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Homework Help: Is this transition allowed or forbidden?

  1. Dec 8, 2016 #1
    1. The problem statement, all variables and given/known data
    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-> 1s2 1S1
    2. Relevant 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*}

    3. 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!
  2. jcsd
  3. Dec 8, 2016 #2
    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?
  4. Dec 8, 2016 #3


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    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.
  5. Dec 9, 2016 #4
    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!
  6. Dec 9, 2016 #5
    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.
  7. Dec 9, 2016 #6


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