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Finding a constant from eigenfunction

  1. Oct 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Since Hamiltonian operator is:

    Ĥ = - (ħ2/(2m))(delta)2 - A/r
    where r = (x2+y2+z2)
    (delta)2 = ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2
    A = a constant

    from Ĥg(r) = Eg(r) form, where:

    g(r) = D e-r/b(1-r/b)

    with b, D as constants, is an EIGENFUNCTION of Ĥ, find the correct b and give the eigenvalue E.

    2. Relevant equations
    g(r) = D e-r/b(1-r/b)

    Ĥ = - (ħ2/(2m))(delta)2 - A/r
    where r = (x2+y2+z2)
    (delta)2 = ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2
    D = a constant

    3. The attempt at a solution
    g(r) = D e-r/b(1-r/b)
    g'(r) = -De-r/b/b - De-r/b/b + Dre-r/b/b2
    = -2De-r/b/b + Dre-r/b/b2
    g''(r) = 3De-r/b/b2 - Dre-r/b/b3

    [delta_g(r)]2 = 5De-r/b/b2 -Dre-r/b/b3 - 4De-r/b/(br)

    Ĥg(r) = Eg(r)
    - (ħ2/(2m))(5De-r/b/b2 -Dre-r/b/b3 - 4De-r/b/(br)) - A/r * (D e-r/b(1-r/b)) = E*D e-r/b (1-r/b)
    ..which is then reduced to:

    -5ħ2/(2mb2) + ħ2r/(2mb3) + 4ħ2/(2mbr) - A/r + A/b = E (1-r/b)....I managed to cancel out constant D and e-r/b, but I am at loss how how to eliminate r?

    Should I equivelate:
    ħ2r/(2mb3) + 4ħ2/(2mbr) - A/r = -E(r/b) ? I tried this way, and still haven't managed to find a way to cancel out r....to make b an independent number...
    With regard to E, if I am not wrong, I think Energy is that of an excited state, NOT ground state, except I haven't figured it out (for I have not gotten b constant), and have no idea which energy level it is....

    Any hints or notices to my mistakes would be appreciated!
     
  2. jcsd
  3. Oct 20, 2014 #2

    stevendaryl

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    [itex]\hat{H} g(r) = E g(r)[/itex]

    gives you an equation of the form:

    [itex]\frac{\alpha}{r} + \beta + \gamma r = 0[/itex]

    where [itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\gamma[/itex] are combinations of your constants: [itex]A, b, E[/itex].

    That equation can only be true for all values of [itex]r[/itex] if each coefficient is equal to zero, separately. So it must be that
    [itex]\alpha = 0[/itex]
    [itex]\beta = 0[/itex]
    [itex]\gamma = 0[/itex]

    So if you figure out what [itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\gamma[/itex] are in terms of your constants, you can get three equations for your two unknowns: [itex]E[/itex] and [itex]b[/itex]. That's too many, but it will turn out that the equations are redundant--the third one is a linear combination of the other two. So you can solve for [itex]E[/itex] and [itex]b[/itex].

    [itex]D[/itex] cancels out, so you can't solve for it using the Schrodinger equation, but instead you have to choose [itex]D[/itex] so that the integral of [itex]|g(r)|^2[/itex] over all space is equal to 1.
     
  4. Oct 20, 2014 #3
    Thank you for your reply! Btw, could you please clarify:

    because, according to what you've said:

    α = 4ħ2/(2mb) - A = 0 --> b = 2ħ2/(Am)
    β = -5ħ2/(2mb2) + A/b = 0
    γ = ħ2/(2mb3) + E/b = 0

    ...but I couldn't derive b from β and γ....does it mean I solved everything from the beginning wrong?

    And

    For the last one, do you mean I need to normalize?
     
    Last edited by a moderator: Oct 20, 2014
  5. Oct 20, 2014 #4

    stevendaryl

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    On the right-hand side of Schrodinger's equation for this case, you have
    [itex]E g(r) = E D e^{-r/b} - (E D r/b) e^{-r/b}[/itex]

    So the equation for [itex]\beta[/itex] should have [itex]E[/itex] on the right-hand side of the equals.

    But you can certainly solve for [itex]E[/itex] using the equation for [itex]\gamma[/itex]. You already know [itex]b[/itex] from your equation for [itex]\alpha[/itex]. Plug that into the equation for [itex]\gamma[/itex]

    Yes. [itex]D[/itex] is a normalization constant.
     
    Last edited: Oct 20, 2014
  6. Oct 20, 2014 #5
    Thank you--I guess I figured out b and E, but may I ask what I should do with β? Can I use it to check b and E I found out?
     
  7. Oct 21, 2014 #6

    stevendaryl

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    When I looked at it, it seemed that the equation for [itex]\beta[/itex] gave the same information as the equation for [itex]\gamma[/itex].
     
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