The Nature of the Cooper Pair - University of Liverpool

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The right hand side of the Schrodinger equation aε(ε + ∆) = −B is a constant. Let this be C. So or � �ωD 0 aε(ε + ∆) = C, aε = C ε + ∆ . a ε ′dε ′ . Substituting this back into the Schrodinger equation, we get Integrating, we find 1 = −B � �ωD 0 dε ′ ε ′ + ∆ . ∆ = �ωD exp(−1/B) = �ωD exp(− 2πkF ). melV0 Superconductivity 45
We have found both aε and ∆. This means that we have found both the wavefunction Ψ and the energy E. In other words, we have solved the Schrodinger equation! The wavefunction is Ψ = �ω �D where k is related to ε by The energy is 0 aε sin kr = ε = �2 k 2 2me �ω �D 0 − E F . 1 ε + ∆ sin kr E = EF − ∆ = E − �ωD exp(− 2πkF ). melV0 Superconductivity 46
Page 1 and 2: Statistical and Low Temperature Phy
Page 3 and 4: When an electron passes in between
Page 5 and 6: Using the small change approximatio
Page 7 and 8: The attractive force is the Coulomb
Page 9 and 10: Recall that the Debye frequency ω
Page 11 and 12: The displacement then gives us the
Page 13 and 14: We shall approximately represent th
Page 15 and 16: We have seen that two electrons mus
Page 17 and 18: ... recall the quantisation conditi
Page 19 and 20: The Schrodinger equation is then of
Page 21 and 22: Here, we just state the solution to
Page 23 and 24: They will start interfering destruc
Page 25 and 26: Using ∆k.ρ = π we can solve for
Page 27 and 28: The bound pair of electrons is the
Page 29 and 30: The number of Cooper pairs is there
Page 31 and 32: So, as in superfluid helium-4, we m
Page 33 and 34: APPENDIX: Solving for the Cooper pa
Page 35 and 36: The attractive potential is very sm
Page 37 and 38: Therefore and sin(kR) = 0, k = nπ
Page 39 and 40: The product of the waves sin(kr) an
Page 41 and 42: Return to the Schrodinger’s equat
Page 43 and 44: So in energy variable, the density
Page 45: However, ε is 2�ω D, then the u

The Nature of the Cooper Pair - University of Liverpool

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