Density of States
Franz Utermohlen
February 13, 2018
Contents
1 Preface
1
2 Summation definition
2.1 Examples of Eqs. 2 and 3 for a fermionic system . . . . . . . . . . . . . . . .
2.2 Example: Ideal Fermi gas . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
3
3 Derivative definition
3.1 Example: Ideal Fermi gas . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4
4 Useful expressions
6
1
Preface
In these notes, the expressions Vd denotes the volume of a d-dimensional sphere of radius R
and Ωd−1 denotes the solid angle a d-dimensional sphere (or equivalently, the surface area of
a d-dimensional sphere of unit radius).1 These expressions are given by
Vd =
π d/2
Rd ,
Γ(1 + d2 )
Ωd−1 =
2π d/2
.
Γ( d2 )
(1)
For example, for d = 1, 2, and 3, these are
d = 1 : V1 = 2R ,
Ω0 = 2 ,
d = 2 : V2 = πR2 ,
Ω1 = 2π ,
4
d = 3 : V3 = πR3 , Ω2 = 4π .
3
1
The d − 1 subscript refers to the fact that the fact that the surface area of a d-dimensional sphere is
(d − 1)-dimensional. It’s also useful to use d − 1 as the subscript, since the surface area of a d-dimensional
sphere of radius R is given by Sd−1 = Ωd−1 Rd−1 .
1
2
Summation definition
Intensive quantities A can generally be expressed in the form
1 X
a(i ) ,
Vd i
A=
(2)
where Vd is the d-dimensional volume of the system, the sum is over all possible singleparticle states i, and i is the energy of the single-particle state i. In the continuum limit
(thermodynamic limit), we can similarly define intensive quantities through
Z ∞
a()g() d ,
(3)
A=
−∞
where g() is called the density of states (DOS). Setting Eqs. 2 and 3 equal to each other,
we obtain
Z ∞
1 X
a(i ) =
a()g() d ,
(4)
Vd i
−∞
which implies that the DOS is given by
g() =
1 X
δ( − i ) ,
Vd i
(5)
as we can verify by inserting this into the right-hand side of Eq. 4:
"
#
Z ∞
Z ∞
1 X
a()
a()g() d =
δ( − i ) d
Vd i
−∞
−∞
Z
1 X ∞
a()δ( − i ) d
=
Vd i −∞
1 X
=
a(i ) . X
Vd i
2.1
Examples of Eqs. 2 and 3 for a fermionic system
As concrete examples of Eqs. 2 and 3, let’s consider a fermionic system, whose distribution
is the Fermi–Dirac distribution
nF () =
1
e(−µ)/kB T
2
+1
.
(6)
We can describe the system’s number density n ≡ N/Vd and energy density u ≡ U/Vd (where
U = Etot ) as
Z ∞
1 X
nF ()g() d ,
(7)
nF (i ) =
n=
Vd i
−∞
Z ∞
1 X
u=
i nF (i ) =
nF ()g() d .
(8)
Vd i
−∞
2.2
Example: Ideal Fermi gas
Consider an ideal, nonrelativistic Fermi gas comprised of electrons of mass m with an energy–
momentum relation
~2 k 2
(k) =
.
(9)
2m
The DOS is given by Eq. 5, which reads
g() =
1 X
δ( − i ) .
Vd i
(10)
We can rewrite the sum over single-particle states as a sum over single-particle momenta
X
X
,
(11)
→2
i
ki
where the factor of 2 is due to the fact that there are two different single-particle states
corresponding to a given momentum ki (spin up and spin down). In the continuum limit,
we use the prescription
Z d
X
d ki
→ Vd
(12)
d
(2π)
k
i
to rewrite Eq. 10 as
Z
g() = 2
dd ki
2
δ( − i )
=
d
(2π)
(2π)d
Z
δ( − 0 ) dd k 0 .
(13)
We can use rotational invariance to rewrite the integration element as dd k 0 = Ωd−1 k 0 d−1 dk 0 ,
so we obtain
Z
Z
2Ωd−1 ∞
2Ωd−1 ∞
d−1
0 0 d−1
0
g() =
δ( − )k
dk =
δ( − 0 )k 0
dk 0 .
d
d
(2π) 0
(2π) 0
Changing the integral to an energy integral through
√
r
0
2m
1
m 0
, dk 0 =
d ,
k0 =
~
~ 20
3
(14)
we obtain
d−1 r
√
Z
2m0
1 m 0
2Ωd−1 ∞
0
g() =
δ( − )
d
d
(2π) 0
~
~ 20
md/2 Ωd−1
= d/2 d d (d−2)/2 θ()
2 π ~
md/2
= d/2−1 d/2 d d (d−2)/2 θ()
2
π ~ Γ( 2 )
md/2 d
= d/2 d/2 d
(d−2)/2 θ() .
d
2 π ~ Γ(1 + 2 )
We thus find that the DOS for an ideal d-dimensional Fermi gas is given by
g() =
3
md/2 d
(d−2)/2 θ() .
d
d/2
d
(2π) ~ Γ(1 + 2 )
(15)
Derivative definition
For a d-dimensional thermodynamic system of volume Vd , the DOS g() is defined by
g() ≡
dn
.
d
(16)
Then, g() d is the number of states per unit volume in the energy interval (, + d).
3.1
Example: Ideal Fermi gas
Consider an ideal, nonrelativistic Fermi gas comprised of electrons of mass m with an energy–
momentum relation
~2 k 2
(k) =
(17)
2m
in a d-dimensional box of volume Vd = Ld . Since all of the energies are nonnegative, we will
multiply g() by the step function θ() to make sure that we only integrate from = 0 to ∞
if we use the DOS inside an integral:
dn
θ() .
d
(18)
dn dk
θ() .
dk d
(19)
g() =
We can rewrite this using the chain rule as
g() =
4
The derivative on the right can simply be obtained from Eq. 17:
√
√
r
dk
2m
d 2m
1 m −1/2
=
= √ =
.
d
d ~
~ 2
2~ (20)
For the other derivative, we need to find an expression for n. In this d-dimensional box of
volume Ld , n is given by
N
(21)
n= d.
L
We need to find N now, which we will find by looking at the problem in d-dimensional
k-space.
For a given value of k, we can consider a corresponding sphere of radius k ≡ |k| in ddimensional k-space whose volume is
Vd (k) =
π d/2
kd .
d
Γ(1 + 2 )
(22)
d
. In the
We now imagine dividing this sphere into cubic cells of length kcell and volume kcell
large k limit (k kcell ), the number of cells that fit inside the sphere is therefore
Ncells =
Vd (k)
.
d
kcell
(23)
Now, we know that we can only have two particles (with opposite spins) of momentum k
per k-space volume (2π/L)d , which means that for this problem we can use
kcell =
2π
L
(24)
to find the number of particles that fit in a d-dimensional k-space sphere of radius k:
d
2Vd (k)
2π d/2 k d
L
Ld k d
.
(25)
N = 2Ncells =
=
=
d
kcell
Γ(1 + d2 ) 2π
2d−1 π d/2 Γ(1 + d2 )
Inserting this into Eq. 21, we find
n=
kd
.
2d−1 π d/2 Γ(1 + d2 )
(26)
We can now compute the k derivative of n:
dn
k d−1 d
= d−1 d/2
.
dk
2 π Γ(1 + d2 )
(27)
In terms of , this is
dn
d
= d−1 d/2
dk
2 π Γ(1 + d2 )
2m
~2
(d−1)/2
=
5
m(d−1)/2 d
(d−1)/2 .
2(d−1)/2 π d/2 ~d−1 Γ(1 + d2 )
(28)
Using this derivative and the one we computed in Eq. 20, we have
r
m −1/2
m(d−1)/2 d
(d−1)/2 1
θ()
g() = (d−1)/2 d/2 d−1
d
~ 2
2
π ~ Γ(1 + 2 )
=
md/2 d
(d−2)/2 θ() .
2d/2 π d/2 ~d Γ(1 + d2 )
We thus find that the DOS for an ideal d-dimensional Fermi gas is given by
g() =
md/2 d
(d−2)/2 θ() ,
d
d/2
d/2
d
2 π ~ Γ(1 + 2 )
(29)
which agrees with Eq. 15.
4
Useful expressions
The DOS is useful for computing thermodynamic quantities, such as the following:
Z ∞
1 X
N
nF/B ()g() d ,
=
nF/B (i ) =
n≡
Vd
Vd i
−∞
Z ∞
U
1 X
u≡
=
i nF/B (i ) =
nF/B ()g() d ,
Vd
Vd i
−∞
(30)
(31)
where U = Etot is the system’s total energy and nF/B () denotes the system’s distribution,
i.e. the Fermi–Dirac distribution nF () or the Bose–Einstein distribution nB ().
For a fermionic system at T = 0, the Fermi–Dirac distribution is just nF () = θ(−)θ( − µ).
The chemical potential at T = 0 is called the Fermi energy:
EF ≡ µ(T = 0) =
~2 kF2
,
2m
so the integral expressions for n and u become
Z EF
n=
g() d ,
0
Z EF
u=
g() d .
0
6
(32)
(33)
(34)