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Can the operator Exp[-I*Pi*L_x/h] be faced as parity?

  1. Nov 7, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem originally asks to evaluate ##exp(\frac{-i\pi L_x}{h})## in a ket |l,m>. So I am wondering if I can treat the operator as a parity operator or if I really have to expand that exponential, maybe in function of ##L_+## and ##L_-##.

    2. The attempt at a solution
    If ##exp(\frac{-i\pi L_x}{h})## operates in |y,z> it returns |-y,-z>. Then, if it operates in |l,m> it would just change the ket's signal.
     
  2. jcsd
  3. Nov 8, 2015 #2

    blue_leaf77

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    Well, proving it for general ##l## might be a tedious task. But if you are only interested in the final answer, think of the fact that ##\exp \left( -\frac{i\pi L_x}{\hbar} \right)## is a rotation operator around ##x## axis through an angle of ##\pi## radian. Now ##|l,m \rangle ## is a state which points in the ##z## direction because it is an eigenstate of ##L_z##. Applying ##\exp \left(- \frac{i\pi L_x}{\hbar} \right)## to this state is the same as rotating this state around ##x## axis through an angle of ##\pi## radian.
     
  4. Nov 8, 2015 #3
    By expanding ##\exp \left( - \frac{i \pi L_x}{\hbar}\right) |l,m\rangle## into a Cartesian basis like ##\int \exp \left( - \frac{i \pi L_x}{\hbar}\right)| x,y,x \rangle \langle x,y,z |l,m\rangle \mathrm{dV}## it would be possible to use the operator in the xyz ket so that it could be simplified into this: ##\int | x,y,x \rangle \langle x,-y,-z |l,m\rangle \mathrm{dV}##. With that said, is there a way to write this answer with the ##|l,m \rangle## ket alone?
     
  5. Nov 8, 2015 #4

    blue_leaf77

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    If you insist that way, you can start from the fact that ##\langle x,y,z | l,m\rangle = Y_{lm}(\theta,\phi)## with ##Y_{lm}## being spherical harmonics defined as
    $$
    Y_{lm}(\theta,\phi) = \sqrt{\frac{(l-m)!}{(l+m)!}} P_l^m(\cos \theta) e^{i m\phi}
    $$
    Now first find out how it will affect ##Y_{lm}(\theta,\phi)## when you make the changes ##y \rightarrow -y## and ##z \rightarrow -z##.
     
  6. Nov 8, 2015 #5
    I'll try that, thank you
     
  7. Nov 8, 2015 #6
    Just to verify the answer, should it be ##(-1)^{(l+m)} |l,m \rangle##?
     
  8. Nov 8, 2015 #7

    blue_leaf77

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    Nope, it's not. It seems that to arrive at that equation you changed ##\theta## only, keeping ##\phi## unchanged.
     
  9. Nov 8, 2015 #8
    How about this? If it isn't, I'll need help.

    ##(-1)^l \frac{(l-m)!}{(l+m)!}|l,-m \rangle##
     
  10. Nov 8, 2015 #9

    blue_leaf77

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    Almost there.
    So, following the aforementioned changes, you should get (I hope)
    $$
    \cos \theta \rightarrow -\cos \theta\\
    \phi \rightarrow 2\pi -\phi
    $$
    Applying these changes to the ##Y_{lm}(\theta,\phi)## yields
    $$
    Y_{lm}(\theta,\phi) = \sqrt{\frac{(l-m)!}{(l+m)!}} P_l^m(\cos \theta) e^{i m\phi} \rightarrow \sqrt{\frac{(l-m)!}{(l+m)!}} P_l^m(-\cos \theta) e^{-i m\phi}
    $$
    Using the parity property of the associated Legendre polynomial ##P_l^m(-x) = (-1)^{l+m} P_l^m(x)## and ##P_l^{-m}(x) = (-1)^m \frac{(l-m)!}{(l+m)!} P_l^m(x)##, the right hand side of the above arrow sign can be manipulated to give
    $$
    (-1)^l \frac{(l+m)!}{(l-m)!} \sqrt{\frac{(l-m)!}{(l+m)!}} P_l^{-m}(\cos \theta) e^{-i m\phi}
    $$
    Now try to manipulate this further such that it becomes proportional to ##Y_{l,-m}(\theta,\phi)##.
     
  11. Nov 8, 2015 #10
    But that is what I did
     
  12. Nov 8, 2015 #11

    blue_leaf77

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    No, it's different from yours. Look what happen when you simplify those fractions with factorial signs.
     
    Last edited: Nov 8, 2015
  13. Nov 8, 2015 #12
    So just ##(-1)^l |l,-m \rangle ## then
     
  14. Nov 8, 2015 #13

    blue_leaf77

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    Yes, it should be that.
     
  15. Nov 8, 2015 #14
    Wow, that was a hell of a evaluation. Thank you for helping me out!
     
  16. Nov 8, 2015 #15

    blue_leaf77

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    You can further check if this is true by applying the second ##\exp \left( \frac{i \pi L_x}{\hbar} \right)## to the last result. You should easily see that it will end up with ##(-1)^{2l} |l,m \rangle##, due to the fact that a double ##\exp \left( \frac{i \pi L_x}{\hbar} \right)##'s is equivalent to a full rotation about x axis. However, for ##l## equal to an half-odd integer, a full rotation results in a negative sign, but this is indeed a characteristic of half-odd integer angular momenta.
     
  17. Nov 8, 2015 #16
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