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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 33B / b / Solution

Spin-one particles are bosons, so their total two-particle state must be symmetric. Write

∣ m, n ⟩ = ∣ m ⟩_{1} ∣ n ⟩_{2}

for

m, n \in {1, 0, - 1}

. A basis of symmetric spin states is

∣1, 1 ⟩, ∣0, 0 ⟩, ∣ - 1, - 1 ⟩, \frac{∣1 , 0 ⟩ + ∣0 , 1 ⟩}{2}, \frac{∣1 , - 1 ⟩ + ∣ - 1 , 1 ⟩}{2}, \frac{∣0 , - 1 ⟩ + ∣ - 1 , 0 ⟩}{2} .

Denote these six states by

χ_{1}^{+}, \dots, χ_{6}^{+}

. A basis of antisymmetric spin states is

χ_{1}^{-} = \frac{∣1 , 0 ⟩ - ∣0 , 1 ⟩}{2}, χ_{2}^{-} = \frac{∣1 , - 1 ⟩ - ∣ - 1 , 1 ⟩}{2}, χ_{3}^{-} = \frac{∣0 , - 1 ⟩ - ∣ - 1 , 0 ⟩}{2} .

This is the six-plus-three decomposition of two identical spin-one bosons.

In the ground orbital both particles occupy

n = 0

, so the spatial wavefunction

Φ_{00} (x_{1}, x_{2}) = ψ_{0} (x_{1}) ψ_{0} (x_{2})

is symmetric. The complete list of ground states is therefore

Φ_{00} χ_{r}^{+}, r = 1, \dots, 6, E = 2 E_{0} .

At the first excited energy, one particle occupies

n = 0

and the other

n = 1

. The normalized symmetric and antisymmetric spatial states are

Φ_{01}^{\pm} (x_{1}, x_{2}) = \frac{ψ _{0} ( x _{1} ) ψ _{1} ( x _{2} ) \pm ψ _{1} ( x _{1} ) ψ _{0} ( x _{2} )}{2} .

Overall bosonic symmetry permits symmetric spatial times symmetric spin and antisymmetric spatial times antisymmetric spin. Hence all first excited states are

Φ_{01}^{+} χ_{r}^{+} (r = 1, \dots, 6), Φ_{01}^{-} χ_{s}^{-} (s = 1, 2, 3), E = E_{0} + E_{1} .

The ground and first excited degeneracies are therefore six and nine.

Solved by gpt-5.6-sol high.

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