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Schrodinger equation problem

  1. Aug 23, 2009 #1
    1. The problem statement, all variables and given/known data

    the ground state wave function for 1-d SHM oscillator is fixed by the diff. eqn a*phi-0=0


    using the expression for the lower operator as a differential operator,a-=(K/2)1/2x-h*([tex]\partial[/tex])/(2pi*(2m)1/2), to find a solution for this differential equation for phi0(x)

    integral(-infinity,infinity)e-y2=pi1/2 to normalize phi0 to the condition integral(-infinity,infinity)abs(phi0)2dx=1
    2. Relevant equations



    3. The attempt at a solution

    a-*phi0=0 =>( K/2)1/2x-h*([tex]\partial[/tex])/(2pi*(2m)1/2)[/SUB]phi0=0

    I don't understand how the three equations a-*phi0=0, integral(-infinity,infinity)abs(phi0)2dx=1,and -infinity,infinity)e-y2=pi1/2 are related;
     
    Last edited: Aug 23, 2009
  2. jcsd
  3. Aug 24, 2009 #2

    gabbagabbahey

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    Yuck!:yuck:

    I think it will help you to better communicate future homework problems if you format your posts a little nicer. That includes using [itex]\LaTeX[/itex] when appropriate, and making an effort to use proper spelling and grammar.

    Here's a nice introduction to using [itex]\LaTeX[/itex] in these forums.

    For example, your post could have looked something like this:

    You can click on any of the above images to see the code that generated them.

    Now, as for your question.... You have a differential equation,

    [tex]\left(\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k}\right)^{1/2}\frac{d}{d x}\right)\varphi_0(x)=0[/tex]

    Just solve it!...You will find that the solution has an unknown constant in front of it....you can then use the normalization condition to find the value of that constant.
     
    Last edited: Aug 24, 2009
  4. Aug 24, 2009 #3
    What am I solving for? \varphi_0(x)? And if I am solving for \varphi_0(x) would I then plug that into my normalization equation to obtain k
     
  5. Aug 24, 2009 #4

    gabbagabbahey

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    Yes. Your problem statement says to 'find a solution for [itex]\varphi_0(x)[/itex]'

    Not quite...try solving for [itex]\varphi_0(x)[/itex] first, worry about the rest later.
     
  6. Aug 24, 2009 #5
    I think I eventually found a solution to for [itex]\varphi_0(x)[/itex]; I won't list my math; [itex]\varphi_0(x)[/itex]=A*exp(-(m)^(1/2)*(k)^(1/2)/h); I would used , [tex]
    \int_{-\infty}^{\infty} |\varphi_0(x)|^2 dx=1
    [/tex]

    to find A?
     
  7. Aug 24, 2009 #6

    gabbagabbahey

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    No, why is there an 'h' and an 'm' in your solution? And why is there no 'x'?

    Make sure you are using the correct expression for the differential operator [itex]a_{-}[/itex] in the x-basis.

    This can either be written as

    [tex]a_{-}\to\left(\frac{m\omega}{2\hbar}\right)^{1/2}x-\left(\frac{\hbar}{2m\omega}\right)^{1/2}\frac{d}{d x}[/tex]

    or

    [tex]a_{-}\to\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{1}{2k}\right)^{1/2}\frac{d}{d x}[/tex]

    with the constant [itex]k[/itex] defined by [itex]k\equiv\frac{m\omega}{\hbar}[/itex]

    Once you find the correct expression for [itex]\varphi_0(x)[/itex], yes.
     
  8. Aug 24, 2009 #7
    Would this expression [tex]
    \int_{-\infty}^{\infty}e^{-y^2}dy=\pi^{1/2}
    [/tex] be useful and helping me find A? Wouldn't I squared this expression and pi^(1/2) becomes Pi?
     
  9. Aug 24, 2009 #8

    gabbagabbahey

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    You still haven't correctly found [itex]\varphi_0(x)[/itex]....do that first, worry about normalizing it later.
     
  10. Aug 24, 2009 #9
    I got [itex]\varphi_0(x)[/itex] to be: [itex]\varphi_0(x)[/itex]=A*exp(-(x^2/2)*(K/2)^(1/2)*((2m)^(1/2)/[tex]\hbar[/tex])
     
  11. Aug 24, 2009 #10

    gabbagabbahey

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    That's closer, but your constants inside the exponential are incorrect. You should be getting
    [tex]\varphi_0(x)=Ae^{-kx^2/2}[/tex]

    Or, in terms of [itex]\omega[/itex],

    [tex]\varphi_0(x)=Ae^{-\frac{m\omega x^2}{2\hbar}}[/tex]

    Are you still using the incorrect expression

    [tex]a_{-}\to\left(\frac{k}{2}\right)^{1/2}x-\left(\frac{\hbar}{2m}\right)^{1/2}\frac{d}{d x}[/tex]

    ???

    If so, refer to post #6 for the correct expressions, or better yet, look them up in your textbook.
     
  12. Aug 24, 2009 #11
    that is the expression given in the textbook except the planck constant is not supposed raised by the 1/2 power
     
  13. Aug 24, 2009 #12

    gabbagabbahey

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    That's odd, which textbook are you using?
     
  14. Aug 24, 2009 #13
    peebles
     
  15. Aug 24, 2009 #14

    gabbagabbahey

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    I found a google book review of that text, which page is this equation on in your copy?
     
  16. Aug 24, 2009 #15
    p.37, eqn 6.34
     
  17. Aug 24, 2009 #16

    gabbagabbahey

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    Okay, I see what he's doing now, he is using a rather odd definition of the raising and lowering operators which results in [itex]\omega=\sqrt{\frac{K}{m}}[/itex]

    In that case, your earlier result,

    [tex]\varphi_0=Ae^{-\frac{\sqrt{mK}x^2}{2\hbar}}[/tex]

    is correct.(I simplified it a little for you)

    Using this result, what do you get when you evaluate the integral [itex]\int_{-\infty}^{\infty}|\varphi_0(x)|^2 dx[/itex] ?
     
  18. Aug 24, 2009 #17

    [tex]\varphi_0^2[/tex]dx=[tex]A^2e^{-\frac{{mK}x^4}{4\hbar^2}}[/tex]dx =1?

    Sorry about latex; I been fumbling with it to a point where I want to put latex to rest for good with a dagger
     
    Last edited: Aug 24, 2009
  19. Aug 24, 2009 #18

    gabbagabbahey

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    No, first off

    [tex](e^u)^2=e^{2u}\neq e^{u^2}[/tex]

    Secondly, you need to actually perform the integration/
     
  20. Aug 24, 2009 #19
    I should have known better; would I normaliased this equation;
    [itex]
    \int_{-\infty}^{\infty}e^{-y^2}=\pi^{1/2}
    [/itex] and y^2 =2u?
     
  21. Aug 24, 2009 #20

    gabbagabbahey

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    First off, what is [itex]|\varphi_0(x)^2|[/itex]?

    Second, what do you get when you integrate it from -infinity to infinity?
     
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