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Two selection rules I can't figure out

  1. Apr 30, 2010 #1
    I am having trouble showing:

    [tex]<4 3 0 | p_z^2 | 4 0 0 > = 0[/tex] and
    [tex]<4 2 1 | x p_y | 3 1 0 > = 0 [/tex]

    I know the usual selection rules for parity, spherical tensors (Wigner-Eckart for x,y,z) but I can't convert these two to anything I know.
    You could also help by telling me how to convert these to combinations of x,y,z or py,pz,px as well..

    This is NOT homework, I am a PhD student and I am trying to self-educate so please don't yell.

    Thanks,
     
  2. jcsd
  3. May 1, 2010 #2
    what is 430 etc referring to?
     
  4. May 1, 2010 #3
    n l m for hydrogen respectively.
     
  5. May 1, 2010 #4
    can you write x^2 etc in terms of spherical coordinates etc? that is basically what u do
     
  6. May 1, 2010 #5

    SpectraCat

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    To expand a bit on angsar's answer, do you know how to use the metric tensor to switch between coordinate frames? That is the most general way to express cartesian coordinates and their derivatives in other arbitrary coordinate systems. Of course spherical coords are common enough that you can just look up the transformations for d/dy, etc. that you need for the matrix elements you gave. However, it is well worth learning how to work with the metric tensor, since you will likely encounter other coordinate systems in the future, where you need to derive things like the gradient operator, Laplacian and volume element for yourself.

    EDIT: Just to be sure, the "pz" and "py" are Cartesian components of momentum, right? Not the short-hand chemist's notation for the l=1, ml=0, and the linear combination of l=1, ml=+1 and l=1, ml=-1 orbitals?
     
    Last edited: May 1, 2010
  7. May 1, 2010 #6
    I think I don't know how to do it in the general way.. pz and py are Cartesian momentums..
    So I think I am trying to write pz^2 xp_y etc.. in terms of spherical tensors, here? But I have no idea how I can do that, as you said. The whole spherical tensor business is a little unclear to me. Can you point out a good introduction?
     
  8. May 1, 2010 #7

    SpectraCat

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    Sure, there is a pretty clear and simple intro http://www.indiana.edu/%7Essiweb/C561/PDFfiles/HAtom2008.pdf" [Broken]. It is a fairly complete development of the H-atom, but the relevant info for you is on pp.212-214. Note that this presentation of the metric tensor is pretty cursory, but I guess that you should be able to extract what you need from the treatment of the gradient operator. Let me know if you have questions/confusion and I will help as best I can.

    Just as an FYI, note that you can also derive expressions for cartesian derivatives (i.e. d/dy and d/dz) in spherical polar coordinates "the hard way" just starting from z=r cos (theta), etc. and using basic calculus. You can see that method in the notes http://www.indiana.edu/%7Essiweb/C561/PDFfiles/Spherical-Harmonics.pdf" [Broken], on pp. 199-201.

    These are lecture notes for a quantum chemistry class taught by a colleague/friend/collaborator of mine who is a professor at Indiana University. The entire set of notes is http://www.indiana.edu/~ssiweb/C561/C561.html" [Broken] if you are interested.

    HTH!
     
    Last edited by a moderator: May 4, 2017
  9. May 2, 2010 #8
    thank you very much spectracat these are indeed very useful...!
     
  10. May 2, 2010 #9

    olgranpappy

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    Hi, maybe I am missing something, but it looks to me like you can just use parity... except you said you already looked at parity?...

    For example, in the first matrix element p_z^2 is invariant under parity, and so is 400, but 430 changes sign. So parity alone shows it is zero.

    Furthermore, p_z^2 can change ell by at most 2, so 3 is too far from 0. Thus again this means the first matrix element is zero.

    Parity seems like it takes care of the other matrix element as well.

    Am I right, or did I miss something? Cheers.
     
  11. May 3, 2010 #10

    SpectraCat

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    Interesting idea, but I don't think I have ever considered the parities of the Cartesian components of the momentum operator. Do you simply say that taking the derivative twice with respect to z preserves the "evenness" or "oddness" of the function, and therefore that pz2 must have even parity? In that case, it is a very nice way to quickly find out if matrix elements of the Cartesian momentum components are zero by symmetry.

    Also, this would mean that the momentum operator in general has odd parity, which is interesting, because the classical velocity also has odd parity. Is it generally true that quantum operators always have the same parities as their classical counterparts? It seems to me that I should know this, but I don't think I ever even asked the question before!
     
  12. May 3, 2010 #11

    DrDu

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    xp_y can be split into a symmetric part, which is related to the quadrupole moment (up to constants, it is [H,xy] ) and an antisymmetric part, which is the z-component of angular momentum L_z. L_z does not make transitions and xy is of even parity.
     
  13. May 3, 2010 #12

    olgranpappy

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    Yes, the momentum operator has odd parity. Cheers.
     
  14. May 3, 2010 #13
    olgranpappy, you are quite right.

    I am sorry I didn't specify this, you can indeed you use parity. I am trying to do an exercise where you calculate these elements by parity, total angular momentum, and angular momentum along +z direction.

    So I know why they are zero because of parity. But for J and Jz, I somehow need to relate the operators to "spherical tensors" I know, to be able to use Wigner-Eckart theorem and the corresponding selection rules.

    I wasn't specific at all, and although I have been checking the references, I still don't see how pz^2 for instance can be related to "z"... Maybe Heisenberg relation, etc.. but I couldn't do it.
     
  15. May 3, 2010 #14
    pz is d/dz, do that one in spherical coordinates
     
  16. May 3, 2010 #15

    olgranpappy

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    Ah, ok. Well, I can help you with the formalism a bit, I think.

    Consider the p_z^2 case and write p_z^2 as
    [tex]
    p^{(1)}_0p^{(1)}_0\;,
    [/tex]
    in a notation where is is apparent that we have a direct product of the zeroth (spherical component) of two vector (rank 1 tensor) operators.

    The point of Clebsch-Gordan coefficients is that they relate various tensors. We can relate this specific product of two vector operators to the sum of three operators as:

    [tex]
    p^{(1)}_0p^{(1)}_0=\sum_{L=0}^2\sum_M \langle L M|1010\rangle X^{(L)}_M=\sum_{L=0}^2\langle L 0|1010\rangle X^{(L)}_0\;,
    [/tex]
    where X^{(L)}_M is the Mth component of a rank L spherical tensor.

    The sum over L goes from 0 to 2 because of the usual angular momentum addition rules 1x1=0+1+2 (in terms of matrix size this read: 3x3=1+3+5).

    As I said in my previous post, the max L is 2. Thus since the initial state ell is zero the final state ell can at most be 2, but it is 3, so the matrix element must be zero.

    Cheers.
     
  17. May 3, 2010 #16

    SpectraCat

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    No, this is not what we are talking about here ... pz is NOT the p0 orbital, it is the z-component of the momentum (that is why I posted all that stuff about momentum earlier). As far as I know, there is no relation between the p0 orbital and [tex]p_z=\frac{\hbar}{i}\frac{\partial}{\partial z}[/tex].
     
  18. May 3, 2010 #17

    olgranpappy

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    What p_0 orbital? I think you may be confused.

    I'm not talking about any orbital. I'm talking about momentum.

    In particular, I'm talking about the z-component of momentum.

    I have simply used a different notation for the z-component of momentum which indicates, as a stated in my post, that it is the 0th component of a rank 1 tensor...
     
  19. May 3, 2010 #18

    SpectraCat

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    Ok .. I think you are right and I am confused, because that is the way I am used to thinking about *angular* momentum, not linear momentum. Of course, since momentum is a 3-vector, there is no reason you couldn't call it a rank 1 tensor as well. I just don't see how the z-component is equivalent to the spherically symmetric component. I am not so used to the tensor formalism, so that is probably what is confusing me. Can you provide a quick explanation, or perhaps a reference where I can study this some more?
     
  20. May 3, 2010 #19

    OK, this was exactly what I was looking for, thank you !
     
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