The free electron gas: Sommerfeld model
• Valence electrons: delocalized ”gas” of free electrons (selectrons of alkalies) which is not perturbed by the ion cores
(ionic radius of N a+ equals 0.97 Å, internuclear separation
in solid Na equals 3.66 Å).
• Constant potential: the variation of the potential due to the
ion cores and the rest electrons is negligible in the interstitial
regions. It is assumed that the net charge associated with
the ion cores and the rest valence electrons is smoothly distributed throughout the volume of the solid (”jellium model”)
giving rise to a constant potential everywhere except at the
surface of the solid. There a potential barrier prevents the
escape of electrons out of the solid. The potential model is
that of a 3D square-well.
• The electrons are treated quantum-mechanically in a meanfield approach
• The electrons obey Pauli’s exclusion principle and FermiDirac statistics
1
References
[1] S. Elliott ”The Physics and Chemistry of Solids”, J. Wiley, Chichester, 1998.
2
For a time-independent potential the stationary-state solutions
of the time-independent Schrödinger equation are the eigenfunctions and eigenenergies:
Hψ(r) = εψ(r)
h̄2 2
∇ ψ(r) + V ψ(r) = εψ(r)
−
2me
The allowed solutions depend upon the boundary conditions, to
which the electrons are subject, and are quantized; i.e. the energies of the allowed states are functions of discrete quantum numbers.
Independent-electron approximation: it works because of the
screening by all other electrons of the interaction between any
two electrons. The free particles (electrons) are in reality ”quasiparticles”, comprising an electron and its associated ”orthogonalization hole”.
H(r1, r2 · · ·) = H(r1) + H(r2) + · · ·
ψ(r1, r2 · · ·) = ψ1(r1) × ψ2(r2) × · · ·
W is the potential well depth or the potential barrier height at
the box boundaries.
is the total energy of an electron; φ = W − is the extra energy
needed for an electron to escape from the solid; work function.
3
Figure 1:
V is the potential energy, assumed constant, set V = 0, in the
interior of the box. It involves the ion-core and averaged electronelectron potential energies.
4
The wavefunctions for the free electron gas
A general solution is:
ψ(r) = N exp(ik · r) = N exp[i(kxx + ky y + kz z)]
1=
Z
ψ(r)∗ψ(r)dr .
The boundary condition for an electron trapped in a potential box
of uniform V is that the wavefunction ψ(r = 0) and ψ(r = L)
goes to zero, i.e. it has a node, at the boundaries of the box
(stricktly valid only for V (0, L) → ∞). This boundary condition
leads to the standing-wave solutions (cf. Appendix 1):
8 1/2
πy
πz
πx

ψ(x, y, z) =
)
sin(n
)
sin(n
)
sin(n
2
3
1
L3
L
L
L


where kx,y,z are the wavevectors of the standing waves
kx,y,z = n1,2,3
π
n1,2,3 = 1, 2 . . .
L
Each allowed k-state can accommodate two electrons with magnetic spin quantum numbers ms = ± 12 .
5
Figure 2:
Dispersion relation for the free electron gas
Substituting the wavefunction in Schrödinger’s equation:
Hψ(x, y, z) = εψ(x, y, z)
h̄2  d2
d2
d2 

−
+ 2 + 2  ψ(x, y, z) = εψ(x, y, z)
2
2me dx
dy
dz


gives for the energy levels of an electron in a 3D potential box:
h̄2 2
h̄2 2
2
2
k + ky + kz ≡
k
ε(k) =
2me x
2me
The function ε(k) is called the dispersion relation. For the
free electron gas ε is a parabolic function of the wavevector k.
The highest occupied states at T = 0K have the Fermi energy
ε = εF =
h̄2 2
2me kF
and their wavevectors terminate on the surface
of the so called Fermi sphere with radius equal to the Fermi
wavevector k = kF.
6
Figure 3:
Figure 4:
7
Density of states of the free electron gas: number of
allowed electron states per energy interval
For macroscopic samples with large size, L is a large number,
therefore the separation in k-space between points corresponding
to allowed states becomes infinitesimal. The allowed physically
distinct standing-wave solutions of the Schrödinger equation for
a 3D potential box occupy the positive octant in k-space (since
ni > 0 and hence ki > 0). The number of allowed electron states
with k-values between k and k + dk is equal to the volume of
the spherical shell with radius k and thickness dk in the positive
octant divided by the volume in k-space, which corresponds to
a single k-point, multiplied by the spin degeneracy gs = 2. The
volume corresponding to a single k-point equals

1  4πk
g(k)dk = 2 

8 Lπ33
g(k)dk =
2




π 3
L .
V k2
dk = 2 dk
π
V 2me
ε(k)dk
π 2 h̄2
where V = L3 is the volume of the box. The density of electron
states as a function of energy g(ε) is given by:
g(ε) = g(k)
8
dk
dε
Figure 5:
dk 1 u
u me
= t ε−1/2
dε h̄ 2
v
V 2me 1 u
u me
g(ε) = 2 2 ε t ε−1/2
π h̄
h̄ 2
v
V  2me 3/2 1/2
g(ε) = 2
ε
2π
h̄2


At T → 0K N electrons fill the N lowest states up to energy
ε = εF :
N=
Z
εF
0
V  2me 3/2 3/2
εF
g(ε)dε =
3π 2 h̄2
2
N = εF g(εF )
3

9

The Fermi energy is determined by the electron density N/V :
εF =
2 
2/3
2
h̄  3π N 


2me
V
.
The Fermi wavevector also depends on the electron density N/V :

kF = 
1/2
2me 
εF
h̄2
10


=
2
1/3
3π N 

V
.
Fermi-Dirac distribution function
The average value of the occupancy of an electron state of energy
ε: hn(ε)i at temperature T is given by the Fermi-Dirac distribution function f (ε):
f (ε) ≡ hn(ε)i =
1
exp[(ε − µ)/kB T ] + 1
• µ is the chemical potential of an electron. At temperature
T → 0K it coincides with the Fermi energy:
µ(T = 0K) ≡ εF
The chemical potential corresponds to the energy at which
f (ε) = 21 .
• At temperature T → 0K f (ε) is a step function. For ε below
µ:
e(ε−µ)/kB T =
1
→ 0, hence f (ε) → 1
exp[(µ − ε)/kB T ]
For ε above µ:
e(ε−µ)/kB T → ∞, hence f (ε) → 0
• For the case where the total number of electrons in a 3D
solid is constant, the chemical potential must decrease with
increasing temperature in order to compensate for the broadening of the distribution.
11
Figure 6:
Appendix 1: Normalization of the wavefunction of an electron
in a 3D box:
Hψ(x, y, z) = εψ(x, y, z)
d
p̂2
p̂x = −ih̄
H=
2me
dx
h̄2 2
−
∇ ψ(x, y, z) = εψ(x, y, z)
2me
Boundary conditions:
ψ(0, 0, 0) = 0
ψ(L, 0, 0) = 0 ψ(0, L, 0) = 0 ψ(0, 0, L) = 0, etc.
Trial wavefunction:
ψ(x, y, z) = A3 sin(kxx) sin(ky y) sin(kz z)
kxL = n1π n1 = 1, 2, . . .
ky L = n2π n2 = 1, 2, . . .
12
kz L = n3π n3 = 1, 2, . . .
kx,y,z = n1,2,3
π
L
Normalization of the wavefunction:
Z
Z
ψ ∗(x, y, z)ψ(x, y, z)dxdydz = 1
∗
ψ (x, y, z)ψ(x, y, z)dxdydz =
L
0
!3
π
A sin ( x)dx
L
2
2
3
Z
3
L
π
6
2


A 0 sin θdθ
=
π
 3
 3
!3
π
L
L
6
∗
ψ (x, y, z)ψ(x, y, z)dxdydz =   A
=   A6 = 1
π
2
2

Z
Z

because
Z
1
x
sin2xdx = − sinx cosx + + C
2
2
The normalization constant equals:
8 1/2

A =
L3
3

13
