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QM 1-D Harmonic Oscillator Eigenfunction Problem

  1. Mar 18, 2014 #1

    tis

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    1. The problem statement, all variables and given/known data
    A particle of mass [itex]m[/itex] moves in a 1-D Harmonic oscillator potential with frequency [itex]\omega.[/itex]
    The second excited state is [itex]\psi_{2}(x) = C(2 \alpha^{2} x^{2} + \lambda) e^{-\frac{1}{2} a^{2} x^{2}}[/itex] with energy eigenvalue [itex]E_{2} = \frac{5}{2} \hbar \omega[/itex].

    [itex]C[/itex] and [itex]\lambda[/itex] are constants. Show that the constant [itex]\lambda[/itex] is not arbitrary.

    NOTE: [itex]\int_{-\infty}^{\infty} e^{-\frac{1}{2} a x^{2}} dx = \sqrt{2\pi / a}, a > 0[/itex].

    2. Relevant equations
    That I can think of: TISE, normalization condition for wavefunctions.


    3. The attempt at a solution
    I assume there's a solution in substituting into the time-independent Schrodinger Equation and solving for [itex]\lambda[/itex], but the equation seems very difficult. And I have no idea how to utilize the hint provided; everything I've tried (normalization condition, etc.) just gives a dependence on C. I've scoured my textbooks with no luck, their only relevant info is using ladder operators and other methods to produce eigenfunctions.

    Just after a point in the right direction. Thanks in advance for your help!
     
  2. jcsd
  3. Mar 19, 2014 #2
    Your idea to substitute this into the TISE seems like a good start. All you will have to do is take derivatives and rearrange some terms, and you will end up with an equation which determines lambda. Give it a try.
     
  4. Mar 20, 2014 #3

    tis

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    Figured it out. Take the ground state of the harmonic oscillator [itex]\psi_{0} = A e^{-\frac{1}{2}\alpha^{2} x^{2}}[/itex] and use the orthogonality condition [itex]\int_{-\infty}^{\infty} \psi_{0}^{*} \psi_{2} \ dx = 0[/itex].

    From there just expand, cancel out A* and C, and solve the integrals with integration by parts and the hint provided. Eventually you come to [itex]0=2\alpha^{2}\frac{\sqrt{\pi}}{2\alpha^{3}}+\lambda\frac{\sqrt{\pi}}{ \alpha }[/itex], hence [itex]\lambda=-1[/itex].
     
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