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Relative intensities of Zeeman components

  1. Jun 19, 2011 #1
    Hi-

    I'm trying to calculate the effect of a magnetic field on some spectral lines. I know that each energy level will split into 2J+1 sub levels (each denoted by Mj). I can calculate the energy level splitting and using the selection rules (dJ = 0, +/-1, dMj = 0, +/-1, dL = 0, +/-1), I can figure out which transitions are allowed electric dipoles. I know that the pi components (dMj = 0) are not emitted parallel to the magnetic field, while the sigma components (dMj = +/- 1) can be seen in both parallel and transverse directions.

    So, I think I can calculate the zeeman splitting alright, but the problem arises when I try to calculate the relative intensities of each of the zeeman components. I found some relevant equations in Condon and Shortley's Theory of Atomic Spectra (Chapter 16, section 4), but I don't think I'm interpreting it right. It's been a while since I had a quantum class, so it's possible I'm forgetting something stupid.

    "For the Zeeman pattern of any line in which there is no change in J, the strengths of the components in transverse observation are proportional to:
    | < alpha J M| P | alpha' J M> |^2 = |<alpha J| P | alpha' J>|^2 M^2 (pi)
    1/4*|<alpha J M | P | alpha' J M -/+ 1>|^2 = |<alpha J| P |alpha' J>|^2 1/4 (J +/- M)(J -/+ M+1) (sigma)

    There are different (but very similar) formulas for delta J = +1, and -1, but I'm not going to type them out here (I can if others would find it useful).

    My problem is when delta Mj = 0, and Mj = 0 as well. Does this mean that since Mj = 0, the pi component is not allowed? To put it another way, when the proportionality factor = zero, is that component forbidden? It doesn't seem to make sense, as you'd be missing the central pi component completely from both the transverse and longitudinal spectra.

    I know there's a selection rule against delta J = 0 when J =0, but i didn't think that prohibited Mj from being zero when delta J = 0.

    Where am I going wrong?

    Thanks,
    C.
     
  2. jcsd
  3. Sep 14, 2011 #2
    Here's an update for anyone who's looking for an answer... It turns out that if the relative intensity = 0, then that Zeeman component does not appear. In effect, it's not allowed. Hope this helps.
     
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