jz can be calculated as:

z

z

z

z

z

z

z

z

BzB

z

BzB

z

z

z

BzB

z

BzB

BzJ

z

BzJz

J-J

JJ

J-J

JJ

J-J

JJ

J-z

J

JJ

BBB

Tk/BJg-J

JJ

Tk/BJg-zB

J

JJ

JzB

J

JJ

Tk/BJg-J

JJ

Tk/BJ-g

Tk/E-J

JJ

Tk/E-J

e

e

e

eJ m

Tk/Bg ,g Let

e

eJg

NN

Jg m

e

e e

e N

N

β

−=

β

−=

β

−=

β

−=

µ

−=

µ

−=

−=

µ

−=

µ

−=

∑

∑

∑

∑

∑

∑

∑

∑∑

β∂∂

α−=

α=><

µ=βµ−=α

µ−=

µ−=><

==

⎥⎦

⎤⎢⎣

⎡ β+β⎟

⎠⎞

⎜⎝⎛ +α−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+

⎟⎠⎞

⎜⎝⎛ +

β∂∂

α−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

β∂∂

α−=

+===−

⋅=++++

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

β∂∂

α−=

β−

β

β−

β

+β−+β

+β−+β

β−

β+β−+β

β−β

β−

β

+β−+β

2coth

21-

212Jcoth

21J

e- e

e e21-

e- e

e e21J

e- e n-e- e n

)1J2n,e a ,e x,a1

a-1xxaxaax(x

e- e

e- en

22

22

)21(J)

21(J

)21(J)

21(J

22)

21(J)

21(J

Jn

1-n2

22

)21(J)

21(J

ll

L

l

Define Brillouin function BJ(x):

We have

x2J1coth

2J1x

2J12Jcoth

2J12J )x(BJ −

++=

B field smallfor )1J(J)(gTk3V

NB M

x.smallfor x3J

1J

x)J2(3

)J4(4J

x)J2(

131x

)J2()1J2(

31

x1

x1

x2J1coth

2J1x

2J12Jcoth

2J12J )x(B

)O(x x31

x1

eeee coth x

1T)Bfor 1K (T TkJBg field, smallFor

)Tk/JB(gB JgVN m

VN M

)Tk/JB(gB Jg J)(B Jm

2B

B

2

2

22

2

J

3x-x

x-xBB

BBJB

BBJB

J

+µ≈∴

+≈

+≈

⎥⎦

⎤⎢⎣

⎡−

++⎟

⎠⎞

⎜⎝⎛ −≈

−++

=

+++≈−+

=

==<<µ

µµ=><=∴

µµ=βα−>=<

L

x

BJ(x) 1

T=∞ Decreasing T

Increasing J

C is the Curie constant,

7. J, S and J can be estimated from Hund’s rule. After J is estimated, g can be calculated with the Lande equation and the p (effective Bohr magnetron number) is known. The estimated value can be compared with experimental value. Notes: (i) Hund’s rule works well for most rare earth (4f electrons), and the calculated p is very close to the measured value. In some cases like Er3+ (Europium) and Sm3+ (Samarium), energy between the j-multiplets is too small and will cause problem in the 2nd order perturbation. (ii) Hund’s rule does not work fine for transition metals (d electrons). In case of rare earth, the 4f electrons are deep inside the ion and well covered by the 5s and 5p shells. This is not the case for the transition metals. The d-electrons are actually extended further out and exposed to the fields from the neighbors (crystal field). This crystal field will affect the LS coupling and modify Hund’s rule in calculating J. (iii) Crystal field will not couple with S, because spin hjas no real space variables in it. However, the crystal field potential will couple with L. It will split the original degenerate l-orbtals >> µB.

21

B

2B

2

2B

B

)]1J(Jg[

numbermagnetron Bohr effective pTk3

pVN

B field smallfor )1J(J)(gTk3V

N BM

+=

=

µ=

+µ≈∂∂

=χ

Susceptibility of paramagnetism follows Curie Law:

TC ≈χ

T

χ

B

2B

2

k3p

VN C µ

=

Example of crystal field splitting (p-orbitals): (iv) Under the crystal field slitting, Lz is not a good quantum number any more. On average over time, <Lz> =0. Therefore, for transition metal, p (with 2 electrons in shell) should be calculated as g[s(s+1)]1/2 = 2[s(s+1)]1/2 instead of g[j(j+1)]1/2 , since L does not contribute to magnetic properties. (v) For splitting of all degenerate orbitals, the crystal field cannot be symmetric, Very often, if the crystal is high symmetric (e.g. cubic), the ions will displaced themselves to produce a non-symmetric crystal potential to quench the angular momentum. This is called Jahn-Teller effect, Pauli paramagnetism 1. Electron has spin, so free electrons demonstrate paramagnetic property, This is known as Pauli paramagnetism. 2. The effect of Pauli paramagnetism is very small, becayse electrons inside the Fermi sphere cannot flip their spins easily when nearly all states are occupied. Only electrons near the Fermi surface can contribute to Pauli paramagnetism. According to Curie Law:

Percentage of electrons that have the freedom to flip spin =T/TF.

Pauli paramagnetism is independent of temperature.

px

py

pz

E

px, py, pz,

px, py

pz

field smallfor TC =χ

FF TC

TT

TC metalfor =⋅=χ∴

3. More quantitative treatment: When there is no field: When an external field B is applied, say, in the ↑ direction, it will lower the energy of the ↑ electrons by µB and raise the energy of the ↓ electrons by µB

E

D(E)

Among of ↑ and ↓ electrons are the same.

EF

E

D(E)

EF Transfer of electrons

µB

µB

2µBUnbalanced magnetic moments

BTk2

N3 E2N3B

)E(BD)N - N( M

)E(BD21 - dED(E)

21 B)-D(E dE

21 N

)E(BD21 dED(E)

21 B)D(E dE

21 N

FB

2

F

2

F2

Pauli

F

E

0

E

B

F

E

0

E

B-

FF

FF

µ=⋅µ=

µ=µ=∴

µ≈µ=

µ+≈µ+=

↓↑

µ↓

µ↑

∫∫

∫∫

This treatment has ignored the spatial effect of magnetic field. In fact, the magnetic field can modify the electron wave function and produces diamagnetism. This diamagnetism is about 1/3 of the above estimated paramagnetism in magnitude: Long range magnetic ordering 1. Long range magnetic ordering is due to exchange field BE from neighbors. In other words, we assume an exchange field between neighbors that gives rise to long range ordering. 2. Magnetic ordered states has higher symmetry and it occurs only at low temperatures when T < Tc. Tc is the critical temperature of the magnetic transition. 3. Three common types of magnetic ordering: 4. It is clear from above schematic drawings that ferromagnetism and ferrimagnetism will give rise to spontaneous magnetisation then ordering occurs at T<Tc. The antiferromagnetism will not produce any magneisation because of the two opposing spin components. When T>Tc there will be no ordering and the material has to be paramagnetic (i.e. the ions should have their own spin at the beginning). Example: Tc for iron (Fe) is 1043K. Iron is actually ferromagnetic possessing ordering and spontaneous magnetisation at room temperature. It is not a magnet because of domain formation.

FB

2

onfreeelectr

FB

2

diaPauli

FB

2

Paulidia

TkN

BM

BTk

N MMmagnetism Total

BTk2

N- M 31- M

µ=

∂∂

=χ

µ=+=∴

µ==

Ferromagnetic ordering

Antiferromagnetic ordering

Ferrimagnetic ordering

Total magnetic moment = 0

Ferromagnetism 1. The exchange field BE is approximated by the average magnetization field M within the sample: BE =λ M where λ is a temperature independent constant. This is known as the mean field approximation. Note that now the exchange field will become stronger as temperature is lowered, because that is what M does according to Curie Law. 2. When T>Tc – Curie Weiss Law and relationship between λ and Tc: If Ba = applied field and χP= paramagnetic susceptibility. 3. When T>Tc- calculation of spontaneous magnetisation: BE is so strong that Ba can be ignored. i.e., B= Ba + BE ~ BE. For simplicity, lwt us consider j=1/2 (2 levels), and g=2.

VkN

CT kV

Nk3

pVNC ,3)p (hence 2g and

21j For

)TT(

C gives theoury groupation renormaliz accurate More

TTfor Law Weiss-Curie TT

C

CT

0CT i.e.

0. B when)ionmagnetisat eoustanspon( finite is M that so tohas ,T T AsCT

C BM

BTCM)

TC-(1

)M B(TC ion approximat field Mean

)TC( )B B(

TC

)B B( M

B

2B

cB

2B

B

2B

2

33.1c

cc

cc

ac

a

a

a

PEa

EaP

λµ=λ=⇒

µ=

µ====

−∝χ

>−

=χ

=λ⇒=λ−

=∞→χ→

λ−==χ⇒

=λ

⇒

λ+=⇒

=χ+=

+χ=

vv

v

v

vv

vv

vv

vvv

2

coth -coth 2

2coth

21-coth g

2coth

21-

212Jcoth

21J M

B

B

⎥⎦⎤

⎢⎣⎡ β

βµ=

⎥⎦⎤

⎢⎣⎡ β

βµ=⎥⎦

⎤⎢⎣

⎡ β+β⎟

⎠⎞

⎜⎝⎛ +α−=

At T=Tc, or , there is no solution because spontaneous magnetization ceased to exist. At small m,

This curve y ~ m/t will have no solution with y=m (except at m=0) when

M)~(B Tk/M tanh )Tk/B2Tk/Bg( Tk/B tanh

2/ tanh eeee

ee2ee

ee)e(e -

eeee2

eeee -

eeee2

BBB

BBBBBBB

B

2/2/

2/2/

B

B

22/2/

B

2/2/

2/2/

B

λλµµ=µ=µ=βµµ=

βµ=

⎥⎦

⎤⎢⎣

⎡+−

µ=

⎥⎦

⎤⎢⎣

⎡−

−+µ=

⎥⎦

⎤⎢⎣

⎡−+

−+

µ=

⎥⎦

⎤⎢⎣

⎡−+

−+

µ=

β−β

β−β

β−β

β−β

β−β

β−β

β−β

β−β

β−β

β−β

β−β

β−β

⎟⎠⎞

⎜⎝⎛=∴

λµ=

λµ⋅

µ=

λµ=

µ=

tm tanh m

)Tk

M TVk

NNMV

tm(

NTVk t and

NMVmLet

B

B

B

2B

B2

B

B

B

m

y 1

t=0.4

t=0.7

t=1

y=tanh (m/t)

y=m

N

VTk t 2

B

cB

λµ=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

tm ~

tmtanh

VkNT 1

NVTk

1 t m tm

B

2B

c2B

cB λµ=⇒=

λµ∴

=⇒=⎟⎠⎞

⎜⎝⎛

This is consistent with the result we derived from the side T>Tc. 4. Spontaneous magnetization neat Tc: As T → Tc, m is small. Expansion of tanh x for small x:

Note that the behavior of M is similar to that of ∆ in the case of syuperconductivity at T ~ Tc.

33.0c

ccB

c3c

22

22

3

3

533

253

42

53

xx

xx

)TT( M givesy ation theorenormaliz accurate More

)T (T TT V

Nm M

)TT(TT3m

)t1(t3m

tm

31

tm m

tmtanhm

3x x

)O(x ! 2

x ! 3

x x

)! 2

x-(1 ) ! 5

x! 3

x x (

) ! 4

x! 2

x 1 2(

) ! 5

x! 3

x x 2(

eeeextanh

−∝

→−∝µ=∴

−=⇒

−=⇒

⎟⎠⎞

⎜⎝⎛−=⇒⎟

⎠⎞

⎜⎝⎛=

−≈

+−+≈

++++≈

+++

+++=

+−

= −

−

LL

L

L

M

T

TT~ c −

5. Low temperature excitations – magnons: Consider N spins coupled to their neighbors:

U is minimum when all spins are parallel (ground state) at T=0: If we consider the j-th spin in antiparallel to the others,

This will raise the system energy by an amount of 8JS2. This excitation energy will be smaller if we allow this antiparallel spin to be shared by all members of the system – formation of magnons. Consider the j-th spin in the system:

Sj is related to its magnetic moment as

n)interactio g(Heisenber SS 2J- U 1ii

N

1i+

=

⋅= ∑vv

…………….

Ground state

NS 2J- U =

[ ]

8JS U 8JS2NJS-

2)2N( 2JS-

SSSS -SSSS 2J- U

20

22

2

1jjj1j1ii

N

1ji1ii

2-j

1i

+=

+=

−−=

⎥⎦

⎤⎢⎣

⎡⋅−⋅⋅+⋅= +−+

+=+

=∑∑

vvvvvvvv

[ ][ ]1j1jj

1jjj1jj

SS S2J-

SSSS 2J- U

+−

+−

+⋅=

⋅+⋅=vvv

vvvv

[ ]

[ ]

)SS(S2J)S(dtd

BSg-)S(dtd B)S(

dtd

B spin th -j on the acting Torque

SS g2J B

asspin -jth on the acting field exchange tneas identified becan {}in termThe

SS g2J - U

Sg-

1j1jjj

j EjB

j j Ejj

j Ej

1j1jB

j E

1j1jB

jj

jBj

+−

+−

+−

+×=⇒

×µ

=⇒×µ=∴

×µ=

+⋅µ

−=

⎭⎬⎫

⎩⎨⎧

+⋅µ

−µ=∴

µ=µ

vvv

h

v

vv

h

vvvvh

vv

vvv

vvv

vv

Assume Sj is not off aligned with the other spins so that Sjx, Sj

y << Sjz. We can ignor

second order terms like SjxSj

y and approximate Sjz as S. The equations of the j-th spin

can the be written as:

u, v are constants, amplitude that measure the maximum deviation of the spin. Substitute these trial solutions into the differential equations:

∴ Sjx and Sj

y are 90o out of phase with equal amplitude. i.e. The spin is precessing circularly about the z-axis:

)]SS(S -)SS(S[2J)S(dtd y

1jy

1jz

jz

1jz

1jy

jx

j +−+− ++=∴h

t)-i(jkayj

t)-i(jkaxj

zj

x1j

x1j

xj

yj

y1j

y1j

yj

xj

ve S

ue S

:solution Trial

0)S(dtd

)]SS( -S2[2JS)S(dtd

)]SS( -S2[2JS)S(dtd

ω

ω

+−

+−

=

=

=

+−=

+=

h

h

iu- v u- vi-v u i-

: vandu for solving satisfy, iscondition thisIf)kacos1(Js4

]kacos1[4Js

0 i]kacos1[4Js

]kacos1[4Jsi

{if only solution trivialnon have v,u

]kacos1[u4Js- vi

:)S(dtdfor equation thefrom ,Similarly

]kacos1[v4Js ui

]ee2[v2Js ui

]eee2[2Js uei

22

yj

ikaika

]tka)1j[(i]tka)1j[(i)tjka(i)tjka(i

=⇒⎭⎬⎫

ω=ωω=ω

−=ω⇒

⎥⎦⎤

⎢⎣⎡ −=ω⇒

=⎥⎥⎥

⎦

⎥

⎢⎢⎢

⎣

⎢

ω−−

−ω

−

−=ω−

−=ω−⇒

−−=ω−⇒

−−=ω−

−

ω−+ω−−ω−ω−

h

h

h

h

h

h

h

h

Spin dynamics is “quantized” into magnons, each of energy . Any spin configuration can be expressed as combination of these magnons. Dispersion relation of magnons: 6. Thermal excitations of magnons: We can derive the thermal properties of magnons from the dispersion relationship. Similar to the case of phonons, magnons are bosons:

ωh

z

S

Top view:

“Normal mode” of spin dynamics

k. smallfor k E

aJsk2 )ka( 8Js~k, small For

kasin 8Js )kacos1(Js4

2

22221

212

∝∴

=⋅ω

=

−=ω

h

h

k

E

π/a -π/a 0

k. smallfor k E 2∝

43421

hh

h

h

h

h

h

h

hh

h

h

h

hhh

h

h

v

v

v

v

v

v

v

v

v

v

22 4A)(0.0587)(4

x0

23

2B

2k

BB

B

Bk0

23

22

Bk

23

22k

23

22

222

22

2

2

2

3

2

3

2

2

2

2222

Bkk

k

k

Bkk

1exdx

Jsa2Tk

4V n

dxTkd , Txk and x Tk

Let

1)Tk/exp(d

Jsa24V

1)Tk/exp(1

Jsa24Vd n

Jsa24V)D(

Jsa221

Jsa22V)D(

Jsa221

2Vk)D(

ddk

2Vk

ddk

V)(2

k4)D(

V)(2dkk4 d)D(

Jsa22 Jsa2

Jsa4 Jska4 dkd aJsk2

:k small of that by band whole the eApproximat1)Tk/exp(

1)D(d )n()D(d n

Nsn

)0(M

M(T)-M(0) M(0)

M

]Ns

n-1 M(0)[

]Ns

magnons ofNumber -1 M(0)[ M(T)

Ns. ofout flipspin 1 toscorrespondmagnon 1 s. ofspin a hasmagnon Each1)Tk/exp(

1 n

π=π=

∞

∞

−⎟⎠⎞

⎜⎝⎛⋅

π=

=ω=ω=ω

−ωω

ω⎟⎠⎞

⎜⎝⎛⋅

π=

−ω⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ω⎟

⎠⎞

⎜⎝⎛⋅

πω=∴

ω⎟⎠⎞

⎜⎝⎛⋅

π=ω⇒

ω⋅

ω⋅

π=ω⇒

ω⋅

π=ω⇒

ωπ=

ωππ

=ω⇒π

π=ωω∴

ω=

ω==

ω⇒=ω

−ωωω=>ω<ωω=

==∆

=

=∴

−ω=><

∫∑

∫

∫∑

∫∫∑

∑

∑

Ferrimagnetic order 1. Example: Magnetite At low temperatures: In genetal:

M

T

TT~ c −

M(0) T ~M 23

∆

re. temperatulowat T ~M(0)

M

)VN(n

Jsa2Tk

nsA

Jsa2Tk

Ns0.0587V

Nsn

M(0)

M

Jsa2TkAV n

23

23

2B

23

2Bk

23

2B

k

∆∴

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛==

∆

⎟⎠⎞

⎜⎝⎛=∴

∑

∑

v

v

Fe3O4

FeO

Fe2O3

{ 23 FeFe ++

43421

N(0) is much smaller than that by considering Fe3O4 as ferromagnetic.

Exchange field on site A: Exchange field on site B: For ferromagnetic order to occur, λ >> µ, ν: Let the Curie constants of sublattice A and B be CA and CB respectively. Mean field theory, when T>Tc:

Ferrimagnetic ordering

A A A

B B B

321

v

321

vv

B sublattice toDue

B

A sublattice toDue

AA M - M B λµ=

321

v

43421

vv

B sublattice toDue

B

A sublattice toDue

AB M M B νλ +−=

43421

vv

43421

vv

A sublattice toDue

AB

B sublattice toDue

BA

M B

M B

λ

λ

−=

−=∴

{

{

⎪⎩

⎪⎨

⎧

=+

=+⇒

⎪⎭

⎪⎬

⎫

−=

−=

=⇒

=−⇒

=

=

−=+=

−=+=

aB

AB

B

aA

BA

A

AaB

B

BaA

A

BAc

BA22

c

cB

Ac

BAAc

AaB

A field Applied

aB

BaA

A field Applied

aA

B T

CM

TC

M

B T

C MTC M

)M B(

TC

M

)M B( T

C M

CCT

0 CCT

0 TCCT

:M andM ofsolution trivial-nonfor ,B ignoring ,TTAt

)M B( T

C )BB (

TCM

)M B( T

C )BB ( TCM

vvv

vvv

vvv

vvv

vvv

vvvvv

vvvvv

λ

λ

λ

λ

λ

λ

λλ

λ

λ

Antiferromangetism 1.

2c

2BAcBA

BA22

BABA

2BA

2

2BABA

a

BA

a2BABA

BA2BA

2

a2BAB

B2BA

2

a2BAA

A2BA

2

TTCCT2-T)CC(

CCTCC2-T)CC(

TCC

-1

TCC2

TCC

B

M M

B )T

CC2T

CC()M M)(

TCC

-(1

B )

TCC

TC

(M) T

CC-(1

B )T

CCT

C(M)

TCC

-(1

−

+=

−+

=

−+

=+

=⇒

−+

=+⇒

⎪⎪⎩

⎪⎪⎨

⎧

−=

−=⇒

λλ

λ

λ

χ

λλ

λλ

λλ

v

vv

vvv

vv

vv

Ferromagnetic ordering Antiferromagnetic ordering

Exchange field: J>0 Exchange field: J<0

T

χ

Tc T

χ

TN

Critical temperature: Curie temperature Tc

Critical temperature: Neel temperature TN

2. Antiferromagnetism is a special case of ferrimagnetism with CA=CB, i.e. Experimentally, 3. When T<TN: Case 1. If Ba ⊥ axis of spin Let M = | MA | = | MB |

N

2N

2

2N

2c

2BAcBA

2Nc

TTC2

TT

CT2-CT2

TTCCT2-T)CC(

C C T T

+=⇒

−=

−

+=

==→

χ

χ

λλ

n)interactioneighbor nearest -next of because Texactly not is ( T

C2 Nθθ

χ−

=

MA

MB

2φ

Ba

0B2 - M4 0ddU

: Uminimize To

B2])2 (21-[1 M-

sinB22 cos M-

)(90 cosB2)2(180 cos M

)MM(BMM

)MM(B)MM-MM(21-

fields) exchange are B and B( )MM(B)MBMB(21- U

a2

a22

a2

oa

o2BAaAB

BAaBAAB

BABAaBBAA

=⇒=

−≈

−=

−⋅−−=

+⋅−⋅=

+⋅−⋅−⋅−=

+⋅−⋅+⋅=

M

M

M

M

φλφ

φφλ

φφλ

φφλ

λ

λλvvvvv

vvvvvvv

vvvvvvvvv

Case 2. If Ba // axis of spin There is no change in U. ∴ χ//(0) = 0

)TC ( 1

B1

M2B

2M B

2Msin

M2B

0MB2 - M4 0ddU

: Uminimize To

B2])2 (21-[1 M-

sinB22 cos M-

)(90 cosB2)2(180 cos M

)MM(BMM

)MM(B)MM-MM(21-

fields) exchange are B and B( )MM(B)MBMB(21- U

Na

a

a

a

a2

a22

a2

oa

o2BAaAB

BAaBAAB

BABAaBBAA

==⋅⋅≈=∴

=⇒

=⇒=

−≈

−=

−⋅−−=

+⋅−⋅=

+⋅−⋅−⋅−=

+⋅−⋅+⋅=

⊥ λλφχ

λφ

φλφ

φφλ

φφλ

φφλ

λ

λλ

M

M

M

vvvvv

vvvvvvv

vvvvvvvvv

MA MB

Ba

T TN

χ

θ

χ⊥

χ//

4. Antiferromagnetic magnons:

From case of ferromagnetism: Now, if lattice A corresponds to even indices (2j) and lattice B to odd indices (2j+1), then S2j

x should have opposite sign with S2j-1x and S2j+1

x. Rewriting the equation for ferromagnetism:

)]SSS2[2iJS)S(dtd

iSSSLet

)]SS -S2[2JS)S(dtd

similarly, and

)SS( )]SSS2[2JS)S(dtd

magnetic)(antiferro )]SS(S -)SS(S[2J)S(dtd

tic)(feromagne )SS(S -)SS(S[2J)S(dtd

1j21j2j2j2

yx

x1j2

x1j2

xj2

yj2

zj2

y1j2

y1j2

yj2

xj2

y1j2

y1j2

zj2

z1j2

z1j2

yj2

xj2rewritten

y1j

y1j

zj

z1j

z1j

yj

xj

++

+−

++

+

+−

+−

+−+−

+−+−

++=∴

+=

−−=

=−−−=⇒

+−−=⎯⎯⎯ →⎯

++=

h

h

h

h

h

iu- v u- vi-v u i-

: vandu for solving satisfy, iscondition thisIf)kacos1(Js4

]kacos1[4Js

0 i]kacos1[4Js

]kacos1[4Jsi

{if only solution trivialnon have v,u

]kacos1[u4Js- vi

:)S(dtdfor equation thefrom ,Similarly

]kacos1[v4Js ui

]ee2[v2Js ui

]eee2[2Js uei

22

yj

ikaika

]tka)1j[(i]tka)1j[(i)tjka(i)tjka(i

=⇒⎭⎬⎫

ω=ωω=ω

−=ω⇒

⎥⎦⎤

⎢⎣⎡ −=ω⇒

=⎥⎥⎥

⎦

⎥

⎢⎢⎢

⎣

⎢

ω−−

−ω

−

−=ω−

−=ω−⇒

−−=ω−⇒

−−=ω−

−

ω−+ω−−ω−ω−

h

h

h

h

h

h

h

h

Corresponding equations for lattice B:

( )

|ka sin|4JS

ka sin4JS

kacos14JS

kacos4JS4JS

0kacos4JS4JS4JS 0 4JSkacos4JS

kacos4JS4JS:solution trivial-non For

]kacosuv[ v

]ue uev2[2iJS- u i-

]kacosvu[4JS- u

] ve veu2[2iJS u i-

become equations aldefferenti Above

u S and u S

:solution Trial

)]SSS2[2iJS)S(dtd

iSSSLet

]SS -S2[2JS)S(dtd

)SS( ]SSS2[2JS)S(dtd

22

2

22

2

222

2

ikaika

ikaika

t]-1)kai[(2j1j2

t]-i[(2j)kaj2

2j2j21j21j2

yx

x2j2

xj2

x1j2

y1j2

zj2

y2j2

yj2

y1j2

x1j2

⎟⎠⎞

⎜⎝⎛=ω⇒

⎟⎠⎞

⎜⎝⎛=ω⇒

−⎟⎠⎞

⎜⎝⎛=ω⇒

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=ω⇒

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −ω⎟⎠⎞

⎜⎝⎛ +ω⇒=

−ω−

+ω

+=ω⇒

++=ω

+=ω⇒

++=ω

==

++−=∴

+=

−−=

=−−=

−

−

ω+++

ω+

++

+++

++

+

+++

+++

h

h

h

hh

hhhhh

hh

h

h

h

h

h

h

Domains 1. Spin in a material with long range magnetic ordering (ferromagnetic, antiferromagnetic etc.) form domains. 2. Reason for domain formation: 3. For small field: domain size will change in accordance to the direction of the magnetic field. Change in domain size can be reversible or irreversible.

k 0

E

π/a -π/a

k. smallfor k E∝

Antiferromagnetism

E

π/a -π/a

k. smallfor k E 2∝

k 0

Ferromagnetism

Higher energy Lower energy

4. For large field: domain magnetization will re-align with the external field. 5. 6. Irreversible boundary displacement and magnetization rotation are the causes of hystersis:

Ba

Ba

Reversible boundary displacement

Irreversible boundary displacement

M

Ba

saturation Magnetization rotation

B

H

Saturation Bs

Coercivity Hc (field needed to reduce B back to 0)

Remanence Br