Density of States · 0.2 0.1 0 State density Number of states 200 Energy, E1 300 400 the top of Figure 9.2 depicts a histogram of diﬀerent energies possible for the particle in

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Density of States 99.1 INTRODUCTION

We have just seen in Chapters 6–8 that one can solve the Schrödingerequation to find the energies and wavefunctions of a particle of interest.We have also alluded to the fact that the model systems we have intro-duced can be used to describe quantum wells, wires, and dots. Now totruly have complete information about a system of interest, we shouldknow all of its possible energies and wavefunctions.

However, our past experience solving the Schrödinger equation showsthat many valid energies and associated wavefunctions exist. Compli-cating this, in certain cases, degeneracies are possible, resulting instates that possess the same energy. As a consequence, the task ofcalculating each and every possible carrier wavefunction as well as itscorresponding energy is impractical to say the least.

Fortunately, rather than solve the Schrödinger equation multipletimes, we can instead find what is referred to as a density of states.This is basically a function that when multiplied by an interval ofenergy, provides the total concentration of available states in thatenergy range. Namely,

Ninterval = ρenergy(E) dE

where Ninterval is the carrier density present in the energy range dE andρenergy(E) is our desired density of states function. Alternatively, onecould write

Ntot =∫E2

E1

ρenergy(E) dE

to represent the total concentration of available states in the systembetween the energies E1 and E2.

Obtaining ρenergy(E) is accomplished through what is referred to as adensity-of-states calculation. We will illustrate this below and in laterparts of the chapter. Apart from giving us a better handle on the statedistribution of our system, the resulting DOS is important for many sub-sequent calculations, ranging from estimating the occupancy of statesto calculating optical transition probabilities and/or transition ratesupon absorbing and emitting light.

In what follows, we briefly recap model system solutions to theSchrödinger equation seen earlier in Chapter 7. This involves solutionsfor a particle in a one-, two-, or three-dimensional box, from which wewill estimate the associated density of states. We will then calculate thedensity of states associated with model bulk systems, quantum wells,wires, and dots. This will be complemented by subsequent calculationsof what is referred to as the joint density of states, which ultimatelyrelates to the absorption spectrum of these systems.

203

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204 Chapter 9 DENSITY OF STATES

Particle in a One-Dimensional Box

Let us start with the particle in a one-dimensional box, which we saw ear-lier in Chapter 7. Recall that we had a potential that was zero inside thebox and infinite outside. We found that the wavefunctions and energiesof the confined particle were (Equation 7.7)

ψn =√

2a

sinnπx

a

and (Equation 7.8)

E = n2h2

8meffa2= �

2k2

2meff,

where n = 1, 2, 3, . . . and k = nπ/a, with a the length of the box.Table 9.1 lists the first 10 energies of the particle. The first eight energylevels are also illustrated schematically in Figure 9.1. Note that we cango on listing these energies and wavefunctions indefinitely, since nolimit exists to the value that n can take. But clearly this is impractical.

We will now calculate the density of states for this system to illustratehow we can begin wrapping our arms around the problem of describingall of these states in a concise fashion. Before beginning, note that therewill be a slight difference here with what we will see later on. In thissection, we are calculating a density of energy levels. In the next section,the density that we derive refers to a carrier density. As a consequence,there will be an extra factor of two in those latter expressions to accountfor spin degeneracy, stemming from the Pauli principle.

Table 9.1 Energies of aparticle in a one-dimensionalbox

n Energy

1 E1 = h2/8meffa2

2 4E1

3 9E1

4 16E1

5 25E1

6 36E1

7 49E1

8 64E1

9 81E1

10 100E1

n = 1n = 2

n = 3

n = 4

n = 7

n = 6

n = 5

n = 8

E

Figure 9.1 The first eight energylevels of a particle in a one-dimensional box. The spacingbetween states is to scale.

At this point, associated with each value of n is an energy. (In latersections, we will utilize k in our calculations since there exists a1:1 correspondence between k and energy.) Let us define an energydensity as

g(E) = �n�E

∼ dndE

with units of number per unit energy. Note that the approximationdn/dE improves as our box becomes larger (large a) such that webegin to see a near-continuous distribution of energies with the dis-creteness due to quantization becoming less pronounced. Since E =n2h2/8meffa2, we find that n = √

8meffa2/h2√

E , whereupon evaluatingdn/dE gives

g(E) ∼ 12

√8meffa2

h2

1√E

.

We see that the energy density becomes sparser as E increases. Thisis apparent from the function’s inverse square root dependence. Yourintuition should also tell you this from simply looking at Table 9.1 or atFigure 9.1, where the spacing between states grows noticeably largeras E increases.

We can now obtain an approximate density of states with units of num-ber per unit energy per unit length by dividing our previous expressionby the length of the box, a. Thus, ρenergy = g(E)/a results in

ρenergy, 1D box =√

2m0

h2

1√E

. (9.1)

Although not exact, we can illustrate Equation 9.1 for ourselves usingthe energies listed in Table 9.1 and shown in Figure 9.1. Specifically,

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INTRODUCTION 205

1.0

0.8

0.6

0.4

0.2

00 100 200

Energy, E1

300 400

0 100

0.2

0.1

0

Stat

e d

ensi

tyN

um

ber

of

stat

es

200Energy, E1

300 400

the top of Figure 9.2 depicts a histogram of different energies possiblefor the particle in a one-dimensional box. Next, an effective probabilitydensity that mimics ρenergy, 1D box can be created by dividing each valueof the histogram by the average “distance" in energy to the next state.On doing this, we see that the resulting probability density decays inagreement with one’s intuition and with Equation 9.1.

Figure 9.2 Histogram of thestates of a particle in aone-dimensional box and theassociated probability densityobtained by weighting each valueof the histogram by the average“distance" in energy to the nexthigher or lower energy state.

Particle in a Two-Dimensional Box

Let us now consider a particle in a two-dimensional box. This problemwas also solved in Chapter 7 and assumes that a particle is confinedto a square box with dimensions a within the (x, y) plane. The poten-tial is zero inside the box and infinite outside. Solving the Schrödingerequation leads to the following wavefunctions (Equation 7.10):

ψnx ,ny = 2a

sinnxπx

asin

nyπya

with corresponding energies (Equation 7.11)

E = n2h2

8meffa2= �

2k2

2meff

where n2 = n2x + n2

y (nx = 1, 2, 3, . . . ; ny = 1, 2, 3, . . . ) and k2 = k2x + k2

y(kx = nxπ/a, ky = nyπ/a). Table 9.2 list the first 10 energies of the par-ticle and Figure 9.3 illustrates the first eight energy levels with theirdegeneracies.

Table 9.2 Energies of a particlein a two-dimensional box

nx ny Energy

1 1 E11 = 2h2/8meffa2

1 2 2.5E11

2 12 2 4E11

1 3 5E11

3 12 3 6.5E11

3 21 4 8.5E11

4 13 3 9E11

2 4 10E11

4 23 4 12.5E11

4 31 5 13E11

5 1

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At this point, as with the previous one-dimensional box example, theenergy density associated with a unit change in energy is characterizedby �n. Since E ∝ n2

x + n2y , where nx and ny are independent indices, we

essentially have circles of constant energy in the (nx, ny ) plane with

a radius n =√

n2x + n2

y . This is illustrated in Figure 9.4. Thus a smallchange in radius, �n, leads to a slightly larger circle (radius n + �n)with a correspondingly larger energy.

E

(1,1)

(1,2) (2,1)

(1,3) (3,1)

(2,2)

(2,3) (3,2)

(1,4) (4,1)

(3,3)

(2,4) (4,2)

Figure 9.3 Depiction of the firsteight energies of a particle in atwo-dimensional box, includingdegeneracies. The numbersrepresent the indices nx and ny .The spacing between states is toscale.

We are interested in the states encompassed by these two circles,as represented by the area of the resulting annulus. As before, if thedimension of the box, a, becomes large, we may approximate �n → dn.We will assume that this holds from here on.

n

n + Δn

nx

ny

Figure 9.4 Constant-energyperimeters described by theindices nx , ny and the radius

n =√

n2x + n2

y for a particle in atwo-dimensional box.

Now, the area of the annulus is 2πn dn (we will see a derivation of thisshortly). It represents the total number of states present for a givenchange dn in energy. Since nx > 0 and ny > 0, we are only interested inthe positive quadrant of the circle shown in Figure 9.4. As a conse-quence, the area of specific interest to us is 1

4 (2πn dn). Then if g(E) isour energy density with units of number per unit energy, we can write

g2D box(E) dE ∼ 14 (2πn dn)

such that

g2D box(E) ∼ π

2n

dndE

.

Since E = n2h2/8meffa2, n = √8meffa2/h2

√E . As a consequence,

dndE

= 12

√8meffa2

h2

1√E

and we find that

g2D box(E) ∼ 2πmeffa2

h2.

Finally, we can define a density of states ρenergy, 2D box = g2D box(E)/a2,with units of number per unit energy per unit area by dividing g2D box(E)

by the physical area of the box. This results in

ρenergy, 2D box = 2πmeff

h2, (9.2)

which is a constant. In principle, one could have deduced this fromlooking at Table 9.2 or Figure 9.3, although it is not immediatelyobvious.

Particle in a Three-Dimensional Box

Finally, let us consider a particle in a three-dimensional box. We didnot work this problem out in Chapter 7 but instead listed the resultingwavefunctions and energies. However, as stated before, the reader canreadily derive these results by applying the general strategies for find-ing the wavefunctions and energies of a particle in a box. The resultingwavefunctions and energies are (Equation 7.12)

ψnx ,ny ,nz =(

2a

)3/2

sinnxπx

asin

nyπya

sinnzπz

a

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INTRODUCTION 207

and (Equation 7.13)

Enx ,ny ,nz = h2n2

8ma2= �

2k2

2m,

where n2 = n2x + n2

y + n2z (nx = 1, 2, 3, . . . ; ny = 1, 2, 3, . . . ; nz = 1, 2, 3, . . . ),

k2 = k2x + k2

y + k2z (kx = nxπ/a, ky = nyπ/a, kz = nzπ/a) and a is the

length of the box along all three sides. Table 9.3 lists the first 10 ener-gies of a particle in a three-dimensional box. The first eight energies areillustrated in Figure 9.5 along with their degeneracies.

Now, just as with the one- and two-dimensional examples earlier,we will find the energy density associated with a unit change inenergy characterized by �n. Since E ∝ n2

x + n2y + n2

z , where nx, ny , nzare all independent indices, we consider a sphere with a radius n =√

n2x + n2

y + n2z , possessing a constant-energy surface. This is illustrated

in Figure 9.6. Thus, for a given change in radius, �n, we encompassstates represented by the volume of a thin shell between two spheres:one with radius n and the other with radius n + �n. Furthermore, if abecomes large, we may approximate �n → dn, which we will assumefrom here on.

The volume of the shell is 4πn2 dn (we will see a derivation of thisshortly). It represents the total number of states present for a givenchange dn in energy. Next, since nx > 0, ny > 0, and nz > 0, we onlyconsider the positive quadrant of the sphere’s top hemisphere. As aconsequence, the volume of interest to us is 1

8 (4πn2 dn). Finally, if wedefine g3D box(E) as our energy density with units of number per unitenergy, we can write

g3D box(E) dE ∼ 18

(4πn2 dn)

such that

g3D box(E) ∼ πn2

2

(dndE

).

Figure 9.5 Depiction of the firsteight energy levels of a particle ina three-dimensional box, includingdegeneracies. The numbersrepresent the indices nx , ny , nz .The spacing between states is toscale.

Next, since E = n2h2/8meffa2 we have n2 = (8meffa2/h2)E and, as aconsequence,

dndE

= 12

√8meffa2

h2

1√E

.

Table 9.3 Energies of a particle in athree-dimensional box

nx ny nz Energy

1 1 1 E111 = 3h2/8meffa2

1 1 2 2E111

1 2 12 1 11 2 2 3E111

2 1 22 2 11 1 3 3.667E111

1 3 13 1 12 2 2 4E111

1 2 3 4.667E111

1 3 22 1 32 3 13 1 23 2 12 2 3 5.667E111

2 3 23 2 21 1 4 6E111

1 4 14 1 11 3 3 6.333E111

3 1 33 3 11 2 4 7E111

1 4 22 1 42 4 14 1 24 2 1

E

(1,1,1)

(1,1,2) (1,2,1) (2,1,1)

(1,2,2) (2,1,2) (2,2,1)

(1,1,3) (1,3,1) (3,1,1)

(2,2,2)

(1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) (3,2,1)

(2,2,2) (2,3,2) (3,2,2)

(1,1,4) (1,4,1) (4,1,1)

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We thus find that

g3D box(E) ∼ π

4

(8meffa2

h2

)3/2 √E .

We now define a density of states by dividing g3D box(E) by the physicalvolume of the box. This gives ρenergy, 3D box = g3D box/a3, with units ofnumber per unit energy per unit volume. The result

ρenergy, 3D box = π

4

(8meff

h2

)3/2 √E (9.3)

possesses a characteristic√

E dependence. As a consequence, at higherenergies, there are more available states in the system. In principle,one could have seen this trend by looking at Table 9.3 or Figure 9.5.However, it is not obvious and highlights the usefulness of the density-of-states calculation.

nx

ny

nz

n

n + Δn

Figure 9.6 Constant-energyspherical surfaces described bythe indices nx , ny , nz and the

radius n =√

n2x + n2

y + n2z for a

particle in a three-dimensionalbox.

9.2 DENSITY OF STATES FOR BULK MATERIALS,WELLS, WIRES, AND DOTS

Let us now switch to systems more relevant to us. Namely, we want tocalculate the density of states for a model bulk system, quantum well,quantum wire, and quantum dot. We have already illustrated the basicapproach in our earlier particle-in-a-box examples. However, there willbe some slight differences that the reader will notice and should keepin mind.

For more information about these density-of-states calculations, thereader may consult the references cited in the Further Reading.

9.2.1 Bulk Density of States

First, let us derive the bulk density of states. There exist a number ofapproaches for doing this. In fact, we have just used one in the previousexamples of a particle in a three-dimensional box. The reader may verifythis as an exercise, noting that the only difference is a factor of two thataccounts for spin degeneracy.

In this section, we will demonstrate two common strategies, since oneusually sees one or the other but not both. Either method leads to thesame result. However, one may be conceptually easier to remember.In our first approximation, consider a sphere in k-space. Note that theactual geometry does not matter and, in fact, we just did this calculationassuming a cube. Furthermore, while we will assume periodic boundaryconditions, the previous example of a particle in a three-dimensionalbox did not (in fact, it assumed static boundary conditions).

Associated with this sphere is a volume

Vk = 43πk3,

where k is our “radius" and k2 = k2x + k2

y + k2z . In general,

kx = 2πnx

Lx, where nx = 0, ±1, ±2, ±3, . . . ,

ky = 2πny

Ly, where ny = 0, ±1, ±2, ±3, . . . ,

kz = 2πnz

Lz, where nz = 0, ±1, ±2, ±3, . . . .

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DENSITY OF STATES FOR BULK MATERIALS, WELLS, WIRES, AND DOTS 209

These k values arise from the periodic boundary conditions imposedon the carrier’s wavefunction. We saw this earlier in Chapter 7 whenwe talked about the free particle problem. Namely, the free carrierwavefunction is represented by a traveling wave. As a consequence, itsatisfies periodic boundary conditions in the extended solid and leadsto the particular forms of kx, ky , and kz shown. By contrast, the factorof 2 in the numerator is absent when static boundary conditions areassumed. Notice also that in a cubic solid, Lx = Ly = Lz = Na, where Nis the number of unit cells along a given direction and a represents aninteratomic spacing.

The above volume contains many energies, since

E = �2k2

2meff= p2

2meff.

Furthermore, every point on the sphere’s surface possesses the sameenergy. Spheres with smaller radii therefore have points on their sur-faces with equivalent, but correspondingly smaller, energies. This isillustrated in Figure 9.7.

k

E1E2E3

kx

ky

kz

Figure 9.7 Constant-energyspherical surfaces in k-space.

Different radii (k =√

k2x + k2

y + k2z )

lead to different constant-energysurfaces denoted by E1, E2, and E3here, for example.

Figure 9.8 Volume (shaded)associated with a given state ink-space. Two views are shown: atop view onto the (kx , ky ) planeand a three-dimensional side view.

We now define a state by the smallest nonzero volume it possesses ink-space. This occurs when kx = 2π/Lx, ky = 2π/Ly , and kz = 2π/Lz, so that

Vstate = kxkykz = 8π3

LxLyLz.

The concept is illustrated in Figure 9.8. Thus, within our imaginedspherical volume of k-space, the total number of states present is

N1 = Vk

Vstate=

43πk3

kxkykz= k3

6π2LxLyLz.

kx

ky

kz

kx

ky

Top

Side

(kx, ky, kz)

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Next, when dealing with electrons and holes, we must consider spindegeneracy, since two carriers, possessing opposite spin, can occupythe same state. As a consequence, we multiply the above expression by2 to obtain

N2 = 2N1 = k3

3π2LxLyLz.

This represents the total number of available states for carriers,accounting for spin.

We now define a density of states per unit volume, ρ = N2/LxLyLz, withunits of number per unit volume. This results in

ρ = k3

3π2.

Finally, considering an energy density, ρenergy = dρ/dE , with unitsnumber per unit energy per unit volume, we obtain

ρenergy = 13π2

ddE

(2meffE

�2

)3/2

,

which simplifies to

ρenergy = 12π2

(2meff

�2

)3/2 √E . (9.4)

This is our desired density-of-states expression for a bulk three-dimensional solid. Note that the function possesses a characteristicsquare root energy dependence.

Optional: An Alternative Derivation of the Bulk Density of States

This section is optional and the reader can skip it if desired. We discussthis alternative derivation because it is common to see a system’s den-sity of states derived, but not along with the various other approachesthat exist.

We can alternatively derive the bulk density of states by starting witha uniform radial probability density in k-space,

w(k) = 4πk2.

The corresponding (sphere) shell volume element is then

Vshell = w(k) dk

= 4πk2 dk.

The latter expression can be derived by taking the volume differencebetween two spheres in k-space, one of radius k and the other of radiusk + dk. In this case, we have

Vshell = V (k + dk) − V (k)

= 43π(k + dk)3 − 4

3πk3,

which when expanded gives

Vshell = 43π(k3 + k2 dk + 2k2 dk + 2k dk2 + k dk2 + dk3 − k3)

= 43π(3k2 dk + 3k dk2 + dk3).

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Keeping terms only up to order dk, since dk2 and dk3 are much smallervalues, yields

Vshell = 4πk2dk

as our desired shell volume. Now, the point of finding the shell volumeelement between two spheres is that the density-of-states calculationaims to find the number of available states per unit energy difference.This unit change is then dk, since there exists a direct connectionbetween E and k.

Next, we have previously seen that the “volume" of a given state is

Vstate = kxkykz = 8π3

LxLyLz.

The number of states present in the shell is therefore the ratio N1 =Vshell/Vstate:

N1 = k2

2π2LxLyLz dk.

If spin degeneracy is considered, the result is multiplied by 2 to obtain

N2 = 2N1 = k2

π2LxLyLz dk.

At this point, we can identify

g(k) = k2

π2LxLyLz

as the probability density per state with units of probability per unit k,since

N2 = g(k) dk.

Furthermore, because of the 1:1 correspondence between k and E , wecan make an additional link to the desired energy probability densityg(E) through

g(k) dk = g(E) dE .

As a consequence,

g(E) = g(k)dkdE

.

Recalling that

k =√

2meffE�2

anddkdE

= 12

√2meff

�2

1√E

then gives

g(E) = LxLyLz

2π2

(2meff

�2

)3/2 √E ,

as the expression for the number of available states in the system perunit energy. Dividing this result by the real space volume LxLyLz yieldsour desired density of states, ρenergy(E) = g(E)/LxLyLz, with units ofnumber per unit energy per unit volume:

ρenergy(E) = 12π2

(2meff

�2

)3/2 √E .

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The reader will notice that it is identical to what we found earlier(Equation 9.4).

9.2.2 Quantum Well Density of States

The density of states of a quantum well can also be evaluated. As withthe bulk density-of-states evaluation, there are various approaches fordoing this. We begin with one and leave the second derivation as anoptional section that the reader can skip if desired.

First consider a circular area with radius k =√

k2x + k2

y . Note that sym-metric circular areas are considered, since two degrees of freedom existwithin the (x, y) plane. The one direction of confinement that occursalong the z direction is excluded. The perimeters of the circles shownin Figure 9.9 therefore represent constant-energy perimeters much likethe constant-energy surfaces seen in the earlier bulk example.

E1E2E3

k

kx

ky

Figure 9.9 Constant-energyperimeters in k-space. The

associated radius is k =√

k2x + k2

y .Perimeters characterized bysmaller radii have correspondinglysmaller energies and are denotedby E1, E2, and E3 here as anexample.

The associated circular area in k-space is then

Ak = πk2,

and encompasses many states having different energies. A given statewithin this circle occupies an area of

Astate = kxky ,

with kx = 2π/Lx and ky = 2π/Ly , as illustrated in Figure 9.10. Recallthat Lx = Ly = Na, where N is the number of unit cells along a givendirection and a represents an interatomic spacing. Thus,

Astate = (2π)2

LxLy.

kx

ky

(kx, ky)

Figure 9.10 Area (shaded)associated with a given state ink-space for a two-dimensionalsystem.

The total number of states encompassed by this circular area istherefore N1 = Ak/Astate, resulting in

N1 = k2

4πLxLy .

If we account for spin degeneracy, this value is further multiplied by 2,

N2 = 2N1 = k2

2πLxLy ,

giving the total number of available states for carriers, including spin.At this point, we can define an area density

ρ = N2

LxLy= k2

2π= meffE

π�2

with units of number per unit area, since k = √2meffE/�2. Our desired

energy density is then ρenergy = dρ/dE and yields

ρenergy = meff

π�2, (9.5)

with units of number per unit energy per unit area. Notice that it is aconstant.

Notice also that this density of states in (x, y) accompanies states asso-ciated with each value of kz (or nz). As a consequence, each kz (or nz)

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value is accompanied by a “subband" and one generally expresses thisthrough

ρenergy(E) = meff

π�2

∑nz

�(E − Enz ), (9.6)

where nz is the index associated with the confinement energy along thez direction and �(E − Enz ) is the Heaviside unit step function, defined by

�(E − Enz ) =

⎧⎪⎨⎪⎩

0 if E < Enz (9.7)

1 if E > Enz . (9.8)

Optional: An Alternative Derivation of the Quantum WellDensity of States

As before, this section is optional and the reader may skip it if desired.We can alternatively derive the above density-of-states expression.

First, consider a uniform radial probability density within a plane ofk-space. It has the form

w(k) = 2πk

such that a differential area Aannulus, can be defined as

Aannulus = w(k) dk

= 2πk dk.

The latter expression can be derived by simply taking the difference inarea between two circles having radii differing by dk. What results is

Aannulus = A(k + dk) − A(k)

= π(k + dk)2 − πk2

= π(k2 + 2k dk + dk2 − k2)

= π(2k dk + dk2).

For small dk, we can ignore dk2. This yields

Aannulus = 2πk dk.

We are now interested in calculating the number of states within thisdifferential area, since it ultimately represents the number of availablestates per unit energy. Since the smallest nonzero “area" occupied by astate is

Astate = kxky = 4π2

LxLy,

the number of states present in the annulus is

N1 = Aannulus

Astate= k dk

2πLxLy .

To account for spin degeneracy, we multiply this value by 2 to obtain

N2 = 2N1 = k dkπ

LxLy .

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This represents the number of states available to carriers in a ring ofthickness dk, accounting for spin degeneracy.

At this point, we identify

g(k) = kπ

LxLy

as a probability density with units of probability per unit k, since

N2 = g(k) dk.

An equivalence with the analogous energy probability density g(E) thenoccurs because of the 1:1 correspondence between k and E :

g(k) dk = g(E) dE .

Solving for g(E) then yields

g(E) = g(k)dkdE

where

k =√

2meffE�2

anddkdE

= 12

√2meff

�2

1√E

.

We thus have

g(E) =(

meff

π�2

)LxLy .

This energy density describes the number of available states per unitenergy. Finally, dividing by LxLy results in our desired density of states,

ρenergy = g(E)

LxLy= meff

π�2,

with units of number per unit energy per unit area. This is identical toEquation 9.5. Since an analogous expression exists for every kz (or nz)value, we subsequently generalize this result to

ρenergy(E) = meff

π�2

∑nz

�(E − Enz ).

where �(E − Enz ) is the Heaviside unit step function.

9.2.3 Nanowire Density of States

The derivation of the nanowire density of states proceeds in an identicalmanner. The only change is the different dimensionality. Whereas wediscussed volumes and areas for bulk systems and quantum wells, werefer to lengths here. As before, there are also alternative derivations.One of these is presented in an optional section and can be skipped ifdesired.

For a nanowire, consider a symmetric line about the origin in k-spacehaving length 2k:

Lk = 2k.

The associated width occupied by a given state is

Lstate = k

�

�

�

�

�

�

�

�


where k represents any one of three directions in k-space: kx, ky , or kz.For convenience, choose the z direction. This will represent the singledegree of freedom for carriers in the wire. We then have possible kzvalues of

kz = 2πnz

Lz, with nz = 0, ±1, ±2, ±3, . . . ,

where Lz = Na, N represents the number of unit cells along the z direc-tion, and a is an interatomic spacing. The smallest nonzero lengthoccurs when nz = 1. As a consequence, the number of states foundwithin Lk is

N1 = Lk

Lstate= k

πLz.

If spin degeneracy is considered,

N2 = 2N1 = 2kπ

Lz

and describes the total number of available states for carriers.We now define a density

ρ = N2

Lz= 2k

π= 2

π

√2meffE

�2

that describes the number of states per unit length, including spin. Thisleads to an expression for the DOS defined as ρenergy = dρ/dE :

ρenergy(E) = 2π

(dkdE

), (9.9)

giving

ρenergy = 1π

√2meff

�2

1√E

. (9.10)

Equation 9.10 is our desired expression, with units of number per unitenergy per unit length. More generally, since this distribution is asso-ciated with confined energies along the other two directions, y and z,we write

ρenergy(E) = 1π

√2meff

�2

∑nx ,ny

1√E − Enx ,ny

�(E − Enx ,ny ), (9.11)

where Enx ,ny are the confinement energies associated with the x and ydirections and �(E − Enx ,ny ) is the Heaviside unit step function. Noticethe characteristic inverse square root dependence of the nanowire one-dimensional density of states.

Optional: An Alternative Derivation of the Nanowire Density of States

Alternatively, we can rederive the nanowire density-of-states expressionby simply considering a differential length element dk between k + dkand k, with the length occupied by a given state being kz = 2π/Lz. Thenumber of states present is then

N1 = dkkz

= dk2π

Lz.

�

�

�

�

�

�

�

�


If spin degeneracy is considered, this expression is multiplied by two,giving

N2 = 2N1 = dkπ

Lz.

We now define a probability density having the form

g(k) = Lz

π,

since N2 = g(k)dk. Furthermore, from the equivalence

g(k) dk = g(E) dE ,

we find that

g(E) = g(k)dkdE

with

k =√

2meffE�2

anddkdE

= 12

√2meff

�2

1√E

.

This yields

g(E) = Lz

2π

√2meff

�2

1√E

as our density, with units of number per unit energy. Finally, we candefine a density of states

ρenergy = 2g(E)

Lz,

where the extra factor of 2 accounts for the energy degeneracy betweenpositive and negative k values. Note that we took this into account inour first derivation since we considered a symmetric length 2k aboutthe origin. We then have

ρenergy = 1π

√2meff

�2

1√E

,

which is identical to Equation 9.10. As before, the density of states canbe generalized to

ρenergy(E) = 1π

√2meff

�2

∑nx ,ny

1√E − Enx ,ny

�(E − Enx ,ny ),

since the distribution is associated with each confined energy Enx ,ny .

9.2.4 Quantum Dot Density of States

Finally, in a quantum dot, the density of states is just a series of deltafunctions, given that all three dimensions exhibit carrier confinement:

ρenergy(E) = 2δ(E − Enx ,ny ,nz ). (9.12)

In Equation 9.12, Enx ,ny ,nz are the confined energies of the carrier, char-acterized by the indices nx, ny , nz. The factor of 2 accounts for spindegeneracy. To generalize the expression, we write

ρenergy = 2∑

nx ,ny ,nz

δ(E − Enx ,ny ,nz ), (9.13)

which accounts for all of the confined states in the system.

�

�

�

�

�

�

�

�

POPULATION OF THE CONDUCTION BAND 217

9.3 POPULATION OF THE CONDUCTION BAND

At this point, we have just calculated the density of states for three-,two-, one-, and zero-dimensional systems. While we have explicitlyconsidered conduction band electrons in these derivations, such cal-culations also apply to holes in the valence band. The topic of bandswill be discussed shortly in Chapter 10. Thus, in principle, with someslight modifications, one also has the valence band density of states forall systems of interest.

In either case, whether for the electron or the hole, the above density-of-states expressions just tell us the density of available states. Theysay nothing about whether or not such states are occupied. For this,we need the probability P(E) that an electron or hole resides in a givenstate with an energy E . This will therefore be the focus of the currentsection and is also our first application of the density-of-states functionto find where most carriers reside in a given band. Through this, wewill determine both the carrier concentration and the position of theso-called Fermi level in each system.

9.3.1 Bulk

Let us first consider a bulk solid. We begin by evaluating the occupationof states in the conduction band, followed by the same calculation forholes in the valence band.

Conduction Band

For the conduction band, we have the following expression for the num-ber of occupied states at a given energy per unit volume (alternatively,the concentration of electrons at a given energy):

ne(E) = Pe(E)ρenergy(E) dE .

In this expression, Pe(E) is the probability that an electron possesses agiven energy E and is called the Fermi–Dirac distribution. It is one ofthree distribution functions that we will encounter throughout this text:

Pe(E) = 1

1 + e(E−EF )/kT. (9.14)

There are several things to note. First, EF refers to the Fermi level,which should be distinguished from the Fermi energy seen earlier inChapter 3. The two are equal only at 0 K. Next, note that the Fermi levelEF is sometimes referred to as the “chemical potential μ". However, wewill stick to EF here for convenience. The Fermi level or chemical poten-tial is also the energy where Pe(E) = 0.5 (i.e., the probability of a carrierhaving this energy is 50%) and, as such, represents an important ref-erence energy. Finally, Figure 9.11 shows that the product of Pe(E)

and ρenergy(E) implies that most electrons reside near the conductionband edge.

The total concentration of electrons in the conduction band, nc , isthen the integral of ne(E) over all available energies:

nc =∫∞

Ec

Pe(E)ρenergy(E) dE .

Notice that the lower limit of the integral is Ec , which represents thestarting energy of the conduction band.

�

�

�

�

�

�

�

�


Ef

re(E)rh(E)

Ph(E)Pe(E)

EcEv E

Next, we have previously found for bulk materials (Equation 9.4) that

ρenergy = 12π2

(2meff

�2

)3/2 √E .

Note that since we now explicitly refer to the electron in the conduction

Figure 9.11 Conduction bandand valence band occupation ofstates. The shaded regions denotewhere the carriers are primarilylocated in each band.

band, we replace meff with me to denote the electron’s effective massthere. We also modify the expression to have a nonzero origin to accountfor the conduction band starting energy. We thus find

ρenergy = 12π2

(2me

�2

)3/2 √E − Ec .

Inserting this into our integral then yields

nc =∫∞

Ec

(1

1 + e(E−EF )/kT

) (1

2π2

) (2me

�2

)3/2 √E − Ec dE

or

nc = A∫∞

Ec

(1

1 + e(E−EF )/kT

) √E − Ec dE (9.15)

where

A = 12π2

(2me

�2

)3/2

.

We can subsequently simplify this as follows:

nc = A∫∞

Ec

(1

1 + e(E−Ec )/kT e(Ec−EF )/kT

) √E − Ec dE ,

and let η = (E − Ec)/kT as well as μ = (−Ec + EF )/kT to obtain

nc = A(kT )3/2∫∞

Ec

√η

1 + eη−μdη.

The resulting integral has been solved and can be looked up. It is calledthe incomplete Fermi–Dirac integral (Goano 1993, 1995) and is definedas follows:

F1/2(μ, c) =∫∞

c

√η

1 + eη−μdη. (9.16)

However, to stay instructive, let us just consider the situation whereE − EF � kT . In this case, the exponential term in the denominator ofEquation 9.15 dominates, so that

1

1 + e(E−EF )/kT� e−(E−EF )/kT .

�

�

�

�

�

�

�

�


We therefore have

nc � A∫∞

Ec

e−(E−EF )/kT√

E − Ec dE ,

which we can again modify to simplify things

nc � A∫∞

Ec

e−[(E−Ec )+(Ec−EF )]/kT√

E − Ec dE

� Ae−(Ec−EF )/kT∫∞

Ec

e−(E−Ec )/kT√

E − Ec dE .

At this point, changing variables by letting x = (E − Ec)/kT , E = Ec +xkT , and dE = kT dx as well as changing the limits of integration, gives

nc � A(kT )3/2e−(Ec−EF )/kT∫∞

0e−x√

x dx.

To evaluate the integral here, we note that the gamma function (n) isdefined as

(n) =∫∞

0e−xxn−1dx. (9.17)

Values of are tabulated in the literature (see, e.g., Abramowitz andStegun 1972; Beyer 1991). In our case, we have

(32

). As a consequence,

nc � A(kT )3/2e−(Ec−EF )/kT (

32

),

and we recall that A = (1/2π2)(2me/�2)3/2. Consolidating terms by let-

ting Nc = A(kT )3/2(

32

)then results in

nc � Nce−(Ec−EF )/kT (9.18)

and expresses the concentration of carriers in the bulk semiconductorconduction band.

Valence Band

We repeat the same calculation for the valence band. The approach issimilar, but requires a few changes since we are now dealing with holes.As before, we first need a function Ph(E) describing the probability thata hole possesses a given energy. Likewise, we need the previous bulkdensity of states (Equation 9.4).

Conceptually, the number of holes at a given energy per unit volume(i.e., the concentration of holes at a given energy) is

nh(E) = Ph(E)ρenergy(E) dE

where the the probability that a hole occupies a given state can beexpressed in terms of the absence of an electron. Since we have pre-viously invoked the Fermi–Dirac distribution to express the probabilitythat an electron occupies a given state, we can find an analogous prob-ability for an electron not occupying it by subtracting Pe(E) from 1. Inthis regard, the absence of an electron in a given valence band stateis the same as having a hole occupy it. As a consequence, we writePh(E) = 1 − Pe(E) to obtain

Ph(E) = 1 − 1

1 + e(E−EF )/kT. (9.19)

�

�

�

�

�

�

�

�


Next, the hole density of states can be written as

ρenergy = 12π2

(2mh

�2

)3/2 √Ev − E ,

which is slightly different from the corresponding electron density ofstates. Note that we have used mh to denote the hole’s effective mass.The different square root term also accounts for the fact that hole ener-gies increase in the opposite direction to those of electrons since theireffective mass has a different sign. We alluded to this earlier in Chapter 7and will show it in Chapter 10. Note that Ev is the valence band startingenergy and, in many instances, is defined as the zero energy of the x axisin Figure 9.11. As a consequence, hole energies become larger towardsmore negative values in this picture. Figure 9.11 likewise illustratesthat most holes live near the valence band edge.

The total hole concentration is therefore the following integral overall energies:

nv =∫Ev

−∞Ph(E)ρenergy(E) dE

= B∫Ev

−∞

(1 − 1

1 + e(E−EF )/kT

) √Ev − E dE

where

B = 12π2

(2mh

�2

)3/2

.

Generally speaking, since E < EF in the valence band (recall that in thisscheme, where we are using a single energy axis for both electrons andholes, hole energies increase towards larger negative values), we write

nv = B∫Ev

−∞

(1 − 1

1 + e−(EF −E)/kT

) √Ev − E dE .

We can now approximate the term in parentheses through the binomialexpansion, keeping only the first two terms:

11 + x

� 1 − x + x2 − x3 + x4 + · · · .

We therefore find that

1

1 + e−(EF −E)/kT� 1 − e−(EF −E)/kT + · · · .

On substituting this into nv, we obtain

nv � B∫Ev

−∞e−(EF −E)/kT

√Ev − E dE .

Changing variables, by letting x = (Ev − E)/kT , E = Ev − xkT , and dE =−kT dx, and remembering to change the limits of integration, then gives

nv � B∫0

−∞e−[(EF −Ev)+(Ev−E)]/kT

√kTx(−kT ) dx

� B(kT )3/2e−(EF −Ev)/kT∫∞

0e−x√

x dx.

The last integral is again the gamma function (

32

). As a consequence,

we find

nv � B(kT )3/2e−(EF −Ev)/kT (

32

).

�

�

�

�

�

�

�

�


Letting Nv = B(kT )3/2(

32

)then yields

nv � Nve−(EF −Ev)/kT , (9.20)

which describes the hole concentration in the valence band.

Fermi Level

We can now determine the location of the Fermi level EF from nv and nc ,assuming that the material is intrinsic (i.e., not deliberately doped withimpurities to have extra electrons or holes in it). We will also assumethat the material is not being excited optically, so that the systemremains in thermal equilibrium. Under these conditions, the followingequivalence holds, since associated with every electron is a hole:

nc = nv,

Nce−(Ec−EF )/kT = Nve−(EF −Ev)/kT ,

A(kT )3/2(

32

)e−(Ec−EF )/kT = B(kT )3/2

(32

)e−(EF −Ev)/kT ,

12π2

(2me

�2

)3/2

e−(Ec−EF )/kT = 12π2

(2mh

�2

)3/2

e−(EF −Ev)/kT .

Solving for EF then yields

EF = Ec + Ev

2+ 3

4kT ln

(mh

me

), (9.21)

which says that at T = 0 K, the Fermi level of an intrinsic semiconduc-tor at equilibrium occurs halfway between the conduction band and thevalence band. However, a weak temperature dependence exists, movingthe Fermi level closer to the conduction band when mh > me. Alterna-tively, if me > mh, the Fermi level moves closer to the valence band.To a good approximation, the Fermi level lies midway between the twobands, as illustrated in Figure 9.12.

Ec

Ev

EF

E

Figure 9.12 Position of the Fermilevel relative to the conductionand valence band edges of anintrinsic semiconductor. At 0 K, itlies exactly halfway between thetwo bands.

9.3.2 Quantum Well

Let us repeat these calculations for a quantum well. The procedure isidentical, and the reader can jump ahead to the results if desired.

Conduction Band

We start with our derived quantum well density of states (Equation 9.6),

ρenergy = meff

π�2

∑ne

�(E − Ene ),

where ne is the index associated with quantization along the z direc-tion. Let us consider only the lowest ne = 1 subband, as illustrated inFigure 9.13. Note further that since we are dealing with an electron inthe conduction band, we replace meff with me. As a consequence,

ρenergy = me

π�2.

�

�

�

�

�

�

�

�


E

r Well

1st subband

2nd subband

3rd subband

EneEc

1st subband

2nd subband

3rd subband

EvEnh

Figure 9.13 First subband(shaded) of the conduction andvalence bands in a quantum well.For comparison purposes, thedashed lines represent the bulkdensity of states for both theelectron and the hole.

Next, recall from the previous bulk example that the concentration ofstates for a given unit energy difference can be expressed as


The total concentration of electrons in the first subband is therefore theintegral over all available energies:

nc =∫∞

Ene

Pe(E)ρenergy(E) dE

where the lower limit to the integral appears because the energy of thefirst subband begins at Ene not Ec . See Figure 9.13. Note that Pe(E) =(1 + e(E−EF )/kT )−1 is again our Fermi–Dirac distribution. We thereforefind that

nc = me

π�2

∫∞

Ene

1

1 + e(E−EF )/kTdE .

To be instructive, we continue with an analytical evaluation of theintegral by assuming that E − EF � kT . This leads to

nc � me

π�2

∫∞

Ene

e−(E−EF )/kT dE ,

or

nc � mekTπ�2

e−(Ene −EF )/kT , (9.22)

which is our desired carrier concentration in the first subband.

Valence Band

Let us repeat the above calculation to find the associated hole densityof states. As with the conduction band, we need the probability thatthe hole occupies a given state within the valence band. The requiredprobability distribution is then Ph(E), which can be expressed in termsof the absence of an electron in a given valence band state (i.e., throughPh(E) = 1 − Pe(E)). We obtain

Ph(E) = 1 − 1

1 + e(E−EF )/kT.

The concentration of states occupied at a given unit energy differenceis then

nh(E) = Ph(E)ρenergy(E) dE ,

where we consider only the first hole subband (nh = 1) in Equation 9.6:

ρenergy = mh

π�2.

�

�

�

�

�

�

�

�


Notice that mh is used instead of meff since we are explicitly consideringa hole in the valence band.

Since the total concentration of holes in the first subband is theintegral over all energies, we find that

nv =∫Enh

−∞Ph(E)ρenergy(E) dE .

The integral’s upper limit is Enh since the subband begins there and notat the bulk valence band edge Ev (see Figure 9.13). One therefore has

nv = mh

π�2

∫Enh

−∞

(1 − 1

1 + e(E−EF )/kT

)dE

and since E < EF in the valence band, we alternatively write

nv = mh

π�2

∫Enh

−∞

(1 − 1

1 + e−(EF −E)/kT

)dE .

Applying the binomial expansion to the term in parentheses

1

1 + e−(EF −E)/kT� 1 − e−(EF −E)/kT + · · · ,

then yields

nv � mh

π�2

∫Enh

−∞e−(EF −E)/kT dE .

This ultimately results in our desired expression for the valence bandhole concentration:

nv � mhkTπ�2

e−(EF −Enh)/kT . (9.23)

Fermi Level

Finally, as in the bulk example, let us evaluate the Fermi level of thequantum well. Again we assume that the material is intrinsic and thatit remains at equilibrium. In this case,

nc = nv,

mekTπ�2

e−(Ene −EF )/kT = mhkTπ�2

e−(EF −Enh)/kT .

Eliminating common terms and solving for EF then gives

EF =(

Ene + Enh

2

)+ kT

2ln

(mh

me

). (9.24)

This shows that the Fermi level of an intrinsic two-dimensional mate-rial at equilibrium occurs halfway between its conduction band andvalence band. The slight temperature dependence means that the Fermilevel moves closer to the conduction (valence) band with increasingtemperature if mh > me (mh < me).

9.3.3 Nanowire

Finally, let us repeat the calculation one last time for a one-dimensionalsystem. The procedure is the same as with the bulk and quantum wellexamples above. The reader can therefore skip ahead to the results ifdesired.

�

�

�

�

�

�

�

�


Conduction Band

Let us start with the conduction band and use the Fermi–Dirac distribu-tion in conjunction with the following one-dimensional density of states(Equation 9.11):

ρenergy = 1π

√2me

�2

∑nx,e,ny,e

1√E − Enx,e,ny,e

�(E − Enx,e,ny,e ).

nx,e and ny,e are integer indices associated with carrier confinementalong the two confined directions x and y of the wire and meff is replacedby me to indicate an electron in the conduction band. We also con-sider only the lowest subband with energies starting at Enx,e,ny,e withnx,e = ny,e = 1. This is illustrated in Figure 9.14.

Figure 9.14 First subband(shaded) of both the conductionband and valence band of aone-dimensional system. Forcomparison purposes, the dashedlines represent the bulk electronand hole density of states.

The concentration of states at a given unit energy difference istherefore


The associated carrier concentration within the first subband is thenthe integral over all possible energies:

nc =∫∞

Enx,e ,ny,e

Pe(E)ρenergy(E) dE

= 1π

√2me

�2

∫∞

Enx,e ,ny,e

1

1 + e(E−EF )/kT

1√E − Enx,e,ny,e

dE .

If E − EF � kT , then

nc � 1π

√2me

�2

∫∞

Enx,e ,ny,e

e−(E−EF )/kT 1√E − Enx,e,ny,e

dE

� 1π

√2me

�2

∫∞

Enx,e ,ny,e

exp[− (E − Enx,e,ny,e ) + (Enx,e,ny,e − EF )

kT

]

× 1√E − Enx,e,ny,e

dE

� 1π

√2me

�2exp

(−Enx,e,ny,e − EF

kT

)

×∫∞

Enx,e ,ny,e

exp(

−E − Enx,e,ny,e

kT

)1√

E − Enx,e,ny,e

dE ,

which we can simplify by letting x = (E − Enx,e,ny,e )/kT , E = Enx,e,ny,e +xkT , and dE = kT dx, remembering to change the limits of integration.

E

r Wire

Enx,e,ny,eEc

1st subband2nd subband 3rd subband

1st subband2nd subband

3rd subband

Enx,h,ny,h Ev

�

�

�

�

�

�

�

�


This yields

nc �√

kTπ

√2me

�2exp


kT

) ∫∞

0e−xx−1/2 dx.

The last integral is again the gamma function, in this case (

12

). As a

consequence, the desired expression for the electron concentration ina one-dimensional material is

nc � 1π

√2mekT

�2exp


kT

)

(12

). (9.25)

Valence Band

A similar calculation can be done for the holes in the valence band. Frombefore, the hole probability distribution function is

Ph(E) = 1 − 1

1 + e(E−EF )/kT,

with a corresponding one-dimensional hole density of states (Equation9.11)

ρenergy = 1π

√2mh

�2

∑nx,h,ny,h

1√Enx,h,ny,h − E

�(Enx,h,ny,h − E).

Note the use of mh instead of meff for the hole’s effective mass. Theindices nx,h and ny,h represent integers associated with carrier con-finement along the x and y directions of the system. Notice also theslight modification to the square root term and to the Heaviside unitstep function. These changes account for hole energies increasing inthe opposite direction to those of electrons due to their different effec-tive mass signs. We alluded to this earlier in Chapter 7 and will see whyin Chapter 10.

Considering only the lowest-energy band, we have

ρenergy = 1π

√2mh

�2

1√Enx,h,ny,h − E

,

whereupon the concentration of states for a given unit energy differ-ence is

nh(E) = Ph(E)ρenergy(E) dE .

The total carrier concentration is then the integral over all energies inthe subband. We thus have

nv =∫Enx,h ,ny,h

−∞Ph(E)ρenergy(E) dE

= 1π

√2mh

�2

∫Enx,h ,ny,h

−∞

(1 − 1

1 + e(E−EF )/kT

)1√

Enx,h,ny,h − EdE .

where the upper limit reflects the fact that the subband begins atEnx,h,ny,h and not at Ev like in the bulk case. Since E EF , we may alsowrite

nv = 1π

√2mh

�2

∫Enx,h ,ny,h

−∞

(1 − 1

1 + e−(EF −E)/kT

)1√


�

�

�

�

�

�

�

�


At this point, using the binomial expansion, keeping only the first twoterms,

1

1 + e−(EF −E)/kT� 1 − e−(EF −E)/kT ,

gives

nv � 1π

√2mh

�2

∫Enx,h ,ny,h

−∞e−(EF −E)/kT 1√

Enx,h,ny,h − EdE

� 1π

√2mh

�2

∫Enx,h ,ny,h

−∞exp

[− (EF − Enx,h,ny,h) + (Enx,h,ny,h − E)

kT

]

× 1√Enx,h,ny,h − E

dE

� 1π

√2mh

�2exp

(−EF − Enx,h,ny,h

kT

)

×∫Enx,h ,ny,h

−∞exp

(−Enx,h,ny,h − E

kT

)1√


To simplify things, let x = (Enx,h,ny,h − E)/kT , E = Enx,h,ny,h − xkT , anddE = −kT dx, remembering to change the limits of integration. Thisyields

nv � 1π

√2mh

�2(kT )1/2 exp


kT

)∫∞

0e−xx−1/2 dx,

where the last integral is again the gamma function (

12

). The final

expression for the total hole concentration in the valence band istherefore

nv � 1π

√2mhkT

�2exp


kT

)

(12

). (9.26)

Fermi Level

Finally, we can find the Fermi level position in the same fashion as donepreviously for two- and three-dimensional materials. Namely, assumingthat the material is intrinsic and remains at equilibrium, we have

nc = nv,

1π

√2mekT

�2

(12

)exp


kT

)= 1

π

√2mhkT

�2

(12

)

× exp


kT

).

Solving for EF then gives

EF = Enx,e,ny,e + Enx,h,ny,h

2+ kT

4ln

(mh

me

)(9.27)

where we again find that the Fermi level lies midway between the con-duction band and the valence band of the system. There is a slighttemperature dependence, with EF moving closer to the conduction bandedge with increasing temperature provided that mh > me.

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JOINT DENSITY OF STATES 227

9.4 QUASI-FERMI LEVELS

As a brief aside, we have just found the Fermi levels of various intrinsicsystems under equilibrium. However, there are many instances wherethis is not the case. For example, the material could be illuminated so asto promote electrons from the valence band into the conduction band.This results in a nonequilibrium population of electrons and holes. If,however, we assume that both types of carriers achieve equilibriumamong themselves before any potential relaxation processes, then wecan refer to a quasi-Fermi level for both species. Thus, instead of refer-ring to a single (common) Fermi level, we have EF ,e and EF ,h for electronsand holes, respectively, and only at equilibrium are they equal,

EF ,e = EF ,h = EF . (9.28)

Furthermore, if we were to look at the occupation probability for statesin the conduction band under these conditions, we would use a slightlyaltered version of Pe(E), namely

Pe(E) = 1

1 + e(E−EF ,e)/kT(9.29)

where the quasi-Fermi level EF ,e is used instead of EF . Likewise, wewould use

Ph(E) = 1 − 1

1 + e(E−EF ,h)/kT(9.30)

for holes.

9.5 JOINT DENSITY OF STATES

To conclude our derivation and discussion of the use of the density-of-states function, let us calculate the joint density of states for both theconduction band and valence band of low-dimensional systems. Ourprimary motivation for doing this is that interband transitions occurbetween the two owing to the absorption and emission of light. Thiswill be the subject of Chapter ??. As a consequence, we wish to con-solidate our prior conduction band and valence band density-of-statescalculations to yield the abovementioned joint density of states, whichultimately represents the absorption spectrum of the material. Namely,we will find that it is proportional to the system’s absorption coefficientα, which is a topic first introduced in Chapter 5.

9.5.1 Bulk Case

First, let us redo our density-of-states calculation. This time, we willconsider the conduction and valence bands in tandem. As before, wecan perform the derivation is a number of ways. Let us just take the firstapproach used earlier where we began our density-of-states calculationwith a sphere in k-space.

The volume of the sphere is

Vk = 43πk3.

where k =√

k2x + k2

y + k2z . Along the same lines, the corresponding

volume of a given state is Vstate = kxkykz = 8π3/LxLyLz. The number of

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states enclosed by the sphere is therefore

N1 = Vk

Vstate= k3

6π2LxLyLz.

We now multiply N1 by 2 to account for spin degeneracy,

N2 = 2N1 = k3

3π2LxLyLz,

and consider a volume density

ρ = N2

LxLyLz= k3

3π2

with units of number per unit volume. Since we are dealing withinterband transitions between states, our expression for k will differfrom before. Pictorially, what occur in k-space are “vertical” transitionsbetween the conduction band and valence band. This is depicted inFigure 9.15. In addition, recall from Chapter 7 that relevant (bulk)envelope function selection rules led to a �k = 0 requirement for opticaltransitions.

Figure 9.15 Pictorial descriptionof the conduction and valencebands in k-space as well as avertical transition between themupon the absorption of light.

Let us therefore call the initial valence band k-value kn and the finalconduction band k-value kk. The associated energy of the conductionband edge is then

Ecb = Ec + �2k2

k

2me,

where Ec is the bulk conduction band edge energy and me is theelectron’s effective mass. For the valence band hole, we have

Evb = Ev − �2k2

n

2mh,

where Ev is the valence band edge energy and mh is the hole’s effectivemass.

Next, the transition must observe conservation of momentum (p =�k). Thus, the total final and initial momenta are equal. We thus have

pk = pn + pphoton

�kk = �kn + �kphoton

E

k

Ec

Ev

kk

kn

Eg

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�

�


where pphoton represents the momentum carried by the photon. As aconsequence,

kk = kn + kphoton. (9.31)

We will see in Chapter ?? that kk and kn for interband optical transitionsare both proportional to π/a, where a represents an interatomic spac-ing. By contrast, kphoton ∝ 1/λ, where λ is the wavelength of light. As aconsequence, λ � a and we find that

kk � kn. (9.32)

This is why optical transitions between the valence band and conduc-tion band in k-space are often said to be vertical.

Next, let us use this information to obtain an expression for the opticaltransition energy, from which we will obtain a relationship between kand E . Namely,

E = Ecb − Evb

= (Ec − Ev) + �2

2

(k2

k

me+ k2

n

mh

).

Note that the band gap Eg is basically the energy difference between theconduction band edge and the valence band energy. It is analogous tothe HOMO–LUMO gap in molecular systems. As a consequence, Eg = Ec −Ev. In addition, we have just seen from the conservation of momentumthat kk � kn ≡ k. This results in

E = Eg + �2k2

2

(1

me+ 1

mh

),

whereupon defining a reduced mass

1μ

= 1me

+ 1mh

yields

E = Eg + �2k2

2μ. (9.33)

Solving for k then gives

k =√

2μ(E − Eg)

�2(9.34)

and if we return to our density ρ = k3/3π2, we find that

ρ = 13π2

(2μ(E − Eg)

�2

)3/2

.

The desired bulk joint density of states is then ρjoint = dρ/dE , yielding

ρjoint = 12π2

(2μ

�2

)3/2 √E − Eg, (9.35)

which has a characteristic square root energy dependence. This isillustrated in Figure 9.16. It is also similar to the experimental bulkabsorption spectrum of many semiconductors, although this lineshapeis sometimes modified by additional Coulomb interactions between theelectron and hole.

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Eg E

renergy Bulk

E

renergy Well

E

renergy Dot

E

renergy Wire

9.5.2 Quantum Well

We can now repeat the same calculation for a quantum well. As before,there are a number of strategies for doing this. Let us just take the firstapproach, where we begin with a circular area in k-space,

Ak = πk2.

The corresponding area occupied by a given state is

Figure 9.16 Summary of the jointdensity of states for a bulksystem, a quantum well, aquantum wire, and a quantum dot.

Astate = kxky

where kx = 2π/Lx and ky = 2π/Ly . The number of states encompassedby the circular area is therefore

N1 = Ak

Astate= k2

4πLxLy .

We multiply N1 by 2 to account for spin degeneracy,

N2 = 2N1 = k2

2πLxLy ,

and subsequently define a density

ρ = N2

LxLy= k2

2π,

with units of number per unit area.

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At this point, let us find an expression for k that accounts for theinterband transition energy. Recall that the energy of the conductionband electron is

Ecb = E′c + �

2k2k

2me,

while that of the hole is

Evb = E′v − �

2k2n

2mh.

Note that instead of Ec and Ev, as seen in our earlier bulk example, wehave E

′c and E

′v. This is because these energies now contain contributions

from quantization along the well’s z direction, leading to discrete kzvalues for the electron and hole:

kze = πnze

a,

kzh = πnzh

a.

In both expressions, a is the thickness of the quantum well, with nze =1, 2, 3, . . . and nzh = 1, 2, 3, . . . . So, for both the electron and hole, Ec andEv have the following offsets:

E′c = Ec + �

2k2ze

2me,

E′v = Ev − �

2k2zh

2mh.

The transition energy is then

E = Ecb − Evb

= (E′c − E

′v) + �

2

2

(k2

k

me+ k2

n

mh

),

where E′c − E

′v is an effective confinement energy Enz,e,nz,h . Note also that

kk � kn ≡ k due to momentum conservation. This results in

E = Enz,e,nz,h + �2k2

2

(1

me+ 1

mh

).

We can again define a reduced mass 1/μ = 1/me + 1/mh to obtain

E = Enz,e,nz,h + �2k2

2μ.

Solving for k then yields

k =√

2μ(E − Enz,e,nz,h)

�2

and returning to our density gives

ρ = k2

2π= μ

π

(E − Enz,e,nz,h)

�2.

At this point, we define the joint density of states as ρjoint = dρ/dE toobtain

ρjoint = μ

π�2. (9.36)

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Since this distribution is associated with each value of nz, we cangeneralize the expression to

ρjoint = μ

π�2

∑nz,e,nz,h

�(E − Enz,e,nz,h). (9.37)

The resulting joint density of states is illustrated schematically inFigure 9.16 and represents the linear absorption spectrum of thequantum well. As with the bulk example earlier, this lineshape isoften modified by additional Coulomb interactions between the electronand hole.

9.5.3 Nanowire

Finally, let us repeat the above calculation for a nanowire. Consider asymmetric length in k-space about the origin,

Lk = 2k.

The width occupied by a given state is then kz = 2π/Lz, so that thenumber of states contained within Lk is

N1 = Lk

kz= k

πLz.

The result is multiplied by 2 to account for spin degeneracy

N2 = 2N1 = 2kπ

Lz.

We now define a density

ρ = N2

Lx= 2k

π,

where we require an expression relating k to the transition energy inorder to evaluate our final joint density of states.

In the case of a one-dimensional wire, the energy of the conductionband electron is

Ecb = E′c + �

2k2k

2me,

while for the hole it is

Evb = E′v − �

2k2n

2mh.

E′c and E

′v are again the effective electron and hole state energies off-

set from their bulk values by radial confinement along the x and ydirections:

E′c = Ec + �

2k2xy,e

2me,

E′v = Ev −

�2k2

xy,h

2mh.

In the expressions above, kxy,e =√

k2x,e + k2

y,e and kxy,h =√

k2x,h + k2

y,h,

where kx,e = nx,eπ/Lx, ky,e = ny,eπ/Ly , kx,h = nx,hπ/Lx, ky,h = ny,hπ/Ly

�

�

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�

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�


with nx,e = 1, 2, 3, . . . , ny,e = 1, 2, 3, . . . , nx,h = 1, 2, 3, . . . , and ny,h =1, 2, 3, . . . .

The transition energy is then

E = Ecb − Evb

= Enx,e,ny,e,nx,h,ny,h + �2

2

(k2

k

me+ k2

n

mh

)

with an effective transition energy Enx,e,ny,e,nx,h,ny,h = E′c − E

′v and kk �

kn ≡ k owing to momentum conservation. We therefore have

E = Enx,e,ny,e,nx,h,ny,h + �2k2

2

(1

me+ 1

mh

).

Defining a reduced mass 1/μ = 1/me + 1/mh, yields

E = Enx,e,ny,e,nx,h,ny,h + �2k2

2μ

and solving for k results in

k =√

2μ(E − Enx,e,ny,e,nx,h,ny,h)

�2.

On returning to our density, we find that

ρ = 2kπ

= 2π

(2μ

�2

)1/2 √E − Enx,e,ny,e,nx,h,ny,h .

We can thus define a joint density of states ρjoint = dρ/dE , to obtain

ρjoint = 1π

(2μ

�2

)1/2 1√E − Enx,e,ny,e,nx,h,ny,h

. (9.38)

This is our desired one-dimensional joint density of states and, asbefore, is associated with the values nx,e, ny,e, nx,h, and ny,h due to car-rier confinement along the x and y directions of the wire. The expressioncan be generalized to

ρjoint = 1π

√2μ

�2

∑nx,e,ny,e,nx,h,ny,h

1√E − Enx,e,ny,e,nx,h,ny,h

�(E − Enx,e,ny,e,nx,h,ny,h).

(9.39)

Equation 9.39 is illustrated schematically in Figure 9.16. In principle,it resembles the actual linear absorption spectrum of the system. How-ever, in practice, additional electron-hole Coulomb interactions oftenmodify the nanowire absorption spectrum.

9.5.4 Quantum Dot

Finally, for a quantum dot, there is no need to do this exercise since thejoint density of states is a series of discrete atomic-like transitions. The

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zero-dimensional joint-density-of-states expression is therefore

ρjoint = 2δ(E − Enx,e,ny,e,nz,e,nx,h,ny,h,nz,h) (9.40)

where the energy Enx,e,ny,e,nz,e,nx,h,ny,h,nz,h contains confinement contribu-tions along the x, y, and z directions for both the electron and hole.Note that, as in the previous cases of the quantum well and nanowire,we implicitly assume that we are operating in the strong confinementregime with both the electron and hole confined independently. We

then have

Enx,e,ny,e,nz,e,nx,h,ny,h,nzh= Eg +

(�

2k2x,e

2me+ �

2k2y,e

2me+ �

2k2z,e

2me

)

+(

�2k2

x,h

2mh+

�2k2

y,h

2mh+

�2k2

z,h

2mh

)

where kx,e = nx,eπ/Lx, ky,e = ny,eπ/Ly , kz,e = nz,eπ/Lz, kx,h = nx,hπ/Lx,ky,h = ny,hπ/Ly , and kz,h = nz,hπ/Lz (nx,e = 1, 2, 3, . . . , ny,e = 1, 2, 3, . . . ,nz,e = 1, 2, 3, . . . , nx,h = 1, 2, 3, . . . , ny,h = 1, 2, 3, . . . and nz,h = 1, 2, 3, . . . ).This can be generalized to the following linear combination thataccounts for all states present:

ρjoint = 2∑

nx,e,ny,e,nz,e,nx,h,ny,h,nz,h

δ(E − Enx,e,ny,e,nz,e,nx,h,ny,h,nz,h). (9.41)

The resulting joint density of states is illustrated schematically inFigure 9.16. In principle, it resembles the actual linear absorptionspectrum of the dots. In practice, though, the inhomogeneous size dis-tribution of nanocrystal ensembles often causes significant broadeningof these transitions.

9.6 SUMMARY

This concludes our introduction to the density of states and to twoearly uses: (a) evaluating the occupation of states in bands and(b) evaluating the accompanying joint density of states. The latteris related to the absorption spectrum of the material and is there-fore important in our quest to better understand interband transi-tions in low-dimensional systems. In the next chapter, we discussbands and illustrate how they develop owing to the periodic poten-tial experienced by carriers in actual solids. We follow this with a briefdiscussion of time-dependent perturbation theory (Chapter ??). Thisultimately allows us to quantitatively model interband transitions inChapter ??.

9.7 THOUGHT PROBLEMS

9.1 Density of states: bulk

Provide a numerical evaluation of ρenergy for electrons in a bulksolid with me = m0 and with an energy 100 meV above the sys-tem’s conduction band edge. Express your answer in units ofeV−1 cm−3.

9.2 Density of states: bulk

Estimate the total number of states present per unit energyin a microscopic a = 1 μm CdTe cube. The effective electronmass me = 0.1m0 and the electron energy of interest is 100 meV

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THOUGHT PROBLEMS 235

above the bulk conduction band edge. Express your result inunits of eV−1.

9.3 Effective conduction and valence band densityof states: bulk

In Equations 9.18 and 9.20,

nc � Nce−(Ec−EF )/kT

nv � Nve−(EF −Ev)/kT ,

the prefactors Nc and Nv are often referred to as the effec-tive conduction and valence band density of states. Providenumerical values for both in GaAs, where me = 0.067m0 andmh = 0.64m0 at 300 K. Express you answers in units of cm−3.

9.4 Density of states: quantum well

Numerically evaluate ρenergy for electrons in a quantum wellwith me = m0 and with an energy 100 meV above Ene, the firstsubband edge. Consider only the lowest subband. Express youranswer in units of eV−1 cm−2.

9.5 Density of states: quantum well

Numerically evaluate the prefactors in Equations 9.22 and 9.23:

nc � mekT

π�2e−(Ene −EF )/kT

nv � mhkT

π�2e−(EF −Enh )/kT .

They are referred to as critical densities. Consider GaAs, whereme = 0.067m0 and mh = 0.64m0 at 300 K. Express all youanswers in units of cm−2.

9.6 Density of states: nanowire

Provide a numerical evaluation of ρenergy for electrons in ananowire with me = m0 and with an energy 100 meV aboveEnx ,e ,ny ,e , the first subband edge. Consider only the lowest

subband. Express your answer in units of eV−1 cm−1.

9.7 Density of states: nanowire

Evaluate numerical values for the prefactors in Equations 9.25and 9.26:

nc = 1π

√2mekT

�2exp

(−Enx ,e ,ny ,e − EF

kT

)

(12

)

=√

2mekT

π�2exp

(−Enx ,e ,ny ,e − EF

kT

),

nv = 1π

√2mhkT

�2exp

(−

EF − Enx ,h,ny ,h

kT

)

(12

)

=√

2mhkT

π�2exp

(−

EF − Enx ,h,ny ,h

kT

).

Consider GaAs where me = 0.067m0 and mh = 0.64m0 at 300 K.Express all you answers in units of cm−1.

9.8 Fermi–Dirac distribution

Consider the Fermi–Dirac distribution

f (E ) = 1

1 + e(E−EF )/kT.

Plot f (E ) for three different temperatures, say 10 K, 300 K, and1000 K. Assume EF = 0.5 eV. Show for yourself that the widthof the distribution from where f (E ) ∼ 1 to where f (E ) ∼ 0 isapproximately 4kT . Thus, at room temperature, the width ofthe Fermi–Dirac distribution is of order 10 meV.

9.9 Fermi–Dirac distribution

The Fermi–Dirac distribution for electrons is

fe(E ) = 1

1 + e(E−EF )/kT

and describes the probability that an electron occupies a givenstate. For holes, we have seen that one writes

fh(E ) = 1 − 1

1 + e(E−EF )/kT,

where this can be rationalized since the second term just reflectsthe probability that an electron occupies a given state and 1minus this is the probability that it remains unoccupied (andhence occupied by a hole). In some cases, however, one seeswritten

fh(E ) = 1

1 + e(EF −E )/kT.

Show that the two expressions are identical. Then show in thelimit where E − EF � kT for electrons and EF − E � kT for holesthat the following Boltzmann approximations exist for both fe(E )

and fh(E ):

fe(E ) � e−(E−EF )/kT

fh(E ) � e−(EF −E )/kT .

9.10 Fermi level: bulk

Calculate the Fermi level of silicon at 0 K, 10 K, 77 K, 300 K, and600 K. Note that Eg = Ec + Ev and assume that me = 1.08m0 andmh = 0.55m0. Leave the answer in terms of Eg or, if you desire,look up the actual value of Eg and express all you answers inunits of eV.

9.11 Fermi level: bulk

Evaluate the position of the Fermi level in bulk GaAs underequilibrium conditions (i.e. no optical excitation) relative to itsvalence band edge. Assume that me = 0.067m0, mh = 0.64m0,and Eg = 1.42 eV at 300 K.

9.12 Fermi–Dirac integral

Show that the concentration of either electrons and holes in abulk semiconductor is given by the following expression:

nc(v) = 1

2π2

(2me(h)kT

�2

)3/2 ∫∞0

√ηeμe(h)−η

1 + eμe(h)−ηdη

where the last integral is the Fermi–Dirac integral F1/2(μe(h))

written in such a way as to make its numerical evaluation moreconvenient. me(h) is the electron (hole) effective mass, η is avariable of integration and μe(h) = (EF − Ec(v))/kT .

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9.13 Quasi-Fermi level: bulk, nonequilibriumconditions

Provide an estimate for the quasi-Fermi level of electrons inthe conduction band of a CdS nanowire, modeled as a bulksystem. Recall from Problem 9.12 that one can express the con-duction band electron concentration in terms of the Fermi–Diracintegral as

nc = 1

2π2

(2mekT

�2

)3/2 ∫∞0

√ηeμe−η

1 + eμe−ηdη,

where η is a variable of integration and μe = (E′F − Ec )/kT , with

Ec the conduction band edge and E′F the desired quasi-Fermi

level under nonequilibrium conditions. Assume an intrinsicmaterial where all excess carriers are photogenerated throughpulsed laser excitation. Recall from Chapter 5 that the numberof carriers generated in such an experiment is

n =(

Iσhν

)τp ,

where I is the laser intensity, σ is the nanowire absorption crosssection, hν is the photon energy, and τp is the laser pulse width.

To carry out the numerical evaluation of E′F , use the following

parameters:

me = 0.2m0,

T = 300 K,

I = 1012 W/m2,

σ = 3.13 × 10−10 cm2,

387 nm excitation,

τp = 150 fs,

a nanowire radius r = 7 nm,

a nanowire length of l = 10 μm.

9.14 Alternative bulk concentration

Show that the concentration of electrons in a bulk semiconductorcan be described by the expression

nc = 14

(2mekT

π�2

)3/2eμe ,

where μe = (EF ,e − Ec )/kT and EF ,e is the quasi-Fermi level ofcarriers in the conduction band. Assume E − EF � kT and makean appropriate choice of variables. Hint: let z = √

η, whereη = (E − Ec )/kT . This approach has been used by Agarwal et al.(2005).

9.15 Quasi-Fermi level: bulk, nonequilibriumconditions

Evaluate the electron quasi-Fermi level in a bulk system underdifferent photogenerated carrier concentrations nc . Show thatunder low pump fluences where carrier densities are relativelysmall, the Fermi–Dirac distribution can be approximated by theBoltzmann distribution. Conversely, show that at high pump flu-ences where nc is large, it is best to not make this approximation.First show that when E − EF ,e � kT ,

nc = 14

(2mekT

π�2

)3/2eμe ,

where μe = (EF ,e − Ec )/kT and EF ,e is the quasi-Fermi level ofelectrons under illumination. Next, solve for μe and compare it

with the same value found numerically by evaluating

nc = 1

2π2

(2mekT

�2

)3/2 ∫∞0

√ηeμe−η

1 + eμe−ηdη,

where the integral is that of the full Fermi–Dirac function and η

is a variable of integration. Consider the specific case of bulkGaAs, where me = 0.067m0 and mh = 0.64m0, both at 300 K,and with excitation intensities leading to either nc = 1020 m−3

or nc = 1025 m−3.

9.16 Peak concentration

Consider the bulk density-of-states expression for electronsderived in the main text,

ρenergy = 1

2π2

(2me

�2

)3/2 √E − Ec ,

with Ec the conduction band energy. Assume that E − EF � kTin the Fermi–Dirac distribution. Find the energy associated withthe peak of the resulting electron distribution function ne =ρenergy(E )f (E ).

9.17 Fermi energy

Consider the electron concentration of the conduction band in abulk semiconductor. Show that at T = 0 K, the associated Fermienergy is

EF ,e = �2

2me(3π2nc )2/3.

Hint: Consider the behavior of the Fermi–Dirac distribution at0 K for E > EF ,e versus E < EF ,e. Note also that the resultingexpression was previously seen in Chapter 3 when discussingmetals.

9.18 Bulk quasi-Fermi level: temperature dependence

In Problem 9.15, we compared two expressions for the carrierdensity in the conduction band of a bulk semiconductor:

nc = 14

(2mekT

π�2

)3/2eμe

and

nc = 1

2π2

(2mekT

�2

)3/2 ∫∞0

√ηeμe−η

1 + eμe−ηdη,

where μe = (EF ,e − Ec )/kT and EF ,e is the quasi-Fermi level ofelectrons under illumination. For the specific case of GaAs whereme = 0.067m0 and nc = 1024 m−3, compare the quasi-Fermilevels from both expressions at 300 K and at 3000 K. Assumethat me is temperature-independent. Explain any observedtrends.

9.19 Quantum well carrier concentration

Evaluate the total carrier concentration nc in a given conductionband subband of a quantum well. This time, however, do notassume E − EF � kT . Consider the integral

∫1

1 + ex dx = x − ln(1 + ex )

and assume lower and upper limits of E1 and E2, respectively.

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FURTHER READING 237

9.8 REFERENCES

Abramowitz M, Stegun IA (1972) Handbook of MathematicalFunctions. Dover Publications, New York, 1972.

Agarwal R, Barrelet CJ, Lieber CM (2005) Lasing in singlecadmium sulfide nanowire optical cavities. Nano Lett. 5, 917(see Supporting Information).

Beyer WH (1991) Standard Mathematical Tables and Formulae,29th edn. CRC Press, Boca Raton, FL.

Goano M (1993) Series expansion of the Fermi–Dirac integralFj (x) over the entire domain of real j and x . Solid-StateElectron. 36, 217.

Goano M (1995) Algorithm 745: computation of the completeand incomplete Fermi–Dirac integral. ACM Trans. Math. Soft.21, 221.

9.9 FURTHER READING

Chuang SL (1995) Physics of Optoelectronic Devices. Wiley,New York.

Davies JH (1998) The Physics of Low-DimensionalSemiconductors: An Introduction. Cambridge University Press,Cambridge, UK.

Fox M (2010) Optical Properties of Solids, 2nd edn. OxfordUniversity Press, Oxford, UK.

Liboff RL (2003) Introductory Quantum Mechanics, 4th edn.Addison-Wesley, San Francisco.

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Documents

Density of States · 0.2 0.1 0 State density Number of states 200 Energy, E1 300 400 the top of Figure 9.2 depicts a histogram of diﬀerent energies possible for the particle in