1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: QM variational principle
  1. Variational Principle (Replies: 3)

  2. Variational Principle (Replies: 4)

  3. QM Variation Method (Replies: 3)

  4. Variational Principle (Replies: 7)

Loading...