WKB Approximation with V-Shaped well

  • #1
117
7

Homework Statement



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?

Homework 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

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!
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,965
3,322
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]
If ##p(x)## represents the momentum, should you be dividing by ##\hbar##?
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]?##
It should be the expression that follows from solving the energy equation ##\frac{p^2}{2m} + V = E## for ##p##.

So my question is: what is my turning point?
[itex] \frac{\sqrt{2m(\frac {V_0}{a}x-V_0-E))}}{\hbar} =0 \rightarrow x=a [/itex]
Is this turning point correct?
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:

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