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!

WKB Approximation with V-Shaped well

  1. May 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Good day all!
    I'm studying for finals and i'd like to know how to do this problem (its not homework):

    "Using the WKB method, find the bound state energies [itex] E_n [/itex] of a particle of mass m in a V-shaped potential well:

    -V_0 (1- \begin{vmatrix}
    \end{vmatrix})\: \: -a<x<a\\
    0 \: \: \: if \: \: \: \: \: \: x\notin (-a,a)

    \end{Bmatrix} [/itex] where [itex]V_0[/itex] and [itex]a[/itex] are positive constants.

    What is the energy of the highest bound state?

    2. Relevant equations

    WKB approximation for "potential with no vertical walls":

    [itex] \int_{x_1}^{x_2} p(x)dx=(n-\frac{1}{2})\pi \hbar [/itex]

    Here [itex]x_1,x_2 [/itex] are the classical "turning points" i.e to find them I set [itex]p(x)=0[/itex] and solve for x

    3. The attempt at a solution

    Well, if its a bound state, then [itex] E < V [/itex], so the wave function is "in" the well.
    [tex] p(x)=\frac{\sqrt{2m(V-E)}}{\hbar}=\frac{\sqrt{2m(-V_0(1-|\frac{x}{a}|)-E))}}{\hbar} [/tex]
    Thus the integral is:

    [itex] \int_{x_1}^{x_2} \frac{\sqrt{2m(V-E)}}{\hbar}=\int_{x_1}^{x_2} \frac{\sqrt{2m(-V_0(1-|\frac{x}{a}|)-E))}}{\hbar} dx=(n-\frac{1}{2})\pi \hbar [/itex]

    So a few question arise:

    Is my p(x) term correct? In this case is it: [tex]p(x)=\frac{\sqrt{2m(V-E)}}{\hbar}[/tex] OR [tex]p(x)=\frac{\sqrt{2m(E-V)}}{\hbar} [/tex]?

    Also, the well is symmetrical, so I can just to the integral:

    [tex]2 \int_{0}^{x_2} \frac{\sqrt{2m(-V_0(1-|\frac{x}{a}|)-E))}}{\hbar} dx[/tex] So my question is: what is my turning point?
    If I look just at the positive x values, I can drop the absolute value sign in the potential and I get:

    [itex] V(x)= \frac {V_0}{a}x-V_0 [/itex] for [itex] 0<x<a [/itex]. Thus my momentum becomes:

    [itex] \frac{\sqrt{2m(\frac {V_0}{a}x-V_0-E))}}{\hbar}[/itex]. Setting this equal to zero to find the left most turning point gives:

    [itex] \frac{\sqrt{2m(\frac {V_0}{a}x-V_0-E))}}{\hbar} =0 \rightarrow x=a [/itex]

    Is this turning point correct? Am I on the right track? Thanks!
  2. jcsd
  3. May 8, 2016 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    If ##p(x)## represents the momentum, should you be dividing by ##\hbar##?
    It should be the expression that follows from solving the energy equation ##\frac{p^2}{2m} + V = E## for ##p##.

    I don't think you solved for ##x## correctly. But, first, you want to make sure the left side is the correct expression for ##p##. A sketch of V(x) with a line drawn for E can help you see that the turning point cannot be ##x = a## in general.
    Last edited: May 8, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: WKB Approximation with V-Shaped well