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SOLID-STATE PHYSICS II 2007
O. Entin-Wohlman
1.
SEMICONDUCTORS
The energy gap. When one plots the electronic density of states of a certain material vs.
the energy, one observes that there are regions on the energy axis in which the density of
states vanishes. The electronic spectrum can then be viewed as ‘bands’ separated by ‘gaps’.
This is a consequence of the structure of the material (which may be, for example, a crystal
or an amorphous system). Another property of the material at hand is the concentration
of electrons. This concentration determines the Fermi energy, which is the highest energy
that is occupied at zero temperature. We then encounter, at zero temperature, two possible
situations: either the topmost band is completely filled, or it is just partially filled. In a
metallic system, the Fermi energy is located well within one of the bands, which is then
termed ‘the conduction band’. In an insulator, the topmost occupied band is completely
filled, and is then termed ‘the valence band’. The next highest band, which is completely
empty at zero temperature, is the conduction band of the insulator.
The magnitude of the gap separating the conduction band from the valence one varies.
When it is of order of few eV ’s, the material remains essentially an insulator even at room
temperature,
1 eV ' 1.6 × 10−12 erg ,
kB ' 1.4 × 10−16 erg/K ,
kB T ' 4.2 × 10−14 erg , at room temperature ,
or 1 eV ' 104 Kelvin .
(1.1)
Thus the thermal energy is not sufficient to overcome a gap of order a few eV ’s, but does
suffice when the energy gap is of order of fractions of an eV . In the latter case, the material
is a ‘semiconductor’.
The distinction between a metal and a semiconductor stems from the temperature dependence of their respective electrical conductivities: the conductivity of a metal,
σ=
ne2 τ
,
m
1
(1.2)
decreases as the temperature is raised, since τ , the mean free time in-between collisions of
the electrons, decreases as the temperature increases. Here, e is the electron charge, m is
its mass and n is the electronic concentration. The dependence of τ −1 on the temperature
follows usually a power law, and therefore it is not particularly strong. On the other hand,
the conductivity of a semiconductor increases exponentially as the temperature is raised,
since the number of electrons making the jump over the gap into the conduction band by
virtue of the thermal energy is increasing exponentially. Semiconductors also exhibit the
phenomenon of photoconductivity: the conductivity of a semiconductor increases when light
is shined on it, since the energy gap (if it is sufficiently small) can be overcome by the energy
of the light. Finally, the conductivities of various semiconductors can vary by several orders
of magnitude. In this respect, one usually distinguishes between intrinsic semiconductors,
in which the conductivity is dominated by the electrons which are excited thermally into the
conductance band, and extrinsic semiconductors, whose electronic behavior is determined
by the electrons contributed to the conduction band by impurities.
One should keep in mind that the value of the energy gap itself depends on the temperature:
the gap shrinks as the temperature is raised (by a few percents). This is because (i) the band
structure is susceptible to the temperature (the material expands thermally); and (b) the
effect of lattice vibrations (phonons). The energy gap is usually measured by monitoring the
absorbtion of incident radiation, or by studying the temperature dependence of the intrinsic
conductivity. The absorbtion of radiation increases abruptly when its frequency ω is such
that ~ω overcomes the threshold of the gap. However, this is true only for direct transitions,
which occur when the maximum of the valence band is right below the minimum of the
conduction band. When this is not the case, and the maximum of the valence band is not
right below the minimum of the conduction band, the minimal energy transitions are indirect
transitions. Then, a phonon or more are required to supply the missing crystal momentum
between the k−space locations of the minimum and the maximum. The energy of the
phonon is then also absorbed, making the required photon energy less than the threshold.
The band structure of a semiconductor. It is plausible to assume that the number of
electrons in the conduction band, as well as the number of holes in the valence band, of a
typical semiconductor is small; therefore we can approximate the relevant portions of the
2
dispersion relations of the both bands by quadratic forms
³ k2
k2
k2 ´
1
ε(k) = εc + ~2
+ 2 + 3
, for electrons ,
2m1 2m2 2m3
³ k2
k2
k2 ´
1
ε(k) = εv − ~2
+ 2 + 3
, for holes .
2m1 2m2 2m3
(1.3)
Here, εc denotes the bottom of the conduction band and εv denotes the top of the valence
band. In both these expressions, the three ‘masses’, m1 , m2 , and m3 are simply the coefficients of the expansions of the true dispersion relations around the respective maximum
and minimum. Therefore, these are just effective masses, and their values have still to be
determined.
∗ ∗ ∗ exercise: Why there are no linear in k−terms in the dispersion?
∗ ∗ ∗ exercise: Find the density of states for the ellipsoidal pockets described by Eqs. (1.3).
∗ ∗ ∗ solution: The constant energy surfaces of the electrons are ellipsoids. Let us first
determine the volume enclosed by such a constant energy surface, of energy ε. Since energies
in the conductance band are measured relative to εc , the constant energy surface is given by
k2
k2 ´
~2 ³ k12
+ 2 + 3 =1.
(1.4)
ε − εc 2m1 2m2 2m3
p
This is the equation of an ellipsoid, with three axes, given by Ri =
2mi (ε − εc )/~,
i = 1, 2, or 3. The volume in k−space of our ellipsoid is (4π/3)R1 R2 R3 = (4π/3)[2ε −
√
εc ]/~2 )3/2 m1 m2 m3 . We divide this volume by the volume of a unit cell in k−space, 8π 3 ,
and find that the total number of states contained inside a constant surface energy ε is
4π ³ 2(ε − εc ) ´3/2 √
1
Nc (ε) =
m1 m2 m3 3 .
(1.5)
2
3
~
8π
The density of states, Nc (ε), (the number of states per unit volume, per energy, per spin),
is the ε−derivative of Nc (ε) with respect to ε,
∂N (ε)
1
Nc (ε) =
= 3 2
∂ε
~π
r
(ε − εc )m1 m2 m3
.
2
(1.6)
A similar calculation can be carried out for the holes; the result then contains |ε − εv | in
place of ε − εc .
∗ ∗ ∗ exercise: What is the difference (if there is any) between the electronic density of
states of a semiconductor, and that of the free electron gas (which describes rather well a
metal)?
3
The effective masses are measured by the technique of cyclotron resonance. This is accomplished by applying a (constant) magnetic field, B. Let us consider the electron pocket,
described by the dispersion (1.3) (the first equation there). We use the semiclassical equations of motion to describe the motion of the electrons, employing the coordinate scheme in
which the magnetic field and the velocity vectors are decomposed along the three principal
axes of the effective mass tensor,
´
dvi
e³
e
mi
= − v × B = − ²ij` vj B` .
dt
c
c
i
(1.7)
Here, ²ij` is the fully anti-symmetric Levi-Chivita tensor, and repeated indices are summed
over. We now assume an oscillatory solution for the velocity, with frequency ω, such that
d2 vi /dt2 = −ω 2 vi . Then, by taking a second temporal derivative in Eq. (1.7), we obtain
³ e ´2
³ e ´2
1
1
2
−ω vi =
²ij` ²ji1 `1
B` B`1 vi1 = −
²ij` ²i1 j`1
B` B`1 vi1 .
(1.8)
c
mi mj
c
mi mj
There are two possibilities, due to the properties of the Levi-Chivita tensor: either i1 = i
and then ` = `1 or i 6= i1 , in which case i1 = ` and ` = i. Therefore we obtain
³ e ´2 X B 2
³ e ´2 X B B v
i ` `
`
2
2
−ω vi = −
vi
(²ij` ) +
(²ij` )2 .
c
mi mj
c
mi mj
j`
j`
(1.9)
(Here, the summation symbol has been introduced explicitly for clarity.) For example, for
i = 1 Eq. (1.9) becomes
³
´
³ e ´2 B
X
1
ωc2 − ω 2 v1 =
Bi m i v i ,
c m1 m2 m3 i=1,2,3
where we have defined
ωc2
=
³ e ´2 P
mi Bi2
,
m1 m2 m3
c
i=1,2,3
(1.10)
(1.11)
the cyclotron frequency squared. In general, Eq. (1.9) represents three linear homogeneous
equations, and one can easily verify that the determinant of that system of equations vanishes
when the frequency ω equals the cyclotron frequency.
Since for that frequency the electrons’ motion is at resonance with the magnetic field, one
can measure it (from peaks in the absorption spectrum as function of the magnetic field).
One can deduce the cyclotron mass, m∗ ,
s P
2
1
i=1,2,3 mi Bi
P
=
.
m∗
m1 m2 m3 i=1,2,3 Bi2
4
(1.12)
The measurement can be carried out for various orientations of the magnetic field, and
then one can deduce information about the masses mi . Such measurements require that
the mean-free time in-between collisions of the electrons will be larger than the cyclotron
period, so that electron will complete several revolutions before being scattered away. In
other words, the measurement requires rather low temperatures and relatively pure samples.
The number density of charge carriers at thermal equilibrium. Quite generally,
the electrons in the conduction band of a semiconductor are there either because they were
transferred from the valence band (by the thermal energy), or because they come from various impurities present in the system. For the same reasons, there are holes in the valence
band. In any event, the electronic concentration of the electrons in the conduction band,
nc , at thermal equilibrium, is given by
Z
nc (T ) =
∞
dεNc (ε)
εc
1
eβ(ε−µ) + 1
.
(1.13)
Here, β ≡ 1/(kB T ) denotes the inverse temperature, µ is the chemical potential, (it is a
function of the temperature) and Nc (ε) is the density of states, found above [see Eq. (1.6)].
Note that we have taken the width of the conduction band to be unrestricted, whereas our
result for the density of states is valid only in relatively small pockets around the minima
of the conduction band. This approximation will turn out to be justified. Note also the
implicit assumption that the presence of impurities does not affect the density of states.
Employing similar approximations, the density of holes in the valence band, pc , is given by
Z εv
´ Z εv
³
1
1
=
dεNv (ε) β(µ−ε)
.
(1.14)
pc (T ) =
dεNv (ε) 1 − β(ε−µ)
e
+1
e
+1
−∞
−∞
The exact location of the chemical potential in the energy scheme depends on the nature of
the impurities, their number, etc. However, we will make the assumption that the chemical
potential is located somewhere within the energy gap between the conduction and the valence
bands, and that the temperature is not too high, such that
εc − µ À kB T , and µ − εv À kB T .
(1.15)
In other words, all energies, kB T , εc , εv , and the chemical potential µ are well separated,
and not close to each other.
∗ ∗ ∗ exercise: Give numerical estimates for the temperature and the width of the energy
gap for Eqs. (1.15) to hold.
5
The approximation (1.15) allows to replace the Fermi distribution by the Boltzmann one,
1
eβ(ε−µ)
+1
' e−β(ε−µ) , and
1
eβ(µ−ε)
+1
' e−β(µ−ε) .
(1.16)
Then, inserting the condition (1.15) into Eq. (1.13), we perform the resulting integration as
follows
r
Z
m1 m2 m3 ∞ p
dε (ε − εc )e−β(ε−µ)
2
εc
r
Z
m1 m2 m3 ∞ p
−β(εc −µ) 1
=e
dε (ε − εc )e−β(ε−εc )
~3 π 2
2
εc
r
Z εc +kB T p
1
m1 m2 m3
' e−β(εc −µ) 3 2
dε (ε − εc )
~π
2
εc
r
m1 m2 m3 2
−β(εc −µ) 1
=e
(kB T )3/2 ≡ Nc (T )e−β(εc −µ) .
3
2
~π
2
3
1
nc (T ) ' 3 2
~π
An analogous calculation for the hole concentration gives
r
m1 m2 m3 2
−β(µ−εv ) 1
pv (T ) ' e
(kB T )3/2 ≡ Pv (T )e−β(µ−εv ) .
3
2
~π
2
3
(1.17)
(1.18)
(Note that the effective masses, mi , i = 1, 2, or 3 need not to be identical for the valence
and for the conduction bands. Therefore, the temperature-dependent coefficients Nc (T ) and
Pv (T ) need not to be identical.) Although the above calculation does not actually give the
concentration of the charge carriers, since we still do not know the chemical potential, it
does tell us that when one of the concentrations is known, we can find the other. The reason
is that the chemical potential is cancelled when we consider the product,
nc (T )pv (T ) = Nc (T )Pv (T )e−β(εc −εv ) = Nc (T )Pv (T )e−βEgap .
(1.19)
The pure (intrinsic) case. When the semiconductor is very clean, we may assume that
all electrons in the conduction band come from the valence one, and therefore the density of
holes must equal the density of the electrons. We denote the chemical potential of this case
by µi , and the concentration of electrons in the conduction band or of holes in the valence
band by ni (T ). Then, from Eqs. (1.17) and (1.18), we have
e2βµi =
εc + εv kB T Pv (T )
Pv (T ) β(εc +εv )
e
V µi (T ) =
+
ln
.
Nc (T )
2
2
Nc (T )
(1.20)
Hence, since the chemical potential is known, we can compute the electron (or the hole)
concentration. We also note that at zero temperature (where the chemical potential is
6
identical to the Fermi energy), the chemical potential of the pure semiconductor lies exactly
at the middle of the gap. Finally we note that the change of the chemical potential with
the temperature is not large, since the ln-term in Eq. (1.20), being determined by the ratio
of the effective hole masses and the effective electron masses, is of order unity. All these
considerations are valid when the temperature is sufficiently low, so that all our assumptions
above are valid.
The doped (extrinsic) case. When the semiconductor is doped, the impurities contribute
electrons to the conduction band (or holes to the valence band). In that case, nc (T ) 6= pv (T ).
Let us then denote
∆n(T ) = nc (T ) − pv (T ) , and nc (T )pv (T ) = n2i (T ) .
(1.21)
The second relation here is simply another way to present our result Eq. (1.19). It is
straightforward to solve these two equations, and to obtain
r³
∆n ´2
∆n
nc =
+ n2i +
,
2
r³ 2 ´
∆n 2
∆n
+ n2i −
pv =
.
2
2
(1.22)
(We suppress the dependence on the temperature for brevity.) We see that if |∆n| À ni ,
then (depending on the sign of ∆n) either the concentration of the charge carriers is mainly
that of the holes (∆n is negative), and then the semiconductor is of the “p-type”, or the
concentration is mainly that of the electrons (∆n is positive) and then the semiconductor is
of the “n-type” .
We can proceed a bit further with the above considerations, and re-write nc for example,
from Eq. (1.17), in the form
−β(εc −µ)
nc = Nc e
=
=
np
np
Nc Pv e
−βEgap /2
Nc Pv e−βEgap /2
o
×
o
nr N
c −β(− Egap +εc −µ)
2
o
e
Pv
n
o
E
k T
v)
−β(− gap
+εc −µ+ B2 ln P
2
N
c
× e
.
(1.23)
The first factor here is just ni , while the second one, using Eq. (1.20), contains µi (the
chemical potential of the pure semiconductor) in the exponent. Therefore we have found
that
nc = eβ(µ−µi ) ni , pv = e−β(µ−µi ) ni .
7
(1.24)
It follows that
∆n
= 2 sinhβ(µ − µi ) .
ni
(1.25)
In order to proceed farther, we need to calculate the concentration of electrons contributed
by the impurities, or captured by them (and then there is a contribution to the hole concentration). Impurities which contribute electrons to the semiconductor are called ‘donors’,
and those that capture electrons and hence contribute holes are called ‘acceptors’.
Impurity energy levels. Let us first estimate the energy of an electron belonging to a
certain donor. For simplicity it is assumed that each donor contributes just one electron,
for example, an arsenic ion (charge 5e) in a germanium (charge 4e) crystal. One might have
thought that the energy of the extra electron of the arsenic is its ionization energy, 9.81 eV ,
which is huge (as compared to the band structure energies). However, this value is for a
free arsenic ion, while in our case the arsenic ion sits in the germanium medium. This fact
reduces that ‘ionization’ energy drastically, for two main reasons. 1. The ionization energy
is related to the Coulomb energy between the ion and the extra electron. The semiconductor
has a relatively high (static) dielectric constant (of about 10-20), and therefore this energy
is reduced by at least an order of magnitude. The reason for the high dielectric constant is
related to the relatively small values of the energy gaps. In a metal, for example, there is
no gap at all, and the static dielectric constant indeed is infinite. The smaller the energy
gap is, the larger is the dielectric constant. 2. The electron released from the donor into
the conduction band has there [see Eq. [1.3)] a parabolic relation between the energy and
the momentum, but the mass is not the free electron mass. In general, that mass is about
a factor of 10 smaller than the free electron mass; by moving into the conduction band the
electron gains kinetic energy. We may combine these two sources to estimate the ground
state energy using the Bohr radius, a0 ,
Ed =
n me4 o n m∗ 1 o n
o n
o
me4
e2
−3
≡ 2 V
×
'
13.6
eV
×
10
.
a0
~
~2
m ²2
(1.26)
This binding energy should be measured relative to the conduction band; in other words,
the additional electronic level contributed by the donor is at an energy distance Ed below
the conduction band, so that it takes a much smaller energy to excite the electron from the
donor level into the conduction band (as compared to exciting it from the valence band).
8
Although this estimate is done for a single donor, we may assume that all donors contribute
their electrons to the localized level located below the conduction band. This is true as
long as the concentration of donors is such that they do not interact among themselves.
Similar considerations hold for the energy levels of the acceptors; those are located close to
the valence band, so that it is easy for the electrons to be excited from the valence band to
the acceptors’ level, leaving behind them holes in the valence band.
Population of impurity energy levels. An impurity energy level can be in one of the
three following configurations: a. it can host no electrons at all, and then it is empty, and
the corresponding energy is zero; b. it can host an electron of spin up or an electron of spin
down, and then it is singly occupied and its energy is Ed ; or c. it can host two electrons, one
for each spin direction, and then it is doubly-occupied, and its energy is 2Ed + U , where U is
the (repulsive) electrostatic energy it costs to bring two electrons to the same level. Taking
the thermal average we find the thermal equilibrium population of an impurity energy level.
The partition function of our little system, which is coupled to an electron bath of chemical
potential µ and temperature T , is
Z = 1 + 2e−β(Ed −µ) + e−β(2Ed +U −2µ) ,
(1.27)
and the average population of the level is then
hni =
´
1 ³ −β(Ed −µ)
2e
+ 2e−β(2Ed +U −2µ) .
Z
(1.28)
∗ ∗ ∗ exercise: Under which circumstances Eq. (1.28) reproduces the Fermi distribution?
explain.
Usually, the on-level Coulomb repulsive interaction, U , is huge; of order of several eV ’s. The
exponents including it are then almost zero. As a result Eq. (1.28) becomes
hni =
1
1 β(Ed −µ)
e
2
+1
,
(1.29)
so that the concentration of the electrons coming to the conductance band from the donors,
nd , is
nd = Nd hni ,
where Nd is the concentration of the donors.
9
(1.30)
∗ ∗ ∗ exercise: Show that the concentration of holes contributed to the valence band by the
acceptors is given by
pa =
Na
1 β(µ−Ea )
e
2
+1
.
(1.31)
Here, Na is the concentration of acceptors, and Ea is the energy level of a single acceptor,
measured relatively to the top of the valence band.
The charge carrier density of a doped semiconductor. Let us now consider a doped
semiconductor, in which the concentration of donors is Nd , and the concentration of acceptors
is Na , such that Nd > Na . At zero temperature, the valence band is filled completely and
the conduction band is empty. In addition, in each unit volume, Na electrons will ‘fall’ from
the donor energy levels to the acceptor levels and fill them (remember that each acceptor
has, e.g., one electron less than the host material, and each donor has one extra electron).
Thus, at zero temperature, all acceptor levels (whose number per unit volume is Na ) are
filled, the valence band is filled, and there are Nd − Na electrons in the donor levels. As the
temperature is raised, the electrons will be redistributed among all the levels, (namely, the
levels in the valence band, in the conduction band, and those belonging to donors and to
the acceptors), but their total number will remain the same.
At a finite temperature T there are nd electrons on the donor levels and nc electrons in the
conduction band (per unit volume), and pv holes in the valence band, and pa holes on the
acceptor levels. The total number of holes must be equal to the total number of electrons,
Nd − Na = nc + nd − pv − pa .
(1.32)
This equation fixes the chemical potential µ of our system, and allows for a full calculation of
all densities. We note that Eq. (1.32) ensures the neutrality of the system: Nd is the number
of donors, which are charged positively, likewise Na is the number of acceptors, which are
charged negatively. Therefore the total positive charge in the system is Nd + pv + pa , while
the total negative charge of the system is Na + nc + nd (all per unit volume). However, in
order to simplify the calculation, we make the assumption [compare with Eq. (1.15)]
Ed − µ À kB T , and µ − Ea À kB T .
(1.33)
When these approximations are introduced into Eqs. (1.29), (1.30), and (1.31), we find that
nd ' 2Nd e−β(Ed −µ) ¿ Nd , and pa ' 2Na e−β(µ−Ea ) ¿ Na ,
10
(1.34)
such that our condition Eq. (1.32) becomes
∆n ≡ nc − pv = Nd − Na .
(1.35)
In summary, [see Eqs. (1.21) and (1.22)], our set of equations for the charge carrier densities
is
Nd − Na
= 2 sinhβ(µ − µi ) ,
ni
(1.36)
and
 
n 
c
 pv 
=
´2
i1/2 1 h
i
1 h³
Nd − Na + 4n2i
± Nd − Na .
2
2
(1.37)
In the mostly intrinsic case ni À |Nd − Na |, and in the mostly extrinsic one ni ¿ |Nd − Na |.
∗ ∗ ∗ exercise: Show that in the mostly intrinsic case, the charge carrier densities are given
by
 
n 
1
' ni ± (Nd − Na ) ,
 pv 
2
c
(1.38)
while for the mostly extrinsic regime
nc ' Nd − Na , and pv '
n2i
, for Nd > Na ,
Nd − Na
(1.39)
pv ' Na − Nd , and nc '
n2i
, for Na > Nd .
Na − Nd
(1.40)
and
11
2.
THE P-N JUNCTION
Introduction. A p-n junction consists of two semi-infinite semiconductors, which we imagine to fill the entire space. One of them has more donors than acceptors, and so it is
negatively charged, (it is the n-part of the junction) and the other has more acceptors than
donors, and so it is positively charged (and hence it is the p-member of the junction). These
two semiconductors are attached to one another at x = 0, such that along the x−axis the
combined system is negatively charged for x > 0, and is positively charged at x < 0, while
along the other two directions y and z it is homogeneous (namely, its properties do not vary
with position). A very crude model to describe the system will be to assume that




N , x>0
 0 , x>0
d
Nd (x) =
, Na (x) =
.
 0 , x<0
 Na , x < 0 
(2.1)
The non-uniformity in impurity concentrations induces non-uniformity in the density of
the charge carriers (the electrons and the holes). It is plausible to expect that this nonuniformity is significant at and around x = 0 and decays far away as |x| → ∞. Hence, there
is formed a ‘layer’ around x = 0, which is called the ‘depletion layer’, where the electron and
hole concentration depends on x. In order to determine the properties of the depletion layer,
we can view the abrupt change in the impurity concentration, as described by Eq. (2.1), as
causing some electrostatic potential, φ(x), which we need to determine in a self-consistent
way, since the charge carriers which are moving around will tend to screen any electrostatic
potential as best as they can.
Screening in a free electron gas. Suppose we have free electrons (that have only kinetic
energy) moving against a positive background, so that the entire system is neutral. This
describes a simple metal. Now let us introduce into this system a positively-charged particle,
at a given position, and hold it there firmly. This charge will attract the electrons to it,
and will thus create in its neighborhood a surplus of negative charge, which will reduce (i.e.,
‘screen’) its positive charge.
The relevant equations that govern this behavior are from electrostatics. Firstly, we have
Poisson’s equation connecting the charge density of the positive ‘extra’ charge (which was
12
introduced into our neutral system) and the potential it creates,
−∇2 φext (r) = 4πρext (r) .
(2.2)
Secondly, we have Poisson’s equation connecting the ‘true’ charge density, formed in the
system after all the re-shuffling of the electrons, with the potential it creates
−∇2 φ(r) = 4πρ(r) .
(2.3)
Now, the true charge density ρ(r), consists of the extra charge density inserted into the
system, and the charge density it induces because it attracts electrons,
ρ(r) = ρext (r) + ρind (r) .
(2.4)
To these (exact) electrostatic equations, we add a semi-classical approximation: we find
a relation between the induced charge density and the (full) potential which the electrons
‘feel’. This approximation, known as the Thomas-Fermi approximation, is carried out as
follows. At equilibrium, the chemical potential, µ, in the system fixes the electron density,
n,
n(µ) =
X
k
1
eβ(Ek −µ)
+1
,
(2.5)
where β = 1/(kB T ) is the inverse of the temperature. When there is a constant electrostatic
potential, φ, acting on the electrons, we may just add it to µ, i.e., µ → µ + eφ. When
the electrostatic potential, φ(r), varies sufficiently smoothly, we nonetheless add it to the
chemical potential. The result, according to Eq. (2.5), is that the electron density is changed,
n(µ) → n(µ+eφ(r)). The change in the electron density, times −e, is just the induced charge
density. Thus we have
ρind (r) = −e[n(µ + eφ(r)) − n(µ)] ' −e2
∂n
φ(r) .
∂µ
(2.6)
Note that the derivative is to be found for φ = 0.
∗ ∗ ∗ exercise: Calculate explicitly the quantity e2 ∂n
for the free electron gas in which
∂µ
Ek = ~2 k 2 /2m, estimate its numerical value, and in particular determine its dimensions.
Let us now write our equations in Fourier space. Equations (2.2) and (2.3) become
φext (q) =
4π
4π ext
ρ (q) , φ(q) = 2 ρ(q) ,
2
q
q
13
(2.7)
and Eqs. (2.4) and (2.6) are
h
∂n i
ρ(q) = ρext (q) + ρind (q) , ρind (q) = −e2
φ(q) .
∂µ
(2.8)
Simple algebraic manipulations give us the relation between the true potential in the system,
and the extra charge or extra potential that have caused this potential (in the initially neutral
system)
φ(q) =
φext (q)
4πρext (q)
h
i=
.
2 + 4πe2 ∂n
q
2 ∂n
e
1 + 4π
∂µ
q2
∂µ
(2.9)
We deduce from this calculation the dielectric function, ε(q), of the free electron gas in the
Thomas-Fermi approximation,
ε(q) = 1 +
4π h 2 ∂n i
e
.
q2
∂µ
(2.10)
For example, when the extra charge density is simply a (positive) point charge, ρext (r) =
eδ(r) then ρext (q) = e, and the potential it creates is obviously φext (r) = e/r, which in
Fourier space becomes φext (q) = 4πe/q 2 . The true potential created in the system is φ,
which according to Eq. (2.9) is given in this example by
φ(q) =
∂n
4πe
, κ2 ≡ 4πe2
.
2
+κ
∂µ
(2.11)
e
φ(r) = e−κr ,
r
(2.12)
q2
Transforming back to real space,
so that instead of the long-range Coulomb potential, that decays at long distances as 1/r,
the potential screened by the free electron gas decays exponentially. The exercise above
will tell us how good this screening is, namely, how far (in space) extends the the potential
(2.12).
∗ ∗ ∗ exercise: Work out explicitly the integral leading to the result (2.12).
Self-consistent determination of the electrostatic potential in a p-n junction.
Coming back to our p-n junction, we will calculate the electrostatic potential created by
the junction (or rather, by the depletion layer) employing the same type of approximation
14
as the Thomas-Fermi approximation used above. Namely, we simply shift the chemical
potential in Eqs. (1.17) and (1.18) by eφ(x), leading to
nc (x) = Nc (T )e−β(εc −eφ(x)−µ) , pv (x) = Pv (T )e−β(µ+eφ(x)−εv ) .
(2.13)
Confining ourselves to the completely extrinsic case, in which the charge carrier density is
entirely determined by the impurity concentration, we write down the charge carrier densities
far away from the junction
nc (∞) ≡ Nd = Nc (T )e−β(εc −eφ(∞)−µ) , pv (−∞) ≡ Na = Pv (T )e−β(µ+eφ(−∞)−εv ) .
(2.14)
These two equations give
eφ(∞) + µ = εc + kB T ln
Nd
,
Nc (T )
−eφ(−∞) − µ = −εv + kB T ln
Na
,
Pv (T )
(2.15)
so that we know the total electrostatic potential drop across the junction is
eφ(∞) − eφ(−∞) = Egap + kB T ln
Nd Na
.
Nc (T )Pv (T )
(2.16)
The spatial variation of the electrostatic potential is determined, as usual, by the Poisson
equation
−∇2 φ = −
4πρ(x)
d2 φ
=
,
2
dx
²
(2.17)
in which ² is the static dielectric constant of the semiconductor. (The static dielectric
constant of the host material of the p-n junction takes into account all static screening
effects which are there before the material turned into a p-n junction.) The charge density
giving rise to the spatially-varying electrostatic potential is
h
i
ρ(x) = e Nd (x) − Na (x) + pv (x) − nc (x) .
(2.18)
The meaning of the various factors here is simple. In the square brackets we have the number
of completely ionized donors (positive charge density) + the number of completely ionized
acceptors (negative charge density) + hole concentration (pv ) + electron concentration (negative charge, nc ).
In principle, we have now all that is needed to find the electrostatic potential, since we can
just insert Eqs. (2.1) and (2.13) into Eq. (2.18), and then in Poisson’s equation (2.17). However, the result will be an horrendous equation, so let us try to simplify it a bit. Combining
15
Eqs. (2.13) with Eqs. (2.14) we find
nc (x) = Nd e−βe(φ(∞)−φ(x)) , pv (x) = Na e−βe(φ(x)−φ(−∞)) ,
(2.19)
from which we see clearly that as x → ±∞, and the respective exponentials approach
unity, the electron and hole concentrations approach the concentration of the donors and
the acceptors, respectively. We may thus say that at x larger than a certain distance, say dn ,
Nd (x) − nc (x) → 0, while for x smaller than a certain distance, say −dp , Na (x) − pv (x) → 0.
Therefore, a plausible approximate expression for the charge density ρ(x) will be


0,




 eN ,
d
ρ(x) =
 −eNa ,





0,






dn > x > 0 , 
.

0 > x > −dp , 




−dp > x .
x > dn ,
(2.20)
In other words, outside the depletion layer, which extends from −dp up to dn , the electron
density cancels the donor density (on the right side) and the hole density cancels the acceptor
density (on the left side).
The solution of Poisson’s equation with the approximate charge density, Eq. (2.20), is now
simple. We have already found the values of φ at ±∞, [see Eq. (2.16)] and we know (from
electrostatics) that φ(x) and its first derivative (which is minus the electric field) should be
continuous. Therefore








φ(∞) ,
´
d
φ(∞) − 2πeN
(x − dn )2 ,
²
³
´
φ(x) =

a

φ(−∞) + 2πeN
(x + dp )2 ,

²




φ(−∞) ,
³
x > dn ,
dn > x > 0 ,









0 > x > −dp , 




−d > x . 
.
(2.21)
p
It remains to apply the boundary conditions at x = 0. First, we look at the electric field at
the origin and require that it will be continuous. This gives us
Nd dn = Na dp .
(2.22)
Second, the continuity of the electrostatic potential itself gives
φ(∞) − φ(−∞) ≡ ∆φ =
16
³ 2πe ´
²
(Na d2p + Nd d2n ) .
(2.23)
We therefore find for the boundaries of the depletion layer the result
s
s
(Na /Nd )²∆φ
(Nd /Na )²∆φ
dn =
, dp =
.
(Nd + Na )2πe
(Nd + Na )2πe
(2.24)
∗ ∗ ∗ exercise: Plot the electrostatic potential across the depletion layer of a p-n junction,
plot the charge carrier density, and estimate the width of the depletion layer as function of
the temperature.
The biased p-n junction–simple considerations. So far, we have considered the p-n
at equilibrium. Let us now examine what happens when the junction is biased by a potential drop, V . In that case the full change of the electrostatic potential is modified as
compared to its equilibrium value. We may try to estimate the effect of the bias by using
the approximation
∆φ ≡ φ(∞) − φ(−∞) = (∆φ)0 − V ,
(2.25)
where (∆φ)0 is the equilibrium value [given by Eq. (2.16)].
One result of the bias voltage is that [see Eq. (2.24)] the depletion layer thickness changes:
the layer shrinks when V > 0 (i.e., V raises the the potential of the p-side), and expands
when V < 0. The other result is that an electrical current will flow across the junction.
Denoting by J the density current and by j the electrical current, we have
je = −eJe , and jh = eJh .
(2.26)
Let us consider first the flow of the holes. Firstly, there are the holes which are generated
in the n-side of the junction. These are called “minority carriers”. There are not many
of them, but once they are in the depletion layer, they will be immediately swallowed into
the p-side by the strong electric field that exists in that layer. This current is called ‘the
hole generation current’ and is not too sensitive to the bias V . Secondly, there is the
current of holes going from the p-side of the junction to the n-side, which is called ‘the
hole recombination current’ (because eventually these holes will combine with the electrons
in the n-side and will disappear). These holes have to overcome the potential barrier, and
therefore we expect that
Jhrec ∝ e−eβ((∆φ)0 −V ) .
17
(2.27)
Since at equilibrium, i.e., when V = 0, we expect the two currents to balance one another,
¯
¯
Jhrec ¯
= Jhgen , we find that
V =0
Jhrec = Jhgen eβeV ,
(2.28)
so that the total current of holes flowing from the p-side of the junction to the n-side is
given by
Jh = Jhrec − Jhgen = Jhgen (eβeV − 1) .
(2.29)
Carrying out the analogous calculation for the electron current, and combining the hole
current with the electron current, the total electric current across the junction is
j = e(Jhgen + Jegen )(eβeV − 1) .
(2.30)
In particular, when V < 0, the current saturates at a (small) negative value, −e(Jhgen +Jegen ),
independent of the voltage. In order to obtain an estimate for that current, we need to carry
out a more detailed calculation.
∗ ∗ ∗ exercise: Plot the current Eq. (2.30) as function of the bias voltage, and discuss the
result.
The biased p-n junction–detailed calculation. Our more elaborate treatment of the
p-n junction begins with the observation that in the presence of both an electric field,
E = −dφ/dx, and density gradients, the current densities obey
Je = −µn nc E − Dn
dnc
,
dx
Jh = µp pv E − Dp
dpv
.
dx
(2.31)
Here, µn and µp are the electron and hole mobilities, respectively, and Dn and Dp are the
diffusion coefficients. The latter are related to the mobilities by the Einstein relations
µn =
eDn
,
kB T
µp =
eDp
.
kB T
(2.32)
∗ ∗ ∗ exercise: Derive the Einstein relation for a degenerate electron gas, and for the non
degenerate one [which is just the one appearing in Eqs. (2.32)].
∗ ∗ ∗ exercise: Prove that the Einstein relations (2.32) ensure that no current is flowing at
thermal equilibrium.
18
We can re-write the mobilities in terms of the the mean free times between collisions and the
effective masses. For example, if there are only electrons of uniform density in the system,
then an electric field will give rise to electrical current, je = −eJe = σE, where σ is the
electrical conductivity. According to Drude theory, σ = ne2 τ col /m, and therefore in our case
eτ col
µn = n ,
mn
eτpcol
µp =
.
mp
(2.33)
We have distinguished here between the mean free time in between collisions of the electrons
(τecol ) and of the holes (τpcol ). We have also assigned different masses, mn and mp , to the
charge carriers, but assumed the same mass for all three principal directions for each of the
species.
At equilibrium, where Je and Jh vanish, one can determine the carrier densities by using
Eqs. (2.31) and (2.32). The solutions are just Eqs. (2.13).
∗ ∗ ∗ exercise: Prove the statement above.
When the junction is biased, the system is no more at equilibrium. In this case we add two
phenomenological equations,
∂Je
∂nc ³ ∂nc ´
−
=
,
∂t
∂t g−r
∂x
∂Jh
∂pv ³ ∂pv ´
−
=
,
∂t
∂t g−r
∂x
(2.34)
which aim to reflect the fact that charge is conserved. In fact, charge is not conserved in
our case, since electrons and holes are created by thermal fluctuations and disappear by
recombining (namely, electrons go back to the valence band). These processes are described
by the first terms on the right hand side of each of Eqs. (2.34). [In the absence of these two
terms, each of Eqs. (2.34) is just the continuity equation for the electrons and the holes,
respectively.] In order to determine the forms of the terms (∂nc /∂t)g−r and (∂pv /∂t)g−r , we
postulate that they act as to restore thermal equilibrium. Hence
³ ∂p ´
³ ∂n ´
nc − n0c
p − p0v
v
c
=−
,
=− v
,
∂t g−r
τn
∂t g−r
τp
(2.35)
where n0c and p0v are the equilibrium values of the densities.
Since our p-n junction is in steady state (namely, it does not evolve in time), we set the
‘true’ time-derivatives in Eqs. (2.34) to be zero, so that
dJe nc − n0c
+
=0,
dx
τn
19
dJh pv − p0v
+
=0.
dx
τp
(2.36)
These equations have to be solved together with Eqs. (2.31). This is not a simple task in
general; however, in the regions where the electric field is almost constant, we may ignore
the terms including it, and combine Eqs. (2.31) and (2.36) to find
Dn
d2 nc
nc − n0c
=
,
dx2
τn
Dp
pv − p0v
d2 pv
=
.
dx2
τp
(2.37)
Inspection of these equations shows that each of them defines a length, the ‘diffusion length’,
L2p = Dp τp .
L2n = Dn τn ,
(2.38)
For example, if the density of holes at some x0 > 0 differs from its (equilibrium) value at
x → +∞, then the second of Eqs. (2.38) tells us that
pv (x) = pv (∞) + e−(x−x0 )/Lp (pv (x0 ) − pv (∞)) .
(2.39)
In other words, at x > Lp , the hole density (on the n-side of the junction) takes almost
its equilibrium value (which is very small, see below). Since holes are created by thermal
fluctuations at a rate p0v /τp , and are not recombined along a distance Lp , the flow of thermally
generated holes per unit area into the depletion layer is about Lp p0v /τp ' (n2i /Nd )Lp /τp .
Hence the saturation current is
Jhgen + Jhgen =
³ n2 ´ L
i
p
Nd τ p
+
³ n2 ´ L
i
n
Na τn
.
(2.40)
∗ ∗ ∗ exercise: Discuss the temperature dependence of the saturation current, Eq. (2.40).
20
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
si ,
i
1
and si = σ ,
2
(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
(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
pi →
pi + A(ri ) .
2m i
2m i
c
21
(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 −
´
e X
e2 X ³ 2 2
2
pi · ri × H +
r
H
−
(r
·
H)
.
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
xi + y i .
∆H = µB H L + g0 S +
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) = −
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.
22
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
∆E0 =
X
e2
2
H
hΨ
|
ri2 |Ψ0 i .
0
12mc2
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
equation in the form
³
´³
i´
Xh
(2)
H + ∆H Ψ0 +
a(1)
Ψ
+
a
Ψ
+
.
.
.
n
n
n
n
³
= E0 +
n
(1)
E0
+
(2)
E0
i´
´³
Xh
(2)
(1)
+ . . . Ψ0 +
an Ψn + an Ψn + . . . .
(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
23
(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
X
(1)
a(1)
n Ψn + E0 Ψ0 .
(3.16)
n
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.
∗ ∗ ∗ 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
24
up to second order in the magnetic field is
∆E0 =µB H · hΨ0 |L + g0 S|Ψ0 i +
+
X
e2
2
H
hΨ
|
(x2i + yi2 )|Ψ0 i
0
2
8mc
i
X |hΨ0 |µB H · (L + g0 S|Ψn i|2
n6=0
E0 − En
.
(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
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.
25
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.
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).
26
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.
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
27
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
e−βγHJz =
Jz =−J
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≡−
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
BJ (x) '
(J + 1)x
.
3J
(3.27)
It therefore follows that
¯
¯
χ¯
single ion
=
(gµB )2 J(J + 1)
, kB T ¿ gµB H .
3
kB T
(3.28)
To obtain the susceptibility of the entire solid, we multiply this susceptibility by the density
of ions in the solid.
28
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
χpar
2J(J + 1) ~2
=−
,
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
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
.
eβµB g0 H/2 + e−βµB g0 H/2
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
F = −kB T ln eβµB g0 H/2 + e−βµB g0 H/2 .
(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.
29
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
,
(3.32)
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
The magnetic moment of the electron gas is
Z
³
M = µB (n− − n+ ) = µB dEN (E)
1
eβ(E−µB H) + 1
−
(3.33)
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.
30
Bibliography
1. N. W. Ashcroft and N. D. Mermin, Solid State Physics, Saunders College, 1975.
31