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Atomic Orbitals ?

  1. Jan 30, 2008 #1
    for n=2 & l=1 we have P orbital for which magnetic quantum no. can take values -1,0,1 , now in p orbital we have 2Px, 2Py &2Pz each corresponding to vale of m like 2Px for m=-1 and so on , is this correct or 2Px can be represented by 0 0r 1 also .
     
  2. jcsd
  3. Jan 31, 2008 #2
    According to some general situations in quantum mechanics I would say (not so sure about) that Px (Py and Pz as well) is likely to correspond to some linear combination of the spatial functions which are eigen functions of the Lz operator (m values are eigen values of this operator).

    Best wishes

    DaTario
     
  4. Jan 31, 2008 #3
    From http://www.3rd1000.com/chem301/chem301b.htm :


    Orbitals with subshell quantum number l = 1 are called p orbitals. Since the magnetic quantum number m can be -1, 0, or +1 when the value of the subshell quantum number l is one, p orbitals come in sets of three. In each set, one of the orbitals is aligned along each of the three mutually perpendicular axes of the atom; these axes are traditionally designated x, y, and z. The three 2p orbitals are correspondingly designated 2px, 2py, and 2pz
     
  5. Jan 31, 2008 #4
    But how nature chooses the particular y and x direction given that magnetic field , for instance, is directed along z axis?

    Another correlated comment: I think that your statement does not give an answer to what was asked. If I understood it well the question was: Is it correct to stabilish the correspondence between m=1 and px, m= 0 and py, and m= -1 and pz (or any other bijection) ?

    Best wishes

    DaTario
     
    Last edited: Jan 31, 2008
  6. Jan 31, 2008 #5
    The direction of the z axis is arbitrary. m = 1 may be assigned to any of Px, Py or Pz. The other two assignments must of course be consistent (orthogonal) with the first. In the labeling of the P prbitals, x, y and z are not related to the initially defined z axis (which may be the cause of some confusion).
     
  7. Jan 31, 2008 #6
    Perhaps
     
    Last edited: Jan 31, 2008
  8. Jan 31, 2008 #7

    reilly

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    Science Advisor

    It's standard to take the z axis as arbitrary unless it is specifically defined. And, peculiar as it sounds, you can call the eigenstates anything you want. There is, however a problem: we would be uncomfortable with M |0> = 1|0>, but as a defining equation it's fine -- could have eigenvalue of -1 for state 1, and 0 for state -1. The normal conventions work quite nicely. Good question.
    Regards,
    Reilly Atkinson
     
  9. Jan 31, 2008 #8
    could you be a little more detailed?

    Thank you in advance,

    best wishes

    DaTario
     
  10. Feb 1, 2008 #9
    I think there's a slightly different convention among chemists and physicists for describing the orbitals.

    The chemists take three mutually perpendicular orbitals along the x, y, and z axes. Physicists take the z orbital, and then mix the chemists' x and y orbitals into combinations which are basicaly x + iy and x - iy. Now they have three states that are symetrical about the z axis and whose spin is +1, 0, and -1.

    A similar thing is done in the d orbitals. The pictures in the chemistry books show orbitals with lobes at 45 degrees, others at 90 degrees, and one with a donut and two lobes. The last one is the one shared with physicists, symetrical about the z axis. The physicists then take mixtures of the chemist's orbitals to get: a pair with spin +/-2, and a pair with spin +/-1 AND a nodal surface through the xy plane (the "equator").
     
  11. Feb 4, 2008 #10
    Very interesting. But, then, according to what you said, the answer to the question is that m = 1 does not correspond to what we call Px (or Py).

    Is it Ok?

    Best wishes

    DaTario
     
  12. Feb 4, 2008 #11

    jtbell

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    Staff: Mentor

    Yes. The m = +1 and m = -1 states are linear combinations of the Px and Py states; and the Px and Py states are linear combinations of the m = +1 and m = -1 states.
     
  13. Feb 4, 2008 #12
    I feel satisfaction due to this learning moment. Thank you.

    Sincerely

    DaTario
     
  14. Feb 4, 2008 #13
    Thanks. I am not used to having people understand my explanations of things.

    marty
     
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