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Homework Help: Normalizing a Wavefunction of a harmonic oscillator

  1. Aug 16, 2008 #1
    1. At a certain time the wavefunction of a one-dimensional harmonic oscillator is

    [tex]\psi[/tex](x) = 3[tex]\phi[/tex]0(x) + 4[tex]\phi[/tex]1(x)

    where [tex]\phi[/tex]0(x) and [tex]\phi[/tex]1(x) are normalized energy eigenfunctions of the ground and first excited states respectively. Normalize the wavefunction and determine the probability of finding the oscillator in the ground state.

    3. I'm not really sure if I'm normalizing the wavefunction correctly, I get the normalizing constant as 1/7. However, when I calculate the probability of the ground state and first state combined they don't equal one. Aren't they supposed to and have I normalized correctly?
  2. jcsd
  3. Aug 16, 2008 #2

    Doc Al

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    Staff: Mentor

    Redo this. What's the definition of normalization? (What must equal 1?)
  4. Aug 16, 2008 #3
    I've got the equation for normalising a wavefunction; one over the square root of the wavefunction squared.

    = ([tex]\int[/tex][3[tex]\phi[/tex]0(x) + 4[tex]\phi[/tex]1(x)]2)1/2 =

    I still get [tex]\sqrt{1/49}[/tex] when I should be getting [tex]\sqrt{1/25}[/tex]; I think I'm squaring the function wrong. The coefficients I get upon squaring are (9 + 12 + 12 + 16), I'm guessing the 12s cancel each other out to leave 9+16=25 but I don't see how. Do I need to know the energy eigenfunctions to calculate this? I think I'm getting myself confused, can you point me in the right direction of squaring the wavefunction? It's just all previous examples I have give the wavefunction experssed as a single function rather than an addition of two. Thanks very much.
  5. Aug 16, 2008 #4


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    show your intermediate steps.

    I think you are forgetting that phi_0 and phi_2 are ortogonal to each other..
  6. Aug 16, 2008 #5

    Doc Al

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    All you need to know is:
    That's the secret. :wink:

    Show your steps, keeping the eigenfunctions in your expansion.
  7. Aug 16, 2008 #6
    psi(x) = 3phi_0(x) + 4phi_1(x)

    => [psi(x)]^2 = [3phi_0(x) + 4phi_1(x)]^2

    => [9(phi_0(x))^2 + 12phi*_0(x)phi_1(x) + 12phi_0(x)phi*_1(x) + 16(phi_1(x))^2

    Is this right?

    When I integrate with limits +-infinity does it give

  8. Aug 16, 2008 #7

    Doc Al

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    Good! Now you've got it.
  9. Aug 16, 2008 #8
    Thanks very much! :D
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