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
Top Related