# WKB Approximation

The WKB (Wentzel-Kramers-Brillouin method) method is a technique for finding approximations to certain differential equations, including the one dimensional Schrödinger equation. It was developed in 1926 by Wenzel, Kramers, and Brillouin, for whom it was named. The logic behind this method is that, as $\hbar\rightarrow 0,\!$ the wavelength, $\lambda=2\pi\hbar/ p,\!$ tends to zero where the potential is smooth and slowly varying. Therefore, $\lambda\!$ can be thought of as a local quantity $\lambda(x).\!$ This is a semiclassical method of solving the Schrödinger equation.

However, for the classical turning points of the potential (i.e., where the energy of the particle becomes equal to the potential), $\lambda\!$ becomes infinite, and we can no longer say that the potential changes slowly. Here, the whole theory is bound to fail. A proper handling of the turning points is the most difficult aspect of the WKB approximation. In this case, the potential at the turning point is approximated as linear and slowly varying (almost constant).

By doing this, we can solve the Schrödinger equation and obtain the wave function in this "region". We can then use this wave function to connect the WKB wave functions at the two sides of the turning point.

## Derivation of the WKB Approximation

For $V(x) = 0\!$ (or constant) the solutions to the Schrödinger equation are simply plane waves of the form $e^{\pm ikx}$. If the potential varies smoothly, and the energy of the particle is fixed, the wave function can be described locally by writing its plane wave form $e^{iu(x)}\!$. If we substitute this form into the Schrödinger equation, then we obtain

$\left (\frac{du}{dx}\right )^2=\left [\frac{p(x)}{\hbar}\right ]^2+i\frac{d^2u}{dx^2},$

where $p(x)=\sqrt{2m[E-V(x)]}.$ Let us now assume that the second derivative on the right is small compared to the rest of the terms and neglect it as a first approximation. Let us call this first approximation to $u(x)\!$ $u_0(x).\!$ The equation satisfied by $u_0(x)\!$ is

$\frac{du_0}{dx}=\pm\frac{p(x)}{\hbar}.$

Solving this equation, we obtain

$u_0(x)=\pm\frac{1}{\hbar}\int_{x_0}^x dx'\,p(x')+C_0.$

We now use this first approximation to solve the full differential equation by an iterative process. We obtain the next approximation by substituting $u_0(x)\!$ into the right-hand side and solving for $u(x)\!$ on the left. If we call the result $u_1(x),\!$ then the equation that we solve is

$\frac{du_1}{dx}=\pm\sqrt{\left [\frac{p(x)}{\hbar}\right ]^2\pm\frac{i}{\hbar}\frac{dp}{dx}}.$

We must assume that $u_1(x)\!$ has the same sign as $u_0(x)\!$ in order for this iterative process to converge, and we must assume that $\frac{dp}{dx}\ll \frac{[p(x)]^2}{\hbar}$ so that the deviation of $u_1(x)\!$ from $u_0(x)\!$ is small. If this is the case, then we may assume that any further corrections introduced in further iterations are even smaller, and that $u_1(x)\!$ is a good approximation to the exact solution. Under this assumption, we may expand the square root on the right-hand side of this equation, obtaining

$\frac{du_1}{dx}\approx\pm\frac{p(x)}{\hbar}+\frac{i}{p(x)}\frac{dp}{dx}.$

Solving this equation, we find that, for the classically accessible region, $E>V(x),\!$ the WKB wave function is

$\psi(x)\approx \frac{C_+}{\sqrt{p(x)}}\exp\left [\frac{i}{\hbar}\int_{x_0}^x p(x')\,dx'\right ]+\frac{C_-}{\sqrt{p(x)}}\exp\left [-\frac{i}{\hbar}\int_{x_0}^x p(x')\,dx'\right ],$

and for the classically forbidden region, $E

$\psi(x)\approx \frac{C_+}{\sqrt{|p(x)|}}\exp\left[\frac{1}{\hbar}\int_{x_0}^x |p(x')|\,dx'\right] + \frac{C_-}{\sqrt{|p(x)|}}\exp\left[- \frac{1}{\hbar}\int_{x_0}^x |p(x')|\,dx'\right].$

This is an exact solution if $V(x)\!$ is constant, otherwise it is a local solution for a locally-defined wavelength. In order for this approximation to be valid, we must assume that the wavelength is slowly varying as a function of $x.\!$ In other words, $\left |\frac{d\lambda}{dx}\right |=\left |\frac{d}{dx}\left(\frac{\hbar}{p(x)}\right)\right | \ll 1.$

Let us now suppose that we let $x_0=a,\!$ where $a\!$ is a classical turning point at a given value of $E\!$ and thereby separates a region where $E > V(x)\!$ from a region where $E < V(x).\!$ Let the classically inaccessible region be to the right of $a.\!$ For $x\!$ sufficiently far away from $a,\!$ the WKB wave function is

$\psi(x)\approx \frac{A}{\sqrt{|p(x)|}}\exp\left[- \frac{1}{\hbar}\int_{a}^x |p(x')|\,dx'\right]+ \frac{B}{\sqrt{|p(x)|}}\exp\left[\frac{1}{\hbar}\int_{a}^x |p(x')|\,dx'\right]$

for $x \gg a, \!$ and

$\psi(x)\approx \frac{C}{\sqrt{p(x)}}\exp\left[- \frac{i}{\hbar}\int_{a}^x p(x')\,dx'\right]+ \frac{D}{\sqrt{p(x)}}\exp\left[\frac{i}{\hbar}\int_{a}^x p(x')\,dx'\right]$

for $x \ll a. \!$

## Patching the Wave Function Near a Classical Turning Point

Note that, at the classical turning point, $p(a) = 0\!$ and the WKB solution diverges, meaning that it is no longer a valid approximation because the true wave function will not exhibit such divergent behavior at the turning points. Thus, around each turning point, we need to splice the two WKB solutions on either side of the turning point with a "patching" function that will straddle each turning point. Because we only need a solution for this function in the vicinity of the turning points, we can approximate the potential as being linear. If we center the turning point at the origin, i.e. we take $a=0,\!$ the potential is approximately

$V(x)\approx E+V'(0)x.\!$

Solving the Schrödinger equation with this linearized potential leads to the Airy equation whose solutions are Airy functions. Our patching wave function, which we denote by $\psi_p(x),\!$ is then:

$\psi_p(x)= C_1 \mathrm{Ai}\left(\alpha x\right)+ C_2 \mathrm{Bi}\left(\alpha x\right)\!$

where $C_1\!$ and $C_2\!$ are complex number coefficients and

$\alpha=\left(\frac{2m}{\hbar^2}V'(0)\right)^{1/3}.$

The key to patching the wavefunction in the region of the turning point is to asymptotically match the patching wavefunction to the wavefunctions far away from the classical turning point. In the vicinity of the classical turning point,

$p^2=2m(-V'(0)x)\Rightarrow 2p\frac{dp}{dx}=-2mV'(0)\Rightarrow \frac{dp}{dx}=-\frac{m}{p}V'(0)$

Since the region of applicability of the WKB approximation is

$1\gg\frac{1}{2\pi}\left|\frac{d\lambda}{dx}\right|,$

we find that, near the turning point,

$|p|^3 \gg \hbar m|V'(0)|,$

or

$|x|\gg \tfrac{1}{2}\hbar^{2/3}|mV'(0)|^{-1/3}.$

This implies that the width of the region around the classical turning point vanishes as $\hbar^{2/3}.$ Thus, we can come as close to the turning point as we wish with the WKB approximation by taking the limit as $\hbar\!$ approaches zero, as long as the distance from the classical turning point is much greater than $\hbar^{2/3}.$ Thus, by extending the patching function towards the singularity in the WKB wave function, while simultaneously extending the WKB wave function toward the classical turning point, it is possible to match the asymptotic forms of the wave functions from the two regions, which are then used to patch them together.

For this purpose, the following asymptotic forms of the Airy functions $\text{Ai}(z)\!$ and $\text{Bi}(z)\!$ as $z\rightarrow\pm\infty$ will be useful:

$z\rightarrow \infty: \mathrm{Ai}(z)\rightarrow \frac{1}{2\sqrt{\pi}}z^{-\frac{1}{4}}e^{-\frac{2}{3}|z|^{\frac{3}{2}}}$
$z\rightarrow \infty: \mathrm{Bi}(z)\rightarrow \frac{1}{\sqrt{\pi}}z^{-\frac{1}{4}}e^{\frac{2}{3}|z|^{\frac{3}{2}}}$
$z\rightarrow -\infty: \mathrm{Ai}(z)\rightarrow \frac{1}{\sqrt{\pi}}z^{-\frac{1}{4}}\cos\left(\frac{2}{3}|z|^{\frac{3}{2}}-\frac{\pi}{4}\right)$
$z\rightarrow -\infty: \mathrm{Bi}(z)\rightarrow -\frac{1}{\sqrt{\pi}}z^{-\frac{1}{4}}\sin\left(\frac{2}{3}|z|^{\frac{3}{2}}-\frac{\pi}{4}\right)$

Noticing that in the vicinity of the turning point (for negative $x\!$),

$\frac{1}{\hbar}\int_{x}^0p(x')\,dx'=\sqrt{\frac{2mV'(0)}{\hbar^2}} \int_x^0\sqrt{-x'}\,dx'=\tfrac{2}{3}\sqrt{\frac{2mV'(0)}{\hbar^2}}|x|^{3/2}=\tfrac{2}{3}|\alpha x|^{3/2}$

and

$\frac{1}{\sqrt{p(x)}}=\left(2mV'(0)\right)^{-1/4}|x|^{-1/4},$

it becomes apparent that the WKB approximation to the wave function is the same as the patching function in the asymptotic limit. This must be the case, since, as $\hbar\rightarrow 0,$ the region of invalidity of the semiclassical wave function in the vicinity of the turning point shrinks, while the solution of the linarized potential problem depends only on the accuracy of the linearity of the potential, and not on $\hbar.$ The two regions must therefore overlap.

For example, one can take $x\sim \hbar^{1/3}$ and then take the limit $\hbar\rightarrow 0.$ The semiclassical solution must hold as we are always in the region of its validity and so must the solution of the linearized potential problem. Note that the argument of the Airy functions at $x\sim \hbar^{1/3}$ goes to $\pm\infty$, which is why we need their asymptotic expansion.

For a classical turning point $x=a\!$ that separates a classically inaccessible region $x>a\!$ from a classically accessible region $x we compare the WKB wave functions with the asymptotic expressions for the Airy functions and find that the WKB wave functions on the two sides of this turning point are connected as follows:

$\frac{2A}{\sqrt{k(x)}}\cos\left (\int_{x}^{a}k(x')\,dx'-\frac{\pi}{4}\right )-\frac{B}{\sqrt{k(x)}}\sin\left (\int_{x}^{a}k(x')\,dx'-\frac{\pi}{4}\right ) \longleftrightarrow \frac{A}{\sqrt{|k(x)|}}\exp\left (-\int_{a}^{x}|k(x')|\,dx'\right )+\frac{B}{\sqrt{|k(x)|}}\exp\left (\int_{a}^{x}|k(x')|\,dx'\right ),$

where $k(x)=\frac{p(x)}{\hbar}.$ Similarly, for a classical turning point $x=b,\!$ that separates a classically accessible region $x>b\!$ from a classically inaccessible region $x we find that the connection between the WKB wave functions on either side of the turning point is

$\frac{2A}{\sqrt{k(x)}}\cos\left (\int_{b}^{x}k(x')\,dx'-\frac{\pi}{4}\right )-\frac{B}{\sqrt{k(x)}}\sin\left (\int_{b}^{x}k(x')\,dx'-\frac{\pi}{4}\right ) \longleftrightarrow \frac{A}{\sqrt{|k(x)|}}\exp\left (-\int_{x}^{b}|k(x')|\,dx'\right )+\frac{B}{\sqrt{|k(x)|}}\exp\left (\int_{x}^{b}|k(x')|\,dx'\right ).$

## Bound States Within the WKB Approximation

We may use the WKB approximation to find the bound state energies of a potential, if any. We will show that, in fact, we recover the Bohr-Sommerfeld quantization rule from the old quantum theory. We will have, in effect, shown that the old quantum theory is in fact a semiclassical approximation to the modern quantum theory that we have been studying thus far.

Let us consider a potential with no rigid walls. Let us also label the boundaries of the classically accessible region for a given energy $E\!$ as $b\!$ and $a,\!$ such that the accessible region is $b To the left of this region, the wave function of a bound state must have the form,

$\psi(x)=\frac{A}{\sqrt{|p(x)|}}\exp\left (-\frac{1}{\hbar}\int_x^b |p(x')|\,dx'\right ),$

in order for it to exponentially decay to zero as $x\rightarrow -\infty.$ Using the connection formulas derived earlier, we find that the wave function inside the classically accessible region is

$\psi(x)=\frac{2A}{\sqrt{|p(x)|}}\cos\left (\frac{1}{\hbar}\int_b^x p(x')\,dx'-\frac{\pi}{4}\right ).$

This may be rewritten as

$\psi(x)=\frac{2A}{\sqrt{|p(x)|}}\sin\left (\frac{1}{\hbar}\int_b^a p(x')\,dx'\right )\cos\left (\frac{1}{\hbar}\int_x^a p(x')\,dx'-\frac{\pi}{4}\right )-\frac{2A}{\sqrt{|p(x)|}}\cos\left (\frac{1}{\hbar}\int_b^a p(x')\,dx'\right )\sin\left (\frac{1}{\hbar}\int_x^a p(x')\,dx'-\frac{\pi}{4}\right ).$

Again using the connection formulas, we see that the coefficient of the second term must vanish in order for the wave function for $x>a\!$ to decay exponentially to zero as $x\rightarrow\infty;$ i.e.,

$\cos\left (\frac{1}{\hbar}\int_b^a p(x')\,dx'\right )=0.$

This is satisfied if

$\int_b^a p(x)\,dx=\left (n+\tfrac{1}{2}\right )\pi\hbar.$

A similar treatment for the case where one of $b\!$ or $a\!$ is a rigid wall rather than a simple classical turning point yields the condition,

$\int_b^a p(x)\,dx = \left(n+\tfrac{3}{4}\right)\pi\hbar.$

If both $b\!$ and $a\!$ are rigid walls, then we obtain

$\int_b^a p(x)\,dx = n\pi\hbar.$

## Transmission and Reflection Through a Barrier

The WKB approximation can also be used to calculate transmission and reflection coefficients through a potential barrier. In the presence of such a barrier, the wave function will have the form,

$\psi(x)= \begin{cases} \frac{A}{\sqrt{|p(x)|}}\exp\left [\frac{i}{\hbar}\int_{a}^{x} p(x')\,dx'\right ]+\frac{B}{\sqrt{|p(x)|}}\exp\left [-\frac{i}{\hbar}\int_{a}^{x} p(x')\,dx'\right ], & xb, \end{cases}$

where, as usual, $a\!$ and $b\!$ are the classical turning points for the barrier. If we now use the connection formulas derived earlier, we arrive at the relation,

$\left [ \begin{matrix} A \\ B \end{matrix} \right ]= \tfrac{1}{2}\left [ \begin{matrix} 2\theta+\frac{1}{2\theta} & i\left (2\theta-\frac{1}{2\theta}\right ) \\ -i\left (2\theta-\frac{1}{2\theta}\right ) & 2\theta+\frac{1}{2\theta} \end{matrix} \right ] \left [ \begin{matrix} F \\ G \end{matrix} \right ],$

where

$\theta=\exp\left [\int_a^b |p(x)|\,dx\right ].$

The transmission coefficient is then defined in terms of the momentum right at the barrier, which will be the same on both sides, and is thus simply given by

$T=\left |\frac{F}{A}\right |^2.$

If we now assume that there is no wave incident from the right, then $G=0\!$ and

$T=\frac{4}{\left (2\theta+\frac{1}{2\theta}\right )^2}.$

If the barrier is high and broad, then $\theta\ll 1,$ so that we may approximate the transmission coefficient as

$T\approx\frac{1}{\theta^2}=\exp\left [-2\int_a^b |p(x)|\,dx\right ].$

## Problems

(1) Consider the potential,

$V(x) = V_{0}\ln\left (\frac{x}{a}\right ),\,x>0,$

where V0 and a are constants. Assume that there is an infinite wall at $x=0.\!$ Determine the bound states of this potential within the WKB approximation. Show that the spacing between the levels is independent of mass.

(2) Calculate the transmission probability for an alpha particle to tunnel through the Coulomb barrier of a nucleus of an atom with atomic number $Z.\!$ The potential has the form,

$V(r)=\frac{2z_{1}e^{2}}{r},$

for $r>a\!$ and is constant for $r (see figure). In doing so, you will have calculated the (approximate) probability for a nucleus to undergo alpha decay. Note that the expression that you will obtain contains an exponential factor that is independent of $a;\!$ this is known as the Gamow factor.