WKB Approximation with V-Shaped well

[DeldotB]

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 E_n of a particle of mass m in a V-shaped potential well:

V(x)= <br /> \begin{Bmatrix}<br /> -V_0 (1- \begin{vmatrix}<br /> \frac{x}{a}<br /> \end{vmatrix})\: \: -a&lt;x&lt;a\\ <br /> 0 \: \: \: if \: \: \: \: \: \: x\notin (-a,a)<br /> <br /> \end{Bmatrix} where V_0 and a are positive constants.

What is the energy of the highest bound state?

Homework Equations

WKB approximation for "potential with no vertical walls":

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

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

The Attempt at a Solution

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

\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

So a few question arise:

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

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

2 \int_{0}^{x_2} \frac{\sqrt{2m(-V_0(1-|\frac{x}{a}|)-E))}}{\hbar} dx 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:

V(x)= \frac {V_0}{a}x-V_0 for 0&lt;x&lt;a. Thus my momentum becomes:

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

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

Is this turning point correct? Am I on the right track? Thanks!

 

[TSny]

the wave function is "in" the well.
p(x)=\frac{\sqrt{2m(V-E)}}{\hbar}=\frac{\sqrt{2m(-V_0(1-|\frac{x}{a}|)-E))}}{\hbar}

If $p(x)$ represents the momentum, should you be dividing by $\hbar$?

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

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?
\frac{\sqrt{2m(\frac {V_0}{a}x-V_0-E))}}{\hbar} =0 \rightarrow x=a
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.