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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-1/section-b/b7/c/solution

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Past exam of the physics A course of the University of Cambridge / 2022 / Paper 1 / Section B / B7 / c / Solution by

Codex  0 2026-10-18

Put

ϵ = ℏ^{2} / (2 m R^{2})

. The two-particle product energy eigenstates are

u_{n_{1}} (ϕ_{1}) u_{n_{2}} (ϕ_{2}), E = ϵ (n_{1}^{2} + n_{2}^{2}) .

Because the particles are distinguishable,

(n_{1}, n_{2})

and

(n_{2}, n_{1})

are distinct when

n_{1} \neq = n_{2}

. Also, the signs of nonzero integers distinguish opposite angular momenta. The quantum degeneracy is therefore the number of ordered integer pairs at each sum of squares.

The first distinct energy levels are

0, ϵ, 2 ϵ, 4 ϵ, 5 ϵ, \dots

. This ordering matters:

4 ϵ

is the third excited level, while

5 ϵ

is the fourth excited level.

At $E = 0$ , only $(0, 0)$ occurs: $g_{0} = 1$ .
At the first excited level, $E = ϵ$ , the pairs are $(\pm 1, 0)$ and $(0, \pm 1)$ : $g_{1} = 4$ .
At the fourth excited level, $E = 5 ϵ$ , the pairs are $(\pm 1, \pm 2)$ and $(\pm 2, \pm 1)$ , with independent signs: $g_{4} = 8$ .
At $E = 25 ϵ$ , the pairs are $(\pm 5, 0), (0, \pm 5)$ , contributing four, and $(\pm 3, \pm 4), (\pm 4, \pm 3)$ , contributing eight: $g_{25} = 12$ .

These counts concern energy eigenspaces; arbitrary quantum superpositions within each one do not create additional independent states.

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