1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Quantum Harmonic Oscillator Problem

  1. Feb 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Substitute [tex] \psi = Ne^{-ax^2} [/tex] into the position-space energy eigenvalue equation and determine the value of the constant a that makes this function an eigenfunction. What is the corresponding energy eigenvalue?

    2. Relevant equations
    [tex] \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \langle x | E \rangle + \frac{1}{2} m \omega^2 x^2 \langle x| E \rangle = E \langle x | E \rangle [/tex]

    3. The attempt at a solution

    So, initially I tried to solve for a by plugging in [tex] \psi [/tex] but I got a nasty quadratic [tex]\frac{2 \hbar^2 x^2}{m} a^2 - \frac{\hbar^2}{m} a + \left( E - \frac{1}{2} m \omega^2 x^2 \right) = 0 [/tex]that didn't really seem right. I then did some research and found a similar problem where the book stopped at a similar quadratic equation (the problem was for the first excited state) and said that "the x^2 terms must cancel.

    Why is that? I guess I haven't seen the energy eigenstates depend on x before... So I suppose that gives reason to believe that the terms with x will negate each other?

    Following that recipe I arrived at the same value of a as they do [tex] a = \frac{m \omega}{2 \hbar} [/tex] and I also arrive at the expected energy (Since this given psi is of the form of the ground state harmonic oscillator) [tex] E = \frac{\hbar \omega}{2} [/tex].

    Thanks for any help with understanding this deeper.
    Last edited: Feb 15, 2017
  2. jcsd
  3. Feb 15, 2017 #2


    User Avatar
    Science Advisor
    Homework Helper

    Because that quadratic equation must hold for any value of ##x##.
  4. Feb 15, 2017 #3
    Ok! I think I see. the rest of the terms are constant, so if x varies the zero on the right must still obviously be a 0. Only way this happens is if the two x^2 terms are equal, right?
  5. Feb 15, 2017 #4


    User Avatar
    Science Advisor
    Homework Helper

  6. Feb 15, 2017 #5
    Thank you so much! That is kind of a subtle thing to notice! Makes total sense now though, thanks again!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted