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Homework Help: QM variational principle

  1. Dec 27, 2007 #1
    [SOLVED] QM variational principle

    1. The problem statement, all variables and given/known data
    In order to use the variational principle to estimate the ground-state energy of the one-dimensional potential V(x) = Kx^4, where K is a constant, which of the following wave functions would be a better trial wave function:

    1) [tex]\psi(x) = e^{-\alpha x^2}[/tex]

    2) [tex]\psi(x) = x e^{-\alpha x^2}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    The potential is symmetric, so we know that the wavefunctions have to be even or odd. The main difference is that the second has psi(0)=0 and is odd while the first one has psi(0)=1 and is even. So I am not sure why one of these is better than the other.
    Last edited: Dec 27, 2007
  2. jcsd
  3. Dec 27, 2007 #2


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    Science Advisor
    Homework Helper

    Generally, the more nodes in a wave function (points where psi=0), the higher the energy. So the first test function would be good to approximate the ground state, the second better for a first excited state. Remind yourself what the wavefunctions look like for the harmonic oscillator.
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