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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:

    [itex]V(x)=
    \begin{Bmatrix}
    -V_0 (1- \begin{vmatrix}
    \frac{x}{a}
    \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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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
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