Fundamentals of Magnetism – 2 J. M. D. Coey

School of Physics and CRANN, Trinity College Dublin

Ireland.

4. Magnetism of the electron

5.  The multi-electron atom

6.  The single-electron atom

7.  Ions in solids

www.tcd.ie/Physics/MagnetismComments and corrections please: jcoey@tcd.ie

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Lecture 2 covers the origin of magnetism in solids, in the spin and orbital moments of the electron. Paramagnetism of non-interacting electrons is discussed in the localized limit. The multi-electron atom is analysed, and the influence of the local crystalline environment and ligand field on its paramagnetism is explained. Finally, magnetic ordering is discussed, in terms of molecular field theory.

An elementary knowledge of atomic physics, and quantum mechanics is assumed.

4. Magnetism of the electron

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Orbital and Spin Moment

Magnetism in solids is due to the angular momentum of the atomic electrons.

Two contributions to the electron moment:

•  Orbital motion about the nucleus

•  Spin- the intrinsic (rest frame) angular m momentum.

(a) (b)

← ←

(a) (b)

← ←

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Einstein-de Hass Experiment

Demonstrates the relation between magnetism and angular momentum.

A ferromagnetic iron rod is suspended on a torsion fibre.

The field in the solenoid is reversed, switching the direction of magnetization of the rod.

An angular impulse is delivered due to the reversal of the angular momentum of the electrons- conservation of angular momentum.

Fe has 26 electrons, moment per Fe is that of 2.2e

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Paradigm shift was needed to explain 1)  Amperian Currents2)  Weiss field 3)  Bohr - van Leewen theorem.

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Bohr-van Leeuwen Theorem

A famous and disconcerting result od classical statistical mechanics;

At any finite temperature, and in all finite electric or magnetic fields, the net magnetization of a collection of electrons in thermal equilibrium vanishes identically!

Origin of Magnetism

1930 Solvay conference

At this point it seems that the whole of chemistry and much of physics is understood in principle. The problem is that the equations are much to difficult to solve….. P. A. M. Dirac

Dirac and Heisenberg in red B. Cabrera in yellow

(

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The electron

The magnetic properties of molecules and solids derive essentially from the magnetism of their electrons. (Nuclei also possess magnetic moments, but they are ≈ 1000 times smaller).

An electron is a point particle with:mass me = 9.109 10-31 kgcharge -e = -1.602 10-19 Cintrinsic angular momentum (spin) ½ħ = 0.527 10-34 J s

On an atomic scale, magnetism is always associated with angular momentum. Charge is negative, hence the angular momentum and magnetic moment are oppositely directed

(a) (b)

← ←

Orbital moment Spin

m

l I

The same magnetic moment, the Bohr Magneton,

µB = eħ/2me = 9.27 10-24 Am2

is associated with ½ħ of spin angular momentum or ħ of orbital angular momentum

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Orbital moment

Circulating current is I; I = -e/τ = -ev/2πr

The moment* is m = IA m = -evr/2

Bohr: orbital angular momentum ℓ is quantized in units of ħ; h is Planck�s constant = 6.626 10-34 J s; ħ = h/2π = 1.055 10-34 J s. |ℓ| = nħ

Orbital angular momentum: ℓ = mer x v Units: J s

Orbital quantum number ℓ, ℓz= mlħ ml = 0,±1,±2,...,±ℓ so mz = -ml(eħ/2me)The Bohr model provides us with the natural unit of magnetic moment

Bohr magneton µB = (eħ/2me) µB = 9.274 10-24 A m2 mz = mlµB

In general m = γℓ γ = gyromagnetic ratio Orbital motion γ = -e/2me

* Derivation can be generalized to noncircular orbits: m = IA for any planar orbit. IEEE Santander 2017

g-factor; Bohr radius; energy scale

The g-factor is defined as the ratio of magnitude of m in units of µB to magnitude of ℓ in units of ħ.

g = 1 for orbital motion

The Bohr model also provides us with a natural unit of length, the Bohr radius

a0 = 4πε0ħ2/mee2 a0 = 52.92 pm

and a natural unit of energy, the Rydberg R0

R0 = (m/2ħ2)(e2/4πε0)2 R0 = 13.606 eV

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Spin moment

Spin is a relativistic effect.Spin angular momentum sSpin quantum number s s = ½ for electronsSpin magnetic quantum number ms ms = ±½ for electrons

sz = msħ ms= ±½ for electrons

For spin moments of electrons we have gyromagnetic ratio and g-factor:γ = -e/me g ≈ 2

m = -(e/me)smz = -(e/me)msħ = ±µB

More accurately, after higher order corrections: g = 2.0023 mz = 1.00116µB

m = - (µB/ħ)(ℓ + 2s)

An electron will usually have both orbital and spin angular momentum

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Quantized mechanics of spin

In quantum mechanics, we represent physical observables by operators – differential or matrix.

e.g. momentum p = -iħ∇; energy p2/2me = -ħ2∇2/2me

n magnetic basis states ⇒ n x n Hermitian matrix, Aij=A*ji

Spin operator (for s = ½ )

s = σħ/2

Pauli spin matrices

Electron: s = ½ ⇒ ms=±½ i.e spin down and spin up statesRepresented by column vectors: |↓〉 = |↑〉 = s |↑〉 = - ½ħ |↑〉 ; s|↓〉 = ½ħ|↓〉

Eigenvalues of s2: s(s+1)ħ2

The fundamental property of angular momentum in QM is that the operators satisfy the commutation relations:

or

Where [A,B] = AB - BA and [A,B] = 0 ⇒ A and B�s eigenvalues can be measured simultaneously [s2,sz] = 0

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Quantized spin angular momentum of the electron

-1/2

1/2

msS

z

g√[s(s+1)]ħ2

H 1/2

1/2

s = ½

-

-

2µ0µBH

-ħ/2

ħ/2

The electrons have only two eigenstates, �spin up�(↑, ms = -½) and �spin down� (↓, ms = ½), which correspond to two possible orientations of the spin moment relative to the applied field.

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Populations of the energy levels are given by Boltzmann statistics; ∝ exp{-Ei/kΒT}. The thermodynamic average 〈m〉 is evaluated from these Boltzmann populations.

〈m 〉 = [µBexp(x) - µBexp(-x)][exp(x) + exp(-x)]

where x = µ0µBH/kBT.

〈m 〉 = µBtanh(x)

Note that to approach saturation x 2At T = 300 K, µ0H. = 900 TAt T = 1K , µ0H. = 3 K.

Useful conversion 1 TµB = 0.672 (µB/kB)

Spin magnetization of localized electrons

0 2 4 6 80

0.2

0.4

0.6 ∞

0.8 1/2

1.0

2

,

x

Linear susceptibilitySlope 1

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In small fields, tanh(x) ≈ x, hence the susceptibility

χ = N〈m 〉/H (N is no of electrons m-3)

χ = µ0NµB2/kBT

This is the famous Curie law for susceptibility, which varies as T-1.

In other terms χ = C/T, where C = µ0NµB

2/kB

C is the Curie constant with dimensions of temperature; Assuming an electron density N of 6 1028 m-3 gives a Curie constant C ≈ 0.5 K. The Curie law susceptibility at room temperature is of order 10-3.

Curie-law susceptibility of localized electrons

T

1/χ

Slope C

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Electrons in a field; paramagnetic resonance

1/2

1/2

ms S

s = ½

-

gµ0µBH

At room temperature there is a very slight difference in thermal populations of the two spin states (hence the very small spin susceptibility of 10-3). The relative population difference is x = gµ0µBH/2kBT At resonance, energy is absorbed from the rf field until the populations are equalized.

The resonance condition is hν = gµ0µBH ν/µ0H = gµB/h [= geħ/2meh = e/2�me]

Spin resonance frequency f = ν/2π is 28 GHz T-1

hf

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m = γl [γ = -e/me]

Γ = m x B

Γ = dl/dt (Newton’s law)

dm /dt = γ m x B

= γ ex ey ez

mx my mz

0 0 Bz

Solution is m(t) = m ( sinθ cosωLt, sinθ sinωLt, cosθ ) where ωL = γBz

Magnetic moment precesses at the Larmor precession frequency fL = γB/2π

dM/dt = γM x B – αeM x dM/dt 28 GHz T-1 for spin

Γ = m x B

m

BZ

dl/dt

dmx/dt = γmyBz dmy/dt = -γmxBz dmz/dt = 0

θ

Electrons in a field - Larmor precession

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Free electrons follow cyclotron orbits in a magnetic field.Electron has velocity v then it experiences a Lorentz force

F = -ev × B

The electron executes circular motion about the direction of B (tracing a helical path if v|| ≠ 0)

Cyclotron frequency fc = v⊥/2πr

fc = eB/2πme

Electrons in cyclotron orbits radiate at the cyclotron frequency

Example: — Microwave oven

Since γe = -(e/me), the cyclotron and Larmor and epr frequencies are all the same for electrons; 28.0 GHz T-1

Electrons in a field – Cyclotron resonance

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D↑,↓(E)

EE

↓

↑ EE

↓

↑

H

±µ0µBH

EF

The splitting is really very small, ~ 10-5 of the bandwidth in a field of 1 T.

M = µB(N↑ - N↓)/V Note M is magnetic moment per unit volumeAt T = 0, the change in population in each band is ΔN = ½ D(EF)µ0µBH

M = 2µB ΔN = D(EF)µ0µB2H The dimensionless susceptibility χ = M/H

χPauli = D(EF)µ0µB2 It is ~ 10-5 and independent of T

Pauli susceptibility

We now show the ↑ and ↓ density of states separately. They split in a field B = µ0H

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Landau diamagnetism

Free electron model was used by Landau to calculate the orbital diamagnetism of conduction electrons. The result is:

exactly one third of the Pauli susceptibility, and opposite in sign.

The real band structure is taken into account in an approximate way by renormalizing the electron mass. Replace me by an effective mass m*

Then χL = -(1/3)(me/m*) χP

In some semimetals such as graphite or bismuth, m* can be 0.01 me, hence the diamagnetism of the conduction electrons may sometimes be the dominant contribution to the susceptibility. (χL = -4 10-4 for graphite)

In the free-electron model, D(EF) = (3/2)n/EF Hence χPauli = {3nµ0µB

2/2EF }[1 + cT2 + ….] (Compare Curie law nµ0µB2/kBT)

The ratio of electronic specific heat coefficient to Pauli susceptibility in the nearly-free, independent electron approximation should be a constant R.

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Susceptibility of electrons; summary

Curie Pauli Landau

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Density of states in other dimension

D(�) ∝ �1/2

D(�) = constant

D(�) ∝ �-1/2

Discreet levels

3-d solid

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Spin-Orbit Coupling

Spin and angular momentum coupled to create total angular momentum j. J = l + s m = γj

From the electron�s point of view, the nucleus revolves round it with speed v ⇒ current loop. It is a relativistic effect

I = Zev/2πrwhich produces a magnetic field µ0I/2r at the centre

Bso = µ0 Zev/4πr2 [~10 T for B or C]

E = - m.B Eso = - µBBso

Since r a0/Z and mevr ≈ ħ Eso -µ0µB2Z4/4πa0

3

The spin – orbit Hamiltonian for a single electron is of the form:in general Hso = (1/2me

2c2r)dV/dr l.s

Here the two ħs have been assimilated into λ, making it an energy (c.f. exchange)

67 3.1 Orbital and spin moments

only be integral. Bold symbols represent angular momentum vectors, whichhave units of h.

Ze

Bso

Spin–orbit interaction fromthe viewpoint of theelectron.

From the electron’s geocentric point of view, the nucleus revolves around itwith speed v. The motion is equivalent to a current loop In = Zev/2πr , whichcreates a magnetic field µ0In/2r (2.7) at the centre. The spin–orbit interaction isdue to this magnetic field, Bso = µ0Zev/4πr2, acting on the intrinsic magneticmoment of the electron. The electron’s magnetic moments associated with ℓ

and s are oppositely aligned. The interaction energy (2.73) εso = −µBBso canbe written approximately in terms of the Bohr2 magneton and the Bohr radius,since r ≃ a0/Z for an inner electron and r = na0 for an outer electron, andmevr ≈ h. In the former case,

εso ≈ −µ0µ2BZ4

4πa30

. (3.12)

The Z variation means that the spin–orbit interaction, while weak for lightelements becomes much more important for heavy elements and especiallyfor inner shells. The associated magnetic field is of order 10 T for boronor carbon. The correct version of the spin–orbit interaction, resulting from arelativistic calculation, is given in §3.3.3. The expression (3.12) is modified bya factor 2. The interaction for a single electron is represented by the spin–orbitHamiltonian

Hso = λl · s, (3.13)

where λ is the spin–orbit coupling energy. l and s are dimensionless operators– the h2 has been absorbed into λ, thus giving it dimensions of energy.

3.1.4 Quantum mechanics of angular momentum

The Bohr model is an oversimplification of the quantum theory of angularmomentum. In quantum mechanics, physical observables are represented bydifferential operators or matrix operators, which we denote by bold symbolswith a hat. For example, momentum is represented by p = −ih∇ and kineticenergy by p2/2m = −h2∇2/2m. The allowed values of a physical observableare given by the eigenvalues, λi , of the equation Oψ i = λiψ i , where O is theoperator and ψ i are the eigenfunctions, which represent the possible observablestates of the system. The eigenvalues are determined by solving the equation|O − λI | = 0 where |· · ·| denotes a determinant and I is the identity matrix.

The angular momentum operator is l = r × p, with components

l = −ih(y∂/∂z − z∂/∂y)ex − ih(z∂/∂x − x∂/∂z)ey − ih(x∂/∂y − y∂/∂x)ez.

(3.14)

2 We usually approximate the spin moment of the electron as 1 µB . Strictly, this equation shouldbe εso = −gµBmsBso.

j

l

s

Orbital angular momentum

The orbital angular momentum operators also satisfy the commutation rules: l x l = Iħl and [l2,lz] = 0

Spherical polar coordinates x = r sinθ cosφ y = r sinθ sinφ z = r cosθ

Nucleus Ze

Electron -e

67 3.1 Orbital and spin moments

only be integral. Bold symbols represent angular momentum vectors, whichhave units of h.

Ze

Bso

Spin–orbit interaction fromthe viewpoint of theelectron.

From the electron’s geocentric point of view, the nucleus revolves around itwith speed v. The motion is equivalent to a current loop In = Zev/2πr , whichcreates a magnetic field µ0In/2r (2.7) at the centre. The spin–orbit interaction isdue to this magnetic field, Bso = µ0Zev/4πr2, acting on the intrinsic magneticmoment of the electron. The electron’s magnetic moments associated with ℓ

and s are oppositely aligned. The interaction energy (2.73) εso = −µBBso canbe written approximately in terms of the Bohr2 magneton and the Bohr radius,since r ≃ a0/Z for an inner electron and r = na0 for an outer electron, andmevr ≈ h. In the former case,

εso ≈ −µ0µ2BZ4

4πa30

. (3.12)

The Z variation means that the spin–orbit interaction, while weak for lightelements becomes much more important for heavy elements and especiallyfor inner shells. The associated magnetic field is of order 10 T for boronor carbon. The correct version of the spin–orbit interaction, resulting from arelativistic calculation, is given in §3.3.3. The expression (3.12) is modified bya factor 2. The interaction for a single electron is represented by the spin–orbitHamiltonian

Hso = λl · s, (3.13)

where λ is the spin–orbit coupling energy. l and s are dimensionless operators– the h2 has been absorbed into λ, thus giving it dimensions of energy.

3.1.4 Quantum mechanics of angular momentum

The Bohr model is an oversimplification of the quantum theory of angularmomentum. In quantum mechanics, physical observables are represented bydifferential operators or matrix operators, which we denote by bold symbolswith a hat. For example, momentum is represented by p = −ih∇ and kineticenergy by p2/2m = −h2∇2/2m. The allowed values of a physical observableare given by the eigenvalues, λi , of the equation Oψ i = λiψ i , where O is theoperator and ψ i are the eigenfunctions, which represent the possible observablestates of the system. The eigenvalues are determined by solving the equation|O − λI | = 0 where |· · ·| denotes a determinant and I is the identity matrix.

The angular momentum operator is l = r × p, with components

l = −ih(y∂/∂z − z∂/∂y)ex − ih(z∂/∂x − x∂/∂z)ey − ih(x∂/∂y − y∂/∂x)ez.

(3.14)

2 We usually approximate the spin moment of the electron as 1 µB . Strictly, this equation shouldbe εso = −gµBmsBso.

l = r x p =

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5. The Single-electron Atom

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Orbital angular momentum operators

l =1 case

ml = 1, 0, -1 corresponds to the eigenvectors

lx, ly and lz operators can be represented by the matrices:

where

Eigenvalues of l2:

l(l+1)ħ2 l is the orbital angular momentum quantum number

√[l(l+1)] " ml "

z

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Solution of Schrodinger’s equation

Schrodinger�s equation:

Satisfied by the wavefunctions:

Where:

And the combined angular parts are

(Vnl are Laguerre polynomials V0

1=1)

(Legendre polynomials)

Normalized spherical harmonics:

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One-electron hydrogenic states

The three quantum number n ,l ml denote an orbital.

Orbitals are denoted nxml, x = s, p, d, f... for l = 0,1,2, 3,...

Each orbital can accommodate at most two electrons* (ms=±1/2)

*The Pauli exclusion principle: No two electrons can have the same four quantum numbers.⇒  Two electrons in the same orbital must have opposite spin.

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

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Single-electron orbitals

s electrons P electrons d electrons

101 4.2 The many-electron atom

0.6

0.5

0.4

0.3

0.2

0.1

0.00 5 10 15 20 25 30

10

21 20

32 31 3043 42 41 40

ρ = r /a0

ρ2 |Rnl(ρ

)|2

Figure 4.3

Radial probabilitydistributions deduced fromthe radial parts of the wavefunction R(r ). The curvesare labelled with thevalues of n and ℓ.

becomes

H0 =!

i

[−(h2/2me)∇2 − Ze2/4πϵ0ri] +!

i<j

e2/4πϵ0rij. (4.7)

The sum over many pairs of interacting particles makes this an intractableanalytical problem.2 An appropriate way of dealing with the extra Coulombinteractions is to suppose that each electron experiences the central potential ofsome different, spherically symmetric charge distribution. The potential withmany electrons is no longer a simple Coulomb well and the degeneracy ofthe energy of electrons with different ℓ-values but the same principal quantumnumber n is lifted. The 4s shell, for example, then turns out to be lower inenergy than the 3d shell, the energy change depends on filling. The sensein which the shells are filled defines the shape of the periodic table. Thepotentials can be determined self-consistently. This is known as the Hartree–Fochapproximation.

n 1 2 3 4 5 6

1s 2s 3s 4s 5s 6s

2p 3p 4p 6p

3d 4d 6d

4f 5f 6f

5g 6g

5d

5p

Sequence of shell filling fora many electron atom.

When several electrons are present on the same atom, at most two of them,with opposite spin, can occupy the same orbital. The ions of interest in mag-netism generally follow the L–S coupling scheme, where individual spin andorbital angular momenta add3 to give resultant quantum numbers (here S andL ≥ 0):

S =!

si, MS =!

msi, L =!

ℓi , ML =!

mℓi .

2 There is a very complicated analytical solution to the three-body problem, but there are usuallymany more than three particles involved in atoms.

3 When spin-orbit coupling is very strong, as it is in the actinides, it is appropriate to first coupleli and si for each electron, to form ji , and then to couple these total angular momenta. This isthe j−j coupling scheme.

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6. The Multi-electron Atom

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The many-electron atom

Hartree-Fock approximation

•  No longer a simple Coulomb potential.

•  l degeneracy is lifted.

•  Solution: Suppose that each electron experiences the potential of a different spherically-symmetric potential.

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4 Be 9.01 2 + 2s0

12Mg 24.21 2 + 3s0

2 He 4.00

10Ne 20.18

24Cr 52.00 3 + 3d3

312

19K 38.21 1 + 4s0

11Na 22.99 1 + 3s0

3 Li 6.94 1 + 2s0

37Rb 85.47 1 + 5s0

55Cs 132.9 1 + 6s0

38Sr 87.62 2 + 5s0

56Ba 137.3 2 + 6s0

59Pr 140.9 3 + 4f2

1 H 1.00

5 B 10.81

9 F 19.00

17Cl 35.45

35Br 79.90

21Sc 44.96 3 + 3d0

22Ti 47.88 4 + 3d0

23V 50.94 3 + 3d2

26Fe 55.85 3 + 3d5

1043

27Co 58.93 2 + 3d7

1390

28Ni 58.69 2 + 3d8

629

29Cu 63.55 2 + 3d9

30Zn 65.39 2 + 3d10

31Ga 69.72 3 + 3d10

14Si 28.09

32Ge 72.61

33As 74.92

34Se 78.96

6 C 12.01

7 N 14.01

15P 30.97

16S 32.07

18Ar 39.95

39Y 88.91 2 + 4d0

40Zr 91.22 4 + 4d0

41Nb 92.91 5 + 4d0

42Mo 95.94 5 + 4d1

43Tc 97.9

44Ru 101.1 3 + 4d5

45Rh 102.4 3 + 4d6

46Pd 106.4 2 + 4d8

47Ag 107.9 1 + 4d10

48Cd 112.4 2 + 4d10

49In 114.8 3 + 4d10

50Sn 118.7 4 + 4d10

51Sb 121.8

52Te 127.6

53I 126.9

57La 138.9 3 + 4f0

72Hf 178.5 4 + 5d0

73Ta 180.9 5 + 5d0

74W 183.8 6 + 5d0

75Re 186.2 4 + 5d3

76Os 190.2 3 + 5d5

77Ir 192.2 4 + 5d5

78Pt 195.1 2 + 5d8

79Au 197.0 1 + 5d10

61Pm 145

70Yb 173.0 3 + 4f13

71Lu 175.0 3 + 4f14

90Th 232.0 4 + 5f0

91Pa 231.0 5 + 5f0

92U 238.0 4 + 5f2

87Fr 223

88Ra 226.0 2 + 7s0

89Ac 227.0 3 + 5f0

62Sm 150.4 3 + 4f5 105

66Dy 162.5 3 + 4f9

179 85

67Ho 164.9 3 + 4f10

132 20

68Er 167.3 3 + 4f11

85 20

58Ce 140.1 4 + 4f0 13

Ferromagnet TC > 290K

Antiferromagnet with TN > 290K

8 O 16.00

35

65Tb 158.9 3 + 4f8

229 221

64Gd 157.3 3 + 4f7

292

63Eu 152.0 2 + 4f7

90

60Nd 144.2 3 + 4f3

19

66Dy 162.5 3 + 4f9

179 85

Atomic symbolAtomic Number

Typical ionic changeAtomic weight

Antiferromagnetic TN(K) Ferromagnetic TC(K)

Antiferromagnet/Ferromagnet with TN/TC < 290 K

Metal

Radioactive

80Hg 200.6 2 + 5d10

93Np 238.0 5 + 5f2

94Pu 244

95Am 243

96Cm 247

97Bk 247

98Cf 251

99Es 252

100Fm 257

101Md 258

102No 259

103Lr 260

36Kr 83.80

54Xe 83.80

81Tl 204.4 3 + 5d10

82Pb 207.2 4 + 5d10

83Bi 209.0

84Po 209

85At 210

86Rn 222

Nonmetal Diamagnet

Paramagnet

BOLD Magnetic atom

25Mn 55.85 2 + 3d5

96

20Ca 40.08 2 + 4s0

13Al 26.98 3 + 2p6

69Tm 168.9 3 + 4f12

56

Magnetism of free atoms

79 out of the 103 first elements are magnetic as free atoms Moments << ZµB

Addition of angular momenta

J

L

SJ = L + S ⎢L-S⎢ ≤ J ≤ ⎢L+S⎢

Different J-states are termed multiplets, denoted by;

2S+1XJ

X = S, P, D, F, ... for L = 0,1,2,3,... Hund�s rules To determine the ground-state of a multi-electron atom/ion. 1) Maximize S 2)  Maximize L consistent with S. 3)  Couple L and S to form J.

•  Less than half full shell J = L-S •  More than half full shell J = L+S

First add the orbital and spin momenta li and si to form L and S. Then couple them to give the total J

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Fe3+ 3d5

2 1 0 -1 -2

↑ ↑ ↑ ↑ ↑ S = 5/2 L = 0 J = 5/2

6S5/2

Note; Maximizing S is equivalent to maximizing Ms = Σmsi , since Ms � S

Hund�s rules; examples

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2 1 0 -1 -2

↑ ↑ ↑ ↑ ↓

↑ ↓

Co2+ 3d7

S = 3/2 L = 3 J = 9/2

4F9/2

Note; Maximizing L is equivalent to maximizing ML = Σmli , since ML � L

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2 1 0 -1 -2

↑ ↑ ↑ ↓

↑ ↓

↑ ↓

Ni2+ 3d8 S = 1 L = 3 J = 4

3F4

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Ce3+ 4f1 ↑

S = 1/2 L = 3 J = 5/2

2F5/2

3 2 1 0 -1 -2 -3

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Nd3+ 4f3

S = 3/2 L = 6 J = 9/2

4I9/2

↑ ↑ ↑

3 2 1 0 -1 -2 -3

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Dy3+ 4f9 ↑ ↑ ↑ ↑ ↑ ↑

↓ ↑ ↓

3 2 1 0 -1 -2 -3S = 5/2 L = 5 J = 15/2

6H15/2

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Hund’s rules 3d and 4f

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Spin-Orbit Coupling

Hso=ΛL.S Λ Is atomic the spin-orbit coupling constant

Λ  > 0 for the 1st half of the 3d or 4f series.Λ  < 0 for the 2nd half of the 3d or 4f series. (for Hund’s 3rd rule)

Compare single-electron atom case: Hso = λl.s

Λ = ± λ/2S

L.S = (1/2)(J2 - L2 - S2) = (ħ2/2)[J(J+1)-L(L+1)-S(S+1)]

ion Λ

3d1 Ti3+ 124 3d2 Ti2+ 88 3d3 V2+ 82 3d4 Cr2+ 85 3d6 Fe2+ -164 3d7 Co2+ -272 3d8 Ni2+ -493

(K)

The magnetic moment of an ion is represented by the expression m = - (L + 2S)µB/"

The Zeeman Hamiltonian for the magnetic moment in a field B along ez is H Zeeman = –m.B

H Zeeman = (µB/")(Lz + 2Sz)Bz

For a particular J-multiplet the matrix elements of L + 2S are proportional to those of J (Wigner Eckart theorem)

〈LSJMJ| L + 2S |LSJMJ〉 = gJ〈LSJMJ| J |LSJMJ〉 gJ is the Landé g-factor

(L + 2S) = gJJ

mz = gJJzµB/"

H Zeeman = gJJzB(µB/")

H ZeemanψLSJM = gJ µB MJB ψLSJM

2S+1LJ

MJ = J

MJ =-J

Zeeman Interaction

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L

SJ

m

z

S

The vector model of the atom, including magnetic moments. First project m onto J. J then precesses around z.

The g-factor for the atom or ion is the ratio of the component of magnetic moment along J in units of µB to the magnitude of the angular momentum in units of ".

gJJ = (L + 2S) Take scalar product with JgJ = -(m.J/µB)/(J2/") = -m.J("/µB)/[(J(J + 1)]

but m.J = -(µB/"){(L + 2S).(L + S)} -(µB/"){(L2 + 3L.S + 2S2)} -(µB/"){(L2 + 2S2 + (3/2)(J2 - L2 - S2)} since J2 = L2 + S2

+ 2 L.S

-(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)}

henceg = 3/2 + {S(S+1) - L(L+1)}/2J(J+1) Check; gS = 2 , gL=1

J2 = J(J + 1)" 2; Jz = MJ"

Landé g-factor

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Co2+ free ion

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Paramagnetic susceptibility - Brillouin theory

Curie law X = C / T C is Curie’s constant.Units: Kelvin, K.Typical values ~ 1K

The thermodynamic average of the moment:

B = Bz E = - m.B

⇒

Using the identities:

and the fact that X = n〈m 〉/H

we find

(n is the number density of atoms/ions)

X = µ0nmeff2µB

2 / 3kBT

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Energy levels of an ion with J = 5/2 in an applied field

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4f ions

J is a good quantum number IEEE Santander 2017

3d ions

S is a good quantum number L is ‘quenched’

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Magnetization curve - Brillouin theory

To calculate the complete magnetization curve, set y = gµBµ0H/kBT,then 〈m 〉 = gµB�/�y[lnΣ-J

J exp{MJy} [d(ln z)/dy = (1/z) dz/dy]

The sum over the energy levels must be evaluated; it can be written asexp(Jy) {1 + r + r2 + .........r2J} where r = exp{-y}

The sum of a geometric progression (1 + r + r2+ .... + rn) = (rn+1 - 1)/(r - 1) ∴ Σ-J

J 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}

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Paramagnetism - Brillouin theory

setting x = Jy, we obtain

where BJ(x) is the Brillouin function { }

This reduces to 〈m 〉 = µB tanh(x) in the limit J = ½, g = 2.and 〈m 〉 = L(x) is the Langevin function {coth x - 1/x} in the large-J limit.

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Comparison of the Brillouin functions for s = ½, J = 2 and the Langevin function (J = ∞)

x = g JµBB/kBT

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Magnetization curves for paramagnetic ions

Reduced magnetization curves of three paramagnetic salts, compared with Brillouin function predictions

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Experimental confirmation

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Mean field theory of ferromagnetism

Weiss (1907) supposed that in addition to any externally applied field H, there is an internal �molecular� field in a ferromagnet proportional to its magnetization.

Hi = nWM

Hi must be immense in a ferromagnet like iron to be able to induce a significant fraction of saturation at room temperature; nW ≈10 - 1000. The origin of these huge fields remained a mystery until Heisenberg introduced the idea of the exchange interaction in 1928.

Magnetization is given by the Brillouin function,

<m> = mBJ(x) where now x = µ0mHi/kBT.

The magnetization at zero temperature, M0 = Nm where m = gµBJ. At nonzero temperature Ms = N<m>

PY3P04 2017

In zero external field, we have Ms/M0 = BJ(x) (1)Also by eliminating Hi from the expressions for Hi (Hi = nWMs) and x (Hi = kBTx/µ0m); since M0= Nm

Ms/M0 = (NkBT/µ0M02nW)x

which can be rewritten in terms of the Curie constant C = µ0Ng2µB2J(J+1)/3kB

Ms/M0 = [T(J+1)/3CJnW]x (2)The simultaneous solution of (1) and (2) is found graphically, or they can be solved numerically.

(2)

(1)

0 2 4 6

1

< m >/ m

T > TC T = TC T < TC

x

Graphical solution of (1) and (2) to find the spontaneous magnetization Ms when T < TC.

in blue for T < TC, T = TC and T > TC.

Mean field theory of ferromagnetism

PY3P04 2017 55

At the Curie temperature, the slope of (2) is equal to the slope at the origin of the Brillouin function

For small x. BJ(x) [(J+1)/3J]x + ...

hence TC = nWC

where the Curie constant C = µ0Ng2µB2J(J+1)/3kB.

In practice, TC is used to determine nW.

The spontaneous magnetization for nickel, together with the theoretical curve for S = ½ from the mea field theory. The theoretical curve is scaled to give correct values at either end.

Mean field theory of ferromagnetism

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Exchange

Heisenberg and Dirac explained the origin of magnetism in solids in terms of the exchange interaction. It is short-range interaction, coupling the spins of nearest-neighbour atoms.It is not due to magnetic dipolar interactions; those fields are ~ 1 T, and far too weak.Nor is there any real ‘molecular field’ inside the sample. Since ∇.B = 0, it would be observed outside as well. It is due to interelectronic Coulomb interactions, subject to the symmetry constraints of quantum mechanics.

Interatomic exchange HHeis = -2 J Si.Sj

J > 0; ferromagnetic interatomic interaction J < 0; antiferromagnetic interatomic interaction

Relation between J and Hi Hi = -2(ΣjJSj).Si -µ0Hi mi = -µ0HigµBSi = -µ0nWMgµBSi

TC = µ0NnWg2µB2S(S+1)/3kB = 2ZJ S(S+1)/3kB

Number of nearest-neighbours in Σi

♦ Filled electronic shells are not magnetic (the spins are paired; ms = ±1/2)♦ Only partly-filled shells may possess a magnetic moment♦  The magnetic moment is given by m = gJµBJ, where "J represents the total angular

momentum. For a given configuration the values of J and gj in the ground state are given by Hund�s rules

��✻✻★✻✻��

When the ion is embedded in a solid, the crystal field interaction is important. This is the electrostatic Coulomb interaction of an ion with its surroundings.The third point is modified:

♦ Orbital angular momentum for 3d ions is quenched. The spin only moment is m ≈ gµBS, with g = 2.♦ Magnetocrystalline anisotropy appears, making certain crystallographic axes easy directions of magnetization.

Summary – so far

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7. Ions in Solids

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SUMMER SCHOOL

!

Energy scales

0.01 0.1 1 meV 10 100 1 eV 10

0.1 1 K 10 100 1000 104 105

3d 4f

4f 3d

UIntratomic exchange

JInteratomic exchange

Λ Spin-orbit interaction

Eσ πLigand-field interaction

EZ 1T Zeeman interaction. of electron

EZ 1T Zeeman interaction. of atom

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The Hamiltonian is now

H = H 0 + H so+ H cf+ H Z

Typical magnitudes of energy terms (in K)

H 0 H so H cf H Z in 1 T

3d 1 - 5 104 102 -103 104 1

4f 1 - 6 105 1 - 5 103 ≈3 102 1

H so must be considered before H 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.

Magnitudes of the interactions

Hi = H0 + Hso + Hcf + HZ

Coulomb interactions |L,S〉

spin-orbit interaction ΛL.S |J〉

Zeeman interaction gµBB.J/ħ |MJ〉 ion Λ

3d1 Ti3+ 124 3d2 Ti2+ 88 3d3 V2+ 82 3d4 Cr2+ 85 3d6 Fe2+ -164 3d7 Co2+ -272 3d8 Ni2+ -493

4f1 Ce3+ 920 4f2 Pr3+ 540 4f3 Nd3+ 430 4f5 Sm3+ 350 4f8 Tb3+ -410 4f9 Dy3+ -550 4f10 Ho3+ -780 4f11 Er3+ -1170 4f12 Tm3+ -1900 4f13 Yb3+ -4140

H 0 H so H cf H Z in 1 T

3d 1 - 5 104 102 -103 104 1

4f 1 - 6 105 1 - 5 103 ≈3 102 1

Crystal field interaction ∫ρ0(r)ϕcf(r)d3r

Magnitudes of the interactions

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Co2+

Co0 Gd

Gd Co Gd Co

As metallic atoms or ions the transition metals occupy one third of the volume of the rare earths.

Gd3+ (105 pm) (75 pm)

3d and 4f compared

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Localized and delocalized electrons

LOCALIZED MAGNETISM DELOCALIZED MAGNETISMIntegral number of 3d or 4f electrons Nonintegral number of unpaired spinson the ion core; Integral number of unpaired spins; per atom. Discreet energy levels. Spin-polarized energy bands with strong

correlations.Ni2+ 3d8 m = 2 µB Ni 3d9.44s0.6 m = 0.6 µB

ψ ≈ exp(-r/a0) ψ ≈ exp(-ik.r)Boltzmann statistics Fermi-Dirac statistics

4f metals   localized electrons4f compounds localized electrons3d compounds localized/delocalized electrons3d metals delocalized electrons.

Above the ferromagnetic Curie temperature, the moments do not disappear, they just become disordered in a paramagnetic state when T > TC.

3d

3d9

8

r

ψ

Ε

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Roct = (21/2 -1)rO = 58 pm Rtet = ((3/2)1/2 - 1)rO = 32 pm

Oxides

Oxides are usually insulating. Structures are based on dense-packed O2- arrays, with cations in interstitial sites.

Octahedral and tetrahedral sites are common in transition metal oxides and other compounds.

Both have cubic symmetry if undistorted

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4-fold tetrahedral

pm 6-fold octahedral

pm 6-fold octahedral

pm 12-fold substitutional

pm

Mg2+ 53 Cr4+ 3d2 55 Ti3+ 3d1 67 Ca2+ 134

Zn2+ 60 Mn4+ 3d3 53 V3+ 3d2 64 Sr2+ 144

Al3+ 42 Cr3+ 3d3 62 Ba2+ 161

Fe3+ 3d5 52 Mn2+ 3d5 83 Mn3+ 3d4 65 Pb2+ 149

Fe2+ 3d6 78 (61) Fe3+ 3d5 64 Y3+ 119

Co2+ 3d7 75 (65) Co3+ 3d6 61 (56) La3+ 136

Ni2+ 3d8 69 Ni3+ 3d7 60 Gd3+ 122

Cation radii in oxides: low spin values are in parentheses.

The radius of the O2- anion is 140 pmIEEE Santander 2017

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To demonstrate quenching of orbital angular momentum, consider the 2p states ψ0, ψ1, ψ-1

corresponding to l = 1, ml = 0, ±1. ψ0 = R(r) cos θψ±1 = R(r) sin θ exp {±ιφ}

The functions are eigenstates in the central potential V (r) but they are not eigenstates of Hcf. Suppose the oxygens can be represented by point charges q at their centres, then for the octahedron,

H cf = Vcf = D(x4 +y4 +z4 - 3y2z2 -3z2x2 -3x2y2)where D ≈ eq/4πεoa6. But ψ±1 are not eigenfunctions of Vcf, e.g. ∫ψi

*VcfψjdV≠ δij, where i,j = -1, 0, 1.We seek linear combinations that are eigenfunctions, namely

ψ0 = R(r) cos θ = zR(r) = pz(1/√2)(ψ1 + ψ-1)= R�(r)sinθcosφ = yR(r) = py(1/√2)(ψ1 - ψ-1)= R�(r)sinθsinφ = xR(r) = px

q

,,X Y Z ,,X Y Z ,,X Y Z

z

θ r

φ y x

Orbital moment quenching is a cubic crystal field

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The 2p eigenfunctions are degenerate in an undistorted cubic environment

ψ0 = R(r) cos θ = zR(r) = pz(1/√2)(ψ1 + ψ-1) = R�(r)sinθcosφ = yR(r) = px(1/√2)(ψ1 - ψ-1) = R�(r)sinθsinφ = xR(r) = py

Note that the z-component of angular momentum; lz = i"/∂φ is zero for these wavefunctions. Hence the orbital angular momentum is quenched.

The same is true of the 3d eigenfunctions, which are

dxy = (1/√2)(ψ2 - ψ-2) = R�(r)sin2θsin2φ ≈ xyR(r) dyz = (1/√2)(ψ1 - ψ-1) = R�(r)sinθcosθsinφ ≈ yzR(r) t2g orbitalsdzx = (1/√2)(ψ1 + ψ-1) = R�(r)sinθcosθcosφ ≈ zxR(r)dx

2-y

2 = (1/√2)(ψ2 + ψ-2) = R�(r)sin2θcos2φ ≈ (x2-y2)R(r) eg orbitalsd3z

2-r

2 = ψ0 = R�(r)(3cos2θ � 1) ≈ (3z2-r2)R(r) The 3d eigenfunctions split into a set of three and a set of two in an undistorted cubic environment

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.

q

x y, px py pz

dxy dyz dzx

d x2-y2 , d z 2

y x

z

n =1l = 0 s orbital

n =2l = 1 p orbitals

n =3 l = 2 d orbitals

t2g

eg

Orbitals in a cubic crystal field

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yx

z

yx

z

yx

z

yx

z

yx

z2p

3d

4s

t2g

eg

dσ

dπ

cf splitting

hybridization

Orbitals in the crystal field

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Crystal-field theory regards the splitting of the 3d orbitals in octahedral oxygen, for example, as an electrostatic interaction with neighbouring point charges (oxygen anions). In reality the 3d and 2p orbitals of oxygen overlap to form a partially covalent bond. The oxygens bonding to the 3d metals are the ligands. The overlap is greater for the eg than the t2g orbitals in octahedral coordination.The overlap leads to mixed wavefunctions, producing bonding and antibonding orbitals, whose splitting increases with overlap. The hybridized orbitals are

φ = αψ2p+βψ3d where α2 + β2 = 1.For 3d ions the splitting is usually 1- 2eV, with the ionic and covalent contributions being of comparable magnitude The spectrochemical series is the sequence of ligands in order of effectiveness at producing crystal/ligand field splitting.

Br-<Cl-<F-<OH-<CO2-3<O2-<H2O<NH3<SO2-

3<NO-2<S2-<CN-

The bond is mostly ionic at the beginning of the series and covalent at the end.

Covalency is stronger in tetrahedral coordination but the crystal field splitting isΔtet = (3/5)Δoct

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σ�bond

π�bond

+

+

++

+

+

+

+

+

+

+

–

–

–

––

–

–

–

–

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Ligand field interaction

One-electron energy diagrams

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Lower symmetry

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The Jahn-Teller effect

• 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 ε, the energy change is

δE = -Aε + Bε2. where the first term is the crystal field stabilization energy and the second term is the increased elastic energy. • The Jahn-Teller distortion may be static or dynamic.

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High and low spin states An ion is in a high spin state or a low spin state depending on whether the Coulomb interaction UH leading to Hund�s first rule (maximize S) is greater than or less than the crystal field splitting Δcf.

Δcf.

Δcf.

UH > Δcf. UH < Δcf.

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Crystal Field Hamiltonian

Charge distribution of the ion potential created by the crystal

structural parameters IEEE Santander 2017

The approximation made so far is terrible.It ignores the screening of the potential by the outer shells of the 4f ion for example, and also the covalent contribution. But it captures the symmetry of the problem. We proceed with it, but treat the crystal field coefficients as empirical parameters.

It is useful to expand the charge distribution of a central 4f ion in terms of the 2n-pole moments of the charge distribution, n = 2, 4, 6

The quadrupole moment

The hexadecapole moment

The 64-pole moment

Rare earth quadrupole moments

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Single-ion anisotropy

Single-ion anisotropy is due to the electrostatic crystal field interaction + spin-orbit interaction. The 4f charge distribution ρ0 (r) interacts with the crystal field potential ϕcf(r) to stabilizes some particular orbitals; spin-orbit interaction -ΛL.S then leads to magnetic moment alignment along some specific directions in the crystal.

The leading term in the crystal field interaction is

where A20 is the uniaxial second-order crystal field parameter, which described the

electric field gradient created by the crystal which interacts with the 4f quadrupole moment. Compare εa = K1sin2θThe crystal field interaction can be expressed in terms of angular momentum operators, using the Wigner-Eckart theorem

Stevens operatorscf coefficientIEEE Santander 2017

Here and θn is different for each 4f ion, proportional to the 2n-pole moment

Q2 = 2 θ2〈r4f2〉 Q4 = 8 θ4〈r4f

4〉 Q6 = 16 θ6〈r4f6〉

Anm ~ γnm parameterises the crystal field produced by the lattice.

NB. Q2 0 for J (or L) � 1

Q4 0 for J (or L) � 2

Q6 0 for J (or L) � 3

The Stevens operators are tabulated, as well as which ones feature in each point symmetry

e.g. The leading term in any uniaxial site is the one in O20

The complete second order (uniaxial) cf Hamiltonian is

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125 4.4 Ions in solids; crystal-field interactions

Table 4.10. Data on the rare-earth ions. The operators O0n are evaluated at T = 0 K , where J z = J

θ 2 θ 2⟨r2⟩O02 θ 4 θ 4⟨r4⟩O0

4 θ 6 θ 6⟨r6⟩O06

J (10−2) a20 (10−4) a4

0 (10−6) a60 G γ s Cmol

Ce3+ 5/2 −5.714 −0.748 63.5 1.51 0.18 −1/3 8.0Pr3+ 4 −2.101 −0.713 −7.346 −2.12 60.99 5.89 0.80 −1/2 16.0Nd3+ 9/2 −0.643 −0.258 −2.911 −1.28 −37.99 −8.63 1.84 −1/4 16.4Sm3+ 5/2 4.127 0.398 25.012 0.34 4.46 −5 0.9Gd3+ 7/2 − 15.75 1 78.8Tb3+ 6 −1.010 −0.548 1.224 1.20 −1.12 −1.28 10.50 2/3 118.2Dy3+ 15/2 −0.635 −0.521 −0.592 −1.46 1.04 5.64 7.08 1/2 141.7Ho3+ 8 −0.222 −0.199 −0.333 −1.00 −1.29 −10.03 4.50 2/5 140.7Er3+ 15/2 0.254 0.190 0.444 0.92 2.07 8.98 2.55 1/3 114.8Tm3+ 6 1.010 0.454 1.633 1.14 −5.61 −4.05 1.17 2/7 71.5Yb3+ 7/2 3.175 0.435 −17.316 −0.79 1.48 0.73 0.29 1/4 25.7

lowest, and the susceptibility in a magnetic field applied along the crystal-fieldaxis is given by (4.14). If D is positive, the MJ = 0 singlet is the ground state.The crystal field appears to have destroyed the moment. The initial suscepti-bility is zero, although the induced moment increases quadratically with field.However, when the magnetic field is applied perpendicular to the crystal-fieldaxis, there is a large susceptibility. What this means is that there is an easy axisof magnetization in the first case, but an easy plane in the second.

A20 < 0

A20 > 0

Examples of atomicconfigurations that producepositive and negativeelectric field gradients atthe central site.

Single-ion anisotropy is the major source of anisotropy in hard ferromagneticmaterials. The tendency for the magnetic moment to lie along specific crystalaxes generally makes the susceptibility a tensor rather than a scalar quantity.There are three principal axes, for which the induced moment lies parallel tothe applied magnetic field.

In general, the details of the response of the system to an applied fieldare obtained by diagonalizing the Hamiltonian Hcf + Hz, and evaluating ⟨m⟩from (4.11). In a series of isostructural rare-earth metals or compounds, Am

n isroughly constant. The leading term in the anisotropy just depends on the productof A0

2 and θ2⟨r24f ⟩J 2

z for the rare-earth in question. Both θ2, and the quadrupolemoment change sign at each quarter-shell filling (Fig. 4.18, Table 4.10). WhenA0

2 is positive, ions such as Nd3+, which have a negative quadrupole momentand an oblate charge distribution, will exhibit easy axis anisotropy, whereasions such as Sm3+, which have a positive quadrupole moment and a prolatecharge distribution, will exhibit hard axis anisotropy.

Consider a rare-earth, for which the spin-orbit interaction stabilizes a(2J + 1)-fold degenerate multiplet |J,MJ ⟩. The crystal field defines the z

axis and splits out the MJ states, subject to Kramers theorem.In the figure, Sm3+ has a |±5/2⟩ ground state, so the moment has its max-

imum positive or negative projection along Oz, which precesses as indicatedby the vector model. The average moment lies along the z-direction, and

125 4.4 Ions in solids; crystal-field interactions

Table 4.10. Data on the rare-earth ions. The operators O0n are evaluated at T = 0 K , where J z = J

θ 2 θ 2⟨r2⟩O02 θ 4 θ 4⟨r4⟩O0

4 θ 6 θ 6⟨r6⟩O06

J (10−2) a20 (10−4) a4

0 (10−6) a60 G γ s Cmol

Ce3+ 5/2 −5.714 −0.748 63.5 1.51 0.18 −1/3 8.0Pr3+ 4 −2.101 −0.713 −7.346 −2.12 60.99 5.89 0.80 −1/2 16.0Nd3+ 9/2 −0.643 −0.258 −2.911 −1.28 −37.99 −8.63 1.84 −1/4 16.4Sm3+ 5/2 4.127 0.398 25.012 0.34 4.46 −5 0.9Gd3+ 7/2 − 15.75 1 78.8Tb3+ 6 −1.010 −0.548 1.224 1.20 −1.12 −1.28 10.50 2/3 118.2Dy3+ 15/2 −0.635 −0.521 −0.592 −1.46 1.04 5.64 7.08 1/2 141.7Ho3+ 8 −0.222 −0.199 −0.333 −1.00 −1.29 −10.03 4.50 2/5 140.7Er3+ 15/2 0.254 0.190 0.444 0.92 2.07 8.98 2.55 1/3 114.8Tm3+ 6 1.010 0.454 1.633 1.14 −5.61 −4.05 1.17 2/7 71.5Yb3+ 7/2 3.175 0.435 −17.316 −0.79 1.48 0.73 0.29 1/4 25.7

lowest, and the susceptibility in a magnetic field applied along the crystal-fieldaxis is given by (4.14). If D is positive, the MJ = 0 singlet is the ground state.The crystal field appears to have destroyed the moment. The initial suscepti-bility is zero, although the induced moment increases quadratically with field.However, when the magnetic field is applied perpendicular to the crystal-fieldaxis, there is a large susceptibility. What this means is that there is an easy axisof magnetization in the first case, but an easy plane in the second.

A20 < 0

A20 > 0

Examples of atomicconfigurations that producepositive and negativeelectric field gradients atthe central site.

Single-ion anisotropy is the major source of anisotropy in hard ferromagneticmaterials. The tendency for the magnetic moment to lie along specific crystalaxes generally makes the susceptibility a tensor rather than a scalar quantity.There are three principal axes, for which the induced moment lies parallel tothe applied magnetic field.

In general, the details of the response of the system to an applied fieldare obtained by diagonalizing the Hamiltonian Hcf + Hz, and evaluating ⟨m⟩from (4.11). In a series of isostructural rare-earth metals or compounds, Am

n isroughly constant. The leading term in the anisotropy just depends on the productof A0

2 and θ2⟨r24f ⟩J 2

z for the rare-earth in question. Both θ2, and the quadrupolemoment change sign at each quarter-shell filling (Fig. 4.18, Table 4.10). WhenA0

2 is positive, ions such as Nd3+, which have a negative quadrupole momentand an oblate charge distribution, will exhibit easy axis anisotropy, whereasions such as Sm3+, which have a positive quadrupole moment and a prolatecharge distribution, will exhibit hard axis anisotropy.

Consider a rare-earth, for which the spin-orbit interaction stabilizes a(2J + 1)-fold degenerate multiplet |J,MJ ⟩. The crystal field defines the z

axis and splits out the MJ states, subject to Kramers theorem.In the figure, Sm3+ has a |±5/2⟩ ground state, so the moment has its max-

imum positive or negative projection along Oz, which precesses as indicatedby the vector model. The average moment lies along the z-direction, and

The cf Hamiltonian for a site with cubic symmetry is

For 3d ions only the fourth-order terms exist; (l = 2)

Kramer�s theorem

It follows from time-reversal symmetry that the cf energy levels of any ion with an odd number of electrons, and therefore half-integral angular momentum, must be at least 2-fold degenerate. These are the |±MJ〉 Kramers doublets.

When J is integral, there will be a |0〉 singlet (with no magnetic moment) and a series of doublets.

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For 3d ions the second-order term

Is often simplified to H cf = DSz2 Typically D � 1 K

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Summary 2

You should now know;

Ø Magnetism in solids is inextricably connected to angular momentumØ  It arises from quantized spin and orbital angular momenta of the electronsØ Curie law paramagnetic susceptibility of localized electronsØ  Pauli susceptibility of delocalized electronsØ Hund’s rules for calculating the ground state of a free ionØ Crystal fields in solids tend to quench orbital angilar momentum; spin

survivesØ Magnetic anisotropy � sources and magnitudes.Ø Magnetic exchange interactionsØ  Ferrromagnetic, ferrimagnetic and antiferromagnetic order.

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Thank you !

IEEE Santander 2017

SUMMER SCHOOL

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Energy scales

0.01 0.1 1 meV 10 100 1 eV 10

0.1 1 K 10 100 1000 104 105

3d 4f

4f 3d

UHIntratomic exchange(Hund’s first rule) J

Interatomic exchange

Λ Spin-orbit interaction

Eσ πLigand-field interaction

EZ 1T Zeeman interaction. of electron

EZ 1T Zeeman interaction. of atom

IEEE Santander 2017

Magnetic dipolar interactions

Ems

Room temperature