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Angular Momentum squared operator (L^2) eigenvalues?

  1. Dec 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the eigenvalues of the angular-momentum-squared operator (L2) for hydrogen 2s and 2px orbitals...

    2. Relevant equations
    Ψ2s = A (2-r/a0)e-r/(2a0)
    Ψ2px = B (r/a0)e-r/(2a0)
    3. The attempt at a solution
    If I am not wrong, is the use of L2 in eigenfunction L2Ψ = ħ2 l(l+1) Ψ?

    ...so I wonder if eigenvalues are l(l+1) from this part, ħ2 l(l+1) Ψ, as in:

    for Ψ2s, because for s orbital, l = 0--> l(l+1) = 0(0+1) = 0
    for Ψ2px, because for p orbital, l = 1--> l(l+1) = 1(1+1) = 2 ?
     
  2. jcsd
  3. Dec 4, 2014 #2

    DrClaude

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    Staff: Mentor

    That's only the radial part. You are missing the angular part, which is essential ;)

    How is an eigenvalue defined?
     
  4. Dec 4, 2014 #3
    OMG, I was like, why does the equation look funny----the correct one for Ψ2px would be B (r/a0)e-r/(2a0)sinθcosφ, right?

    Also, as how an eigenvalue is defined....well, from one previous eigenfunction example I solved, the one involving Hamiltonian operator, HΨ = EΨ, E is the eigenvalue?
    So, is the eigenvalue for Ψ2px2?
     
  5. Dec 4, 2014 #4

    DrClaude

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    Staff: Mentor

    Right.

    Yes. For an operator ##\hat{A}##, ##\hat{A} \psi = a \psi## where ##a## is a scalar means that ##\psi## is an eigenfunction of ##\hat{A}## and ##a## its corresponding eigenvalue.

    Yes.

    I don't know what the level of the question is, so I don't know how simple the answer can be. If this is an advanced problem, you probably are expected to actually derive the eigenvalue.
     
  6. Dec 4, 2014 #5
    Thank you! I am quite surprised myself at how simple the answer turns out to be.....but still thank you.
     
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