Is this transition allowed or forbidden?

[Poirot]

Homework Statement

Under the multi-electron electric dipole (E1) selections rules, state whether this transition is allowed or forbidden and state which selection rule(s) has been violated.
for He: 1s1p ¹P₁-> 1s^{2 1}S₁

Homework Equations

For E1:
\begin{eqnarray*} Parity \ changes: (-1)^l\\
\Delta L = 0, \pm 1 (0\rightarrow{}0 \ not \ allowed)\\
\Delta J = 0, \pm 1 (0\rightarrow{}0 \ not \ allowed)\\
\Delta S = 0\end{eqnarray*}

The Attempt at a Solution

The initial state has: L=1, 2S+1=1 so S=0, J=1
Final state has: L=0, 2S+1=1 so S=0, J=0

so in theory this is allowed because ΔL=-1, ΔS=0 and ΔJ=-1, but I'm not sure how to implement the parity condition? Is it that because the electron is initially in 1p, l=1 and then ends in 1s^2 so l=0 so this transition is odd and parity is fine?
I have a feeling this is physically impossible because I don't think 1p exists? I'm quite lost here so any help would be great!

 

[eys_physics]

The parity of the state is ##(-1)^L##, i.e. negative for the initial state and positive for the final state.
You can also this in terms of the individual electron configurations
For the initial state: ##(-1)^0\times (-1)^1=-1##, in a similar way you should be able to get that the final state has positive parity.
I don't see why the 1p state cannot exist? What is your arguments for this?

 

[TSny]

The parity of the state is ##(-1)^L##, i.e. negative for the initial state and positive for the final state.
You can also this in terms of the individual electron configurations
For the initial state: ##(-1)^0\times (-1)^1=-1##, in a similar way you should be able to get that the final state has positive parity.
I don't see why the 1p state cannot exist? What is your arguments for this?

1p would imply a p-state for n = 1, which doesn't exist. I think there must have been a misprint in the original statement of the problem. Maybe the configuration for the initial state was meant to be 1s2p.

 

[Poirot]

Do you do (-1)^l for each electron in the shell? or just the l for that shell, so for example if you had:
1s² 2s² 2p 3s
would the parity calculation be: (-1)⁰(-1)⁰(-1)¹(-1)⁰ = -1

or is it per electron?

And as for the 1s1p, I think it must be a typo.

Thanks for your replies!

 

[eys_physics]

Yes, probably it should be 2p. First I was not aware of the restrictions on the n quantum since I am not an expert in atomic physics.

Yes, you are correct regarding the parity. Total parity of a state is the product of the parities of the individual states.
This is easy to understand from the definition of parity. If the parity is -1 you have $$\phi(-\mathbf{r})=-\phi(-\mathbf{r})$$ and if parity is 1, $$\phi(\mathbf{r})=\phi(-\mathbf{r})$$ for the wave function.

 

[TSny]

1s² 2s² 2p 3s
would the parity calculation be: (-1)⁰(-1)⁰(-1)¹(-1)⁰ = -1

or is it per electron?

It's per electron. Thus 1s²2s²2p² is even parity.
https://en.wikipedia.org/wiki/Term_symbol#Term_symbol_parity