QM variational principle

  • Thread starter ehrenfest
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[SOLVED] QM variational principle

Homework Statement


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]

Homework Equations





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.
 
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Answers and Replies

  • #2
Dick
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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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