Two particle wavefunctions of helium

[Jdraper]

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(r₁)Ψ_1s(r₂) + Ψ_1s(r₁)Ψ_3p(r₂))(|V>|Λ> - |V>|Λ>)

½(Ψ_3p(r₁)Ψ_1s(r₂) - Ψ_1s(r₁)Ψ_3p(r₂))(|V>|Λ> + |V>|Λ>)

1/√2 (Ψ_3p(r₁)Ψ_1s(r₂) - Ψ_1s(r₁)Ψ_3p(r₂))(|Λ>|Λ>)

1/√2 (Ψ_3p(r₁)Ψ_1s(r₂) - Ψ_1s(r₁)Ψ_3p(r₂))((|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

 

[DrClaude]

½(Ψ_3p(r₁)Ψ_1s(r₂) + Ψ_1s(r₁)Ψ_3p(r₂))(|V>|Λ> - |V>|Λ>)

½(Ψ_3p(r₁)Ψ_1s(r₂) - Ψ_1s(r₁)Ψ_3p(r₂))(|V>|Λ> + |V>|Λ>)

I guess you mean (|V>|Λ> - |Λ>|V>) and (|V>|Λ> + |Λ>|V>) for the spin states.

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?

That sounds right. There is a mistake in the answer, see post #4 below.

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.

What is meant by "two transitions" and "one transition" in the solution is the number of possible transitions for the original state to a lower-energy one. A better way to say it would be to ask "How many different states can each these initial states decay into?"

Why do you think that there is more orbital angular momentum in the triplet case? Do you know spectroscopic notation? If yes, write the term symbols for these states and the lower energy ones and apply selectrion rules.

 

[Jdraper]

Sorry, this makes total sense now, transitioning to the 1s state would mean a full 'shell' and thus S=0 has to apply. Therefore this can only occur with the S=0 singlet state so it doesn't violate this rule. The S=1 triplet state however cannot do this as it would violate this rule. Both the singlet and triplet states can transition to the lower 2s because of the shell isn't full and thus S=0 does not apply. Is this thinking correct?

I was confusing myself when i spoke about orbital angular momentum, i meant to talk about spin and the conditions that apply to a full shell.

Thanks for your help. John

 

[DrClaude]

Sorry, this makes total sense now, transitioning to the 1s state would mean a full 'shell' and thus S=0 has to apply. Therefore this can only occur with the S=0 singlet state so it doesn't violate this rule. The S=1 triplet state however cannot do this as it would violate this rule. Both the singlet and triplet states can transition to the lower 2s because of the shell isn't full and thus S=0 does not apply. Is this thinking correct?

What you say is correct, but it is better to consider are the possible term symbols associated with the electronic configurations 1s² and 1s2s. In the former case, only a singlet state is possible, while the other admits a single and a triplet state. Since $\Delta S=0$, there is only one possibility for the triplet state.

I just now realize that there is an error in the solution. The answer doesn't mention the decay to 1s3s, which is also allowed.

 

[Jdraper]

I'm trying to write these term symbols out as you suggested.

1s² has a term symbol ¹S₀.

However i thought term symbols weren't applicable to electronic configurations with two empty shells? (Hunds first rule)

I'll try anway

³L₁ (Unsure what L should be.

I'm guesing you were trying to highlight is the fact that in the 1s² state S=0 and because of the transition rules (ΔS=0) only the singlet state can make this transition, as it had S=0, and therefore ΔS=0 is not violated.

1s2s however can have a spin of either S=0 or S=1, because the electrons are in two shells and thus aren't bound by the pauli exclusion principle. So this triplet can make this transition.

This was useful and did help my understanding, Thanks.

 

[DrClaude]

However i thought term symbols weren't applicable to electronic configurations with two empty shells? (Hunds first rule)

Hund's rules allow to find the ground state for equivalent electrons. Terms symbols can be used as long as Russell-Saunders coupling applies.

³L₁ (Unsure what L should be.

How is L defined?

 

[Jdraper]

Ok, didn't realize that.

Have done a bit of work relevant to that in the past hour;

L is the sum of the angular momentum of the electrons, L_i. Since the 2 electrons are both in S orbitals (S : L_i=0) then the value of L=0.

Thanks for the help :)

 

[Jdraper]

Sorry forgot to mention so therefore

³S₁