 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2019/3/d/solution

For independent

χ, χ^{*}

, the derivatives needed in the

χ^{*}

\frac{\partial L _{2}}{\partial χ ^{*}} = \frac{i ℏ}{2} \overset{χ}{˙} - μ (χ + χ^{*}), \frac{\partial L _{2}}{\partial χ ˙ ^{*}} = - \frac{i ℏ}{2} χ, \frac{\partial L _{2}}{\partial ( \nabla χ ^{*} )} = - \frac{ℏ ^{2}}{2 M} \nabla χ .

Consequently the two conjugate field equations are

i ℏ \overset{χ}{˙} = - \frac{ℏ ^{2}}{2 M} \nabla^{2} χ + μ (χ + χ^{*}), - i ℏ \overset{χ}{˙}^{*} = - \frac{ℏ ^{2}}{2 M} \nabla^{2} χ^{*} + μ (χ + χ^{*}) .

Write

χ = χ_{1} + i χ_{2}

with real components, and put

A = ℏ^{2} / (2 M)

. Separating real and imaginary parts gives

- ℏ \overset{χ}{˙}_{2} = (- A \nabla^{2} + 2 μ) χ_{1}, ℏ \overset{χ}{˙}_{1} = - A \nabla^{2} χ_{2} .

For each real field take the real part of a complex plane wave amplitude,

χ_{j} = Re [q_{j} e^{i (k \cdot r - ω t)}]

. With

ϵ_{k} = ℏ^{2} ∣ k ∣^{2} / (2 M)

the two equations become

i ℏ ω q_{2} = (ϵ_{k} + 2 μ) q_{1}, - i ℏ ω q_{1} = ϵ_{k} q_{2} .

A nonzero normal mode requires their determinant to vanish, so

ℏ^{2} ω^{2} = ϵ_{k} (ϵ_{k} + 2 μ) .

ω (k) = \frac{1}{ℏ} ϵ_{k} (ϵ_{k} + 2 μ) = \frac{∣ k ∣ ^{2}}{2 M} (\frac{ℏ ^{2} ∣ k ∣ ^{2}}{2 M} + 2 μ) .

There is also the negative-frequency conjugate solution, as required for real component fields. Since

μ > 0

ω^{2} \geq 0

, consistent with the stability of the chosen uniform ground state.

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