Appendix F: Density of Electron States

Appendix F: Density of Electron States F.1

THREE-DIMENSIONAL SOLID

Consider a three-dimensional lattice in the form of a cube of side L and volume V. Imposing the cyclic boundary condition on a crystalline solid shows that there is one k-state in a volume of (2 p)3/V. So, the number of electron states in an elemental volume d3k is given by Ne ðkÞd3 k ¼ 2

V ð2pÞ3

d3 k

(F.1)

Ne(k) d3k is the number of electron states lying between k and k + dk. Here the factor of 2 takes account of the spin degeneracy of the electron states. Let the volume element d3k lie between two constant-energy surfaces with energies Ek and Ek + dEk, as shown in Fig. F.1. The elemental volume d3k can then be written as ^  dk d3 k ¼ dSEk  dk ¼ dSEk n

(F.2)

^ is a unit vector perpendicular to dSEk, a small element of surface with constant energy Ek. If dk? is the component where n ^, then of the wave vector dk in the direction of n d3 k ¼ dSEk dk?

(F.3)

3

In this approximation, Ne(k) d k can also be interpreted as the number of electron states lying between energies Ek and Ek + dEk, which is represented as Ne(Ek) dEk. So, Eq. (F.1) can be written as

FIG. F.1 The two constant energy surfaces with energies Ek and Ek + dEk in the k-space of a three-dimensional solid. d3k is an elemental volume lying between the two constant-energy surfaces having wave vectors k and k + dk. dSEk is an elemental surface on the constant-energy surface with energy Ek and wave vector k.

605

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Appendix F

Ne ðEk Þ dEk ¼ 2

ð

V ð2pÞ3

d3 k ¼ 2 SEk

V ð2pÞ3

ð dSEk dk?

(F.4)

SEk

The integral dSEk is over the surface of constant energy Ek. The energy derivative is given by dEk ¼

∂Ek ∂Ek ∂Ek dkx + dky + dkz ∂kx ∂ky ∂kz

(F.5)

¼ ðrk Ek Þ  dk ¼ jrk Ek jdk? From the above equation one can write dk? ¼

dEk jrk E k j

Substituting Eq. (F.6) into Eq. (F.4), one can write Ne ðEk ÞdEk ¼ 2

ð

V ð2pÞ3

SE k

dSEk dEk jrk E k j

Therefore, the density of electron states per unit energy Ne(Ek) is given by ð V dSEk Ne ðEk Þ ¼ 2 3 jrk Ek j ð2pÞ

(F.6)

(F.7)

(F.8)

SEk

Hence the density of electron states per unit energy per unit volume becomes ð 2 dSEk g e ð Ek Þ ¼ 3 r j ð2pÞ k Ek j

(F.9)

SEk

Eqs. (F.8), (F.9) are the general expressions for any type of constant energy surface. In the free-electron approximation, the energy of an electron is given by Ek ¼

ħ2 k2 2me

(F.10)

Therefore, jrk Ek j ¼

ħ2 k me

(F.11)

The area of the constant-energy surface SEkis 4 p k2. Substituting Eq. (F.11) and the value of SEk into Eq. (F.8), one can write   V 4pk2 V 2me 3=2 1=2 ¼ Ek (F.12) Ne ðEk Þ ¼ 2 ð2pÞ3 ħ2 k=me 2p2 ħ2 which is the same expression as obtained in Eq. (9.22). Therefore, the density of electron states per unit energy per unit volume becomes   1 2me 3=2 1=2 Ek (F.13) ge ð Ek Þ ¼ 2 2p ħ2 It is worthwhile to point out here that real crystals are anisotropic, which gives rise to nonspherical constant-energy surfaces. Therefore, the evaluation of Ne(Ek) involves the ab initio calculation of constant-energy surfaces in a crystalline solid.

F.2

TWO-DIMENSIONAL SOLID

Consider a two-dimensional crystalline solid having a closed surface with area A0. To find the density of electron states, one can proceed in the same way as in a three-dimensional solid. Fig. F.2 shows an elemental area d2k between vectors

Appendix F

607

FIG. F.2 The two constant-energy contours with energies Ek and Ek + dEk in the k-space of a two-dimensional solid. d2k is an elemental surface lying between the two constant-energy contours having wave vectors k and k + dk. dlEk is an elemental portion of the constant-energy contour with energy Ek and wave vector k.

k and k + dk and bounded by contours of constant energy Ek and Ek + dEk. The number of electron states in a unit area in the k-space is A0/(2 p)2. Therefore, the number of electron states in d2k is given by ð A0 Ne ðEk ÞdEk ¼ 2 d2 k ð2pÞ2 l Ek ð (F.14) A0 dl dk ¼2 E ? k ð2pÞ2 l Ek

Here the integration is along the contour of constant energy Ek. As the energy is a function of wave vector, therefore, dEk ¼

∂Ek ∂Ek dkx + dky ∂kx ∂ky

¼ ðrk Ek Þ  dk ¼ jrk Ek jdk? From the above equation one can write dk? ¼

dEk j r k Ek j

Substituting the value of dk? from the above equation into Eq. (F.14), we get ð A0 dlEk dEk Ne ðEk ÞdEk ¼ 2 ð2pÞ2 jrk Ek j

(F.15)

(F.16)

lEk

Therefore, the density of electron states per unit energy is given by ð A0 dlEk N e ð Ek Þ ¼ 2 2 r j ð2pÞ k Ek j

(F.17)

l Ek

The density of electron states per unit energy per unit area is given by ð 1 dlEk ge ð Ek Þ ¼ 2 ð2pÞ2 jrk Ek j l Ek

(F.18)

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Appendix F

In the case of a two-dimensional free-electron gas, the electron energy and its derivative are given by Eqs. (F.10), (F.11), respectively. The circumference of a circle with constant energy Ek is lEk ¼ 2 p k. Substituting these values into Eq. (F.17), we get N e ð Ek Þ ¼ 2

A0

2pk me A0 ¼ pħ2 ð2pÞ ħ k=me 2 2

(F.19)

which is the same result as obtained in Eq. (9.44). Hence the density of electron states per unit energy per unit area becomes g e ð Ek Þ ¼

me pħ2

(F.20)

In a one-dimensional solid the contour of integration reduces to two points k. Therefore, the free electron density of states become   L 1 L 2me 1=2 1=2 ¼ Ek (F.21) N e ð Ek Þ ¼ 2 2p ħ2 k=me 2p ħ2 which is the same expression as Eq. (9.51). Hence the density of electron states per unit energy per unit length becomes   1 2me 1=2 1=2 g e ð Ek Þ ¼ Ek (F.22) 2p ħ2