Density of States 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 1 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 1 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 1 Z 1 1 X a()g() d = a() δ( − ) d V i −∞ −∞ d i 1 X Z 1 = 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 1 n = n ( ) = n ()g() d ; (7) V F i F d i −∞ 1 X Z 1 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 1 Z 1 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 p 2m0 1r m k0 = ; dk0 = d0 ; (14) ~ ~ 20 3 we obtain p Z 1 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 1 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: p p dk d 2m 2m 1rm = = p = −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 ≡ jkj 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 1 n ≡ = n ( ) = n ()g() d ; (30) V V F=B i F=B d d i −∞ U 1 X Z 1 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 6.