# QM variational principle

1. Dec 27, 2007

### ehrenfest

[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) $$\psi(x) = e^{-\alpha x^2}$$

2) $$\psi(x) = x e^{-\alpha x^2}$$

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. Dec 27, 2007

### Dick

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.