 Solution

ID: past-exam-of-the-options-ii-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/j2/a/solution

r_{0} = q E / (m ω^{2})

and

y = r - r_{0}

\frac{1}{2} m ω^{2} ∣ r ∣^{2} - q E \cdot r = \frac{1}{2} m ω^{2} ∣ y ∣^{2} - \frac{q ^{2} ∣ E ∣ ^{2}}{2 m ω ^{2}} .

Translation leaves the Laplacian unchanged, and the shifted three-dimensional quantum harmonic oscillator separates into three one-dimensional oscillators. Their energies add, giving

E_{N} = (N + \frac{3}{2}) ℏ ω - \frac{q ^{2} ∣ E ∣ ^{2}}{2 m ω ^{2}}, N = n_{1} + n_{2} + n_{3} .

The energy-level degeneracy is the dimension of the energy eigenspace, or number of linearly independent states with that energy. Counting nonnegative integer triples of total

N

d_{N} = (2 N + 2) = \frac{( N + 1 ) ( N + 2 )}{2} .

The constant electric field displaces the oscillator and shifts all levels by the same amount, preserving their degeneracies.

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