Theorie der Kondensierten Materie I WS 16/ Chemical potential in the BCS state (40 Punkte)

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1 Karlsruhe Institute for Technology Institut für Theorie der Kondensierten Materie Theorie der Kondensierten Materie I WS 6/7 Prof Dr A Shnirman Blatt PD Dr B Narozhny, T Ludwig Lösungsvorschlag Chemical potential in the BCS state (4 Punte We consider at first the number of particles in the System for a given chemical potential, ie N = c σ c σ = c c + c c, ( σ where we used that the sum over momentum is symmetric Now, we express the electron creation/annihilation c σ /c σ operators by the creation and annihilation operators for Bogoliubov particles α σ α σ We get N = v + (u v (α α + α α + u v (α α + α α ( Since we consider T = it is = BCS BCS and, since α σ BCS =, we obtain N = v = ( (3 + ( ɛ µ = dɛ ν(ɛ (4 + (ɛ µ Now, following the idea of the Bohr-Sommerfeld-approximation That is, we define b(ɛ = ɛ dɛ ν(ɛ and perform a partial integration to obtain µ+ ωd N = dɛ b(ɛ µ ω D ɛ ( ɛ µ + (ɛ µ, (5 where we restricted the interval of integration to [µ ω D, µ + ω D ], since v should vanish outside of this interval for phonon mediated Cooper-pairing Now, approximating b(ɛ b(µ + b (µ(ɛ µ + b (µ(ɛ µ +, (6

2 we obtain N = b(µ + b (µ b(µ b (µ ln x ω D ( + x 3 }{{} ω D +( ω D ( ω D + arsinh ( ωd (7 (8 Next, we have to write b(µ and b (µ in a more useful form We use the density of states of free electrons in three dimensions, ie ν(ɛ ɛ Thus, b(µ = µ dɛ α ν(ɛ = 3 α µ 3, (9 b (µ = α µ ( b (µ = α µ ( with some constant α To determine α, note that ν b (µ α = ν µ So for N we obtain, ( ( 4 N = ν µ 3 + µ ln ωd ( Now, we have to solve for µ Although it is possible do solve exactly for µ it is instructive to go for the approximate solution, which is also applicable in more general cases For that purpose, note that µ Thus the second term in parentheses is only a small correction We mae use of this by rearranging the equation as µ = 3 N 3 ( 4ν 8 µ ln ωd, (3 which can now be ßolvedïteratively We obtain as zeroth order contribution µ ( = 3 4ν N, (4 which is the same as in the absence of superconductivity The first iteration yields µ ( = 3 N ( ( ν ωd ln, (5 4ν 3 N where the second term on the right hand side is the leading correction to the chemical potential due to superconductivity Thermodynamics of BCS theory (6 Punte (a From statistical mechanics, the entropy of a free Fermion-gas (with dispersion relation E is nown to be S = B [f(e ln f(e + ( f(e ln( f(e ], (6 σ

3 where f(e is the fermi distribution Since a system in its ground state has no entropy (ie no contribution from BCS, and excitations in the superconductor (Bogoliubov quasi-particles are free Fermions, the entropy is given by Eq(6 with the quasi-particle dispersion E = + (b Knowing the entropy, it is straightforward to determine the heat capacity by ( S C V (T = T V = B T ( f V f ln f }{{} βe = ( E E T + f, (9 E E where in the second step, we have rewritten the temperature derivative ( f in V terms of f E (c The self-consistency equation for (T reads = g V + β + (7 (8 ( To find T c, we have to loo for a solution to the self-consistency equation with (T c =, ie = g V β c ( Going from the sum over to the integral over, we obtain = g ν d βc, ( or after substitution β c / x, = g ν βc ω D x x ( γ = g ν ln β c ω D, (3 b T c = γ ω De gν, (4 where γ is the Euler constant (note: γ, 3 If the temperature T is just below T c, then 36 [T c (T c T ], (5 which can be found numerically Alternatively, one might proceed as follows: Starting again from Eq ( but now with, we have in integral form = g ν d + β + (6

4 Now, using Eq ( to replace the on the left hand side, and then subtracting β in the integrand on both sides we obtain [ ] β c β [ ωd d = d + β + β (7 The lhs can be handled as before, whereas for the rhs we extend the integral to (justified, since the integrand approaches zero fast enough for > B T and ω D B T and use the identity x = 4 x = ( + + x, (8 to obtain ln T = 4β ( d T c ( + (β (β + + ( + (β + (9 Now, we can expand for β, which yields, ln T = 4β ( ( n n β d T c [ ] n+ (3 = n= ( + (β + = 4β ( ( n n β d ( + (n+ [ ( ] n+ (3 = n= + = 4 = n= ( n ( + n+ β (+ ( n β [ + x n+ (3 ] Using the identity ( + x = x (n + n (n ( + x n (n and the integral we obtain and therefore, we get ln T c T = ( β ( n+ n=, (33 ( + x n = arctan x, (34 + x [ + x ] n+ = n (n!! (n!! (n!! (n!!, (35 ( n+ ( + = }{{ n+ } ζ(n+,, (36 where ζ(z, q is the Hurwitz-Zeta function Now, eeping only the lowest order in, ie the n = term in the sum, we obtain ln T c T = ζ(3, 8 β, (37 ]

5 further approximating ln T ( c T = ln + T c T T c T T T, (38 yields 8 = ζ(3, B T (Tc T 36 B Tc (T c T (39 where we also approximated T (T c T T c (T c T (d To determine the universal jump of the heat capacity, we insert (39 into (9 and assume to be very close to T c, such that is negligible in E, whereas the derivative is still important due to its behaviour We obtain and further, by using f C V (T c = d ( ν T + ν δ( 6 ( BT δ (, we find f, (4 C V (T c = ν B T c + (36 B ν T c = C V (T c + + C V, (4 3 and finally, we obtain C V C V (T c + 43, (4 which is the universal jump in heat capacity for the BCS theory

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