The Nature of the Cooper Pair - University of Liverpool

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If V (r) is constant, the solution is of the form ψ(r) = sin(kr). However, V (r) is not constant. It becomes zero beyond r = l. We can try a sum of sine functions: Ψ(r) = � ak sin(kr), where a k are unknown constants that we must find. k We now come the the important part. The need for this sum means that energy states with many wavevectors k must be used to make up the wavefunction. However, we are not allowed to use k values below the Fermi energy, because those states are fully occupied. Therefore, we must impose the condition that k must be more than k F , the wavevector at the Fermi energy. The detailed steps to solve the Schrodinger equation are given in the appendix. Superconductivity 19
Here, we just state the solution to the Schrodinger equation. The wavefunction is Ψ = �ω �D 0 aε sin kr = �ω �D 0 1 ε + ∆ sin kr where k is in steps of π/R, R being roughly the radius of a sphere similar in size to the metal. (R corresponds to the length L of the cube that is commonly used for electrons in metal.) ε is the energy above the Fermi energy. k is related to ε by − EF . 2me The energy of the electron pair is ε = �2 k 2 E = E F − ∆. ∆ is called the binding energy, and is given by ∆ = �ωD exp(− 2πkF ). melV0 Superconductivity 20
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: The Schrodinger equation is then of
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 and 46: However, ε is 2�ω D, then the u
Page 47 and 48: We have found both aε and ∆. Thi

The Nature of the Cooper Pair - University of Liverpool

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