 Solution

ID: past-exam-of-the-condensed-matter-physics-major-option-course-of-the-university-of-oxford/2016/8/b/solution

Let

m, q, n_{s}

denote pair mass, charge and number density, respectively. A uniform-phase condensate in the stated gauge has zero canonical momentum

m v_{s} + q A = 0

. Therefore

j_{s} = n_{s} q v_{s} = - \frac{n _{s} q ^{2}}{m} A, \nabla \times j_{s} = - \frac{n _{s} q ^{2}}{m} B .

For static fields,

\nabla \times B = μ_{0} j_{s}

and

\nabla \cdot B = 0

. Taking a curl and using the vector identity for a double curl gives

- \nabla^{2} B = - \frac{μ _{0} n _{s} q ^{2}}{m} B, \nabla^{2} B = \frac{B}{λ ^{2}}, λ = \frac{m}{μ _{0} n _{s} q ^{2}} .

For electronic pairs,

m ≃ 2 m_{e}

and

q = - 2 e

, while

n_{s}

is the pair density, not twice that density. In a general gauge the canonical momentum is

ℏ\nabla θ

; taking the curl away from vortices gives the same London equation.

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