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Variational method approximation for half-space linear potential

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    This was a test question I just had, and I'm fairly certain I got it wrong. I'm confused as to what I did wrong, though. We were told that our potential was infinite when x<0, and Cx where x>0. We were asked to approximate the ground state potential using the variational method with a test function [tex]xe^{- \alpha x}[/tex].

    2. Relevant equations

    The variational method states that

    [tex]\frac{\langle \psi \mid \hat{H} \mid \Psi \rangle}{\langle \psi \mid \Psi \rangle} \ge E_0.[/tex]


    3. The attempt at a solution

    I calculate

    [tex] \langle \psi \mid \hat{H} \mid \Psi \rangle = \frac{3\hbar^2}{8m} + \frac{c}{4 \alpha^3}.[/tex]

    and

    [tex]\langle \psi \mid \Psi \rangle = \frac{1}{4 \alpha^3}.[/tex]

    In total this gives me

    [tex]\frac{\langle \psi \mid \hat{H} \mid \Psi \rangle}{\langle \psi \mid \Psi \rangle} = \frac{3 \hbar^2 \alpha^3}{2m} + c.[/tex]

    Now assuming I had gotten this far correctly, I was a little confused where to go from here. Presumable I need to minimize this with respect to alpha. However, this would mean that alpha will be negative and unbounded, correct? Or do I need to assume that alpha is greater than 0. In that case, wouldn't the minimum energy just be c? This seems wrong to me. Any advice would be appreciated.
     
  2. jcsd
  3. Dec 6, 2011 #2
    Those should all be the same bras and kets up there. Can't go back and edit for some reason.
     
  4. Jan 30, 2012 #3
    I believe that your integrals were a little off.

    I find that the integrals should be

    hbar^2/(8m*alpha), 3/(8*alpha^4), (4*alpha^3)^(-1)

    After you simplify those, your next step is always to take the derivative with respect to alpha and set to zero. You were right to be alarmed by an unbounded, negative energy. That should not occur.
     
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