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Simple Harmonic Oscillator (time independant Schrodingers)

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Particle mass m is confined by a one dimensional simple harmonic oscillator potential V(x)=Cx2, where x is the displaecment from equilibrium and C is a constant

    By substitution into time-independant schrodingers with the potential show that
    [tex]\psi[/tex](x)=Axe-ax2
    is a possible spatial wavefunction for this particle provided constant a has a certain value.
    Find a in term os C, m, h bar (\h)
    What the corresponding energy eigen value

    3. The attempt at a solution

    [tex]\psi[/tex](x)= Axe-ax2
    d/dx [tex]\psi[/tex](x)= [A - 2Aax2]*e-ax2
    d2/d2x [tex]\psi[/tex](x)= [4a2Ax3 - 6aAx]*e-ax2

    so shrodingers:


    [tex]\frac{-hbar^2}{2m}[/tex]*[4a2x2 - 6a] + Cx2 = E

    Ive gotten to this bit but i dont understand what to do, i have read the 4 page solution in Eisburg&Resnick and spoke to my lecturer, but my lecturer has said that i dont need to do the long winded complete solution and that i is also not necessary to evaluate the normalisation constant A.
    I just dont know how to show that the psi above is a possible wavefunction

    help or advice on how to do this problem would be much appreciated

    ps I have tried looking at and doing some other example but they all had the potential as V(x)= 1/2 Cx2, this seemed to be the standard example. I also tried using the substitution w2= (c/m) and this would give a potential V(x)= mw2x2, but i didnt know if this was right as the question ask for a in terms of C (and m and hbar)
     
    Last edited: Nov 10, 2009
  2. jcsd
  3. Nov 10, 2009 #2

    lanedance

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    so i think you're pretty much there, so assuming you have subbed into the TISE correctly (which looks ok to me...), then how can you "choose" a, so that the above equation i satisfied?
     
  4. Nov 10, 2009 #3
    well the thing is i dont know when it would be satisfied other than the LHS=RHS but i dont know how i would get E on the LHS, unless i made the sub that E=hbar*w
     
  5. Nov 10, 2009 #4

    lanedance

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    I don't see a w anywhere... but there should be one value of a to satisfy TISE for all x, and what if the TISE is telling you the energy of the given state....
     
  6. Nov 10, 2009 #5
    As lanedance said, you're nearly there. You do know that E is a constant (but don't know it yet), i.e. it is independent of x.
     
  7. Nov 10, 2009 #6
    thats what im having trouble with - knowing when it is satisfied

    is the LHS supposed to equal the RHS?
    i tried a=[tex]\sqrt{\frac{mc}{hbar}}[/tex]

    and i got

    Cx2(1-2hbar) + [tex]\frac{3x(hbar)\sqrt{hbar*mC}}{m}[/tex] = E

    but i dont know what to do with it...is my choice for a even right?
    How do you choose the value, is it a guess? or is there some method?
     
  8. Nov 10, 2009 #7

    lanedance

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    no, i don't think so, the Energy of a eigenstate in the TISE, should be a constant scalar value (independent of x), this should help you choose a to remove any x dependence
     
    Last edited: Nov 10, 2009
  9. Nov 10, 2009 #8
    ok i tried this

    [tex]\frac{(hbar)^2}{2m}[/tex]*[6a - 4ax2 + [tex]\frac{2m}{(hbar)^2}[/tex]Cx2] = E

    6a + x2*[[tex]\frac{2m}{(hbar)^2}[/tex]Cx2 - 4a] = [tex]\frac{2m}{(hbar^2}[/tex]Cx^2]E

    i tried a= mC/2(hbar)2

    this gave me

    [tex]\frac{6mC}{2(hbar)^2}[/tex] + x2[[tex]\frac{2mC}{(hbar)^2}[/tex] - [tex]\frac{4mC}{2(hbar)^2}[/tex]] = [tex]\frac{2m}{(hbar)^2}[/tex]Cx^2]E

    =[tex]\frac{6mC}{2(hbar)^2}[/tex] = [tex]\frac{2m}{(hbar)^2}[/tex]Cx^2]E

    E=[tex]\frac{3C}{2}[/tex]
     
  10. Nov 10, 2009 #9

    lanedance

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    i can't follow your equations i the last post... but i think you've got the idea
     
  11. Nov 10, 2009 #10

    lanedance

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    also do you know you can write longer equations in tex? eg.

    [tex] \frac{\hbar^2}{2m}(4a^2x^2 - 6a) + Cx^2 = E [/tex]
     
  12. Nov 10, 2009 #11
    sorry made a mistake there, this is what i meant

    [tex]\frac{(hbar)^2}{2m}[/tex]*[6a - 4ax2 + [tex]\frac{2m}{(hbar)^2}[/tex]Cx2] = E

    6a + x2*[[tex]\frac{2m}{(hbar)^2}[/tex]Cx2 - 4a] = [tex]\frac{2m}{(hbar)^2}[/tex]E

    i tried a= mC/2(hbar)2

    this gave me

    [tex]\frac{6mC}{2(hbar)^2}[/tex] + x2[[tex]\frac{2mC}{(hbar)^2}[/tex] - [tex]\frac{4mC}{2(hbar)^2}[/tex]] = [tex]\frac{2m}{(hbar)^2}[/tex]E

    =[tex]\frac{6mC}{2(hbar)^2}[/tex] = [tex]\frac{2m}{(hbar)^2}[/tex]E

    E=[tex]\frac{3C}{2}[/tex]
     
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