The Nature of the Cooper Pair - University of Liverpool

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Then suppose that two electrons at the Fermi level E F start pairing up. When they pair up, their energy falls by ∆, to 2E F − ∆. Next, imagine that more electrons pair up. As more and more electrons pair up, the Fermi sea gets depleted, and the ”sea” level falls. (This is not to be confused with the Fermi energy, which is still defined as E F .) The electron pair would always have the same energy of 2E F − ∆, because they have the same wavefunction Ψ. This goes on until the sea level has fallen by ∆/2, to E F − ∆/2. A pair of electrons at this level would have already have a total energy of 2E F − ∆. There is no advantage of forming a bound pair, which also have the same energy. So the electrons stop pairing up. Superconductivity 27
The number of Cooper pairs is therefore equal to half of the number of electrons with ∆/2 of the Fermi energy. This is given by 1 2 × 2g(E F )∆/2. Recall from the notes on electrons in metal that g(EF ) = 3N . 4EF So the number of Cooper pairs is N ′ = g(E F ) ∆ 2 3N∆ = . 8EF We can estimate the distance between Cooper pairs by dividing the volume of the metal V by N ′ to get the volume occupied by one Cooper pair. Taking this to be a cube, the side of the cube would give an estimate of the distance D between adjacent pairs: D = Superconductivity 28 � V N ′ �1/3
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: The bound pair of electrons is the
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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