Solution manual - Chapter 2 Jens Zamanian March 9, 2014 2.1 The Free and Independent Electron Gas in Two Dimensions 2 2 a) In two dimensions each state takes up an see Fig. 1. The two dimensional Ground-Stat<' uf area lhl' 4⇡ /L in k-space,35 !<, imeo""'nal k-sp:u:c of the fo rm J. , = L Noh: th:ot the:: per point is Jl"l t he n:•lwn.: r<:r point is t:! n L r . • • • • • • • • k, • • • • • • • f k. That latter volume (see Figure 2.2) is j ust t2n L) 3 • We therefore Figure 1: Two dimensional k-space. Each state takes up an area (2⇡/L)2 . region of k-space tlf \'Olumc: n will contain nv n equivalent of the Fermi sphere is a circular area with radius kF . This area must contain all the (2. 17) electrons. The= number (1rr, L):J lfn3 of states in the Fermi circle is the area of the circle multiplied by the the number of states per area multiplied by a factor 2 to account for spin. We hence have f k. equivalently. that the number of allowed k-valucs per unit L2 = N lalso known itS the k-sp:.1cc density lcvd sl2⇡k isF2just 4⇡ 2 and hence Jl (1) r (2.18) p 2⇡N = 2⇡n, (2) L2 where,regions in 2D, it makes sense(to talk of electrons per area instead of per volume and all deal with k-sp;H:e so large I about ponumber ints) and so regular 2 we hence use n = N/L . ) that to all intents and purposes (2.17) and (2.1 S) can be regarded b) In 3D rs is the radius of a sphere which has the same volume as the volume per conduction l begin to applyelectron. these important formulas In 2D it mustcounting be defined as the radius shortly. of a circle which has the same area as the area sume the electrons non interacting Gill build upL the N-clcctron perare conduction electron. Forwe N electrons in an ⇥ L solid we have g1t3 . kF = on placing electro ns into the allowed one-electronL2le,•els we have j ust 1 = = ⇡rs2(as it does exclusion principle plays a vital role in thts construction N n e states many dcctron atoms} : we may place at most one electron ctron level. The one-electron levels are specified by the wave vectors ection of the electron's spin along an arbitrary axis, 1which can take values f• / 2 or - lr 2. Therefore associated w1th each allowed wave electronic levels, one for each direction of tl1e electron's spin. ng up the N-electron ground sla te we begin by placing two electrons (3) which gives 1 rs = p . ⇡n (4) c) In two dimensions one is often interested in calculating the sum of a function over all states. We may write X 2L2 X 2 F (k) = F (k) k (5) 4⇡ 2 k k where the first factor 2 is due to the possible spin states and where k = 4⇡ 2 /L2 . We then get that Z X 2 X k dk F (k) = 2 F (k) ! F (k) 2 , (6) L2 4⇡ 2 2⇡ k k in the limit L ! 1. Often one is interested in functions F that only depends on k through the energy E(k) = ~2 k 2 /(2m), i.e. F (k) = F (E(k)) = F (k). Switching to cylindrical coordinates we have Z Z 1 Z 2⇡ Z dk 1 1 1 I= F (E(k)) = dkkF (E(k)) d = dkkF (E(k)) . (7) 4⇡ 2 2⇡ 2 0 ⇡ 0 0 Now we wish to switch variables to E = ~2 k 2 /(2m). The di↵erential is found as dE = So that I= m ⇡~2 Z ~2 kdk m 1 ) kdk = dEF (E) = 0 where the density of states is ⇢ Z 1 m dE. ~2 (8) dEg(E)F (E), (9) 1 m ⇡~2 E >0 (10) 0 E < 0. R1 d) One is often interested in integrals like 1 dEH(E)f (E) where f (E) is the Fermi-Dirac distribution. In the book they derive the Sommerfeld expansion which is g(E) = Z 1 dEH(E)f (E) = 1 Z µ dEH(E) + 1 1  2j X d j 1 dE 2j H 1 ✏=µ Z The integral to calculate n in two dimensions is given by Z 1 n= dEg(E)f (E), 1 1 dE (E µ)2j (2j)! ✓ @f (E) @E ◆ . (11) (12) 1 where g(E) is the density of state as above. Since this is a constant only the first term in the expansion survives. We then have Z µ Z µ m µm n= dEg(E) = dE 2 = , (13) 2 ⇡~ ⇡~ 1 0 which yields n⇡~2 k 2 ~2 ⇡ k 2 ~2 = F = F , (14) m 2⇡m 2m where we have used (2) to express n in terms of the Fermi wave vector kF . The last term is identified as the Fermi energy so we hence have µ= µ = EF , 2 (15) independently of temperature! Although this is very close to the exact result, there is a small correction to this and the Sommerfeld expansion does not give the exact result in this case. e) If we instead directly use (12) to calculate n we have Z 1 Z 1 m 1 m 1 n= dE (E µ)/k T = dE (E µ) , (16) B ⇡~2 0 ⇡~2 0 e +1 e +1 where we have defined = 1/(kB T ) for convenience. This integral can easily be solved using for example tables, but as we are physicist we do it the real way. First we switch variables to x = e(E µ) which yields dE = dx/( x) 1 The integral is then Z 1 1 1 dx . (17) x(x + 1) µ e This is now solved by the standard method of separating the fraction. If we try to write this as two separate brackets we get 1 A B = + , (18) x(x + 1) x x+1 which gives A + Ax + Bx = 1 which gives us A = 1 and B = A = 1. The integral is now easily solved  Z ⇣ ⌘ 1 1 1 1 1 1 dx = [ln x ln(x + 1)]e µ = µ + kB T ln 1 + e µ/(kB T ) . (19) x x+1 e µ Inserting this into Eq. (16) we get ⇣ µ + kB T ln 1 + e µ/(kB T ) ⌘ = EF . (20) f ) Typically we have kB T ⌧ EF and then we must also have µ kB T (try the limit µ and see that this lead to nonsens (EF ! 0). For µ ⇠ EF kB T we get ⌘ k T EF µ kB T ⇣ B = ln 1 + e µ/(kB T ) ⇡ e µ/(kB T ) , µ µ µ kB T (21) which is very small. The reason why the Sommerfeld expansion did not work perfectly is because the function g(E) has a discontinuity at E = 0 where it jumps down to zero. This means that the function is not analytical and cannot be Taylor expanded exactly with no amount of coefficients. 2.3 The Classical Limit of Fermi-Dirac Statistics a) In the classical limit Furthermore, we know that The density is now n = = = 1A f (E) ⇡ e (E µ)/(kB T ) . (22) 1 4⇡rs3 = . n 3 calculated from the distribution function as p Z 1 Z m3/2 2eµ/(kB T ) 1 p dE Ee E/(kB T ) dEg(E)f (E) = 3 ⇡2 ~ 1 0 p  Z m3/2 2eµ/(kB T ) (kB T )3/2 1 p y = E/(kB T ) = dy ye dE = kB T dy ~3 ⇡ 2 0 p p m3/2 2eµ/(kB T ) (kB T )3/2 ⇡ m3/2 eµ/(kB T ) (kB T )3/2 = , ~3 ⇡ 2 2 21/2 ~3 ⇡ 3/2 slightly shorter solution is obtained by instead using the variable substitution z = e 3 (E (23) y2 = (24) µ) . where we have looked up the integral using a table (it is actually related to the -function if you are interested). Inserting this into Eq. (23) we get rs3 = 3~3 ⇡ 1/2 23/2 m3/2 eµ/(kB T ) (k or rs = b) Note that this shows that if e µ/(3kB T ) 1/3 1/6 e 3 ⇡ (2kB T )1/2 µ/(kB T ) ✓ rs 3/2 (25) . (26) BT ) ~ 1 we also have ~2 2mkB T ◆1/2 . (27) To see the significance of this we define the thermal velocity vT by vT2 = We then have ✓ rs ~2 m2 vT2 2 mvT 2 = kB T or 2kB T . m ◆1/2 (28) = ~ , mvT (29) i.e. this says that the typical distance between the electrons rs much be much larger than the (thermal) de Broglie wavelength of the particle. c) In terms of the Bohr radius a0 = ~2 /(me2 ) [in SI-units it is a0 = 4⇡✏0 ~2 /(me2 )] we have ✓ rs a0 me4 2~2 kB T ◆1/2 . (30) If we now insert the number in CGS units (SI-units in brackets, but not that you need to use a0 in SI-units in that case) m ~ = = kB = e = 9.1095 ⇥ 10 28 1.3807 ⇥ 10 16 1.05459 ⇥ 10 4.80324 ⇥ 10 (9.1095 ⇥ 10 g 27 erg · s erg/K 10 esu 31 kg) (1.05459 ⇥ 10 (1.3807 ⇥ 10 (1.60219 ⇥ 10 (31) 34 Js) (32) J/K) (33) C), (34) 23 19 where 1erg = 1gcm2 /s2 is what energy is measured in in the CGS-system and 1esu = 1g1/2 cm3/2 /s is the unit of charge. We then get rs a0 ✓ 157888 K T ◆1/2 ⇡ ✓ 105 K T ◆1/2 . (35) Note that in metals rs /a0 ⇠ 1 and we have Fermi-Dirac statistics up to quite high temperatures. d) To solve this (which involves a bit of algebra) we use n = kF3 3⇡ 2 EF = k B TF = (36) ~2 kF2 2m (37) Now I’ll give you the start m3 m3 n m3 3⇡ 2 n = = = use kF = (2mkB TF )1/2 /~ . . . 4⇡ 3 ~3 4⇡ 3 ~3 n 4⇡ 3 ~3 kF3 4 (38)