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Hamiltonian of the Quantum Harmonic Oscillator-Eigenfunction & Eigenvalue

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the equation below is an eigenfunction for the Quantum Harmonic Oscillator Hamiltonian and find its corresponding eigenvalue.


    2. Relevant equations
    u1(q)=A*q*exp((-q[itex]^{2}[/itex])/2)


    3. The attempt at a solution
    Ok, so I know that the Quantum Harmonic Oscillator Hamiltonian (H[itex]_{QHO}[/itex]) is :
    (H[itex]_{QHO}[/itex])=[itex]\frac{1}{2}[/itex][itex]\hbar[/itex]ω(((-d^2)/(dq^2))+q^2) and I know that:
    (H[itex]_{QHO}[/itex])u1(q)=Eu1(q)

    but how do I show that it's an eigenfunction? Simply subbing it into the eqn doesn't appear to help.

    Many thanks in advance.
     
  2. jcsd
  3. Feb 5, 2012 #2

    vela

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    It should. Either that or you need to solve the differential equation, which is a much harder task.

    Show us what you got when you plugged u1 into the equation.
     
  4. Feb 5, 2012 #3
    Ok, so here goes:

    HQHO*U1(q)=E*U1(q)

    [itex]\frac{1}{2}[/itex][itex]\hbar[/itex]ω([itex]\frac{-d^{2}}{dq^{2}}[/itex]+q[itex]^{2}[/itex]).A.q exp([itex]\frac{-q^{2}}{2}[/itex])=E*U1(q)

    [itex]\frac{A}{2}[/itex][itex]\hbar[/itex]ω(-[itex]\frac{d}{dq}[/itex][itex]\frac{d}{dq}[/itex](q.exp([itex]\frac{-q^{2}}{2}[/itex]))+(q[itex]^{3}[/itex])exp([itex]\frac{-q^{2}}{2}[/itex]))=E*U1(q)

    which eventually comes to:

    (q[itex]^{3}[/itex]+q[itex]^{2}[/itex]-2q-1)exp([itex]\frac{-q^{2}}{2}[/itex])[itex]\frac{A}{2}[/itex][itex]\hbar[/itex]ω=E*U1(q)

    So does this mean that : (q[itex]^{3}[/itex])+q[itex]^{2}[/itex]-2q-1) is the corresponding eigenvalue?

    Is my method correct?

    Many thanks.
     
  5. Feb 5, 2012 #4

    vela

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    You must have calculated the second derivative incorrectly. You should get
    $$u_1''(q) = (q^3-3q)e^{-q^2/q}.$$
     
  6. Feb 5, 2012 #5
    Thanks, my mistake. So I now have:

    [itex]\frac{\hbarω}{2}[/itex]A.exp([itex]\frac{-q^{2}}{2}[/itex])(-q[itex]^{3}[/itex]-q[itex]^{2}[/itex]+3q)=E.U[itex]_{1}[/itex](q).

    So is -q[itex]^{3}[/itex]-q[itex]^{2}[/itex]+3q the corresponding eigenvalue? Or can I simplify even further?

    Many thanks.
     
  7. Feb 5, 2012 #6

    vela

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    That's still not correct. The eigenvalue is a constant. It can't depend on q.
     
  8. Feb 5, 2012 #7
    I think I've figured it out now. Many thanks Vela
     
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