 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-1/section-b/b8/b/solution

cos^{2} ϕ = 1/2 + (e^{2 i ϕ} + e^{- 2 i ϕ}) /4

. With the normalized

u_{m}

from part (a), this gives

∣Ψ (0)⟩ = \frac{2}{3} ∣ m = 0 ⟩ + \frac{1}{6} ∣ m = 2 ⟩ + \frac{1}{6} ∣ m = - 2 ⟩ .

The coefficient squares sum to one, also verifying the stated normalization. The Born rule therefore gives angular momentum outcomes

L_{z} = 0, + 2ℏ, - 2ℏ with probabilities \frac{2}{3}, \frac{1}{6}, \frac{1}{6} .

For an energy measurement the two distinct

m = \pm 2

states belong to the same energy eigenspace. Sum their probabilities, rather than adding their amplitudes, to obtain

E = 0 with probability \frac{2}{3}, E = \frac{2 ℏ ^{2}}{I} with probability \frac{1}{3} .

Their relative probabilities are

4 : 1 : 1

for the angular momenta and

2 : 1

for the energies. In particular, zero mean angular momentum does not imply zero rotational energy.

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