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Schrodinger's Equation in 1D

  1. Aug 13, 2010 #1
    1. The problem statement, all variables and given/known data
    An electron in a one dimensional crystal is bound by:

    [tex]U(x) = \frac{-\overline{h}^{2}x^{2}}{mL^{2}\left(L^{2}-x^{2}\right)}[/tex]
    for
    [tex]\left|x\right| < L[/tex]

    and
    [tex]x = infinity[/tex]
    for
    [tex]\left|x\right| \geq L[/tex]

    Show that a stationary state for the electron in the potential well
    [tex]\psi(x) = A\left(1-\frac{x^{2}}{L^{2}}\right)[/tex]
    satisfies the Schrodinger's Equation

    and find E

    2. Relevant equations

    [tex]\frac{-\overline{h}^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}+U(x)\psi = E\psi[/tex]

    3. The attempt at a solution

    from Schrodinger's:

    [tex]\frac{d^{2}\psi}{dx^{2}} = \frac{-2m}{\overline{h}^{2}}\left(\frac{\overline{h}^{2}x^{2}}{mL^{2}\left(L^{2}-x^{2}\right)}+E\right)\psi[/tex]

    and from the guess solution:

    [tex]\frac{d^{2}\psi}{dx^{2}} = \frac{-2A}{L^{2}} = \frac{-2}{\left(L^{2}-x^{2}\right)}\psi[/tex]

    and so equating [tex]\frac{d^{2}\psi}{dx^{2}}[/tex],
    I deduced that it satisfies the Schrodinger's equation but only when E = 0 and x = L.

    Am I right?

    I am also concerned because the potential, U, is negative 'inside' the well...
     
  2. jcsd
  3. Aug 14, 2010 #2
    It can't be a stationary state if it only satisfies the Schrodinger eqn at a single 'x' point. I think you need to double check your work on this problem, since you made an incorrect assumption.

    I suggest simplifying the two terms you list in the parenthesis of the first eqn in your attempt at a solution. Convert it to one giant fraction. Then try to find an energy that will remove the 'x'-dependence in the numerator of that fraction. You will want to keep the x-dependence in the denominator since it needs to equal your 2nd equation. You want this equality to be applied to any value of 'x' and not just one specific value like before.
     
  4. Aug 14, 2010 #3
    Hi nickjer

    After putting it as a single fraction, I got:

    [tex]
    \frac{d^{2}\psi}{dx^{2}} = \left(\frac{-2}{\left(L^{2}-x^{2}\right)}\frac{x^{2}\left(\overline{h}^{2}-mE\right)+mL^{4}E}{L^{2}\overline{h}^{2}}\right)\psi
    [/tex]

    I just cannot figure out how to remove the x dependencies in the numerator.

    If I equate this with
    [tex]
    \frac{d^{2}\psi}{dx^{2}} = \frac{-2}{\left(L^{2}-x^{2}\right)}\psi
    [/tex]

    I can get a value for E, however it depends on x, which i think is wrong...

    Maybe, just maybe, is the question wrong and
    [tex]
    \psi(x) = A\left(1-\frac{x^{2}}{L^{2}}\right)
    [/tex]
    is not a possible solution to the situation?
     
  5. Aug 14, 2010 #4
    Double check your math for the numerator. I believe you are missing something. Also, once you correct the numerator, then you see the x^2 has a coefficient next to it in parenthesis. Very similar to what you have now (although yours is slightly wrong). Since you are solving for the energy that makes this stationary state satisfy Schrodinger's eqn, then you are free to choose an E that will make this coefficient in parenthesis go to 0. Then you will have 0*x^2 = 0. So you cancelled out the x-dependence.
     
  6. Aug 14, 2010 #5
    If I'm right this time, I was missing L2

    And so I equate 0 = hbar2 - L2mE

    and also, after eliminating x2, I equate to the guess solution,

    and both of them give me a consistent answer:

    [tex]E = \frac{\overline{h}^{2}}{mL^{2}}[/tex]

    Now, I think this is right!

    If it is, thanks very much :smile:
     
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