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Solutions to a quantum harmonic oscillator - desperate for help!

  1. Aug 12, 2012 #1
    I need to find the value σ for which:

    ψ0(x) = (2πσ)-1/4 exp(-x2/4σ)

    is a solution for the Schrodinger equation

    I know the equation for the QHO is:

    Eψ = (P2/2m)ψ + 1/2*mw2x2ψ

    I've tried normalizing the wavefunction but I end up with a σ/σ term :(

    Any help would be greatly appreciated :)
     
  2. jcsd
  3. Aug 12, 2012 #2
    You aren't using the right form of the Schrodinger Equation. Try plugging it into the form with the derivatives:

    [tex]
    E\Psi = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\Psi + \frac{1}{2}m\omega^{2}x^{2}\Psi[/tex]

    Solve the resulting equation

    edit - wrote the wrong equation.
     
  4. Aug 12, 2012 #3
    that is the equation I was using I just put P as the momentum operator :( I'm confused! Do I enter the wavefuntion into schrodinger and then solve for what?

    <3 xoxo
     
  5. Aug 12, 2012 #4
    Are you given the general expression for [itex] E_n [/itex] ?? You are given the expression for [itex] \psi_n [/itex] for [itex] n = 0 [/itex]. So take the necessary derivatives, plug in the expression for E_0 and recall that this solution must be valid for all x. Do you know a specific value for x that might make this much easier?

    edit - the general expression for E isn't really necessary now that I think about it. You'll still be able to solve for [itex] \sigma [/itex] in terms of E
     
  6. Aug 12, 2012 #5
    I can find σ in terms of E but the question says give the value for it that makes ψ0 (the one I originally posted) a solution to the schrodinger

    Then part II asks me to find the energy associated by putting the wavefunction ψ0 into schrodinger like you said, so I'm guessing there must be an actual value or σ?

    I'm really confuzzled! :(

    All I'm given is the function ψ0 like I originally posted, nothing else.

    <3 xoxo
     
  7. Aug 12, 2012 #6
    Huh, I'm confused then as well, typically when a question asks you what value for a constant makes it a solution to an equation the thing to do is plug it into said equation. Could you possibly write out the entire problem? Or is it in a certain textbook? If so, what text and what's the problem number.
     
  8. Aug 12, 2012 #7
    Part I

    Find the value σ0 for which:

    ψ0(x) = (2πσ)-1/4 exp(-x2/4σ)

    is a solution to the Schrodinger equation

    Part II

    What is the associated energy in part I? [hint: insert the given wave function into the schrodinger equation]

    That's it word for word.

    Thank you so much for helping I really appreciate it! <3 xoxoxo
     
  9. Aug 12, 2012 #8
    when you substitute ψ into schrodingers equation
    Hψ=Eψ, you know E is a constant (independent of x)
    but when you calculate Hψ you'll get terms proportional
    to x^2. you need to pick σ so that those terms go away.
     
  10. Aug 12, 2012 #9
    I put the wavefunction in and took the 2nd derivative and all that I cancelled a load of common terms and ended up with

    E = (-h(bar)2/2m)*(x2)/(4σ2) + 1/2*(mω2x2)

    Is this right? I don't get how to pick a value so all the x's disappear?

    <3 Cheers guys xoxox
     
  11. Aug 12, 2012 #10
    qbert, that would mean [itex] \sigma [/itex] is a function of x^2. This is definitely not the case.
     
  12. Aug 12, 2012 #11
    Bowenwww, You should end up with the following:

    [tex] E = \frac{\hbar^{2}}{4m\sigma} - \frac{\hbar^{2}x^{2}}{8m\sigma} + \frac{1}{2}m\omega^{2}x^{2} [/tex]
     
  13. Aug 12, 2012 #12
    more like:
    [tex]E = \frac{1}{4}\frac{\hbar^2}{m \sigma}-\frac{1}{8}\frac{\hbar^2x^2}{m\sigma^2}+\frac{1}{2} m \omega^2 x^2[/tex]

    Clever-Name, picking σ to make the x^2 terms 0 does not
    mean that σ is a function of x. in fact, if you just do it
    you'll see precisely what σ has to be!
     
  14. Aug 12, 2012 #13
    So that's how I find out the energy levels, right? But I still don't know how to get the sigma?

    I really appreciate this guys I just need pointing in the right direction ^ _ ^


    <3 xoxoxox
     
  15. Aug 12, 2012 #14
    i told you how to get σ.
     
  16. Aug 12, 2012 #15
    But there's still two unknowns E and sigma - sorry I know I'm being really stupid here; I know how to get to the form you've given I just don't know how to get from there to a value for sigma?

    :( xoxox
     
  17. Aug 12, 2012 #16
    pick σ to make the x^2 terms go away.
     
  18. Aug 12, 2012 #17

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    If [itex]C_1 = (a-2)x^2 +C_2[/itex] where [itex]a[/itex], [itex]C_1[/itex] and [itex]C_2[/itex] are all constants, then [itex]a[/itex] must equal 2 (otherwise [itex]C_1[/itex] would be a function of [itex]x[/itex], not a constant) and [itex]C_1[/itex] must equal [itex]C_2[/itex]
     
  19. Aug 12, 2012 #18
    qbert you're right, my mistake, but your attitude is unnecessary. Bowen is obviously struggling to figure this out and your bitter responses aren't very helpful. I also made a typo in the equation you corrected. The appropriate thing to do is assume I made that typo instead of being so pompous in response.

    Bowen, The last two terms (the ones with x^2) should cancel out based on what [itex] \sigma [/itex] is. In other words, their sum should be zero! Work that out and tell me what your answer for sigma is! i.e. solve this equation:

    [tex]
    -\frac{\hbar^{2}x^{2}}{8m\sigma^{2}} + \frac{1}{2}m\omega^{2}x^{2} = 0 [/tex]
     
  20. Aug 12, 2012 #19
    wow. bitter and pompous and here i thought i was being helpful.
    i'm sorry if i offended anyone, that wasn't my intention.
     
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