Download Report - IISER Pune

Project Report
By Amartya Singh
Project Mentor - Dr. Surjeet Singh
1
Introduction
The key ideas that are the foundation stones of our understanding of magnetism
in matter are described succintly below:
1.1
Magnetic Moments
In magnetism, there is no analogue of electric charge, instead the fundamental
object of consideration is the magnetic moment. Classically, the quantity is
associated with a current loop, and is defined in the following manner:
Suppose a current of magnitude I flows in an infinitesimal loop of area dA,
then the magnetic moment dµ is given by
dµ = IdA
It should be noted that according to the definition of the magnetic moment,
it is independent of the shape of the loop and depends only on the area enclosed
by the loop. Therefore, for a finite sized loop, the magnetic moment µ can be
calculated by summing up all the magnetic moments due to the equal infinitesimal current loops which make up the total area of the given finite loop. We
therefore have,
Z
µ = I dA
1.1.1
Association of Magnetic Moment and Angular Momentum
In classical electromagnetism, a current is explained as originating from the motion of charges. For most purposes this provides a very convenient picture. From
this picture one should expect that there should be a net angular momentum
L due to the movement of charges around the loop. Thus, magnetic moment is
intimately linked with angular momentum.
In atoms, the magnetic moment µ associated with an orbiting electron lies along
the same direction as the angular momentum L of that electron and is proportional to it. Thus, we have
µ
=γ
L
Here γ is known as the gyromagnetic ratio. The relation between the magnetic moment and the angular momentum is demonstrated by the Einstein-de
Haas effect, which is a rotation induced by magnetization. The reverse effect
(magnetization induced by rotation) has also been observed and is known as the
Barnett effect.
1
1.1.2
Magnetic Moment in External Magnetic Field
The Lorentz force F on a charge q moving in a uniform external magnetic field
B is,
F = q(v × B)
This can be carried over to the case of a straight segment of wire of length dr
carrying current I in a uniform magnetic field B to give,
dF = I(dr × B)
Considering an infinitesimal square loop with sides of length dx and dy oriented
perpendicular to the z-direction, the couple on the elementary loop is given by,
dτx = −IdxdyBy
dτy = IdxdyBx
We define the magnetic moment of the loop as dµz = Idxdy. This argument can
be repeated for arbitrary directions of the loop. The couple dτ will then given
by,
dτ = dµ × B
For a finite sized loop, we can integrate the above expression to get,
τ =µ×B
From this expression it is clear that, a magnetic moment µ aligned with a
magnetic field B does not experience any torque. Therefore, if the magnetic
moment µ is rotated at an angle with respect to the magnetic field, the work
done by the applied torque is given by,
Z θ
Z θ
0
0
0
W =
τ dθ =
µB sin θ dθ
0
0
⇒ W = µB(1 − cos θ)
We can therefore write the potential energy U of the magnetic moment as,
U = −µB cos θ
or,
U = −µ · B
1.1.3
Precession
We have seen that the torque experienced by a magnetic moment in an external
magnetic field B is given by,
τ =µ×B
This implies that the magnetic moment tends to align with the magnetic field.
This is in the case where the magnetic moment has no angular momentum.
2
However, we have seen earlier that the two are intimately linked with one another
as,
µ = γL
Torque τ is equal to the rate of change of angular momentum, which therefore
gives us,
dµ
= γ(µ × B)
dt
The change in µ is perpendicular to both µ and B. Therefore, the magnetic field
causes the direction of µ to precess around B.
It will be helpful to consider a specific example where B is along the z direction and µ is initially at an angle of θ to B in the xz plane. Using the above
relation we get the following equations:
µ˙x = γBµy
µ˙y = −γBµx
µ˙z = 0
On solving these differential equations we get:
µx (t) = |µ| sin θ cos(ωL t)
µy (t) = |µ| sin θ sin(ωL t)
The angular frequency ωL in the above equations is known as the Larmor
precession frequency given as,
ωL = γB
1.1.4
Bohr Magneton
In the atomic model proposed by Bohr for a one electron atom, one of the
postulate was that only those orbits exist for which the angular momentum of
the electron revolving about the nucleus is an integral multiple of the constant
h
, where h is the Planck’s constant. Despite the uncomfortable fact
h̄ = 2π
that no underlying reason was given to explain why this was so, the model
was a success since it was able to explain many features of the hydrogen atom,
such as its spectrum, which could not be explained using classical mechanics.
However, even this model was realized to be extremely limited in its applicability,
since it could explain the spectrum for single electron atoms. Still, it was not
clear for some time why the Bohr model could explain the spectrum and the
energy levels of single electron atoms so well. Was it just a lucky coincidence?
A semi-philosophical argument by Louis deBroglie which also reflected on the
Bohr model of the atom led to a radical revision of ideas on the seemingly
separate entities; particles and waves. According to deBroglie every particle
3
exhibits wave-like behaviour, i.e., every particle has a wave associated with it,
which has a wavelength given by,
λ=
h
p
where p is the momentum of the particle. The postulate given by Bohr can now
be stated as; only those orbits are allowed, where the circumference is an integral
multiple of the wavelength λ of the matter wave associated with the electron.
Thus, the allowed orbits are those where stationary waves of the electron exist.
Bohr’s postulate could now be understood by way of the Wave-Particle duality
principle by deBroglie which hinted at a fundamentally different world at the
atomic and the sub-atomic scale
Let us estimate the size of the atomic magnetic moments and also arrive at
the gyromagnetic ratio by making use of the erroneous but simple Bohr model.
Assume that we have the simplest atom with only the proton at the center
and an electron revolving around it (using Bohr model). The current due to
the revolving electron is given by: I = −e/τ where τ = 2πr/v is the orbital
period (here v is the speed and r is the radius of the circular orbit). According
to Bohr’s postulate, the angular momentum of the revolving electron (me vr) in
the ground state must be equal to h̄. Thus we have,
me vr = h̄
⇒r=
h̄
me v
Therefore,
h̄2
−ev 2 me
×
me 2 v 2
2πh̄
eh̄
−µB = −
2me
is the Bohr magneton, defined by,
µ = πr2 I = π
where µB
µB =
eh̄
2me
It takes the value 9.274 × 10−24 Am2
It is important to note here that because of the negative charge of the electron its magnetic moment is antiparallel to its angular momentum.
The gyromagnetic ratio is the ratio of the magnetic moment with the angular momentum. We therefore have for the present case,
γ=
µB
(−eh/4πme )
−e
=
=
h/2π
h/2π
2me
4
1.1.5
Magnetization and Field
So far we have just been talking about single magnetic moments. A magnetic solid consists of a large number of atoms with magnetic moments associated with
them. A relevant quantity in this context is the magnetization M which is
defined as the magnetic moment per unit volume. M is assumed to be a smooth
vector field which is continuous everywhere except at the edges of the solid.
In free space the magnetic field can be described by the two vector fields B
and H which are linearly related,
B = µ0 H
where µ0 = 4π × 10−7 is the permeability of free space. Thus, the two vector
fields B and H are just scaled versions of each other, B is measured in Tesla
(in SI) and H is measured in Am−1 . This reflects their place in experiment,
where, H is the applied field whose strength is determined by the strength of
the applied current and hence the unit of Am−1 . On the other hand, B is the
field measured in a vacuum (or a solid).
In a magnetic solid, the relation between B and H is more complicated and
the two vector fields may be very different in magnitude and direction. The
general vector relationship is,
B = µ0 (H + M)
In the special case where the magnetization M is linearly related to the magnetic
field H, the solid is called a linear magnetic material. For such materials we
have,
M = χH
where χ is a dimensionless quantity called the magnetic susceptibility. The
definition of M means that χ represents the magnetic moment induced by a
magnetic field H per unit volume. We now have the following relation between
B and H,
B = µ0 (1 + χ)H = µ0 µr H
where µr = 1 + χ is the relative permeability of the material.
1.2
Classical Mechanics and Magnetism
We will first consider the effect of an applied magnetic field on a single charge
using classical arguments. This result will then be used to evaluate the magnetization of a system of charges.
The force F on a particle with charge q moving with velocity v in an electric
field E and magnetic field B is,
F = q(E + v × B)
5
This is better known as the Lorentz force.
Now, B = ∇ × A and E = −∇V − ∂A/∂t. Writing F as F = m ddtv , we may
rewrite the Lorentz force law as,
m
dv
= −q∇V − q∂A/∂t + qv × (∇ × A)
dt
We use the vector identity,
v × (∇ × A) = ∇(v.A) − (v.∇)A
to obtain,
m
dv
∂A
+ q(
+ (v.∇)A) = −q∇(V − v.A)
dt
∂t
Here m ddtv is the force on the charged particle measured in a coordinate system
A measures the rate of
that moves with the particle. The partial derivative ∂∂t
change of A at a fixed point in space. We can rewrite the above expression as,
d
(mv + qA) = −q∇(V − v.A)
dt
We therefore define the canonical momentum as,
p = mv + qA
The kinetic energy can be written in terms of the canonical momentum as
(p−q A)2
2m
1.2.1
The Bohr-van Leeuwen Theorem
We want to find the magnetization that is induced by the magnetic field. Now,
according to Lorentz force law the effect of a magnetic field is to always produce
forces on charged particles which are perpendicular to their velocities. Thus no
work is done and therefore the energy of the system does not depend on the applied magnetic field. If the energy of the system does not depend on the applied
magnetic field, then there can be no magnetization.
This idea is contained in the Bohr - van Leeuwen theorem which states
that in a classical system there is no thermal equilibrium magnetization. A qualitative argument for the theorem is as follows; Electrons in a classical system
with an applied field undergo cyclotron orbits in the bulk of the system. These
orbits will precess in the anti-clockwise (or clockwise) direction depending on
whether applied field goes into the plane of the cross-section of the material
or comes out of that plane. These contribute to a net anti-clockwise current.
However, electrons near the surface cannot perform complete loops and instead
make repeated elastic collisions with the surface. They perform what are called
skipping orbits around the sample perimeter. the anti-clockwise (or clockwise)
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current due to the bulk electrons precisely cancels out the clockwise (or anticlockwise) current associated with the skipping orbits of the electrons.
The Bohr - van Leewen theorem is however at odds with experiment since real
systems containing electrons do have a net magnetization. The conclusion that
can be drawn here is that classical mechanics is insufficient to explain magnetic
properties of materials. Instead we must use quantum theory to understand the
magnetic properties of the real materials.
1.3
Basic Quantum Mechanics
The fundamental object in quantum mechanics is the wave function (also
known as the state function) which contains all the information one obtains
about a system from observation. The wavefunction in general is a complex
function.
Born suggested in 1927 that with reference to the propagation of particles |ψ|2
can be interpreted as the probability density. Quantitatively, Born’s postulate states the following (in Cartesian space). The wavefunction for a particle
ψ(x, y, z, t) is such that,
|ψ|2 dxdydz = P dxdydz
where P dxdydz is the probability that measurement of the particle’s position
at time t finds it in the volume element dxdydz about the point x,y,z.
The four basic postulates of quantum mechanics, which when taken with the
Born postulate serve to formalize the rules of quantum mechanics will be introduced now.
Postulate 1
This postulate states the following: To any self-consistent and well-defined observable in physics, such as linear momentum, energy, angular momentum there
corresponds a Hermitian operator (call it Â) such that measurement of A
yields (call these measured values a) which are eigenvalues of Â. That is, the
values, a, are those values for which the equation,
Âψ = aψ
has a solution ψ. The function ψ is called the eigenfunction of  corresponding
to the eigenvalue a. For Hermitian operators, the eigenvalues are real and the
corresponding eigenfunctions are orthonormal.
Postulate 2
The second postulate of quantum mechanics states that, measurement of the
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observable A that yields the value a leaves the system in the state ψa , where
ψa is the eigenfunction of  that corresponds to the eigenvalue a.
Postulate 3
The third postulate of quantum mechanics establishes the existence of the state
function and its relevance to the properties of a system: The state of a system
at any instant of time may be represented by a state or wave function ψ which
is continuous and differentiable. All the information regarding the state of the
system is contained in the wavefunction. Specifically, if a system is in the state
ψ(r, t), the average of any physical observable C relevant to that system at time
t is,
Z
hCi = ψ ∗ Ĉψdr
where dr represents the differential volume.
Postulate 4
The fourth postulate of quantum mechanics specifies the time development of
the state function ψ(r, t): the state function for a system develops in time according to the equation,
ih̄
∂
ψ(r, t) = Ĥψ(r, t)
∂t
This equation is called the time-dependent Schrödinger equation.
A quantity that is extremely useful in Q.M is the commutator. The commutator of two operators  and B̂ is defined by [Â, B̂] = ÂB̂ − B̂Â
If the two operators  and B̂ commute, i.e., [Â, B̂] = 0, they are said to be
compatible and measurement of one does not affect the value of the other in
any way.
An observable whose operator commutes with the Hamiltonian is a conserved
quantity and is known as a constant of the motion or a good quantum
number
1.3.1
Dirac Bra and Ket Notation
This notation is an invaluable tool in calculation. The integral of the product
of two state functions ψ(x) and φ(x) is written in Dirac notation as,
Z ∞
hψ|φi =
ψ ∗ (x)φ(x)dx
−∞
8
If a is any complex number and the functions ψ and φ are such that,
Z ∞
ψ ∗ (x)φ(x)dx < ∞
−∞
then the following rules hold:
hψ|aφi = a hψ|φi
haψ|φi = a∗ hψ|φi
∗
hψ|φi = hφ|ψi
hψ + φ| = hψ| + hφ|
The expectation value of an observable (operator) can be written using this
notation as,
hÂi = hψ|Â|ψi
This notation conveys the asymmetry of the scalar product.
1.3.2
Orbital Angular Momentum
In classical mechanics thhe significance of angular momentum is that it is one
of the fundamental constants of motion (together with linear momentum and
energy) of an isolated system. The counterpart of this statement also holds for
isolated quantum mechanical systems. This conservation principle for angular
momentum stems from the isotropy of space.
Classically, angular momentum of a particle of a particle is a property that
depends on the particle’s linear momentum p and its displacement r from some
prescribed origin. It is given by
L=r×p
The classical Cartesian components of the orbital angular momentum L for
a particle with momentum p = (px ,py ,pz ) at the displacement r = (x,y,z) are
Lx = ypz − zpy
Ly = zpx − xpz
Lz = xpy − ypx
If we replace p̂ by its corresponding gradient operator (in x-basis). There follows
L̂x = ŷ p̂z − ẑ p̂y = −ih̄(y
9
∂
∂
−z )
∂z
∂y
∂
∂
−x )
∂x
∂z
∂
∂
L̂z = x̂p̂y − ŷ p̂x = −ih̄(x
−y )
∂y
∂x
L̂y = ẑ p̂x − x̂p̂z = −ih̄(z
In terms of the three dimensional vector linear momentum operator
p̂ = −ih̄∇
the equations above may be written as the single vector equation
L = −ih̄r × ∇
T he following commutation relations hold for L̂
[L̂i , L̂2 ] = 0
[L̂i , L̂j ] = iijk L̂k
where ijk is the Levi-Civita tensor.
The first relation signifies that we can determine the component of the angular momentum and the total angular momentum simultaneously in a single
measurement. The second relation tells us that no two components of the angular momentum can be determined in a single measurement.
The operator L̂2 has eigenvalues l(l + 1) where l is the angular momentum
quantum number.The eigenfunctions of L̂2 are known as spherical harmonics. Thus, we have
L̂2 |ψi = l(l + 1)|ψi
L̂z has eigenvalues ml . Thus we have
L̂z |ψi = ml |ψi
where ml is known as the magnetic quantum number which can take values
from −l to l.
1.3.3
The Stern - Gerlach Experiment
The solution of the angular part of the Schrödinger equation for a central potential reveals that the z-component of the angular momentum of the system
is quantized in units of h̄. This result does not have any bearing on the energy
eigenvalues of the system as energy quantization follows solely from the solution
of the radial equation.
An experiment done by Stern and Gerlach in 1921 would provide evidence
for the quantization of angular momentum (before the development of wave mechanics! ) and more importantly would lead to the recognition of an additional
10
degree of freedom associated with electrons, which is now known as spin.
Lying at the heart of the Stern-Gerlach experiment is the concept that some
atoms behave like tiny magnets, i.e., they have a magnetic moment associated
with them. We also found that the magnetic moment and the angular momentum are closely linked as,
µ = γL
Considering the Bohr model this becomes,
µ=(
e
)L
2me
p
If L is quantized as we have claimed above as L = h̄ l(l + 1), then we can write
the magnitude of µ as
eh̄ p
µ=(
) l(l + 1)
2me
But we already know that the prefactor in the above equation is the Bohr
magneton µB and it serves as a natural unit of magnetic dipole moment for
atomic systems.
Let us turn now to the study of the Stern-Gerlach experiment. We know that
if we pass a stream of atoms possessing nonzero magnetic moments is directed
to pass through a nonuniform magnetic field, the magnetic forces involved will
split up the stream into a number of substreams , with one substream for each
possible value of the z-component of µ (usually written as µz ). Suppose for sake
of simplicity that the atoms are all in the same state l. Thus, for a given value
of l there can be (2l + 1) possible values of the magnetic quantum number m,
and therefore (2l + 1) possible values of µz . Thus we should expect to see the
beam split into an odd number of components on passing through the magnetic
field. This is the essence of the Stern-Gerlach experiment.
The essential setup is that atoms of mass m are emitted from an oven with
speed v and directed through a region of nonuniform magnetic field of linear
extent x. The entire apparatus is enclosed in a vacuum chamber to minimize
collisions of the atoms with molecules of air.
For atoms of mass m with z-components of µ given by µz , it can be shown
that the vertical deflection w works out to
w=
dB
x
(R + x/2)µz (
)
mv 2
dz
On performing this now-historic experiment with a beam of silver atoms, Stern
and Gerlach observed two components, in contradiction to the above argument.
Resolution of this observation with the theoritical expectation was provided
11
in the form of a postulate by Uhlenbeck and Goudsmith. Their suugestion was that in addition to their orbital angular momenta, electrons possess an
intrinsic spin angular momentum S. In analogy with the orbital angular momentum quantum number l, Uhlenbeck and Goudsmith proposed the existence of an
electron spin quantum number s having magnitude 1/2 (in units of h̄), and, in
analogy to the (2l + 1) possible z-projections of L, there can only be 2s + 1 = 2
possible z-projections of S. The magnitude of the spin angular momentum is
then
p
p
|S| = s(s + 1)h̄ = 3/4h̄
and the possible z-projections are
sz = sh̄ = ±h̄/2
The overall effect of this purely empirical proposal is to add a fourth quantum
number s = ±1/2 to the already exixting quantum numbers hn, l, mi, therefore
leading to a doubling of the number of possible electronic states.
We can now roughly summarize the Stern-Gerlach experiment. In their original
experiment they had used silver atoms. A neutral silver atom has 47 electrons
- odd number. If it is assumed that the electrons with equal and opposite spin
angular momenta cancel their contribution in pairs, and that the remaining
electron is in l = 0 state, then the presence of two beams in the experiment can
be explained since the lone unpaired electron will have sz = ±1/2
The existence of intrinsic electronic angular momentum was later found to be
a natural consequence of reworking Schrödinger’s equation to incorporate relativistic effects.
Just like orbital angular momentum the spin angular momentum is associated with a magnetic moment which can have a component along a particular
axis equal to
µs = −gµB ms
and a magnitude equal to
p
s(s + 1)gµB =
√
3gµB /2
In the above expressions, g is a constant known as the g - factor. According to
Dirac’s theory of the electron the value of g comes out to be approximately 2 (the
exact value is near 2.003). Therefore the component of the intrinsic magnetic
moment of the electron along the z-axis is ≈ ∓µB . The energy of the electron
in an external magnetic field B is therefore
E = −µs .B
⇒ E = gµB ms B
12
The energy levels of an electron therefore split in a magnetic field by an amount
gµB B. This is called Zeeman splitting.
In general the g-factor can take different values in real atoms depending on
the relative contributions of spin and orbital angular momenta.
1.3.4
Pauli Spin Matrices and Spin Angular Momentum Operator
The behaviour of the electron spin can be described by algebra involving three
special matrices first put forth by Pauli. These matrices called the Pauli Spin
Matrices are defined as
0 1
σ̂x =
1 0
0 −i
σ̂y =
i 0
1 0
σ̂y =
0 −1
It will be convenient to think of these as a vector of matrices σ̂ = (σ̂x , σ̂y , σ̂z )
The spin angular momentum operator is defined as
Ŝ =
h̄
σ̂
2
This implies
h̄
0
Ŝx =
1
2
h̄
0
Ŝy =
i
2
h̄
1
Ŝz =
0
2
1
0
−i
0
0
−1
An important point that will be pointed out here is that Sˆz is diagonal, therefore
if the electron points along the z-direction the representation becomes greatly
simplified. The eigenvalues of Sˆz which are denoted by ms take the values ms =
± 21 and the corresponding eigenvectors/eigenstates are
1
| ↑z i =
0
0
| ↓z i =
1
which correspond to the spin pointing parallel and antiparallel to the z-axis
respectively. Therefore
1
Sˆz | ↑z i = | ↑z i
2
13
1
Sˆz | ↓z i = − | ↓z i
2
The eigenstates corresponding to the spin pointing parallel or antiparallel to the
x and y axes are
1
1
| ↑x i = √
1
2
1
1
| ↓x i = √
−1
2
and
1
1
| ↑y i = √
i
2
1
1
| ↓y i = √
−i
2
This two component representation of spin wave functions is known as spinor
representation and the states are referred to as spinors.
The total angular momentum operator is defined by
 ˆ 
Sx
Ŝ =  Sˆy  = iSˆx + jSˆy + kSˆz
Sˆz
where i,j,k are the unit cartesian vectors
2
The operator Ŝ is therefore given by
2
2
2
2
Ŝ = Sˆx + Sˆy + Sˆz
2
2
2
Now, we know that the eigenvalues of Sˆx ,Sˆy ,Sˆz take the same value given by
2
± 12 = 14
Therefore, for any general spin state |ψi
2
2
2
2
1 1 1
3
Ŝ |ψi = (Sˆx + Sˆy + Sˆz )|ψi = ( + + )|ψi = |ψi
4 4 4
4
2
The most important result we have obtained is that the eigenvalue of Ŝ is given
by s(s + 1)
The commutation relation between the spin operators is
[Sˆx , Sˆy ] = iSˆz ]
and other cyclic permutations of the above. Each of the spin operator commutes
2
with Ŝ
[Sˆ2 , Sˆz ] = 0
What this means is, that it is possible to determine the total spin and one of
its components simultaneously.
14
2
Isolated Magnetic Moments
In this section we ignore interactions between magnetic moments on different
atoms, or between magnetic moments and their immediate environments. We
will consider simply the physics of isolated atoms and their interaction with an
applied magnetic field.
2.1
Atom in a Magnetic Field
An electron in an atom has an orbital angular momentum and also spin angular
momentum. Note that if the position of the ith electron in the atom is ri , and
it has momentum pi then the total angular momentum expressed in units of h̄
is given by,
X
h̄L =
ri × pi
i
We can write down the Hamiltonina for the atom as,
Ĥ0 =
Z
X
pi 2
+ Vi )
(
2m
i=1
where we sum over all the electrons of the atom. We assume that the eigenstates
and eigenvalues of the Hamiltonian are known.
If an external magnetic field B is applied to the atom, the Hamiltonian will
be perturbed. Let us now find out what are the perturbation terms.
We know that
B=∇×A
We choose a gauge such that
A(r) =
B×r
2
Remembering that the kinetic energy must be changed to
[pi + eA(ri )]2
2me
We get the perturbed Hamiltonian as
Ĥ =
=
Z
X
[p + eA(ri )]2
( i
+ Vi ) + gµB B.S
2me
i=1
Z
X
pi 2
e2 X
(
(B × ri )2
+ Vi ) + µB (L + gS).B +
2m
8m
e
i=1
i
15
= Ĥ0 + µB (L + gS).B +
e2 X
(B × ri )2
8me i
The dominant perturbation is usually µB (L + gS).B, but it sometimes vanishes
when L and S are 0, i.e., in case of completely filled orbitals that will be discussed
later. This is known as the paramagnetic term. The third term is called the
diamagnetic term.
2.2
Diamagnetism
Diamagnetism is present in all materials, but it is a weak effect. By diamagnetic
behaviour we mean the induction of a magnetic moment that opposes the applied
magnetic field that caused it.
The effect can be illustrated using the quantum mechanical approach. Consider
an atom in an external apllied magnetic field B parallel to the z axis, then B
× ri = B(−yi , xi , 0), implies
(B × ri )2 = B 2 (xi 2 + yi 2 )
Therefore the first order shift in the ground state energy due to the diamagnetic
term is
Z
e2 B 2 X
h0|(xi 2 + yi 2 )|0i
∆E0 =
8me i=1
where |0i represents the ground state wave function. Assuming spherical symmetry of the atom, we have, hxi 2 i = hyi 2 i = 31 hri 2 i. Thus,
∆E0 =
Z
e2 B 2 X
h0|ri 2 |0i
12me i=1
Now, it is really not very useful to talk about a single atom and its diamagnetism,
since what we actually determine by experiment is the magnetic susceptibility
of the material. Thus, let us generalize our conclusion for a single atom to
a solid composed of N ions (each with the same number of electrons) in a
volume V . To determine magnetization M , let us first write down the first law
of thermodynamics as,
dE = T dS − pDV − M dB
The Helmholtz free energy F is defined by,
F = E − TS
implies
dF = dE − T dS − SdT
which gives
dF = −SdT − pdV − M dB
16
Therefore,
∂F
)T,V
∂B
Therefore, the magnetization M for our case at T = 0 is,
M = −(
M =−
Using the relation χ =
Z
∂F
N ∂∆E0
N e2 B X 2
=−
=−
hri i
∂B
V ∂B
6me V i=1
M
H
≈ µ0 M
H (assuming χ << 1), we obtain the result,
χ=−
Z
N e2 µ0 X 2
hri i
V 6me i=1
Thus we can state that diamagnetic susceptibilities are temperature independent (Note: the above expression assumes first-order perturbation theory).
Relatively large and anisotropic diamagnetic susceptibilities are observed in molecules with delocalized π electrons, such as naphthalene and graphite. The π
electrons are mobile and induced currents can run round the edge of of the ring,
producing a large diamagnetic susceptibility which is largest if the magnetic
field is perpendicular to the plene of the ring. The effective ring diameter is
several times larger than an atomic diameter and therefore the effect is larger.
2.3
Paramagnetism
In paramagnetism, an applied magnetic field induces a magnetization which aligns parallel with the applied field which caused it. Paramagnetic behavious is
dominant in atoms with some unpaired electrons, which would lead to a non-zero
magnetic moment. Without an applied magnetic field, these magnetic moments
on neighouring atoms interact only very weakly with each other and can therefore be assumed to be independent. the application of a magnetic field lines them
up, the degree of which depends on the magnitude of the applied magnetic field.
We stated earlier that nn electron in an atom has an orbital angular momentum
and also spin angular momentum, thus in an atom we expect that the magnetic
moment be associated with the total angular momentum. This is denoted as J
and is simply defined as the sum of the orbital angular momentum L and the
spin angular momentum S,
J=L+S
We expect that an increase in magnetic field will tend to line up the spins, and
an increase in the temperature will randomize the spins. It is therefore expected
that the magnetization of a paramagnetic material will depend on the ratio B
T.
17
We will now consider the general quantum mechanical case, where J can take any integer or half-integer value. For that analysis, we need to recall that the
probability of occupying a given state i with energy Ei at a temperature T is
given by,
e−Ei /kB T
pi =
Z
where Z is the single particle partition function, which is used to normalize the
probability, so that,
X
pi = 1
i
But this condition implies that,
Z=
X
e−Ei /kB T
i
Assuming for our system that the magnetic field points in the z-direction, the
partition function is given to be,
J
X
Z=
emJ gJ µB B/kB T
mJ =−J
If we take x = gJ µB B/kB T , we have
PJ
hmJ i =
mJ =−J
mJ emJ x
PJ
mJ =−J
em J x
=
1 ∂Z
Z ∂x
This leads us to,
ngJ µB ∂Z ∂B
Z ∂B ∂x
Notice that the partition function Z is a geometric progression with initial term
a = e−Jx and the multiplicative ratio r = ex . This can be summed using the
formula
N
X
a(1 − rN )
arj−1 =
1−r
j=1
M = ngJ µB hmJ i =
where N is the number of terms in the series which is for our case is N = 2J + 1.
After some algebraic manipulations this leads to
Z=
sinh[(2J + 1)x/2]
sinh[x/2]
which on substituting
y = xJ
yields
M = ngJ µB JBJ (y)
18
The term ngJ µB J is called as the saturation magnetization, denoted by Ms .
BJ (y) is known as the Brillouin function, given by
BJ (y) =
2J + 1
2J + 1
1
y
coth(
y) −
coth
2J
2J
2J
2J
For y << 1 and hence χ << 1, using Maclaurin expansion of coth y,
BJ (y) =
(J + 1)y
+ O(y 3 )
3J
Therefore for low magnetic fields the susceptibility is given by
χ=
M
µ0 M
nµ0 µef f 2
≈
=
H
B
3kB T
This shows that the magnetic susceptibility is inversely proportional to the
temperature. This is known as Curie’s Law. Also, the measurement of χ allows
us to arrive at µef f ,
p
µef f = gJ µB J(J + 1)
where gJ known as Lande g-value is given by,
gJ =
2.4
2.4.1
3 S(S + 1) − L(L + 1)
+
2
2J(J + 1)
Ground State of an Ion and Hund’s Rules
Fine Structure
Uptil now, L and S have been kept separate, since they are independent of
one other. However, they do couple weakly through spin-orbit interaction. This
acts as a perturbation on the states with well defined L and S. Due to this, L
and S are not separately conserved, however the total angular momentum J is
conserved. If the relativistic effects are considered as a perturbation, then one
can consider L2 = L(L + 1) and S2 = S(S + 1) as being conserved. Thus, the
states with L and S are split into a number of levels having different J values,
this is known as the fine structure. Here J takes the values from |L − S| to
L + S. This implies,
J2 = L2 + S2 + 2L.S
It is known that the spin-orbit interraction takes the form λL.S, where λ is
constant. The expected value of this interaction energy is
hλL.Si =
λ
[J(J + 1) − L(L + 1) − S(S + 1)]
2
The precise value that J takes in the absence of spin-orbit interraction is immaterial. Each level is a multiplet of (2L + 1)(2S + 1) states. When the spin-orbit
interaction is added as a perturbation, these multiplets split up into different
19
fine level structure levels which are labeled by J. Each of these levels themselves have a degeneracy of 2J + 1, which can therefore be removed on application
of a magnetic field. The splitting of the different fine structure levels follows a
relationship known as the Lande interval rule. We have,
E(J)−E(J−1) =
λ
λ
[J(J+1)−L(L+1)−S(S+1)]− [(J−1)J−L(L+1)−S(S+1)]
2
2
⇒ E(J) − E(J − 1) = λJ
Therefore the energy level splittign is proportional to J
2.4.2
Hund’s Rules
We have seen above that J takes the values from |L − S| to L + S, but which one
describes the ground state of an ion. The combination of angular momentum
quantum numbers which are found to minimmize energy are determined by
Hund’s rules. These three empirical rules are to staisfied in the order stated
below:
1. Arrange the electronic wavefunction so as to maximize S. This is in accordance with the Pauli exclusion principle, which prevents electrons with
parallel spins being in the same place.
2. Given the wavefunction determined by the first rule, the next step is to
maximize L
3. Finally, the value of J is found using J = |L − S| if the shell is less
than half-filled and J = |L + S| if it is more than half-filled. This third
rule attempts to minimize the spin-orbit energy. (Note: The third rule is
applicable only in certain circumstances. For example, in transition metal
ions the spin-orbit energy is not as significant as some other energy term
such as the crystal field, so that Hund’s rule is disobeyed. However, for
rare earth ions Hund’s third rule works very well.)
Having thus found the values of S, L and J. This ground state can be denoted by
a term symbol of the form 2S+1 LJ , where L is not written as a number, but as
a letter according to the sequence:L = 0 → S, L = 1 → P ,L = 2 → D and so on.
It is important to remember that Hund’s rules lead to a prediction of the ground
state, however they reveal nothing about the excited states. Thus we can estimate the magnetic moment of an ion assuming that only the ground state is
populated. We found earlier that a measurement of the susceptibility enables
us to deduce the effective moment. The effective moment can be expressed in
units of Bohr magneton as
p = µef f /µB
Therefore we should expect to measure
p
p = gJ J(J + 1)
20
Good experimental agreement is found between the predicted and the measured
values of p for 4f ions in the solid state. There exists a discrepancy for Sm
and Eu, but it is due to low-lying excited states, which because of their close
proximity with the ground state in energy, are also populated to some extent,
and therefore cause p to shift from its ground state value. Comparatively, much
poorer agreement is found for 3d ions because of the effect of the local crystal
environment that will be considered later. This effect is not significant for 4f
ions because the partially filled 4f shells lie deep within the ion, beneath the
filled 5s and 5p shells.
3
The Crystal Environment
We just saw that the magnetic moments of many crystals containing rare earths
can be determined by considering the rare earth ions to be completely independent with no interactions with each other and the surroundings. However,
we also noted that this is not the case for crystals containing transition metal ions, where the interactions between these transition metal ions and their
surrounding are large and significant.
3.1
Crystal Fields
Consider the angular dependence of the electron density of the s,p and d orbitals
Only s orbitals are spherically symmetric, the other orbitals have pronounced
angular dependence. This is important since the local environments are often
not spherically symmetric, which leads to varying behaviour of the different
orbitals.
3.1.1
Origin of crystal fields
The crystal field is an electric field originating from the neighbouring atoms in
the crystal. In crystal field theory the neighbouring orbitals are modelled as
negative point charges. The size and the nature of crystal field effects depend
crucially on the symmetry of the local environment. One common environment
in many crystals is the octahedral environment. In this environment, a transition
metal ion sits at the center of an octahedron with commonly an oxygen ion at
each corner. In such a scenario, the crystal field basically arises from the electrostatic repulsion from the negatively charged electrons in the oxygen orbitals.
Figures for the octahedral and tetrahedral environment are shown below.
21
Now, we will concern ourselves primarily with d orbitals which are of significance to transition metal ions. The d orbitals fall into two classes, the t2g orbitals
which are oriented between the x, y and z axes (these are the dxy , dxz , dyz
orbitals) and the eg orbitals which are oriented along the x, y and z axes (these
are the dz2 and the dx2 −y2 orbitals). These are shown below:
22
Suppose we place a cation containing ten d electrons is placed in the center of
a sphere of radius r which is uniformly negatively charged. The d orbitals will
all be degenrate in this case, though the presence of charge will raise the enrgy
of the entire system. If the charge on the sphere is now made to collect into
six discrete point charges, each of which lie on the vertex of an octahedron. In
this case, the total electronic energy of the d orbitals will not change, but the d
orbitals, will not be degenerate anymore.
In an octahedral environment, neighbouring positive charges congregate at the
points (±r, 0, 0), (0, ±r, 0) and (0, 0, ±r). Therefore, the three orbitals dxy , dxz ,
dyz which point between the x, y and z axes will be lowered in energy, but the
dz2 and the dx2 −y2 orbitals which point along the x, y and z axes will be raised
in energy.
In a tetrahedral environment, the crystal field works in the opposite sense. The
orbitals which point along the axes maximmally avoid the charge density associated with the atoms situated on four of the corners of the cube which describe
a tetrahedron. Thus in this case, the two-fold eg levels are lower in energy.
All of the above is summarized in the picture below:
If one is dealing with a transition metal ion in which the 3d is partially filled.
The precise order in which the orbitals fill depends on the relative strengths of
the crystal field and Coulomb repulsion for pairing. The coulomb energy cost of
putting two electrons in the same orbital is known as pairing energy.
If the crystal field energy is lower than the pairing energy (known as the weak
field case), then as the electrons are added to the system, they will first occupy
each orbital singly before any orbital is doubly occupied. If on the other hand,
the crystal field energy is larger is larger than the pairing energy (this is the
strong field case), the electrons will first doubly occupy the lower energy orbitals before filling the higher energy orbitals. The figure below illustrates both
cases:
23
4
Interactions
There are two main interactions that we need to consider in order to understand
how magnetic moments communicate with each other
4.1
Magnetic dipolar interaction
The interaction energy between two magnetic dipoles µ1 and µ2 separated by r
is given by
3
µ0
[µ1 .µ2 − 2 (µ1 .r)(µ2 .r)]
Uint =
3
4πr
r
From the above expression we can infer that Uint depends on the separation
of the magnetic moments and their degree of mutual alignment. Let us try to
arrive at an estimate Uint assuming that µ ≈ µB and they are separated by
r ≈ 1 Angstrom. Using all this information we arrive at the approximation
µ2
≈ 10−23 J
4πr3
Using the approximation E ≈ kT , where k = 1.38 × 10−23 is the Boltzmann’s
constant. Thus the energy is of the order of 1K in temperature. What this means
is that we cannot account for the ordering of the magnetic materials since it
is very weak at room or higher temperatures. However, we cannot ignore this
interaction at temperatures which are of the order of milliKelvin.
4.2
Exchange Interaction
Exchange interactions are the ones which are primarily responsible for long range order. Let us consider the origin of exchange interaction.
Consider a simple system of two electrons with spatial wavefunctions ψa (r1 )
and ψb (r2 ) respectively. The overall wavefunction (including the spin parts)
24
must be antisymmetric under exchange (this is explained beautifully in Lectures on Physics Vol.3 by Richard P Feynman. Thus, the spatial wavefunction can be either symmetric or anti-symmetric under exchange of electrons.
Let us digress a bit and consider the interaction of two spin-half particles, where
the interaction is described by the Hamiltonian,
Ĥ = ASˆa .Sˆb
where Sˆa and Sˆb are the spin operators for the two particles respectively. We
can consider the total spin of the system, defined as
ˆ = Sˆa + Sˆb
Stot
which implies
ˆ )2 = (Sˆa )2 + (Sˆb )2 + 2Sˆa .Sˆb
(Stot
Combining two spin-half particles results in a joint entity with spin quantum
ˆ ) is s(s + 1) which is
number s = 0 or 1. Therefore, the eigenvalue of (Stot
ˆ
a
b
ˆ
consequently 0 or 2. The eigenvalues of S and S are 43 . Therefore we obtain
For s = 1
1
Sˆa .Sˆb =
4
and
for s = 0
3
Sˆa .Sˆb = −
4
ˆ
a
b
ˆ
The Hamiltonian is Ĥ = AS .S for the system. Therefore the system has two
energy levels for s = 0 and s = 1 with energies − 3A
4 and
We
the
the
the
A
4
respectively.
know that the degeneracy of a state is given by 2s + 1, which tells us that
s = 0 state is a singlet and the s = 1 state is a triplet. The z component of
spin,ms , takes the value 0 for the singlet state and the values −1, 0, 1 for
triplet.
Let us now come back top our original argument. We said that the overall
wavefunction must be symmetric under the exchange of electrons. States like
| ↑↑i and | ↓↓i are symmetric under exchange of electrons (where in this representation the first arrow refers to the z component of the spin labeled by a
and the second arrow refers to the z component of the spin labeled b. States
| ↑↓i and | ↓↑i are neither symmetric nor anti-symmetric under exchange of the
two electrons. Therefore, we must take their linear combinations (which must
√
be normalized). The state |↑↓i+|↓↑i
is symmetric under exchange, just like the
2
states | ↑↑i and | ↓↓i and the other state
change.
25
|↑↓i−|↓↑i
√
2
is antisymmetric under ex-
Therefore, we can conclude now that the eigenstate corresponding to eigen√
value s = 0 is |↑↓i−|↓↑i
and the eigenstates corresponding to s = 1 are | ↑↑i,
2
| ↓↓i and
|↑↓i+|↓↑i
√
.
2
Let us denote the antisymmetric singlet state by χS and a symmetric triplet
state by χT . We can write the overall wavefunctions (for singlet and triplet)
now as,
1
ΨS = √ [ψa (r1 )ψb (r2 ) + ψa r2 )ψb (r1 )]χS
2
1
ΨT = √ [ψa (r1 )ψb (r2 ) − ψa r2 )ψb (r1 )]χT
2
The energies of the two states are
Z
ES = ΨS ∗ ĤΨS dr1 dr2
Z
ET =
ΨT ∗ ĤΨT dr1 dr2
Therefore the difference between the two energies is
Z
ES − ET = 2 ψa ∗ (r1 )ψb ∗ (r2 )Ĥψa (r1 )ψb (r2 )dr1 dr2
The difference in energy can be parametrised using S1 .S2 .
Ĥ =
1
(ES + 3ET ) − (ES − ET )S1 .S2
4
We define the exchange constant or exchange integral as
Z
ES − ET
= ψa ∗ (r1 )ψb ∗ (r2 )Ĥψa (r1 )ψb (r2 )dr1 dr2
J=
2
and therefore the spin dependent term of the above Hamiltonian can be written
as,
ˆ = −2JS1 .S2
H spin
If J > 0, ES > ET and the triplet state is favoured. If J < 0, ES < ET and the
singlet state is favoured.
Although we found the above relation considering a system of only two electrons, it was recognized earlier that the interactions of this form apply between
all neighbouring atoms. This motivated the Hamiltonian of the Heisenberg
model:
X
Ĥ = −
Jij Si .Sj
ij
26
4.2.1
Ferromagnetic Interaction
A ferromagnet is a magnetic material that has a spontaneous magnetization
even in the absence of an applied magnetic field. All the magnetic moments
conspire to result in the maximum magnetization possible given the constraint
imposed by the crystal field. This effect is primarily due to exchange interactions.
Thus, the Hamiltonian that is to be solved is
X
Ĥ =
Jij Si .Sj
ij
The exchange constant has to be positive for nearest neighbours to ensure ferromagnetic alignment.
4.2.2
Antiferromagnetic Interaction
For antiferromagnets the situation is exactly opposite, where given the constraint imposed by crystal field the magnetic moments conspire to minimize the
magnetization. In this case the exchange constant has to be negative for nearest
neighbours to ensure antiferromagnetic alignment.
4.2.3
Ising Model
In the Heisenberg model the spins Si are treated as three dimensional vectors
because we allow them to point in any direction in 3-D space.
A related model is the Ising model in which the spins are only allowed to
point up or down, i.e., only the z component is considered. The corresponding
Hamiltonian for this model is,
X
Ĥ = −
JSiz Sjz
ij
5
Frustration
The first frustrated system identified was crystalline ice, which has frozen-in
disorder remaining down to very low temperatures. This property is known as
residual or zero-point entropy. Pauling predicted a special type of proton
disorder that obeys what are called the ice rules. These rules proposed by Bernal and Fowler require that two protons are near to and two further away from
each oxygen ion as shown below.
27
Pauling showed that the ice rules do not lead to order in the proton arrangement
but rather the ground state is macroscopically degenerate, which says that the
number of degenerate or energetically equivalent proton arrangements diverges
exponentially with the size of the sample. Pauling estimated the degeneracy
to be ≈ (−3/2)N/2 where N is the number of water molecules. The zero-point
entropy is therefore S0 = (R/2) log(3/2). The disordered ice-rules proton arrangements was eventually comfirmed by neutron diffraction experiments.
Spin ice
It was noticed that a model of ferromagnetism on the pyrochlore lattice would
exactly map onto the ice model if the magnetic moments were constrained to
point along the axes joining the centers of the two tetrahedra to which it belongs.
28
This seemed rather surprising since we normally do not expect frustration in a
ferromagnet (from experience of triangular and kagome lattice). However, the
ferromagnetic model is compatible with cubic symmetry and has been observed to be approximated by the apparently ferromagnetic pyrochlore material
Ho2 T i2 O7 . Since, this constituted the first simple physical realization of a three
dimensional magnetic analog of ice, the name spin ice was coined to emphasize
the analogy.
Let us now explore the pyrochlore material Ho2 T i2 O7 in more detail. In Ho2 T i2 O7 ,
the Ho3+ occupy a pyrochlore lattice of corner-linked tetrahedra as shown below:
29
Magnetism arises solely from Ho3+ as T i4+ is nonmagnetic. Ho3+ has a particularly large magnetic moment of about 10µB which persists to the lowest temperatures. This large temperature-independent moment is ensured by the surrounding environment of Ho3+ , i.e, the crystal field in the pyrochlore structure.
It should be mentioned here that the pyrochlore structure has cubic symmetry,
i.e., the unit cell of the crystal is a cube which contains a single tetrahedron,
this cell when repeated, generates the pyrochlore structure.
Each tetrahedron of Ho3+ has an oxide ion at its center, what this means is that
two oxide ions lie close to each Ho3+ along the axis that connects the vertex to
the center of the tetrahderon. For a tetrahedron in a cube, it can be shown that
the axes connecting the center of the tetrahedra to the vertices lie along h111i,
h1̄1̄1i, h1̄11̄i and h11̄1̄i for the four vertices respectively. We can also state this
in the following way: Suppose we denote the center of the tetrahedron by G, the
vertex of the tetrahedron at the origin is denoted by O, the other three vertices
are denoted by P , Q and R respectively then (ignoring magnitudes)
~ = h1̄1̄1̄i
GO
~ = h111̄i
GP
~ = h11̄1i
GQ
~ = h1̄11i
GR
30
Coming back to our original discussion, the anisotropic crystallographic environment changes the ground state of Ho3+ such that its magnetic moment has
its maximum possible magnitude and lies parallel to the h111i axis (here we
are making use of the symmetry of the tetrahedron to generally state that the
direction is h111i. Now, for Ho3+ S = 2 and L = 6 and since the shell is more
than half-filled, J = L + S implies J = 8. This state can be denoted as 5 I8 . This
state is split by the local anisotropic crystal field such that the ground state is
an almost pure |J, mJ i = |8, ±8i doublet with h111i quantization axis. The first
excited state is several hundreds of Kelvin above the ground state as has been
revealed by experiment. Thus, at low temperatures, the excited states are not
accessed thermally.
With this information let us now estimate the net magnetic moment for a single
tetrahedron for the case of antiferromagnetic interaction and the ferromagnetic
interaction.
In case of antiferromagnetic interaction, it can be immediately realized by symmetry considerations that the net magetic moment for a tetrahedron will be
zero for the two cases of all spins pointing in or all spins pointing out
For the ferromagnetic case, it is not so straightforward, and we need to consider
two cases and conclude which is the correct configuration for the ferromagnetic
case. The two cases are:
1. 1-in, 3-out (or 3-in, 1-out)
2. 2-in,2-out
For 1-in, 3-out
(~
µnet )1 = C[h111i + h111̄i + h11̄1i + h1̄11i]
= C[h222i]
√
⇒ |(~
µnet )1 | = 12C
where C is a constant which is equal to the magnitude of the magnetic moment
of Ho3+ ion divided by the normalization factor of the axis vectors.
For the 2-in, 2-out case
(~
µnet )2 = C[h111i + h1̄1̄1i + h11̄1i + h1̄11i]
= C[h004i]
√
⇒ |(~
µnet )2 | = 16C
This implies that
|(~
µnet )2 | > |(~
µnet )1 |
31
Thus, ferromagetic interaction would lead to the 2-in, 2-out configuration.
This configuration has since been verified by various experiments and the residual entropy has also been determined experimentally (for Dy2 T i2 O7 ) which is
quite close to Pauling’s estimate for the entropy associated with the extensive
degeneracy of ice: R/2 log(3/2) = 1.68J mol−1 K −1 Finally we explore the behaviour of spin-ice compound in the presence of an applied magnetic field.
We will estimate the magnetization per unit spin when the field is applied along
the three symmetry directions - h100i, h110i and h111i. The application of the
field results in the selection of magnetic order from the macroscopically degenerate ground state, which will result in a component of magnetization along the
field direction.
In the first case, when the field is applied along the h100i all the four spins
on the tetrahedron order so that each has a component of magnetic moment
along the field direction, and the degeneracy is completely broken. The saturated magnetization per unit spin is √13 ≈ 0.578
In the second case when the field is applied along h110i, two of the local Ising
axes are perpendicular to the direction of the applied field, so they will not
order. The only spins that can order are the other two remaining spins. The
maximum net magnetization per spin is √12 . √13 = √16 ≈ 0.408
In the last case when the field is applied along the h111i axis, there is one spin
on each tetrahedron with its anisotropic axis parallel to the field, the others are
only partially ordered. When the field becomes strong enough to overcome the
exchange the two-spins-in, two-spins-out rule will be broken to give an ordered
arrangement of one spin in, three spins out. The saturation magnetization per
spin in this case will be 12 . We see that the degeneracy breaking is a two stage
process and this is borne out in experiment as well.
6
References
1. Magnetism in Condensed Matter - Stephen Blundell, OUP
2. Spin Ice State in Frustrated Magnetic Pyrochlore Materials - Steven T
Bramwell et al, Science, Vol.294
3. Liquid-Gas Critical Behaviour in a Frustrated Pyrochlore Ferromagnet M J Harris, S T Bramwell et al, Physical Review Letters, Vol.81, Number
20
4. Crystal-field interaction in the pyrochlore magnet Ho2 T i2 O7 - S Rosenkranz, A P Ramirez et al, Journal of Applied Physics, Vol.87, Number
9
32