WKB Approximation with V-Shaped well

In summary, the conversation discusses using the WKB method to find the bound state energies of a particle in a V-shaped potential well. The participant asks questions about the validity of their approach, specifically regarding the calculation of the momentum term and the determination of the turning point. They are advised to check their equations and consider using a sketch to visualize the problem.
  • #1
DeldotB
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!
 
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  • #2
DeldotB said:
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:

1. What is the WKB Approximation method?

The WKB (Wentzel-Kramers-Brillouin) Approximation method is a mathematical technique used to approximate the solutions to the Schrodinger equation, which describes the behavior of quantum particles. It is commonly used to solve problems involving potential wells, such as the V-Shaped well.

2. How does the WKB Approximation with V-Shaped well work?

The WKB Approximation with V-Shaped well involves dividing the potential well into two regions: the classically allowed region, where the particle has enough energy to move freely, and the classically forbidden region, where the particle does not have enough energy to move. The solutions to the Schrodinger equation are then approximated separately in each region and matched at the boundaries.

3. What are the advantages of using the WKB Approximation with V-Shaped well?

The WKB Approximation with V-Shaped well is advantageous because it provides a simple and intuitive way to approximate the solutions to the Schrodinger equation in a potential well. It also allows for the calculation of energy levels and wavefunctions for the particle in the well, which can provide valuable insights into its behavior.

4. Are there any limitations to the WKB Approximation with V-Shaped well?

Yes, there are limitations to the WKB Approximation with V-Shaped well. It is most accurate for shallow potential wells and breaks down for deep wells. It also does not take into account quantum effects such as tunneling.

5. How is the WKB Approximation with V-Shaped well used in practical applications?

The WKB Approximation with V-Shaped well is used in various practical applications, such as in the study of electron transport in semiconductor devices and in the analysis of molecular spectra. It is also used in the field of quantum mechanics to provide insights into the behavior of particles in potential wells.

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