 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2010/paper-2/section-b/b9/b/solution

Ψ_{s} = \frac{1}{2 π} ∣0, 0 ⟩, E_{s} = 0, g_{s} = 1 .

For the lowest triplet states take

λ = \pm 1

and

m_{s} = - 1, 0, 1

Ψ_{t, λ, m_{s}} (ϕ_{1}, ϕ_{2}) = \frac{e ^{iλ ϕ_{2}} - e ^{iλ ϕ_{1}}}{2 π 2} ∣1, m_{s} ⟩, E_{t} = ϵ_{R}, g_{t} = 6 .

u_{0}

and

u_{λ}

verifies wavefunction normalization of the antisymmetric spatial factor. At this same energy there are also two excited singlet states, obtained by replacing its minus sign with a plus sign and the spin factor by

∣0, 0 ⟩

. Together they reproduce the eight states counted in part a.

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