Franz Utermohlen
February 13, 2018
Contents
1 Preface 1
2 Summation definition 2 2.1 Examples of Eqs. 2 and 3 for a fermionic system ...... 2 2.2 Example: Ideal Fermi gas ...... 3
3 Derivative definition 4 3.1 Example: Ideal Fermi gas ...... 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
d/2 d/2 π d 2π Vd = d R , Ωd−1 = d . (1) Γ(1 + 2 ) Γ( 2 ) For example, for d = 1, 2, and 3, these are
d = 1 : V1 = 2R, Ω0 = 2 , 2 d = 2 : V2 = πR , Ω1 = 2π , 4 d = 3 : V = πR3 , Ω = 4π . 3 3 2 1The 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 d−1 sphere of radius R is given by Sd−1 = Ωd−1R .
1 2 Summation definition
Intensive quantities A can generally be expressed in the form
1 X A = a( ) , (2) V i d i where Vd is the d-dimensional volume of the system, the sum is over all possible single- particle 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 = a()g() d , (3) −∞ where g() is called the density of states (DOS). Setting Eqs. 2 and 3 equal to each other, we obtain 1 X Z ∞ a( ) = a()g() d , (4) V i d i −∞ which implies that the DOS is given by
1 X g() = δ( − ) , (5) V i d i as we can verify by inserting this into the right-hand side of Eq. 4: " # Z ∞ Z ∞ 1 X a()g() d = a() δ( − ) d V i −∞ −∞ d i 1 X Z ∞ = a()δ( − ) d V i d i −∞ 1 X = a( ) . V i X d 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 1 nF () = . (6) e(−µ)/kB T + 1
2 We can describe the system’s number density n ≡ N/Vd and energy density u ≡ U/Vd (where U = Etot) as 1 X Z ∞ n = n ( ) = n ()g() d , (7) V F i F d i −∞ 1 X Z ∞ u = n ( ) = n ()g() d . (8) V i F i F d i −∞
2.2 Example: Ideal Fermi gas
Consider an ideal, nonrelativistic Fermi gas comprised of electrons of mass m with an energy– momentum relation 2k2 (k) = ~ . (9) 2m The DOS is given by Eq. 5, which reads 1 X g() = δ( − ) . (10) V i d i We can rewrite the sum over single-particle states as a sum over single-particle momenta X X → 2 , (11)
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 → V (12) d (2π)d ki to rewrite Eq. 10 as Z ddk 2 Z g() = 2 δ( − ) i = δ( − 0) ddk0 . (13) i (2π)d (2π)d
d 0 0d−1 0 We can use rotational invariance to rewrite the integration element as d k = Ωd−1k dk , so we obtain Z ∞ Z ∞ 2Ωd−1 0 0d−1 0 2Ωd−1 0 0d−1 0 g() = d δ( − )k dk = d δ( − )k dk . (2π) 0 (2π) 0 Changing the integral to an energy integral through √ 2m0 1r m k0 = , dk0 = d0 , (14) ~ ~ 20 3 we obtain √ Z ∞ 0 d−1 r 2Ωd−1 0 2m 1 m 0 g() = d δ( − ) 0 d (2π) 0 ~ ~ 2 md/2Ω = d−1 (d−2)/2θ() 2d/2πd~d md/2 = (d−2)/2θ() d/2−1 d/2 d d 2 π ~ Γ( 2 ) md/2d = (d−2)/2θ() . d/2 d/2 d d 2 π ~ Γ(1 + 2 ) We thus find that the DOS for an ideal d-dimensional Fermi gas is given by
md/2d g() = (d−2)/2θ() . (15) d/2 d d (2π) ~ Γ(1 + 2 )
3 Derivative definition
For a d-dimensional thermodynamic system of volume Vd, the DOS g() is defined by
dn g() ≡ . (16) d
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 2k2 (k) = ~ (17) 2m d in a d-dimensional box of volume Vd = L . 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 g() = θ() . (18) d We can rewrite this using the chain rule as dn dk g() = θ() . (19) dk d 4 The derivative on the right can simply be obtained from Eq. 17: √ √ dk d 2m 2m 1rm = = √ = −1/2 . (20) d d ~ 2~ ~ 2 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 n = . (21) Ld 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 d- dimensional k-space whose volume is
d/2 π d Vd(k) = d k . (22) Γ(1 + 2 ) d We now imagine dividing this sphere into cubic cells of length kcell and volume kcell. In the large k limit (k kcell), the number of cells that fit inside the sphere is therefore
Vd(k) Ncells = d . (23) kcell 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 2π k = (24) cell L to find the number of particles that fit in a d-dimensional k-space sphere of radius k:
2V (k) 2πd/2kd L d Ldkd N = 2N = d = = . (25) cells d d 2π d−1 d/2 d kcell Γ(1 + 2 ) 2 π Γ(1 + 2 ) Inserting this into Eq. 21, we find kd n = . (26) d−1 d/2 d 2 π Γ(1 + 2 ) We can now compute the k derivative of n: dn kd−1d = . (27) dk d−1 d/2 d 2 π Γ(1 + 2 ) In terms of , this is
dn d 2m(d−1)/2 m(d−1)/2d = = (d−1)/2 . (28) dk d−1 d/2 d 2 (d−1)/2 d/2 d−1 d 2 π Γ(1 + 2 ) ~ 2 π ~ Γ(1 + 2 )
5 Using this derivative and the one we computed in Eq. 20, we have
m(d−1)/2d 1rm g() = (d−1)/2 −1/2θ() (d−1)/2 d/2 d−1 d 2 2 π ~ Γ(1 + 2 ) ~ md/2d = (d−2)/2θ() . d/2 d/2 d d 2 π ~ Γ(1 + 2 ) We thus find that the DOS for an ideal d-dimensional Fermi gas is given by
md/2d g() = (d−2)/2θ() , (29) d/2 d/2 d d 2 π ~ Γ(1 + 2 ) which agrees with Eq. 15.
4 Useful expressions
The DOS is useful for computing thermodynamic quantities, such as the following:
N 1 X Z ∞ n ≡ = n ( ) = n ()g() d , (30) V V F/B i F/B d d i −∞ U 1 X Z ∞ u ≡ = n ( ) = n ()g() d , (31) V V i F/B i F/B d d i −∞ 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:
2k2 E ≡ µ(T = 0) = ~ F , (32) F 2m so the integral expressions for n and u become
Z EF n = g() d , (33) 0 Z EF u = g() d . (34) 0
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