SOLID-STATE PHYSICS II 2007 O. Entin-Wohlman

SOLID-STATE PHYSICS II 2007
O. Entin-Wohlman
1.
2.
3.
DIAMAGNETISM AND PARAMAGNETISM
The interaction of electrons with a uniform magnetic field. A uniform magnetic
field couples to the electronic motion, and to the electron spin. The coupling with the
spin adds to the Hamiltonian the Zeeman interaction
g 0 µB H · S ,
(3.1)
in which H is the magnetic field. Here, S is the total spin of the electrons, i.e.,
S=
X
1
and si = σ ,
2
si ,
i
(3.2)
where σ is the vector of Pauli matrices,

σx = 
σ = x̂σx + ŷσy + ẑσz ,




1
0 −i
1
0
 , σy = 
 , σz = 
 .
0
i
0
0 −1

0
1
(3.3)
In Eq. (3.1), µB is the Bohr magneton µB = e~/2m = 0.927 × 10−20 erg/G and g0 is the
g−factor (Landé factor), which is about 2. The coupling of the magnetic field to the orbital
motion of the electron is described by the vector potential A, such that
H=∇×A .
(3.4)
We shall use the gauge in which ∇ · A = 0. (One can always shift the vector potential by
an arbitrary function ∇χ and make ∇ · A = 0 without changing the magnetic field, which
is the physical quantity). We hence take the vector potential to be
1
A(r) = − r × H .
2
1
(3.5)
(Note that the magnetic field is uniform.) The vector potential modifies the kinetic energy,
making the momentum of the i−th electron, pi , to be pi + (e/c)A(ri ). The kinetic energy
part of the Hamiltonian becomes, in the presence of a uniform magnetic field,
´2
1 X 2
1 X³
e
p →
pi + A(ri ) .
2m i i
2m i
c
(3.6)
It follows from Eqs. (3.1) and (3.6) that the change in the Hamiltonian of the electrons due
to the magnetic field is
∆H = g0 µB H · S −
´
e2 X ³ 2 2
e X
2
.
pi · ri × H +
H
−
(r
·
H)
r
i
i
2mc i
8mc2 i
(3.7)
The terms linear in the magnetic field can be combined together. Since the total electronic
angular momentum of the electrons, L, is
~L =
X
ri × pi ,
(3.8)
³
´
µ B H L + g0 S .
(3.9)
i
the linear terms give
We can therefore write the change in the Hamiltonian in the form
´
³
´
X³
e2
2
2
2
∆H = µB H L + g0 S +
xi + y i .
H
8mc2
i
(3.10)
In writing down the second term here we have assumed that the magnetic field is along
the z−direction. Once we know the modifications of the Hamiltonian in the presence of a
uniform magnetic field, we can find the change in the energy of the system (or the change in
the free energy) and use them in order to compute the magnetic properties of our system.
The magnetic susceptibility. The response of a system to a magnetic field is characterized by its magnetic susceptibility. This quantity is defined as follows. Let us consider a
quantum-mechanical system at zero temperature, and calculate the change in the ground
state energy, E0 , under the application of a magnetic field. Then, the magnetization density
is given by
M(H) = −
2
1 ∂E0 (H)
,
V ∂H
(3.11)
where V is the volume. The susceptibility, χ, is defined by
χ=
∂M
.
∂H
(3.12)
At finite temperatures, where the system is not in the ground state, we have to replace in
the above definitions the ground state energy by the free energy.
Larmor diamagnetism. When a solid consists of ions whose all electronic shells are filled,
the wave function of the ground state is characterized by zero angular momentum (since
such ions are spherically symmetric) and zero spin. In such a case there is no contribution
to the ground state energy from the term linear in H [see Eq. (3.10)], and we are left with
X
e2
2
∆E0 =
ri2 |Ψ0 i .
H
hΨ
|
0
2
12mc
i
(3.13)
The magnetic susceptibility given in Eq. (3.12) is negative, and the material is diamagnetic.
This is dubbed ‘Larmor diamagnetism’. Materials in which the magnetic susceptibility is
negative are called ‘diamagnetic’
since in the presence of a magnetic field their energy
increases, they try to avoid it by directing the induced magnetic moment opposite to the
field.
∗ ∗ ∗ exercise: Explain how Eq. (3.13) is obtained, find an explicit form for the diamagnetic
Larmor susceptibility and estimate its magnitude.
Partially filled shells. A partially filled ion is an ion whose all shells are either completely
filled or completely empty, except for one (the ‘outer’ shell). There are two questions to be
asked: (a) what is the modification of the ground state energy caused by the magnetic field,
and (b) how is the ground state specified.
The first question is somewhat easier. Going back to Eq. (3.10), we use perturbation theory
to find the change in the energy caused by the extra term in the Hamiltonian, ∆H.
The calculation of the change in the energy in perturbation theory is carried out as follows.
The full Hamiltonian is H + ∆H, where ∆H is assumed to be small. The eigen functions
of the part H of the Hamiltonian are denoted Ψn , and their corresponding energies are En .
It is important to remember that the eigen functions form a complete orthonormal basis.
In order to find the correction of the ground state energy, we write the (full) Schrödinger
3
equation in the form
³
´³
i´
Xh
(1)
(2)
H + ∆H Ψ0 +
an Ψn + an Ψn + . . .
n
³
(1)
(2)
= E0 + E0 + E0 + . . .
´³
i´
Xh
(2)
Ψ0 +
a(1)
Ψ
+
a
Ψ
+
.
.
.
.
n
n
n
n
(3.14)
n
(i)
Here, n runs over all eigen values, the coefficients an give the correction of order i (i =
1, 2, . . .) of the eigen functions, (namely, the corrections to the eigen functions are expanded
(i)
in the complete basis formed by the Ψn ) and E0 is the correction of order i of the ground
state energy. The next step is to equate identical orders in Eq. (3.14).
(i)
(i)
At order zero, ∆H = 0, and all an and E0 are zero as well. Equation (3.14) is then
HΨ0 = E0 Ψ0 .
(3.15)
In first order in ∆H, Eq. (3.14) is
H
X
a(1)
n Ψn + ∆HΨ0 = E0
n
X
(1)
a(1)
n Ψn + E0 Ψ0 .
(3.16)
n
When his equation is multiplied on the left by Ψ0 , it gives
(1)
E0 = hΨ0 |∆H|Ψ0 i ,
(3.17)
and when it is multiplied from the left by any other eigen function Ψ` , ` 6= 0 it gives
(1)
a` =
hΨ` |∆H|Ψ0 i
,
E0 − E`
` 6= 0 .
(3.18)
To second order in the perturbation ∆H, Eq. (3.14) gives
H
X
a(2)
n Ψn + ∆H
X
n
a(1)
n Ψn = E0
X
n
(1)
a(2)
n Ψn + E0
n
X
(2)
a(1)
n Ψn + E0 Ψ0 .
(3.19)
n
Multiplying from the left by Ψ0 , we obtain
(2)
E0 =
X
a(1)
n hΨ0 |∆H|Ψn i
n
=
X hΨn |∆H|Ψ0 ihΨ0 |∆H|Ψn i
n6=0
E0 − En
=
X |hΨn |∆H|Ψ0 i|2
n6=0
E0 − En
.
(3.20)
We have used here Eq. (3.18). Obviously, we can use the second-order equation to find
other coefficients in the expansion of the eigen functions, but those are not required for our
purposes.
4
∗ ∗ ∗ exercise: Does the second-order correction to the energy have a definite sign? what
is this sign? what happens to the second-order corrections of energies which are not the
ground state energy?
In our case, ∆H, Eq. (3.10), includes a term linear in the magnetic field, and a term which
is quadratic in the magnetic field. Therefore, the correction to the ground state energy, valid
up to second order in the magnetic field is
X
e2
2
∆E0 =µB H · hΨ0 |L + g0 S|Ψ0 i +
H
hΨ
|
(x2i + yi2 )|Ψ0 i
0
2
8mc
i
X |hΨ0 |µB H · (L + g0 S|Ψn i|2
+
.
E0 − En
n6=0
(3.21)
Hund’s rules. In order to find the magnetic nature of systems made of partially filled ions,
we now need to (a) specify the ground state (in the absence of the magnetic field), (b) insert
the result in Eq. (3.21) to find the change in the energy of the ground state, and (c) take
the second derivative with respect to the magnetic field and find the magnetic susceptibility.
For example, in the case of transition metals, e.g., copper, the outer shell is the d−shell, of
angular momentum ` = 2. [This means that the orbital angular momentum squared of each
electron–the expectation value of L2 , has the value `(` + 1).] The projection of the angular
momentum vector along the z−direction, `z , can take 2` + 1 values,
`z = −`, −` + 1 , . . . ` − 1 , ` .
(3.22)
Hence the d−shell is five-fold degenerate, namely, there are five single-electron wave functions
(or orbitals) corresponding to the d−shell. Each of these orbitals can accumulate two spin
directions, namely it may have sz = ±1/2, and therefore the full degeneracy of the d−shell
is 10. In other words, we can put up to 10 electrons in the d−shell. Copper, for example,
has 9 electrons in that shell.
In general, the number of electrons in the outer shell is n, such that 0 < n < 2(2` + 1). If
these electrons do not interact with each other, then there are many ways to distribute n
electrons on 2(2` + 1) levels. However, the electron-electron interactions, and the spin-orbit
interaction, reduce significantly the number of these different possibilities. This is achieved
according to famous rules (which are in fact only approximate), called the Hund rules. We
shall state these rules without their derivation, assuming that the many-electron eigen states
5
and eigen energies of the ion are characterized by the quantum numbers corresponding to
the total spin of the electrons, S, their total orbital angular momentum, L, and their total
angular momentum, J.
Hund’s first rule. The electronic states with the lowest energy are those with the largest
value of the total spin, such that these states are still consistent with the exclusion principle.
This means that as long as the number of electrons, n, is such that n ≤ 2` + 1, all their
spins are parallel, and S = n/2. When n > 2` + 1, the total spin is reduced.
Hund’s second rule. The electronic states with the lowest energy have the largest possible
P
value of the angular momentum, L = | `z |, which is consistent with Hund’s first rule and
with the exclusion principle.
Hund’s third rule. This rule has to do with the total angular momentum, J. The total
angular momentum takes integral values in the range |L − S| and L + S. Therefore, once S
and L are given, there are still (2L + 1)(2S + 1) many-electron possible states. (Remember
that the degeneracy of a level with a certain J is 2J + 1, since Jz = −J, −J + 1, . . . , J.)
Hund’s third rule uses the spin-orbit interaction to choose the ground state(s) among these
states. The spin-orbit interaction reads λL · S, where λ is the spin-orbit coupling. It turns
out that λ > 0 for shells that are less than half filled and is negative for shell which are more
than half filled. Hund’s third rule tells us that J = |L − S| when n ≤ 2` + 1, because then
the spin-orbit interaction (with λ > 0) reduces the energy, and J = L + S, for n ≥ 2` + 1,
for the same reason(with negative λ).
The ground state of a d−shell ion.
6
n
2
1
0
-1
1
↑
2
↑
↑
3
↑
↑
↑
4
↑
↑
↑
↑
5
↑
↑
↑
↑
6
↑↓
↑
↑
7
↑↓
↑↓
8
↑↓
9
10
-2
S
L
J
1/2
2
3/2
1
3
2
3/2
3
3/2
2
2
0
↑
5/2
0
5/2
↑
↑
2
2
4
↑
↑
↑
3/2
3
9/2
↑↓
↑↓
↑
↑
1
3
4
↑↓
↑↓
↑↓
↑↓
↑
1/2
2
5/2
↑↓
↑↓
↑↓
↑↓
↑↓
0
0
0
∗ ∗ ∗ exercise: Prepare a similar table for the ions with partially filled f −shell (L = 3).
Hund’s three rules determine the ground state(s) of the partially-filled ion. However, that
ground state is still degenerate. Take for example, the case n = 2 in the Table. After
applying Hund’s first and second rules, it has total spin S = 1 and total orbital angular
momentum L = 3. This means that the states with J = 2, 3, and 4 are all possible. This
gives for the case of n = 2 electrons 5 + 7 + 9 = 21 options. (Note that in this case,
(2L + 1)(2S + 1) = 21.) However, Hund’s third rule tells us that the lowest energy is
obtained for J = |L − S| = 2, and therefore, the ground state of a partially-filled d−shell
with two electrons has J = 2 and is 5−fold degenerate.
∗ ∗ ∗ exercise: Repeat this argument and find the degeneracy of all ground states corresponding to the d and f shells.
Now that we have specified the ground state(s) of the ions, we turn to the calculation of the
ground state energy. Here we distinguish between two possibilities: either the ground state
is non degenerate, which happens when J = 0, or it is degenerate. If it is not degenerate,
we may use Eq. (3.21) for the energy. It turns out that hΨ0 |L + g0 S|Ψ0 i = 0 when J = 0,
but hΨn |L + g0 S|Ψ0 i 6= 0 (this will be explained below). Therefore, only the two terms
in Eq. (3.21) which are quadratic in H contribute to the energy. The first one leads to
diamagnetism, as we have found above, and yields the Larmor diamagnetic susceptibility.
7
The second term quadratic in H yields positive magnetization, which means that the material
is paramagnetic. In a paramagnetic material, the application of a magnetic field reduces
the energy, and therefore the material does not try to ‘oppose’ the effect of the magnetic
field, as is the case with a diamagnetic material. We see that partially filled band with
J = 0 can be either paramagnetic or diamagnetic, depending on the competition between
the two H2 −terms in Eq. (3.21). (Note that this is correct as long as one can deduce the
magnetization from the ground state energy alone, namely, when the usual thermal energy
is not enough to excite higher energy states.)
When J 6= 0 the ground state energy is 2J + 1−fold degenerate, and Eq. (3.21) for the
ground state energy cannot be used. The application of the magnetic field removes this
degeneracy, but then we need to diagonalize an (2J + 1) × (2J + 1) matrix, made of the
matrix elements hJLSJz |(Lz + g0 Sz |JLSJz0 i. Luckily enough, there is a theorem, called the
Eckart-Wigner theorem, which states that within the 2J + 1 manifold,
hJLSJz |(Lz + g0 Sz |JLSJz0 i = g(JLS)Jz δJz Jz0 ,
(3.23)
where g(JLS) is a number which depends on the values of J, L, and S. Therefore, (to first
order in the magnetic field H), the ground state energy splits into a ladder-like spectrum of
2J +1 levels. However, since in the absence of the field the ground state energy is degenerate,
we must take into account the entropy in calculating the magnetic susceptibility, in addition
to the energy. In other words, we need to find the free energy.
Curie’s law. The free energy, F, of an ion, whose relevant possible energies are given by
E(Jz ) ≡ γHJz ,
γ = g(JLS)µB ,
(3.24)
is given by
−βF
e
≡
J
X
Jz =−J
−βγHJz
e
eβγH(J+1/2) − e−βγH(J+1/2)
=
.
eβγH/2 − e−βγH/2
(3.25)
The magnetization of such an ion is given by [cf. Eq. (3.11)]
∂F
= γJBJ (βγJH) ,
∂H
³ (2J + 1)x ´
³x´
1
2J + 1
coth
−
coth
is the Brillouin function . (3.26)
where BJ (x) =
2J
2J
2J
2J
M≡−
8
Note that the Brillouin function approaches 1 as x → ∞ (since then the coth approaches
1). This means that when the Zeeman energy γJH is much larger than the thermal energy,
the magnetization of the ion attains its maximal value, γJ. At temperatures such that the
thermal energy is larger than the Zeeman energy, we use the fact that coth(x) '
1
x
+ x3 , to
find
(J + 1)x
.
3J
(3.27)
(gµB )2 J(J + 1)
=
, kB T À gµB H .
3
kB T
(3.28)
BJ (x) '
It therefore follows that
¯
¯
χ¯
single ion
To obtain the susceptibility of the entire solid, we multiply this susceptibility by the density
of ions in the solid.
Equation (3.28) is the Curie’s law. It tells us that partially-filled ions with J 6= 0 are, generally, paramagnetic, and that their inverse susceptibility is proportional to the temperature,
at temperatures which are not too low.
∗ ∗ ∗ exercise: Derive in detail Eqs. (3.25), (3.26), (3.27), and (3.28). Give explicit expressions for the case J = 1/2, and compare with Eq. (3.30) below. Plot the magnetization and
the susceptibility for this specific case, as function of the temperature.
∗∗∗ exercise: Consider an ion with a partially filled shell of total angular momentum J, and
Z additional electrons in filled shells. Show that the ratio of the paramagnetic susceptibility
to the Larmor diamagnetic susceptibility is
2J(J + 1) ~2
χpar
=−
,
χdia
ZkB T mhr2 i
(3.29)
and estimate its magnitude.
In order to clarify the use of the free energy [see Eq. (3.25) above], let us consider the
magnetization of a single spin 1/2, as function of the temperature. A spin half, in the
presence of a magnetic field H, can be either aligned with the field, in which case its energy
is enhanced by µB g0 H/2, or it can be anti parallel to the field, in which case its energy is
reduced by µB g0 H/2. It is hence clear that at zero temperature, the spin will be anti parallel
to the field, namely, it will be magnetized. However, at very high temperatures, that spin
9
has equal probabilities to be aligned or anti aligned with the field, in which case its average
magnetic approaches zero.
At temperature T , the average magnetization of the spin is
M = µB g0
0.5eβµB g0 H/2 − 0.5e−βµB g0 H/2
µB g0
βµB g0 H
=
tanh
.
βµ
g
H/2
−βµ
g
H/2
0
0
B
B
e
+e
2
2
(3.30)
We can re-derive this formula using the definition of the free energy, Eq. (3.25), which in
this case is simply
´
³
βµB g0 H/2
−βµB g0 H/2
.
F = −kB T ln e
+e
(3.31)
It is easy to verify that using this free energy in Eq. (3.11) gives the result (3.30).
Pauli paramagnetism. Here we consider the contribution of the conduction electrons to
the magnetic moment of the crystal. Stated in other words, we consider the (para)magnetism
of metals, whose conduction electrons can be considered as free electron gas.
The magnetic moment of the free electron gas can be obtained as follows. Each electron
has spin half, and therefore its energy is enhanced when it is aligned with the field, and is
reduced when its spin is anti-parallel to the field. All we have to do is to find how many of
the electrons at a temperature T are aligned with the field, and how many of them ar anti
parallel to the field, and take the difference. For simplicity, we assume in this calculation
that the Landé factor g0 is 2.
The number of electrons having a certain energy E at temperature T is given by the Fermi
distribution, f (E) = (eβE + 1)−1 (energies are measure with respect to the chemical potential). The number of energy levels of about the same energy E is given by the density
of states (per unit volume), N (E). The chemical potential of the electrons with their spin
aligned with the field is decreased by µB H, and the chemical potential of those which are
anti parallel to the field is increased by the same amount. Hence, the density of electrons
aligned with the field is
Z
n+ =
dEN (E)
1
eβ(E+µB H)
+1
,
and the density of those which are anti parallel to the field is
Z
1
.
n− = dEN (E) β(E−µ H)
B
e
+1
10
(3.32)
(3.33)
The magnetic moment of the electron gas is
Z
³
M = µB (n− − n+ ) = µB dEN (E)
1
eβ(E−µB H) + 1
−
1
eβ(E+µB H) + 1
Expanding in µB H ¿ EF (where EF is the Fermi energy), we find
Z
³ ∂f ´
2
M ' 2µB H dEN (E) −
.
∂E
´
.
(3.34)
(3.35)
Since minus the derivative of the Fermi energy is very close to a delta-function confining
the energy to be about the Fermi energy, we see that magnetization is simply given by the
density of states at the Fermi energy,
M ' 2µ2B HN (EF ) .
(3.36)
It also follows that the paramagnetic susceptibility of the free electron gas, which is called
Pauli paramagnetism, is essentially independent of the temperature.
∗ ∗ ∗ exercise: Compare Eqs. (3.30) and (3.36), and discuss the similarity and the difference
between the two cases.
11
4.
EXCHANGE INTERACTIONS
The dipolar interaction. The direct dipolar interaction of two magnetic dipoles, m1 and
m2 , separated by a distance r reads
´
1³
U = 3 m1 · m2 − 3(m1 · r̂)(m2 · r̂) .
r
(4.1)
In order to estimate its magnitude, we take m1 ' m2 ' gµB ' e~/mc. Then, using
a0 = ~2 /me2 and α = e2 /(~c) = 1/137,
³ e~ ´2 1
³ a ´3 e2 ³ e~ ´2 1 ³ me2 ´2
0
U'
=
3
mc r
r a0 mc e2 ~2
³ a ´3 e2 ³ 1 ´2 ³ a ´3 ³ 1 ´2
0
0
=
Rd .
=
r a0 137
r
137
(4.2)
Since 1 Ry=13.6 eV and r is about several Bohr radii, this energy is about 10−4 eV (which
amounts to a temperature of a few degrees), and is far too small small to explain the typical
magnetic energies.
∗ ∗ ∗ exercise: What is the preferred direction of two identical magnetic dipoles interacting
via the dipolar interaction? [Answer: The dipolar interaction is minimal when the two
dipoles are parallel to one another and to the radius vector r. If they are perpendicular to
the radius vector, then they prefer to be anti-parallel.]
The exchange energy. Let us consider two electrons interacting via the Coulomb interaction alone (namely, we neglect spin-dependent interactions like the spin-orbit interaction,
etc.). The Hamiltonian reads
H(1, 2) = −
~2 ∂ 2
e2
~2 ∂ 2
−
+
V
(r
)
+
V
(r
)
+
,
1
2
2m ∂r21 2m ∂r22
r12
(4.3)
where V (r) is due to the ions, and r12 = |r1 − r2 |. When there are two independent orbital
states, ψa (r) and ψb (r), the orbital part of the two-electron wave function can be
´
1 ³
Ψ(1, 2) = √ ψa (1)ψb (2) + ψb (1)ψa (2) , singlet ,
2
´
1 ³
Ψ(1, 2) = √ ψa (1)ψb (2) − ψb (1)ψa (2) , triplet .
2
Let us now calculate the total energy of the system. We have
Z Z
Z Z
∗
dr1 dr2 Ψ (1, 2)HΨ(1, 2) =
dr1 dr2 |ψ1 (a)ψb (2)|2 H
Z Z
±
dr1 dr2 ψa∗ (1)ψb∗ (2)ψa (2)ψb (1)H .
12
(4.4)
(4.5)
(We have used here the fact that the Hamiltonian is invariant under the change 1 ↔ 2.)
The integral
Z Z
Jab =
dr1 dr2 ψa∗ (1)ψb∗ (2)ψa (2)ψb (1)H ,
(4.6)
is called the exchange integral. When ψa is orthogonal to ψb , only the Coulomb interaction
contributes to this integral. Hence
Z Z
e2
Jab =
dr1 dr2 ψa∗ (1)ψb∗ (2)ψa (2)ψb (1)
.
r12
(4.7)
Let us denote
ρ(r) = ψa (r)ψb∗ (r) .
Then the exchange integral can be written in the form
Z Z
e2
Jab =
dr1 dr2 ρ∗ (r1 ) ρ(r2 ) .
r12
(4.8)
(4.9)
Let us further denote
Z
φ(r) =
dr0
e2
ρ(r0 ) .
|r − r0 |
(4.10)
By its definition, φ(r) satisfies the Poisson equation,
4φ(r) = −4πe2 ρ(r) .
(4.11)
Using this in Eq. (4.9), we have
1
Jab = −
4πe2
Z
dr(4φ∗ (r))φ(r) .
(4.12)
We can now use Green’s theorem (in other words, integrate by parts). Since the surface
contribution vanishes as |r| → ∞, we obtain
Z
1
dr|∇φ(r)|2 > 0 .
Jab =
4πe2
(4.13)
It follows that the exchange energy is positive.
Returning to the expression for the energy, Eq. (4.5), we see that the spatial integration in
the first term there can be written in the form
Z Z
Z
Z
2
2
dr1 dr2 |ψa (1)ψb (2)| H = dr|ψa (r)| H0 + dr|ψb (r)|2 H0
Z Z
e2
+
dr1 dr2 |ψa (1)ψb (2)|2
≡ Ea + Eb + Kab .
r12
13
(4.14)
The term Kab is called the Coulomb integral. Here, H0 = −(~2 /2m)4 + V (r) is the single
electron part of the Hamiltonian.
In summary, we have found that the energy of the two electron system, which is described
by a spin-independent Hamiltonian is given by Ea + Eb + Kab + Jab when the two electrons
are in the symmetric spatial wave function, and by Ea + Eb + Kab − Jab when they are in
the anti-symmetric one. This property of the two electron system can be written in terms
of spin operators, in the form
1
H = Ea + Eb + Kab − Jab (1 + 4sa · sb ) .
2
(4.15)
In order to prove this statement, we note that
2sa · sb = (sa + sb )2 − s2a − s2b = s2 −
3
.
2
(4.16)
In the singlet state, s = 0, and hence sa · sb = −3/4; in the triplet state, s = 1 (and
s2 = s(s + 1) = 2), and therefore sa · sb = 1/4.
The Heisenberg interaction is based on the above picture. It reads
H=−
X
Jij si · sj .
(4.17)
hiji
This Hamiltonian is the basis for most of the investigations in magnetism. When J is
positive, the ground state of the system is ferromagnetic: the spins are aligned all in the
same direction. When J is negative, the system has an antiferromagnetic order.
Super-exchange. Consider an atom with a single orbital (i.e., a single energy level and a
single wave function). When this level is empty, the energy is zero; if one electron is on the
level, then the energy is ε; and if two electrons occupy the atom, then the energy is 2ε + U ,
where U represents the Coulomb repulsion. We denote the single-electron wave function by
|Ri.
Now consider a molecule made of two such atoms. We denote the wave function of the
electron when it is on the second atom by |R0 i, and allow for quantum mechanical tunneling
processes between the two atoms, such that the single-electron Hamiltonian, denoted h, has
the matrix elements
hR|h|R0 i = hR0 |h|Ri = −t ,
14
(4.18)
in addition to
hR|h|Ri = hR0 |h|R0 i = ε .
(4.19)
∗ ∗ ∗ exercise: Find the eigen energies and the eigen functions of the molecule when the
electrons are spinless.
When the two electrons are in the singlet state, their spatial (symmetric) wave functions are
´
1 ³
0
0
Φ0 = √ |Ri|R i + |R i|Ri , Φ1 = |Ri|Ri , Φ2 = |R0 i|R0 i .
2
(4.20)
The matrix of the Hamiltonian in the singlet subspace is (assuming that hR|R0 i = 0,
hR|Ri = hR0 |R0 i = 1)



H H H
2ε
 00 01 02   √

 
 H10 H11 H12  =  − 2t

  √
H20 H21 H22
− 2t
√ 
− 2t − 2t


2ε + U
0  .

0
2ε + U
√
(4.21)
This Hamiltonian matrix has the eigen energies
U
2ε + U , 2ε + ±
2
r
U2
+ 4t2 ,
4
and therefore the ground state of the singlet state has the energy
r
U
U2
4t2
sing
Eg = 2ε + −
+ 4t2 ' 2ε −
, for t ¿ U .
2
4
U
(4.22)
(4.23)
Namely, the electrostatic energy reduces the total energy of the two electrons. However,
this forces them to have opposite spins. In other words, the electrostatic energy favors the
antiferromagnetic state.
∗ ∗ ∗ exercise: Derive explicitly Eqs. (4.21) and (4.22). Find the exact ground state of
the singlet. Plot the probability to find the two electrons on the same atom in this state as
function of U/t and explain the result.
Curie-Weiss law and ferromagnetism. Let us first re-visit the derivation leading to the
Curie law, Eq. (3.28). When we suppose that each atom behaves like a small magnet of
moment µ, then in the magnetic field H it acquires the energy −µ · H. We further assume
that each atom is independent of its neighbors and can rotate freely under the effect of the
15
temperature. Since the atom is localized, it satisfies the Boltzmann statistics, and therefore
its average moment is given by
R
dΩµeβ µ·H
hµi = R
,
dΩeβ µ·H
(4.24)
where dΩ is the element of the solid angle for rotation. The total magnetization of a system
of N atoms will be Nhµi, and its magnetic susceptibility will be in general a tensor (the
derivative of µi with respect to Hj ). Confining ourselves to cubic symmetry, we find
D ∂µ E
N
1 Nhµ2 i
χ=N
'
hµ · µi =
, for small enough H .
(4.25)
∂H
kB T
3 kB T
This is the Curie law.
∗∗∗ exercise: Find the temperature dependence of the total magnetization of the system
by using Eq. (4.24) in the limit of small magnetic fields, and compare with the results (3.26)
and (3.27).
In many cases the interaction between neighboring atoms cannot be ignored. To account
for this effect approximately, one may introduce an internal magnetic field exerted on each
atom by its neighbors. This field is called the Weiss field (or sometimes molecular field). It
is plausible to assume that the internal field is proportional to the average of the magnetic
moment, and that the proportionality constant reflects the strength of the inter-atom interactions. In other words, λ is related to the exchange coupling J, see Eq. (4.17). Hence, the
internal magnetic field is
HI = λNhµi .
The total field acting on each atom is now H + HI , and therefore we find
R
dΩµeβ µ·(H+HI )
N
Nhµi =N R
'
hµµ(H + HI )i
β µ·(H+HI )
kB T
dΩe
N hµ2 i
'
(H + λNhµi) .
kB T 3
(4.26)
(4.27)
Let us define an effective temperature, Θ,
Θ≡
λNhµ2 i
.
3kB
(4.28)
Then the susceptibility is
χ=N
D ∂µ E
∂H
=
16
Nhµ2 i
.
3kB (T − Θ)
(4.29)
This is the Curie-Weiss law for the magnetic susceptibility of a ferromagnet. In particular, the susceptibility diverges as the temperature approaches the ferromagnetic critical
temperature, (the Curie temperature), which in our approximation is Θ.
In order to show that Θ is indeed the critical temperature of our model, we need to return
to the full calculation of the magnetic moment, as is done for example in Eq. (3.26). There
we have found that the total magnetization, i.e., Nhµi, is given by the Brillouin function,
³
´
³ ´
(2J+1)x
1
x
BJ (x) = 2J+1
coth
−
coth
, with x = H/kB T . For the case J = 1/2 the
2J
2J
2J
2J
Brillouin function reduces to tanh, and hence
Nhµi = Ntanh
³H + H ´
I
kB T
= Ntanh
³ H + λNhµi ´
kB T
.
(4.30)
Let us now consider this equation when the applied magnetic field H is zero. Obviously,
hµi = 0 is a solution for this equation. However, there might be another solution: since
tanhx ' x − x3 /3 for small x, we see that as long as the temperature is less than λN/kB
(which is up to a numerical constant equal to Θ above), we can have a nonzero solution for
the magnetization. In other words, the system sustains a spontaneous magnetic order below
some critical temperature.
Antiferromagnetism. We may view an antiferromagnet as consisting of two sublattices.
At zero temperature, the average magnetizations of the two sublattices are anti parallel to
one another.
At finite temperatures, the internal fields acting on the two sublattices are different. The
internal field acting on an atom belonging to the ‘-’ sublattice is
H− = λNµ+ ,
(4.31)
while the internal field on on an atom of the ‘+’ sublattice is
H+ = λNµ− ,
(4.32)
where now λ is negative.
At sufficiently high temperatures, we can try to perform the same calculation as the one in
Eqs. (4.27), (4.28), and (4.29). This will give us (note that N is the number of atoms in
17
each sub lattice)
N
kB T
N
Nhµ− i '
kB T
Nhµ+ i '
hµ2 i
(H + λNhµ− i) ,
3
hµ2 i
(H + λNhµ+ i) .
3
(4.33)
Adding the two equations, we find that we have exactly the same result as before. However,
in the present case, λ is negative, and therefore it is better to define
ΘN ≡ −
λNhµ2 i
,
3kB
(4.34)
to obtain
χ=N
D ∂µ E
∂H
=
Nhµ2 i
.
3kB (T + ΘN )
(4.35)
The paramagnetic susceptibility of an antiferromagnet is smaller than the one of a ferromagnet. The temperature ΘN is called the Néel temperature.
Spin waves. Let us consider the Heisenberg Hamiltonian for a ferromagnet,
H=−
X
J``0 S` · S`0 .
(4.36)
h``0 i
We now assume that in the ground state of our system, all spins are aligned: they have their
maximal possible value of S z .
S`z |Si` = S|Si` .
(4.37)
The ground state itself is the product,
|0i = |Si1 |Si2 . . . |SiN .
(4.38)
We now introduce the spin-deviation operators, S ± , (dropping the subscript ` for brevity)
S ± = S x ± iS y .
(4.39)
We can show that the application of these operators on the eigen states of S z gives another
eigen state of S z ,
³
´
³
´
+
z
x
y
x z
y
y z
x
z
S S |Si = S (S + iS )|Si = S S + iS + i[S S − iS ] |Si
³
´
= S + S z |Si + S + |Si = (S + 1) S + |Si .
18
(4.40)
In a similar fashion,
³
´
³
´
S z S − |Si = (S − 1) S − |Si .
(4.41)
Using the spin-deviation operators, we re-write the Heisenberg Hamiltonian in the form
H=−
X
³
J``0
S`z S`z0
h``0 i
´
1 + −
− +
+ [S` S`0 + S` S`0 ] .
2
(4.42)
As a result we see that
H|0i = −
X
J``0 S 2 |0i ,
(4.43)
h``0 i
namely, only the product of S z in Eq. (4.42) contributes to the ground state energy.
What are the excitations of this system? One may think that an excited state is a state in
which one spin deviates from being maximally aligned. However, this is not an eigen state,
because the operation of S`− S`+0 will shift the deviation to the nearest neighbor. In order to
specify the excited states, we use the following description. Let us denote by |ni a state
which has n deviations (n < 2S). Operating on this state with S + will reduce n by 1, and
operating with S − will increase it by 1. We see that in some sense S + plays the role of an
annihilation operator, denoted a, and S − plays the role of a creation operator, denoted a† .
In fact, we know that
[S + , S − ] = i[S y , S x ] − i[S x , S y ] = 2S z .
(4.44)
S−
S+
, a† = √
,
a= √
2S z
2S z
(4.45)
So, if we define
we find that the creation and annihilation operators obey the usual commutation law,
[a, a† ] = 1 .
(4.46)
The definition (4.45) is not really proper, since S z is also an operator. However, for large
enough values of S, we can replace S z by its average, S. This means that we may consider
only small deviations from the ground state (and therefore this theory is not valid at all for
spins 1/2).
19
Now the spin deviation states are eigen states of a† a, and we have
a† a|ni = n|ni , a|ni =
√
√
n|n − 1i , a† |ni n + 1|n + 1i ,
(4.47)
and
S z = S − a† a .
(4.48)
Putting all this back into the Hamiltonian, Eq. (4.42), we find
H'−
X
h``0 i
'−
X
´
³
J``0 [S − a†` a` ][S − a†`0 a`0 ] + S[a†` a`0 + a` a†`0 ]
³
´
J``0 S 2 + S[a†` a`0 + a` a†`0 − a†` a` − a†`0 a`0 ] .
(4.49)
h``0 i
The first (non operator) term here is just the ground state, so we drop it when considering
the excitations.
The Hamiltonian now resembles the one of lattice vibrations. As in that case, it is expedient
to introduce the Fourier transforms
r
r
1 X
1 X † −iq·R`
†
iq·R`
a` =
aq e
, a` =
a e
.
N q
N q q
(4.50)
(R` is the radius-vector to site ` of the lattice.) Then, taking into account that in Eq. (4.49)
` and `0 are nearest neighbors and that J``0 depends only on |R` − R`0 | we find
H=
X³X
q
´
2SJ(1 − e−iq·Rnn ) a†q aq ,
(4.51)
Rnn
in which Rnn is the radius-vector to the nearest neighbors.
Since the form (4.51) is exactly the same as encountered in the theory of lattice vibrations,
we can simply use all the know results for this problem as well. Noting that the dispersion
of the spin wave excitations,
ωq =
X
2SJ(1 − e−iq·Rnn ) ' 2SJq 2 a2 ,
(4.52)
Rnn
we can find the average occupation number of each mode,
ha†q aq i =
1
eβωq /kB T
20
−1
,
(4.53)
is given by the Bose distribution. It follows that the temperature dependence of the magP
netization, M (T ), is given by N S − q ha†q aq i.
∗∗∗ exercise: Find an explicit form for the temperature dependence of the magnetization
using the spin wave theory. Find the specific heat of the spin wave excitations.
21