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I Eigenfunction of L2 operator

  1. Apr 6, 2016 #1
    hi, am major new on quantum mechanics. please help me understand. is the real wave function
    Ψ2Px= [Ψ2p+1 +Ψ2p-1]1/2 an eigen function of L2 or Lz?
    if so, how is it?
    and if so kindly explain the values of l and m
    thanks
     
  2. jcsd
  3. Apr 7, 2016 #2

    Simon Bridge

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    Good question - did you apply the operators to find out?

    If I am reading your notation correctly as:
    ##\psi_{2p,x} = \frac{1}{\sqrt{2}}\left(\psi_{2p,+1}+\psi_{2p,-1}\right)##
    This says that the 2p state for, say, a hydrogen atom, is equally likely to involve the electron spin up or spin down.
    The "2" is the energy eigenstate number, the "p" is from "s,p,d,f..." notation, and is the orbital angular momentum state number. You can look these up.
     
  4. Apr 7, 2016 #3
    By all means tell me more.... you are being a major help.
    So how would I go about applying
    And from the explanation you've just given above am guessing l and m are normal quantum number values
     
  5. Apr 7, 2016 #4

    kith

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    Let's say we have an operator [itex]A[/itex], an eigenvalue [itex]a[/itex] and the corresponding eigenfunction [itex]\psi_a[/itex]. How does the eigenvalue equation look like?

    What do you get if you replace [itex]A[/itex] by [itex]H, L^2[/itex] or [itex]L_z[/itex] and [itex]\psi_a[/itex] by [itex]\psi_{n,l,m}[/itex]?
     
  6. Apr 7, 2016 #5
    Oh and the entire equation in the brackets are to the power of 1/2 (square root) not multiplied by the root of 1/2
    Sorry I think there was a font issue there
     
  7. Apr 10, 2016 #6

    Simon Bridge

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    No - I was just assuming someone made a mistake.

    What you wrote was:

    Ψ2Px= [Ψ2p+1 +Ψ2p-1]1/2

    Another interpretation would be: ##\psi_{2px} = \left[ \psi_{2p+1}+\psi_{2p-1}\right]^{1/2}##
    ... but that does not make any sense, though it looks like it can be used to answer the questions.

    ... how would you normally go about applying an operator?
    (By the time you get to see angular momentum in QM, you have met operators and how to apply them: go back through your notes.)

    ... have you looked up "spdf notation" yet?
    https://en.wikipedia.org/wiki/Electron_configuration

    The quantum numbers for an atomic state are n,l,m(,s) - so ##H\psi_{nlm} = E_n\psi_{nlm}##
    ... for hydrogen, ##E_n=-13.6\text{eV}/n^2##. The advantage of doing QM this way is that you don't have to do lots of differentiating and integrating.
    Also see: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qangm.html

    Note: you have received some suggestions and some questions designed to help you.
    If you do not follow suggestions and do not answer questions, we cannot help you.
    We can only point in the right sort of direction - you have to do the work.
     
    Last edited: Apr 10, 2016
  8. Apr 10, 2016 #7
    Yes I have followed the your suggestions and I have checked out the spdf notations...... it has been a great help....thanks alot
     
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