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: Particle in a box

  1. Dec 15, 2011 #1
    1. The problem statement, all variables and given/known data

    V(x) = 0 if [itex]\frac{-L}{2}[/itex]<x<[itex]\frac{L}{2}[/itex] and [itex]\infty[/itex] otherwise.

    Is the wave function [itex]\Psi[/itex] = (2/L)[itex]^{1/2}[/itex] (sin ([itex]\pi[/itex]x/L) an acceptable solution to this? Explain

    2. Relevant equations

    H[itex]\Psi[/itex]= E[itex]\Psi[/itex] , normalization: 1 = [itex]\int[/itex] wavefunction^{2}dx

    3. The attempt at a solution

    My logic is that I have to come up with a wave function that satisfy the boundary condition. Therefore [itex]\Psi[/itex] (x= [itex]\frac{-L}{2}[/itex]) = [itex]\Psi[/itex] (x = [itex]\frac{L}{2}[/itex]) = 0

    My initial answer was that it's not because I was thinking that the wavefunction itself has to be A[/itex]cosine(bx), where b=2n[itex]\pi[/itex]/L. n = 1/4, 3/4 , 5/4 ...

    I am not quite sure if that's correct. Also when I try to calculate the normalization factor (A), it turns out to be 1 = [itex]\frac{A^{2}L}{2}[/itex] + [itex]\frac{A^{2}L}{n\pi}[/itex] sin ([itex]\frac{n\pi}{2}[/itex])
     
  2. jcsd
  3. Dec 15, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    This is right. Does the proposed wave function satisfy these conditions?

    You'll actually get both sine and cosine solutions if you solve the infinite square well problem completely, so just because this wave function has a sine in it isn't reason enough to exclude it.

     
  4. Dec 15, 2011 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think your reasoning is correct. The wave function doesn't vanish at the boundaries. So it's not good. I wouldn't worry about whether the normalization is correct if the boundary conditions aren't correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: QM: Particle in a box
  1. Particle in a box (Replies: 8)

  2. Particle in a box (Replies: 10)

  3. Particle in a box (Replies: 6)

  4. Particle in the box (Replies: 3)

Loading...