PROOFS PROGRAMME

QM-Problem Solving Unit-22 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

  1. Tips
  2. Estimating Bound State Energies
  3. Tunneling Through a Barrier

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                                                                                                                          

                                                                                                                       

The RoadMap

(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
  \begin{equation}
      H=\frac{p^2}{2m} +  V(x), \qquad V(x)=\frac{1}{2}m \omega^2 x^2.
  \end{equation}
  The turning points for energy $E$ are obtained by solving $E=V(x)$ for $x$.
  \begin{eqnarray}
     E =\frac{1}{2} m \omega^2 x^2, \qquad\Rightarrow\qquad x^2  = 
\frac{2E}{m\omega^2}  \qquad \qquad \therefore ~~~~~ x= \pm\sqrt{2E/m\omega^2}
\end{eqnarray}
Thus the two turning points are $\pm a$ with $a=\sqrt{2E/m\omega^2}$. The quantization condition is
\begin{equation}
  2\int_{x_1}^{x_2} p(x) \, dx = (n+1/2) h
\end{equation}
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\\ &=&
\int_{-a}^a \sqrt{2m({\tfrac{1}{2}m \omega^2 a^2$}-\tfrac{1}{2} m\omega^2 x^2)}\,\, dx\nonumber\\ &=&m\omega \int_{-a}^a
\sqrt{a^2-x^2}\, dx \end{eqnarray}

This is a standard integral.  Always try to use a dimensionless integration variable. To integrate, substitute

$$x=a \sin\theta, \quad dx = a \cos\theta d\theta .$$
This gives
\begin{eqnarray} \int_{-a}^a p(x) dx &=& m\omega a^2 \int_{-\pi/2}^{\pi/2}
~\cos^2\theta d\theta\\ &=& \cdots \quad \cdots \nonumber\\ &=& \frac{\pi m
\omega a^2}{2}. \end{eqnarray}
Therefore quantization condition takes the form
\begin{eqnarray} 2 \Big(\frac{\pi m\omega}{2}\Big) a^2 &=& (n+1/2)
$h$ \\ {\pi m\omega}\frac{2E}{m\omega^2} &=& (n+1/2)
$h$ \\ \text{or}\qquad \qquad E&=&(n+1/2)\hbar \omega.
\end{eqnarray}
which is the well known exact result for the harmonic oscillator energy levels. 

Exercises

(1) Using WKB approximation find the energy levels of a particle in a  gravitational potential of the earth. You may assume the acceleration due to gravity to remain constant and that the surface of the earth is a rigid boundary.
(2) 

Find the energy levels of a particle in well
           \[ V(x) = A |x|  \qquad A > 0 \]
using WKB approximation. How does your answer compare with the large $n$ limit
          \[ E_n = (n-1/4)^{\tfrac{2}{3}} \Big( \frac{3\pi}{2\surd 2}\Big)^{\tfrac{2}{3}} \Big( \frac{A^2\hbar^2}{m} \Big)^{\tfrac{1}{3}} \]
of the exact answer. 

(3)

Show that the energy levels of a particle in potential well
          \[ V(x) = A |x|^p  \qquad A>0 \]
using the WKB quantization rule are given by
     \[ E_n = (n+\tfrac{1}{2})^{2p/(p+2)}\left( \frac{2\pi \hbar|A|^{1/p}}  {\sqrt{2m}}\, I_p \right) \]
where $I(p)$ is given by \[ I(p)=2 \int_0^1 \sqrt{1-|x|^p} dx  \]

(4)

Find the energy levels of a particle in a potential well  \[v(x)= A|x|^{1/2}\]
using the WKB approximation.
       \[ E_n= n^{2/5}\Big(\frac{15 \hbar A^2}{8\sqrt{2m}} \Big )^{\frac{2}{5}}\]

           

        Tunneling Through a Barrier                                                                                                                              

The RoadMap

The trasmission probability for particle to tunnel through a high potential barrier  is given by 

\[  T = \exp\Big(-2 \int_{x_1}^{x_2} |p(x)|\, dx\Big) \]

where \(x_1, x_2\) are two turning points, and \(p(x)= \sqrt{2m(E-V(x)}\) is the classical momentum. It must be remebered that this formula is valid for a high barrier. It cannot be used, for example, for a  \(E\) ~ barrier height,  or for a potential well.

Solved Example

Compute the transmission coefficient in the WKB approximation for a potential  barrier $V(x)$ as shown in figure for $E=V_0/2$. 
The given a potential \(V\) 
     $$ V(x) = \begin{cases} V_0 (1+x/L), & -L < x < 0\\ V_0 (1-x/L), & 0 < x <  L \\ 0,  \text{otherwise .}
t\end{cases}$$

The turning point are obtained by solving for $E=V(x)$. From the figure we find that the two turning points are $x_1=-L/2, x_2=L/2$. The transmission coefficient is then given by
\begin{eqnarray} T = \exp\Big(-2 \int_{x_1}^{x_2} |p(x)| \Big) \end{eqnarray}  

where $|p(x)| = \sqrt{2m(V(x)-E)}$. Compute the integral $\int |p(x)|\, dx$: 

\begin{eqnarray} \int_{x_1}^{x_2} |p(x)|\, dx &=& \int_{-L/2}^{L/2}
\sqrt{2mV(x)-E)}\,dx\nonumber\\ &=& \int_{-L/2}^{0} \sqrt{2m(V_0(1+x/L)-E)}\,dx
+ \int_{0}^{L/2} \sqrt{2m((V_0(1-x/L)-E)}\,dx \nonumber\\ &&
\hfill{{\color{blue}V(x) \text{ has  different expressions in intervals} (-L/2,0)
\text{ and } (0,L/2)} }\nonumber\\ &=&2 \int_{0}^{L/2}
\sqrt{2m(V_0(1-x/L)-E)}\,dx \nonumber\\ &=&2 \int_{0}^{L/2} \sqrt{2m(V_0-E
-V_0x/L)-E)}\,dx \nonumber\\ &=&\left.2\times (2/3)\times \sqrt{2m} \Big(V_0 -E
- \frac{V_0 x}{L} \Big)^{3/2}\right|_0^{L/2}\times
\Big(-\frac{L}{V_0}\Big)\nonumber\\ &=&
-\frac{4L}{3V_0}\sqrt{2m}\Big[(V_0-E-V_0/2)^{3/2} -
(V_0-E)^{3/2}\Big]\nonumber\\ &=&-\frac{4L}{3V_0}\sqrt{2m}\Big(
\frac{V_0}{2}\Big)^{3/2}, \qquad\qquad \because E=V_0/2 \nonumber\\ &=&
\frac{2}{3}\sqrt{mV_0L^2} \end{eqnarray}

Therefore the final answer for the transmission coefficient takes the form $$   T = \exp\Big( - \frac{4}{3}\frac{L}{\hbar}\sqrt{mV_0}\Big) $$ 

Exercises

(1) Plot the potential

\[     V(x) = \begin{cases}            0 & \text{for } x < 0 \\               V_0 -f x & \text{for } x \ge 0   \end{cases}\]
and find the transmission coefficient for the potential barrier using WKB approximation for $E=3V_0/4$.\hfill

(2) Find the transmission probability that an alpha particle with energy $E= Ze^2/ 4R$  will come out of a  a nucleus of charge $Z$ and radius $R$. You may assume that the potential seen by the  alpha particle inside the nucleus can be represented by a square well of range $R$. 

Exclude node summary : 

n

Comments

Tanuhree Gope Comments: i solved 2 questions by wkb, its easy, thanks sir... without ur document, i think, its impossible to solve by reading only 2-3 pages.. ME: So, Do you think this format is helpful? Tanushree Gope: yup, very useful, sir... thats what i said in my previous msg..it is good that i can solve question, which is important to crack any competitive exams.
a typo in equation 7 in the above document, h bar should be h, i guess..., sir.. for figure 3, b the graph is of mod x , so, i can take the turning points as E=V|x| where V is constant., then, x=+-E/V. and for figure a, i guess one turning point will be 0 and other, V(x)=Ax.(form is y=mx) so, x = E/A. am i correct, sir?

1) YES It should be $h$.

2) The turning points will be $x=-E/V$ and $x= + E/V$.

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 solved by exact method takes n=1,2,3.. does the same values is for wkb, sir?
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 perturbation is being prepared. It will b uploaded soon.
sir, if i have to calculate transmission coefficient, by wkb, how shud i do that..??
I have already prepared write up for transmission coefficient. It will be posted soon.