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Physical Chemistry, P.W.Atkins, the Schrödinger equation

  1. Oct 15, 2007 #1
    1. The problem statement, all variables and given/known data

    Im doing an A-level project on the Schrödinger equation and am unsure on the mathematics used to obtain the following results:

    The Schrödinger for a particle in no potential field (=0) has the solution:

    psi(x)=e^ikx. i is defined below, I haven't really a clue as to how you get this result.

    Looking at it I would guess that its a double integral to remove the second derivative of psi with respect to x, can anyone help me out here? I know how the rest of the equations goes after and using Eulers relationship to express it in tems of cos kx + i sin kx. You can also re-arrange k to show de Broglie's relationship.

    any help is greatly appreciated

    2. Relevant equations

    [​IMG]

    3. The attempt at a solution

    ??
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 15, 2007 #2

    dextercioby

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    Do you know how to solve 2-nd order ODE-s ?
     
  4. Oct 15, 2007 #3
    ordinary differential equations?

    i can solve second order derivatives but only relatively simple ones. This is a bit above my head. Would you be able to point me in the right direction of what to start looking at to get the correct result?

    Thanks for the reply so far.

    Regards, Simon.
     
    Last edited: Oct 15, 2007
  5. Oct 15, 2007 #4

    Astronuc

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    Staff: Mentor

    What dextercioby was getting at is reforming the differential equation into a workable or recognizable form.

    e.g. take [tex]\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2}\,=\,E\psi[/tex] and rewrite it as

    [tex]\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}\,+\,E\psi = 0[/tex] and then


    [tex]\frac{d^2\psi}{dx^2}\,+\,\frac{2mE}{\hbar^2}\psi = 0[/tex]

    and letting [tex]k^2\,=\,\frac{2mE}{\hbar^2}[/tex], see what that form looks like and what the possible solutions are.
     
  6. Oct 16, 2007 #5
    Ok i think i have got this now:

    [tex]\frac{d^2\psi}{dx^2}\,+\,\frac{2mE}{\hbar^2}\psi = 0[/tex]

    as you said: let:

    [tex]k^2\,=\,\frac{2mE}{\hbar^2}[/tex]

    [tex](1.) \frac{d^2\psi}{dx^2} = -k^2\psi [/tex]

    let:

    [tex]\psi = e^mx[/tex]

    [tex]\frac{d\psi}{dx} = me^{mx}[/tex]

    [tex]\frac{d^2\psi}{dx^2} = m^2e^{mx}[/tex]

    Substitute into (1.):

    [tex]m^2e^{mx} = -k^2e^{mx}[/tex]

    [tex]m^2 = -k^2[/tex]

    [tex]m = -k[/tex]
    if:
    [tex]-k < 0[/tex]
    then:
    [tex]m = ik[/tex]
    so:
    [tex]\psi = e^{mx}[/tex]
    Therefore:
    [tex]\psi = e^{ikx}
    [/tex]

    How does that look? it was a big effort but worth it :D

    only thing im not too sure about is the sign of k at the end of the process, + or -?

    Thanks for the help so far!

    Regards, Si
     
  7. Oct 16, 2007 #6

    Gokul43201

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    There's an error here:
    [tex]m^2 = -k^2 ~\implies m=ik[/tex]
    It doesn't matter if k>0 or k<0.
     
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