Note that some of these solutions are only a rough list of suggestions for what a proper answer might include.

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Osborn Chase
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1 Suprajohtavuus/Superconductivity S, Tentti/Examination (Solutions) Note that some of these solutions are only a rough list of suggestions for what a proper answer might include.. Explain the following (mostly words, possibly some formulas or a figure) (a) intermediate state of type I superconductor (b) mixed state of type II superconductor (c) ac and dc (tasavirta ja vaihtovirta) Josephson effects (a) κ < / 2; depending on geometry, critical field H c may be exceeded locally even at H < H c ; normal and superconducting regions both present; in film perpendicular to field N and S regions alternate; configuration detemined by two competing energy contributions: magnetic field inhomogeneity should be minimized, and so should the amount of N-S interface; figure of H c (T );... (b) κ > / 2; magnetic field penetrates via quantized flux lines, each carrying flux Φ 0 = h/2 e ; hexagonal lattice; motion of flux lines associated with dissipation; figure with H c,2 ;... (c) picture of a Josephson junction; dc: I = I c sin φ where φ is phase difference; ac: φ = 2eV/ h φ(t) = 2eV t/ h supercurrent oscillates with angular frequency 2eV/ h; Show that ˇγ σ ψ 0 = 0 for the BCS ground state ψ 0, which means that ψ 0 is the vacuum state for excitations. It is enough to consider σ =. Some preliminaries first. These should be useful also elsewhere. The BCS ground state is ψ 0 = (u + v a a ) vac () where vac is the vacuum state: a vac = 0. Defining c = u + v a a, we may write ψ 0 = c vac (2) Is the order of c s in the product relevant, or can we freely commute them? Let s see. For l we find c c l = (u + v a a )(u l + v l a l a l ) = u u l + u v l a l a l + u lv a a + v v l a a a l a l = u u l + u l v a a + u v l a l a l + v lv a l a l a a = c l c

2 Thus the operators commute. (Getting to the second to last line requires anticommuting operators in the four-operator term 2*3=6 times (a i a j = a j a i ), which thus eeps the sign intact.) Now to the problem itself. Using the above commutation result, we can isolate from the BCS ground state an arbitrary c = u + v a a factor and bring it to the front of the product. Thus we need to calculate for example γ ψ 0 = γ c c l vac (3) l where γ = u a v a. Now an intermediate result. γ c = (u a v a )(u + v a a ) = u 2 a + u v a a a u v a + v2 a a a = u 2 a + u v a u v a a a u v a v2 a a a = u 2 a + u v a a a = u c a Since there is no c -factor in the product l c l (and [a, c ] = 0, as you may chec), we may move the a operator all the way through: γ ψ 0 = γ c c l vac = u c a c l vac = u c c l a vac = 0 l A similar proof can be given for γ ψ 0 = 0. One can for example isolate c in front of the product and tae it on from there. l 3. In BCS theory, the following equation appears l = g 2L 3 tanh( E ), (4) E 2 B T where E = ξ What is the meaning of E,, and thus of the equation? How can you turn the -sum into a one-dimensional integral? Simplify the integral as much as possible and explain the needed approximations. The equation in question is the so-called gap equation. The E s are excitation energies, E = ξ 2 + 2, where ξ = ɛ µ and is the energy gap. The equation determines (T ). In the lectures we had the following relation (derived as an exercise) d 3 d 2 L 3 g() = (2π) 3 g() = Ω dξ N(ξ)g(). (5) 4π In the gap equation there is no angular dependence in g() and thus the angular integration is trivial. We also notice that the integrand is peaed 2

3 at the Fermi surface ξ = 0, so that we can approximate N(ɛ) N(0). However, the integrand decays lie /ξ, so the integral does not converge at large energies. Thus we need an additional energy a cutoff parameter. This way = gn(0) 2 ɛc ɛ c dξ ξ2 + tanh( ξ2 + 2 ), (6) 2 2 B T The cutoff has a physical meaning, because the attractive interaction should decay away from the Fermi energy, which is not captured in the contact interaction model. (The cutoff is on the order of the maximal phonon energy.) 4. In Ginzburg-Landau theory, the free energy densities in homogeneous superconducting (s) and normal (n) states in a constant external field H are g s = f 0 + α ψ 2 + β 2 ψ 4 (7) g n = f 0 + B2 B H. (8) Minimize these to find ψ and B in equilibrium. Based on these, find an expression for the critical field H c. How do α, β, and hence H c depend on temperature near the critical temperature? The equilibrium energy density in the bul of the superconducting state (far from N-S ir I-S interfaces so that B = 0) is g s = f 0 + α ψ 2 + β 2 ψ 4 = f 0 α2 2β, where the equilibrium value ψ 2 = α/β obtained by minimization of g was inserted. Similarly the equilibrium energy density in the normal state far from interfaces (so that ψ = 0) is g n = f 0 + B2 BH = f 0 2 µ 0H 2, where the minimizing value B = µ 0 H was used. The critical field H c is defined so that these two energy densities are equal 2 µ 0H 2 = α2 2β This gives H = H c = α / µ 0 β. The coefficient β is a positive constant (so that the energy density is bounded from below) and α = a(t T c ), where a > 0, because α must be negative at T < T c so that we can have ψ 2 = α/β > 0. Thus H c is linear in T. N-S interfaces are only stable at H = H c, because otherwise g s > g n or vice versa and the lower-energy phase taes over everywhere. 3

4 5. The behavior of a superconductor in small magnetic field is governed by the London equation B = λ 2 2 B. (9) a) How can this be derived from the Ginzburg-Landau theory and what is the expression for λ? b) What does this equation tell about the penetration of magnetic field into a superconductor? a) For the derivation of λ, consider case ψ =constant. Thus ψ = ψ 0 e iφ, where ψ 0 =constant. Inserting this into the 2nd GL equation: µ 0 B = 2qγψ 2 0( h φ qa) (0) Apply with on the left, and use B = B 2 B = 2 B as well as φ = 0 and B = A, to find which is easy to rearrange to µ 0 2 B = 2q 2 γψ 2 0B () 2 B = q 2 γψ 2 0B (2) Here q 2 γψ0 2 must have units of length 2, because the unit of 2 is length 2. So we identify the length scale as λ = /( q 2 γψ0 2 ). Here we could still insert ψ0 2 = α /β, which is the correct equilibrium value in a homogeneous state. b) λ is the length scale over which the magnetic field B varies (decays to zero) inside a superconductor. Here could be a figure and formulas to show the exponential solution. Formulas to support your memory df = SdT + V H db (3) G = F V H B (4) g = f 0 + α Ψ 2 + β 2 Ψ 4 + γ ( h i qa)ψ 2 + B2 B H (5) ) 2 ( h γ i qa Ψ + αψ + β Ψ 2 Ψ = 0, (6) B = q hγ (Ψ Ψ Ψ Ψ ) 2q 2 γ Ψ 2 A. µ 0 i (7) } {ǎ, ǎ = δ,, {ǎ, ǎ } = {ǎ, ǎ } = 0. (8) 4

5 ǎ = u ˇγ + v ˇγ ǎ = u ˇγ v ˇγ. (9) ˇγ = u ǎ v ǎ ˇγ = u ǎ + v ǎ. (20) ψ 0 = q (u q + v q ǎ q ǎ q ) vac, (2) u 2 q + v 2 q = (22) E = ρ ɛ 0, (23) E = B t, (24) B = 0, (25) B = E ɛ 0 µ 0 t + µ 0j. (26) E = ϕ A (27) t B = A, (28) 5

The Ginzburg-Landau Theory

The Ginzburg-Landau Theory A normal metal s electrical conductivity can be pictured with an electron gas with some scattering off phonons, the quanta of lattice vibrations Thermal energy is also carried