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Transcript
Magnetism of the Localized Electrons on the Atom
1.
The hydrogenic atom and angular momentum
2.
The many-electron atom
3.
Spin-orbit coupling
4.
The Zeeman interaction
5.
Ions in solids
6.
Paramagnetism
Comments and corrections please: [email protected]
Dublin January 2007
1
1. The hydrogenic atom and angular momentum
Consider a single electron in a central potential. A hydrogenic atom is composed of a nucleus of charge
Ze at the origin and an electron at r,!,". First, consider a single electron in a central potential "e =
Ze/4!e 0r
z
2
2
2
! = - (" /2m)# - Ze /4!e 0r
l -e
r
In polar coordinates:
!
#2 = #2/#r2 +(2/r)#/#r + 1/r2{#2/#!2 + cot!$/#! + (1/sin2!)#2/#2"}
Ze
"l
The term in parentheses is -l2. Schrödinger’s equation is ! % = E%
x
The wave function % means that the probability of finding the electron in a small volume dV ar r
is %*(r)%(r)dV. (%* is the complex conjugate of %).
Eigenfunctions of the Schrödinger equation are of the form %(r,!,") = R(r)&(!)'(").
( The angular part &(!)'(") is written as Ylml(!,").
The spherical harmonics Ylml(!,") depend on two integers l, ml, where l is $ 0 and |ml| % l.
'(") = exp(iml") where ml = 0, ±1, ±2 .......
The z-component of orbital angular momentum, represented by the operator lz = -i"#/#",
has eigenvalues <'| lz |'> = ml".
&(!) = Plml(cos!), are the associated Legendre polynomials with l $ |ml|,
so ml = 0, ±1, ±2,....±l.
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2
z
The square of the orbital angular momentum l2 has eigenvalues l(l+1)".
The orbital angular momentum has magnitude )[l(l+1)]" and its projection
along z can have any value from -l" to +l". The quantities lz and l2 can be
measured simultaneously (the operators commute). In the vector model,
The total angular momentum is a vector which precesses around z.
m l"
)[l(l+1)]"
Spherical harmonics.
s
p
d
f
Y00 = )(1/4!)
Y10 = )(3/4*) cos !
Y20 = )(5/16*)(3cos2! - 1)
Y1±1 = ±)(3/8*) sin ! e±i"
Y2±1 = ±)(15/8*) sin! cos! e±i"
Y2±2 = )(15/32*) sin2! e±2i"
Y30 = )(7/16*)(5cos3! - 3cos!) Y3±1 = ±)(21/64*)(5cos2! - 1)sin! e±i"
Y3±3 = ±)(35/64*) sin3! e±3i"
Y3±2 = )(105/32*) sin2!cos! e±2i"
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3
Spherical harmonics.
n =1
l=0
s
n =2
l=1
ml =0
p
ml =±1
n =3
l=2
ml =0
ml =±1
d
ml =±2
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The radial part of the wavefunction R (r)
•The radial part R(r) depends on l and also on n,
the total quantum number; n > l; hence l = 0, 1,
......(n-1).
0.5
10
2
! |Rnl (! )|
0.4
R(r) = Vnl(Zr/na0)exp[-(Zr/na0)]
0.3
"2/me2
= 1. Here a0 = 4!+0
= 52.9 pm is the
first Bohr radius, the basic length scale in
atomic physics.
2
V10
0.6
21 20
0.2
32
31
0.1
30
43
42
41
40
0.0
0
5
10
15
20
25
30
! = r / a0
The energy levels of the 1-electron atom are
E = -Zme4/8h2+02n2 = -ZR0/n2
The quantity R0 = me4/8h2+0 = 13.6 eV is the Rydberg, the basic energy in atomic physics.
For the central Coulomb potential "e, the potential energy V(r) depends only on r, not on ! or ". E
depends only on n.
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The three quantum numbers n, l, ml denote an orbital, a spatial distribution of electronic charge.
Orbitals are denoted nx, x = s, p, d, f for l = 0, 1, 2, 3. Each orbital can accommodate
up to two electrons with spin ms = ±1/2. No two electrons can be in a state with the same four quantum
numbers (Pauli exclusion principle). The hydrogenic orbitals are listed in the table
n
l
ml
ms
No of states
1s
1
0
0
±1/2
2
2s
2
0
0
±1/2
2
2p
2
1
0,±1
±1/2
6
3s
3
0
0
±1/2
2
3p
3
1
0,±1
±1/2
6
3d
3
2
0,±1,±2
±1/2
10
4s
4
0
0
±1/2
2
4p
4
1
0,±1
±1/2
6
4d
4
2
0,±1,±2
±1/2
10
4f
4
3
0,±1,±2,±3
±1/2
14
•The Pauli principle states that no two electrons can have the same four quantum numbers. Each
orbital can be occupied by at most two electrons with opposite spin.
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2. The many-electron atom
Hartree-Foch approximation, each electron sees
a different self-consistent potential Vl(r).
!
E
Coulomb interactions between electrons
n=
n
0
6s
6p
4p
4s
6d
1
2
3
4
5
6
1s
2s
3s
4s
5s
6s
2p
3p
4p
5p
6p
3d
4d
5d
6d
4f
5f
6f
5g
6g
6f
4
3
3d
3p
3s
2
2p
2s
1s
l
s
p
d
f
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Example: The carbon atom
1s22s22p2
There are no options for the
first four electrons. There
must be a pair of them with
opposite spin in each s
orbital.
However, there are 15 ways
of accomodating two
electrons in the p orbitals.
The 15 states fall into three
groups (terms).
Notation: States with L =
0,1,2,3,4,5,6 are S,P,D,F,G,H
2S+1X
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Term
1S
3P
1D
S
J
L
S
(ML, MS)
0
1
2
0
1
0
(0,0)
(1,1)(1,0)(1,-1)(0,1)(0,0)(0.-1)(-1,1)(-1,0)(-1,-1)
(2,0)(1,0)
(2,1)(2,0)(0,0)(-1,0)(-2,0)
L
The terms are widely separated in energy; one is the lowest
Addition of L and S in the vector model
Finally we need to couple the spin and orbital angular momentum
to form a resultant J. J = L + S
L-S % J % L+S
Hund’s rules; A prescription to give the ground state of a multi-electron atom
1) First maximize S for the configuration
2) Then maximize L consistent with that S
3) Finally couple L and S; J = L - S if shell is < half-full; J = L + S if shell is > half-full.
In the example, S = 1, L = 1, J = 0. The ground state of carbon is 3P0, which is nonmagnetic (J = 0).
General notation for multiplets is 2S+1XJ where X = S, P, D ...... for L = 0, 1, 2 .....
In spectroscopy, the energy unit cm-1 is used. Handy conversions are:1 eV , 11605 K and 1 cm-1 , 1.44 K
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Some examples:
Fe3+
3d5
S = 5/2 L = 0
-----| ooooo
J = 5/2
6S
Ni2+
S=1
-----|...oo
J=4
3F
3d8
L=3
5/2
4
Nd3+
4f3
S = 3/2 L = 6
---oooo |ooooooo
4I
J = 9/2
9/2
Dy3+
4f9
S = 5/2 L = 5
-------|..ooooo
6H
J = 15/2
15/2
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3. Spin-orbit coupling
This relatively-weak relativistic interaction is responsible for Hund's third rule. In the multi-electron
atom, the spin-orbit term in the Hamiltonian can be written as
! so = /L.S
/ is > 0 for the first half of the 3d or 4f series and < 0 for the second half. It becomes large in
heavy elements. / is related to the one-electron spin-orbit coupling constant 0 by / = ±0/2S for the
first and second halves of the series. The resultant angular momentum (see above) is
J =L±S
The identity J2 = L2 + S2 + 2 L.S is used to evaluate Hso.The eigenvalues of J2 are J(J + 1) "2 etc,
hence L.S can be calculated.
1
3+
Spin-orbit coupling constants in the
3d and 4f series
L
ion
4f
Ce
920
4f2
Pr3+
540
4f3
Nd3+
430
3d1
Ti3+
124
4f5
Sm3+
350
3d2
Ti2+
88
4f8
Tb3+
-410
3d3
V2+
82
4f9
Dy3+
-550
3d4
Cr2+
85
4f10
Ho3+
-780
3d6
Fe2+
-164
4f11
Er3+
-1170
3d7
Co2+
-272
4f12
Tm3+
-1900
3d8
Ni2+
-493
4f13
Yb3+
-4140
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9
L
S
J
8
7
6
5
4
3
2
1
0
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Application of Hund’s rules to the trivalent rare-earth ions.
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Magnetic properties of free atoms: only the elements marked in bold are nonmagnetic, w ith J = 0
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4. The Zeeman interaction
The magnetic moment of an ion is represented by the term m = (µB/") (L + 2S)
The Zeeman Hamiltonian for the magnetic moment in a field B applied along z is –m.B
! Z = (µB/")(L + 2S).B
z
The vector model of the atom, including
magnetic moments. First project m onto J. J
then precesses around z.
We define the g-factor for the atom or ion as the
ratio of the component of magnetic moment
along J in units of µB to the magnitude of the
angular momentum in units of ".
g = -(m.J/µB)/(J2/") = m.J/J(J + 1)"µB.e
but
S
m
J
S
L
m.J = (µB/"){(L + 2S).(L + S)}
J2 = J(J + 1)"2;
(µB/"){(L2 + 3L.S + 2S2)}
(µB/"){(L2 + 2S2 + (3/2)(J2 - L2 - S2)}
(µB/"){((3/2)J2 – (1/2)L2 + (1/2)S2)}
(µB /"){((3/2)J(J + 1) – (1/2)L(L + 1) + (1/2)S(S + 1)}
Jz = MJ"
hence
g = 3/2 + {S(S+1) - L(L+1)}/2J(J+1)
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The g-factor is also the ratio of the z-component of magnetic moment (in units of µB)
to the z component of angular momentum (in units of ")
mz/Jz = m.J/J2 = gµB/ "
The magnetic Zeeman energy is EZ = –mz B. This is –(mZ /Jz).(Jz B) = (gµB&/")JzB
Hence
EZ = -gµBMJB
MJ
-5 / 2
Note the magnitudes of the energies involved, with g
= 2 and µB = 9.27 10-24 J T-1. The splitting of two
J=5/2
adjacent energy levels is gµBB. For B = 1 T, this is
only ' 2 10-23 J, equivalent to 1.4 K. [kB = 1.38 10-23
J K-1]
-3 / 2
-1 / 2
1/2
3/2
5/2
In a large field at low temperature the moment is
gµBB
saturated at the value -gµBJ Bohr magnetons.
The effect of applying a magnetic field on an ion with J = 5/2.
Susceptibility gives meff2 = g2µB2J(J+1)
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Co2+ free ion
Energy levels of Co2+ ion, 3d7. Note that the
Zeeman splitting is not to scale.
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Data on rare-earth ions
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Data on 3d ions
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Summary:
For free ions:
( Filled electronic shells are not magnetic. A - and a . electron is paired in each orbital
( Only partly-filled shells may possess a magnetic moment
( The magnetic moment is related to the total angular momentum by m = g(µB/ ")J, where J is
the total quantum number given by Hund’s rules.
The third point has to be modified for ions in solids
( Orbital angular momentum for 3d ions is quenched. The spin-only moment is m = gµBS, with
g =2.
( Certain crystallographic directions become easy axes of magnetization – magnetocrystalline
anisotropy.
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5. Ions in solids
When an ion is embedded in a solid, the Coulomb interaction of the charge distribution of the ion
with the potential 1cf created by the surrounding charges is the crystal-field interaction.
! cf = ( 1cf(r)2(r) d3r
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The Hamiltonian now has four terms
! = ! 0 + ! so+ ! cf+ ! Z
Typical magnitudes of energy terms (in K)
!0
! cf
! Z in 1 T
3d 1 - 5 104 102 -103
1 - 104
1
1 - 6 105 1 - 5 103
!3 102
1
4f
! so
! so must be considered before ! cf for 4f ions, and the converse for 3d ions. Hence J is a good
quantum number for 4f ions, but S is a good quantum number for 3d ions.
The 4f electrons are generally localized, and 3d electrons are localized in oxides and other ionic
compounds.
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The most common coordination for 3d ions is 6-fold (octahedral) or 4-fold (tetrahedral). Both have cubic
symmetry, if undistorted. The crystal field can be estimated from a point-charge sum.
q
5.1 One-electron states
Octahedral and tetrahedral sites.
To demonstrate quenching of orbital angular momentum, we consider the l = 1 states %0, %1, %-1
corresponding to ml = 0, ±1.
%0
=
R(r) cos !
%±1
=
R(r) sin ! exp {±3"}
The functions are eigenstates in the central potential V(r) but they are not eigenstates of ! cf. Suppose
the oxygens can be represented by point charges q at their centres, then for the octahedron,
! cf = eVcf = Dq(x4 +y4 +z4 - 3y2z2 -3z2x2 -3x2y2)
where D ' e/4pe oa6. But %±1 are not eigenfunctions of Vcf, e.g. (%i*Vcf%jdV) 4ij, where i,j = -1, 0, 1.
We seek linear combinations that are eigenfunctions, namely
%0
= R(r) cos !
(1/*2)(%1 + %-1)= R’(r)sin!cos"
(1/*2)(%1 - %-1)= R’(r)sin!sin"
= zR(r) = pz
= xR(r) = px
= yR(r) = py
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Note that the z-component of angular momentum; lz" = i"#/#" is zero for these wavefunctions. Hence
the orbital angular momentum is quenched.
The same applies to 3d orbitals; the eigenfunctions there are
dxy = (1/*2)(%2 - %-2) =
dyz = (1/*2)(%1 - %-1) =
dzx = (1/*2)(%1 + %-1) =
dx2-y2 = (1/*2)(%2 + %-2) =
=
d3z2-r2 = %0
R’(r)sin2!sin2"
R’(r)sin!cos!sin"
R’(r)sin!cos!cos"
R’(r)sin2!cos2"
R’(r)(3cos2! 5 1)
The three p-orbitals are degenerate in a
cubic crystal field, whether octahedral
or tetrahedral, whereas the five d-orbitals
split into a group of three t2g and a group
of two eg orbitals
'
'
'
'
'
xyR(r)
yzR(r)
zxR(r)
(x2-y2)R(r)
(3z2-r2)R(r)
t2g orbitals
eg orbitals
dx2-y2, dz2,
eg
dxy,dyz, dzx
px,, py, pz
oct / tet
6Dq
10Dq
dxy,dyz, dzx
oct
t2
t2g
dx2-y2, dz2,
e
tet
Notation; a or b denote a nondegenerate single-electron orbital, e a twofold degenerate orbital and t a
threefold degenerate orbital. Capital letters refer to multi-electron states. a, A are nondegenerate and
symmetric with respect to the principal axis of symmetry (the sign of the wavefunction is unchanged), b.
B are antisymmetric with respect to the principal axis (the sign of the wavefunction changes). Subscripts
g and u indicate whether the wavefunction is symmetric or antisymmetric under inversion. 1 refers to
mirror planes parallel to a symmetry axis, 2 refers to diagonal mirror planes.
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s, p and d orbitals in the crystal field
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As the site symmetry is reduced, the degeneracy of the one-electron
energy levels is raised. For example, a tetragonal extension of the
octahedron along the z-axis will lower pz and raise px and py. The effect
on the d-states is shown below. The degeneracy of the d-levels in
different symmetry is shown in the table.
Px,py
pz
The effect of a tetragonal distortion of octahedral symmetry on the
one-electron energy levels.
The splitting of the 1-electron levels
in different symmetry
.
l
Cubic Tetragonal Trigonal
Rhombohedral
s
1
1
1
1
1
p
2
3
1,2
1,2
1,1,1
d
3
2,3
1,1,1,2
1,2,2
1,1,1,1,1
f
4
1,3,3
1,1,1,2,2
1,1,1,2,2
1,1,1,1,1,1,1
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One-electron energy diagrams
b1g
0
:
e
1/2:
1/20
a1g
3/590
90
b2g
2/590
e
8
7
1/38
eg
2/37
6
4 e
a1
field-free
ion
octahedral
Oh
tetragonal
D4h
trigonal
C3v
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monoclinic
C2
27
z
z
Jahn Teller Effect
x
x
x
y
y
x
x
z
x
z
z
8
8
Dcfse
E
90
4
4
Dublin January 2007
•A system with a single electron
(or hole) in a degenerate level
will tend to distort
spontaneously. The effect is
particularly strong for d4 and d9
ions in octahedral symmetry
(Mn3+, Cu2+) which can lower
their energy by distorting the
crystal environment. This is the
Jahn-Teller effect. If the local
strain is +, the energy change 9
E = -A+ +B+2, where the first
term is the crystal-field
stabilization energy Dcfse and
the second term is the increased
elastic energy.
The J-T distorsion may be static
or dynamic.
28
dx2-y2, dz2
eg
dxy, dyz, dxz
t2g
3/59o
dxy, dyz, dxz
t2
2/59c
2/59t
9o
E
9t
9c
3/59t
3/59c
2/59o
e
t2g
dx2-y2, dz2
dxy, dyz, dxz
eg
dx2-y2, dz2
cubic
tetrahedral
spherical
Dublin January 2007
octahedral
29
T1g
5.2 Many-electron states
A2g
T1g
P
In insulators, the electrons in an unfilled shell interact
strongly with each other giving rise to a series of sharp
energy levels which are determined by the action of the
crystal field on the orbital terms of the free atom. The
spacing of theses levels may be determined by
spectroscopy, and the crystal-field determined.
T2g
F
T1g
T2g
T1g
A2g
E
Orgel Diagrams
!0
These diagrams show the effect of a cubic crystal field
on the Hund’s rule ground state term. Since a half-filled
shell has spherical symmetry, the cases dn and d5+n are
equivalent. Also, since a hole is the absence of an
electron, the cases dn and d10-n are related.
!0
0
d3, d8 octahedral
(d2, d7 tetrahedral)
d2, d7 octahedral
(d3, d8 tetrahedral)
Eg
T2g
E
D
S
A1
A1
T2g
Eg
!0
; Do
0
Dt <
4
9
d , d octahedral
(d1, d6 tetrahedral)
d5 octahedral or tetrahedral
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0
!0
d1, d6 octahedral
(d4, d9 tetrahedral)
30
High-spin and low-spin states
An ion is in a high-spin state or a low spin state, depending on whether the Coulomb interaction U leading to
Hund’s first rule (maximize S) is greater or less than the the crystal-field splitting D.
Consider a 3d6 ion such as Fe3+.
eg
eg
eg
t2g
t2g
eg
D
U
D
t2g
U
t2g
U > D, gives a high-spin state, S = 2 e.g. FeCl2
U < D, gives a low-spin state, S = 0 e.g. Pyrite FeS2
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Tanabe-Sugano diagrams
These show the splitting of the ground state and higher terms by the crystal field. The high-spin < lowspin crossover is seen. Diagrams shown are for d-ions octahedral environments.
d6
1
3
3
3
3
I
G
F
P
3
T2 g
3
A1g
3
T1g
3
Eg
3
Eg ; (t2 g ) (eg )3
H
E
5
3
D
1
A1g (t2 g ) 6
3
T2 g
3
T1g
3
T2 g
3
T1g
5
T2 g ; (t2 g )(eg ) 2
crystal field splitting
Dublin January 2007
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d5
Matching the optical absorption spectrum of Fe3+ - doped Al2O3 with the calculated Tanabe-Sugano energy-level
diagram to determine the cubic crystal field splitting at octahedral sites.
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d2
d7
Note the similarities between the Tanabe-Sugano diagrams for d2 and d7. The differences are associated with the
possible low-spin states for d7 (e.g. Co2+)
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d3
For Cr3+ in Al2O3, the cf parameter Dq/B is 2.8
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d3
For Cr3+ in Al2O3, the cf parameter Dq/B is 2.8
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5.3 Single-ion anisotropy
The electrostatic interaction of the ionic charge distribution 21(r) with the potential 1cf created by the rest of the crysta
gives rise to the crystal field splittings. It is also the source, via spin-orbit coupling, of magnetocrystalline anisotropy.
E = = 1cf 2(r)dr
where
1cf(r) = -(e/4*+0) = {2(R) / |R - r|} dR
The anisotropy energy is therefore
Ea(r) = -(e/4*+0) = {2(r,!") 2(R ) / |R - r|}dr dR
Both the charge distribution 2(r) and the potential 1cf(r) can be expanded in spherical harmonics.
Using the Wigner-Eckart theorem, it is possible to write the corresponding crystal-field Hamiltonian in terms of angular
momentum operators Jx, Jxy Jz J2 which is a particularly useful way to find the energy-levels (eigenvalues). The Hamiltonian
matrix is written in an ML or MJ basis for the 3d transition elements or 4f rare earths respectively. In concise form
! cf = >>n=2,4,6>m=-nn Bnm Ô nm
Crystal field parameters
!n?rn@Anm
Stevens coefficients
Stevens operators
Crystal field coefficients
In a site with uniaxial anisotropy, the leading term is ! cf = B20 Ô20 The Stevens operator Ô 20 is {3Jz2 - J(J+1)}
! cf = B20 {3Jz2 - J(J+1)}
Dublin January 2007
37
9/2
Charge distributions of the rare-earth ions. Those with a positive quadrupole moment (!2 > 0), italic type are
distinguished from those with a negative quadrupole moment (!2 < 0) bold type. Note the quarter-shell changes,
±9/2
±1/2
3+
±3/2
e.g. Nd J = 9/2
±5/2
±7/2
±1/2
±9/2
B2 0 < 0
B20 > 0
Dublin January 2007
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z
6 Paramagnetism
6.1 Langevin theory
2!sin"d"
"
m
H
Dublin January 2007
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39
6.2 Brillouin theory
The general quantum case was treated by Brillouin; m is gµBJ, and x is defined as x = µ0mH/kBT.
There are 2J+1 energy levels Ei = -µ0gµBMJH, with moment mi =gµBMJ where MJ = J. J-1, J-2, … -J.
The sums over the energy levels have 2J+1 terms. Their populations are proportional to exp(-Ei/kT)
a) Susceptibility To calculate the susceptibility, we can take x << 1, because the susceptibility is defined
as the initial slope of the magnetization curve. We expand the exponential as exp(x) = 1 + x + ..,
<m> = >-JJgµBMJ(1 + µ0gµBMJH/kBT)/>-JJ(1 + µ0gµBMJH/kBT) MJ
Recall >-JJ1 = 2J + 1
-5/2
-3/2
>-JJMJ = 0
J=5/2
-1/2
>-JJMJ2 = J(J + 1)(2J + 1)/3
1/2
Hence <m> = µ0g2µB2HJ(J + 1)(2J + 1)/3(2J + 1)kBT
3/2
The relative susceptibility is N<m>/H, where N is the
5/2
-5/2
3
number of atoms/m .
-3/2
Ar = µ0Ng2µB2J(J + 1)/3kBT
H
-1/2
1/2
This is the general form of the Curie law. Again it can be
written Ar = C/T where the Curie constant
3/2
2
2
2
C = µ0Ng µB J(J+1)/3kB or C = µ0Nmeff /3kB where
5/2
meff = gµB)[J(J+1)]. A typical value of C for J = 1, N = 8.1028 m-3 is 3.5 K.
Note that results for the classical limit and S = 1/2 are obtained when J < B (m = gµBJ) and J = 1/2, g =
2. (m = µB) .
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Magnetization Curve
To calculate the complete magnetization curve, set y = µ0gµBH/kBT,
then
<m> = gµB#/#y[ln>-JJ exp{MJy}]
[d(ln z)/dy = (1/z) dz/dy]
The sum over the energy levels must be evaluated; it can be written as
exp(Jy) {1 + r + r2 + .........r2J}
where r = exp{-y}
The sum of a geometric progression (1 + r + r2+ .... + rn) = (rn+1 - 1)/(r - 1)
C >-JJ exp{MJy} = (exp{-(2J+1)y} - 1)exp{Jy}/(exp{-y}-1)
multiply top and bottom by exp{y/2}
= [sinh(2J+1)y/2]/[sinh y/2]
<m> = gµB($/#y)ln{[sinh(2J+1)y/2]/[sinh y/2]}
= gµB/2 {(2J+1)coth(2J+1)y/2 - coth y/2}
setting x = Jy, we obtain <m> = mBJ(x)
1.0
where the Brillouin function BJ(x) =
{(2J+1)/2J}coth(2J+1)x/2J - (1/2J)coth(x/2J).
0.8
1/2
Again, this reduces to the previous equations
in the limits J < B (m = gµBJ) and J = 1/2, g = 2.
2
0.6
B
,
0.4
0.2
Comparison of the Langevin function and the
Brillouin functions for J = 1/2 and J = 2.
0
0
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4
6
8
x
41
7
S = 7/2 (Gd3+)
µB per ion
Magnetic moment,
6
5
S = 5/2 (Fe3+)
4
3
S = 3/2 (Cr3+)
1.30K
2.00K
3.00K
4.21K
Brillouin functions
2
1
0
0
1
2
3
4
-1
BJ /T (TK )
Reduced magnetization curves for three paramagnetic salts, with Brillouin-theory predictions
The theory of localized magnetism gives a good account of magnetically-dilute 3d and 4f salts where the
magnetic moments do not interact with each other. Except in large fields or very low temperatures, the
M(H) response is linear. Fields > 100 T would be needed to approach saturation at room temperature.
The excellence of the theory is illustrated by the fact that data for quite different temperatures superpose
on a single Brillouin curve plotted as a function of x ' H/T
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