Solutions to the problems in Chapter 6 and 7

Transcription

1 Solutions to the probles in Chapter 6 and Pressure of a Feri gas at zero teperature The nuber of electrons N and the internal energy U, inthevoluev,are N = V D(ε)f(ε)dε, U = V εd(ε)f(ε)dε, () The Feri distribution function f(ε) and the density of states (per unit volue) D(ε) are f(ε) = e β(ε µ) +, D(ε) =A 3 ε, A = π 3. () At zero teperature f(ε) =θ( ε), = kf ( /3, k F = 3π n) (3) Introducing the step function and D(ε) =A ε in (), we ay deterine U() and an alternative expression for A: N = V εf A εdε = 3 VAε3/ F A = 3n ε 3/ F, U() = V 5 Aε5/ F = 3 5 N () i.e. U kf V /3 and the pressure at zero teperature is P = U() = U V 3 V = 5 n = (3π ) /3 n 5/3 (5) 5 in sharp contrast to a classical gas, where the zero-teperature pressure is zero. 6. Density of states in low diensions The density of states in k-space is D k =/(π) d in d diensions (6.3) D(ε) = [d k]δ(ε ε k ) D k δ(ε ε k )d k = (π) d δ(ε ε k )d k () Introducing spherical coordinates in d diensions and k = k, then d dk, d = k = πk dk, d = () πk dk, d =3 The factor in the one-diensional case appears because the one-diensional wave vector ay assue both positive and negative values, whereas k per definition. ε k = ε k = k ( ) / k = εk, dk = dε ε k (3) k ) / D(ε) = ( δ(ε ε π k ) ε k (π) (π) 3 εk π δ(ε ε k ) π ε k δ(ε ε k ) ( ) / ε k ( ) / ε k π ε /, d = π, d = 3 π 3 ε/, d =3

2 The probles of st week ( electrons ) 6.5 Feri pancakes Thin layer of Ag: L x = L y = L = 6 ÅandL z = d. The density of electrons is the sae as the density of atos (one conduction electron per Ag ato), i.e. n =5.6 c 3 =.56 Å 3, according to the properties given in the periodic table on the front page of Marder. The wave function of the free electrons should vanish at the boundaries z =andz = d. This condition is fulfilled if assuing the one-electron wave function to be ψ(x, y, z) e i(k x x+k y y) sin (pqz), q = π, p =,,... () d The electron states are characterized by the two-diensional wave vector k = (k x,k y, ) and the band index p. The eigenenergies are (k = k x + k y ): ε p k = (k + p q ) ε = q () At T =(T T F ), as assued iplicitly in the exercise, the occupied states are all those with energies saller than. The nuber of electrons in the pth band, N p, is zero if ε p >. In the opposite case: N p = L kp (π) πk dk = L k p π (3) where k p is the Feri wave nuber of the pth band, i.e. the largest value of k of occupied states in the pth band, as deterined by ε pkp = ( kp + p q ) = ε F kp = ε ( ) F p q = q εf p () ε (a) In the first case d =. Åorq = π/d =.766 Å,and ε = q = ( ) ev =.369 ev (5).677 Assuing the Feri energy to be saller than the lowest energy of the (p =) band, <ε =ε, then only the (p = ) states are occupied, in which case N has to be equal the total nuber N of free electrons: N = L dn = N = L k π k = πd n =.66 Å (6) Introducing this result in (3) we get the Feri energy and the band width W of the occupied states: = ε k = ε [ ] (k /q) + =7.99 ev, W = ε =5.75 ev (7) which is in accordance with our starting assuption of <ε =.9 ev. These results ay be copared with that obtained for bulk Ag: (bulk) = W (bulk) = ( 3π /3 n) =5.5 ev ()

3 Solid state physics II ( electrons ) 3 (b) In the case of d =. Å, q = π/d =.33 Å and ε =.559 ev. In order to deterine in this case, we need to include ore bands than the lowest one. The nuber of bands turns out to be 3, and [( ) ( ) ( )] N = N +N +N 3 πd n = q εf εf + εf + 3 (9) ε ε ε Solving this equation with respect to /ε,weget /ε =.53, which is larger than 3 but saller than, in accordance with the assuption that all electrons are found in the three lowest bands. Hence the results are: =6. ev, W = ε =5.9 ev () 6 p Band energy (ev) Band energy (ev) Wave nuber k [r.l.u.] 6 k 5 p 3 The two first figures show the energy bands as functions of k in the cases (a) and (b), where the unit of k is the length of a reciprocal lattice vector π/a [r.l.u.]. The last figure shows the Feri energy (and the band width) as a function of the nuber of atoic layers of Ag. The crystal structure of Ag is fcc with the lattice paraeter a =.9 Å. This is very nearly the thickness d assued in (a), i.e. this case corresponds to a fil with two atoic layers of Ag atos. Feri energy and band width (ev) k Wave nuber k [r.l.u] 6 W 6 Nuber of Ag layers (d/a) k k Bulk value

4 The probles of st week ( electrons ) Solution to HS s proble Heat capacity of a two-diensional electron gas GaAs/AlGaAs heterostructure (see Marder Section 9.5): n = c, ε = k, =.67 e () The present situation corresponds to the case (a) of the previous proble 6.5 in Marder, i.e. the gas is purely two-diensional in the sense that only the (p =) band needs to be considered, and k is a two-diensional vector with the length k = kx + ky. The ost iportant quantity is the Feri energy, which is deterined by evaluating N at zero teperature: n = N A = D k d k = kf k k F (π) πk dk = k F π k F = πn (a) The sae result is obtained by using that, according to proble 6. or equation (6.35), D(ε) = /π in the two-diensional case: n = εf D(ε) dε = π = nπ or k F = πn (b) Introducing the nubers, and using that k F =Å corresponds to a Feri energy =3. ev when the ass is e,thenweget ( k F =.793 Å.793 ), = 3. ev = 3.5 ev (3).67 This Feri energy corresponds to a Feri teperature T F = /k B =.5 K. ) T = K is uch saller than the Feri teperature and the heat capacity ay be deterined by the leading order expression (6.77) c V = π 3 D()k BT = π 6 T T F k Fk B, as D( )= π = In the case of a saple with the area A =c the result is k F π = k F πk B T F () C V (el) = Ac V = A π 6.5 ( c ) J/K =. 3 J/K (5) The explicit result is C V (el) = Aπ kb T/(3 ). Hence, the sall value of C V is not due to the low electron density n but to the two-diensionality of the syste and the sall effective ass. ) In order to estiate the phonon contribution to the heat capacity we shall use the Debye odel (Section 3.3. in Marder). The Debye teperature of GaAs is Θ D = 3 K (which value is not uch change when soe of the Ga ions are replaced by Al ions). Using (3.7) in Marder and x e x π (e x dx = ) 5 ( ) 3 c V = π T 5 n A k B (T Θ Θ D ) (6) D

5 Solid state physics II ( electrons ) 5 According to Table.5 (page 7) in Marder, GaAs has the zincblende structure ( diaond structure) with the lattice paraeter a = 5.63 Å. In this structure there are atos per unit cell, and the density of atos is n A =/a 3 =. c 3. [The ean atoic ass (periodic table) is ( )/ = 7.3 u, iplying a ass density ρ =5.3 g c 3 ]. Assuing V =c 3 and T = K, the phonon contribution becoes C V (ph) = π ( ) 3 J/K = J/K (7) 3 which is uch larger than the electronic contribution. Utilizing that C V (el) T and C V (ph) T 3, the teperature T at which the two contributions are equal, is deterined by (T in K). 3 T = T 3 T =. K () A ore fair coparison would be to consider a fil of thickness µ, i.e. A =c and V = c 3,inwhichcaseT =. K,ateperaturewithin an accessible range (however, the reduction of the size of the saple akes it ore difficult to deterine the heat capacity). Solutions to the probles in Chapter 7 7. Norals to surfaces r =(x,x,x 3 )= s(t) is the paraetrization of a curve lying within the surface defined by f( r) =ε. Since f( s(t)) is a constant ε, the derivative of this function is : d dt f( s(t)) = α f ds α x α dt = f d s dt = () Because s(t) ay be any arbitrary curve lying within the surface, the sae is true for the curve tangent d s(t)/dt, and () is only generally valid if f is noral to the surface. 7.3 Van Hove singularities (a) The proble becoes the sae as the one considered in proble 7. if aking the replaceents k r and ε n k f( r), hence k ε n k is perpendicular to the energy surface defined by ε n k = ε. (b) The energy is assued to be ε n k = ε ax k, and in the two-diensional case D(ε) = [d k]δ(ε ε n k )= (π) πk δ(ε ε ax + k ) dk = δ(ε ε π ax + k )d(k )= () π θ(ε ax ε) The density of states is zero if ε>ε ax and /(π) whenε<ε ax. (c) In the three diensional case, the result is D(ε) = [d k]δ(ε ε n k )= (π) 3 πk δ(ε ε ax + k ) dk = π kδ(ε ε ax + k )d(k )= εax π εθ(ε ax ε) ()