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# PSU-QM-22001 ---- WKB Approximation

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Summary
Problem Solving Unit-1

Bound State Energies, Tunneling Through a Barrier
In this unit use of the WKB approximation for estimating bound state energies and transmission through a barrier is explained through short examples and solved problems. A couple of  exercise sets containing practice problems on direct application of the WKB results are included. No prior knowledge of WKB approximation is required.

This unit is designed to be self contained and learning to solve problem from this unit does not require use of any other text book or reference material.  Details of derivations of WKB formula  are not needed and are not included here.

Contents

Users are welcome to post comments, questions at the end of the page and also to contribute problems.

It is assumed that you already familiar with the main results of WKB approximation. For purposes of this set an understanding of the` derivation of the formula is not required,

Objectives
The aims of problem solving unit are as follows.

1.   to understand the standard formula for
(i)  bound state energies in one dimension
(ii) transmission coefficient through a barrier
2.   to learn how to apply them to potential problems
3.   to compute bound state energy spectrum and
4.   to find transmission coefficients from a potential barrier.

Tips for Turning Points

A few potentials are drawn in figure below. The turning points are

marked as open circles for energies indicated by  dashed horizontal lines.

(1)   For potential in Fig. 1, there are two turning points for all energies

(2)  The potential in Fig. 2 (a) has two turning points for some energies and four turning points for some other energies. Fig 2(b) shows a barrier with two turning points for energies $E>0$ and  below the barrier  height $V_0$. Fig2(b)  shows a well with bounded motion with two turning points for energy $E <0$ and  unbounded motion for  $E>0$.

Fig. 2

(3)  In the above two examples the turning  points are obtained by solving  $V(x)=E$. If the potential is piece wise continuous, this may or may not be the case. For example, for one of the two potentials in Fig 3, the turning points are not obtained by solving $E=V(x)$.\\

Fig 3

Estimating Bound State Energies

(1) In a classical description the total energy of a particle is,
$E = \frac{p^2}{2m} + V(x)$
and the particle can move only in regions where $E \ge V(x)$, and this region is called classical region. The turning points of a particle are the boundary points of the classical region.

(2) For a potential with two turning points, the energy eigenvalues in the WKB approximation are given by the quantization rule
$2 \int_{x_1}^{x_2} p_\text{c}(x) dx = (n+\tfrac{1}{2}) h \label{EQ1.1}$
where $h$ is Planck's constant and $p_\text{c}(x)$ is the classical momentum function given by $p_\text{c} = \sqrt{2m(E-V(x))}$
Do not miss a factor of 2 in front of the integral in the left hand side of \eqref{EQ1.1}. The bound state energies are obtained by solving (\ref{EQ1.1}) for $E$.

(3) Note that the WKB quantization rule is written in several equivalent forms such as
$\oint_{x_1}^{x_2} p_\text{c}(x) dx = (n+\tfrac{1}{2}) h, \label{EQ1.3}$    $\frac{1}{2\pi} \oint_{x_1}^{x_2} p_{c}(x) dx = (n+\tfrac{1}{2}) \hbar \label{EQ1.4}$ Note that in the (\ref{EQ1.3}), the integral, $\oint ...\,dx$, is over one full period, over the motion from $x_1$ to $x_2$ and back. While considering integral from $x_2$ to $x_1$ the sign of momentum is to be reversed. In (\ref{EQ1.4}) the appearance of $\hbar$ in the right hand side, rather than $h$, should be noted.

### Solved Example

Problem : Show that the WKB approximation gives exact answer for energy eigenvalues of a harmonic oscillator.
Solution :  For harmonic oscillator

H=\frac{p^2}{2m} +  V(x), \qquad V(x)=\frac{1}{2}m \omega^2 x^2.

The turning points for energy $E$ are obtained by solving $E=V(x)$ for $x$.
\begin{eqnarray}
\end{eqnarray}
Thus the two turning points are $\pm a$ with $a=\sqrt{2E/m\omega^2}$. The quantization condition is

2\int_{x_1}^{x_2} p(x) \, dx = (n+1/2) h

In this case $x_1=-a, x_2=a$  and we compute

\begin{eqnarray} 2\int_{x_1}^{x_2} p(x) \, dx &=& \int_{-a}^a \sqrt{2m(
E-\tfrac{1}{2} m\omega^2 x^2)}\, dx\\ &&        \text{write energy in terms of the turning point ~} a\nonumber\\ &=&

### this are the turning points

yup , these are the turning points of fig 3 b. for fig. 3 a. turning points will be 0, E/A if my potential is Ax.is this correct?

### allowed values of energy

allowed values of energy solved by exact method takes n=1,2,3.. does the same values is for wkb, sir?

### sir, i want perturbation

sir, i want perturbation questions as like wkb.. there are 2-3 questions in every competitive exams!!! so, it will be helpful ..

### The problem solving unit on

The problem solving unit on perturbation is being prepared. It will b uploaded soon.

### sir, if i have to calculate

sir, if i have to calculate transmission coefficient, by wkb, how shud i do that..??

### I have already prepared write

I have already prepared write up for transmission coefficient. It will be posted soon.