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## Homework Statement

The two-particle wavefunctions are given:

Where |V> is the spin down state and |Λ> is the spin down state

½(Ψ

_{3p}(r

_{1})Ψ

_{1s}(r

_{2}) + Ψ

_{1s}(r

_{1})Ψ

_{3p}(r

_{2}))(|V>|Λ> - |V>|Λ>)

½(Ψ

_{3p}(r

_{1})Ψ

_{1s}(r

_{2}) - Ψ

_{1s}(r

_{1})Ψ

_{3p}(r

_{2}))(|V>|Λ> + |V>|Λ>)

1/√2 (Ψ

_{3p}(r

_{1})Ψ

_{1s}(r

_{2}) - Ψ

_{1s}(r

_{1})Ψ

_{3p}(r

_{2}))(|Λ>|Λ>)

1/√2 (Ψ

_{3p}(r

_{1})Ψ

_{1s}(r

_{2}) - Ψ

_{1s}(r

_{1})Ψ

_{3p}(r

_{2}))((|V>|V>)

For each of the four two-particle states written down in part (iii), state by how many dipole allowed transitions the state can decay into a lower energy state. Give your reasoning.

## Homework Equations

## The Attempt at a Solution

Here is the answer from the solutions given.

The key is that only one electron can change state and this is the excited electron which can go from 3p to the 2s or the 1s [1]. The 3p to 1s transition is forbidden for the three S=1 states as spin can’t change and the ground state cannot have S=1 due to particle exchange symmetry[1]. Thus the first state can decay by two transitions [1] and the other three by one transition [1].

My question is are these solutions wrong?

My first thought was that it would take the first s=0 singlet state 1 transitions to decay as it doesn't have to change orbital angular momentum. The second three triplet S=1 states would surely take 2 transitions to decay as they have to lose orbital angular momentum.

Thanks, John