# WKB Approximation

The WKB Approximation, named after scientists Wentzel–Kramers–Brillouin, is a method to approximate solutions to a time-independent linear differential equation or in this case, the Schrödinger Equation. Its principal applications are for calculating bound-state energies and tunneling rates through potential barriers. The WKB Approximation is most often applied to 1D problems, but also works for 3D spherically symmetric problems. As a general overview, the wavefunction is assumed to be an exponential function with either amplitude or phase taken to be slowly changing relative to the de Broglie wavelength $$λ$$. It is then semi-classically expanded.

## Solving the Schrödinger Equation

The WKB Approximation states that the wavefunction to the Schrödinger Equation take the form of simple plane waves when at a constant potential $$U$$ (i.e., acts  like a free particle).

$\psi(x)=Ae^{\pm ikx}$

where

$k=\frac{2\pi}{\lambda}=\sqrt{\frac{2m(E-U)}{\hbar^2}}=constant\label{56.1}$

If the potential changes slowly with x $$(U \rightarrow U(x))$$, the solution of the Schrödinger equation is:

$\psi(x)=Ae^{\pm i\phi(x)}\label{56.2}$

where $$\phi (x)=xk(x)$$. For a case with constant potential U, $$\phi (x)=\pm kx$$. Thus, phase changes linearly with $$x$$. For a slowly varying U, $$\phi (x)$$ varies slowly from the linear case $$\pm kx$$.

The classical turning point is defined as the point at which the potential energy U is approximately equal to total energy E $$(U \approx E)$$ and the kinetic energy equals zero. This occurs because the mass stops and reverses its velocity is zer0. It is an inflection point that marks the boundaries between regions where a classical particle is allowed and where it is not, as well as where two wavefunctions must be properly matched. Figure $$\PageIndex{1}$$: A classical particle would be confined to the region where $$E \ge U(x)$$, which is also viewed as the area in between the turning points.

If E > U, a classical particle has a non-zero kinetic energy and is allowed to move freely. If U is a constant, the solution to the one-dimensional Schrödinger equation is:

$\psi(x)=Ae^{\pm ikx},\label{56.3}$

in which the wavefunction is oscillatory with constant wavelength λ and constant amplitude A. $$k(x)$$ is defined as:

$k(x)=\sqrt{\frac{2m(E-U(x))}{\hbar^2}}\label{56.4}$

If E < U, the solution to the Schrödinger equation for a constant U is:

$\psi(x)=Ae^{\pm\kappa x}. \label{56.5}$

In these regions, a classical particle would not be allowed, but there is a finite probability that a particle can pass through a potential energy barrier in quantum mechanics. The quantum particle is described as ‘tunnelling’, which is important in determining the rates of chemical reactions, particularly at lower temperatures. $$k(x)$$ is defined as:

$k(x)=-i\sqrt{\frac{2m(U(x)-E)}{\hbar^2}} = -i\kappa(x).\label{56.6}$

If U(x) is not a constant, but instead varies very slowly on a distance scale of λ, then it is reasonable to suppose that ψ remains practically sinusoidal, except that the wavelength and amplitude change slowly with x.

## WKB Approximation

Substituting in the normalized version of Equation \ref{56.2}, the Schrödinger Equation:

$\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+U(x)\psi(x)=E\psi(x)\label{56.7}$

becomes

$i\frac{\partial^2\phi}{\partial x^2}-\left(\frac{\partial\phi}{\partial x}\right)^2+(k(x))^2=0.\label{56.8}$

The WKB Approximation assumes that the potentials, $$k(x)$$ and $$\phi (x)$$ are slowly varying.

The 0th order WKB Approximation assumes: $$\frac{\partial^2\phi}{\partial x^2} \ge 0.\label{56.9}$$

Thus,

$\left(\frac{\partial\phi_0}{\partial x}\right)^2=(k(x))^2\label{56.10}$

Solving,

$\phi_0(x)=\pm\int k(x)dx+C_0\label{56.11}$

and substituting $$\phi_0(x)$$ into Equation \ref{56.2},

$\psi(x)=e^{i(\pm\int k(x)dx+C_0)}.\label{56.12}$

To obtain a more accurate solution, we manipulate Equation \ref{56.8} to solve for $$\phi(x)$$.

$i\frac{\partial^2\phi}{\partial x^2}-\left(\frac{\partial\phi}{\partial x}\right)^2+(k(x))^2=0$

$\left(\frac{\partial\phi}{\partial x}\right)^2=(k(x))^2+i\frac{\partial^2\phi}{\partial x^2}$

$\frac{\partial\phi}{\partial x}=\sqrt{(k(x))^2+i\frac{\partial^2\phi}{\partial x^2}}$

So,

$\phi(x)=\pm\int\sqrt{(k(x))^2+i\frac{\partial^2\phi}{\partial x^2}}dx+C_1\label{56.13}$

The 1st order WKB Approximation follows the assumption of Equation \ref{56.10} from the 0th order solution. Taking its square root, we find that:

$\frac{\partial\phi_0}{\partial x}=\pm k(x).\label{56.14}$

Taking the derivative on both sides with respect to x, we find that:

$\frac{\partial^2\phi_0}{\partial x^2}=\pm \frac{\partial}{\partial x}k(x).\label{56.15}$

Solving,

$\phi_1(x)=\pm\int\sqrt{(k(x))^2\pm i\frac{\partial}{\partial x}k(x)}dx+C_1\label{56.16}$

and substituting $$\phi_1(x)$$ into Equation \ref{56.2},

$\psi(x)=e^{i\left( \pm\int\sqrt{(k(x))^2\pm i\frac{\partial}{\partial x}k(x)}dx+C_1\right)}.\label{56.17}$

Example $$\PageIndex{1}$$

Determine the tunneling probability $$T$$ at a finite width potential barrier. Figure $$\PageIndex{2}$$: A classical particle moves through Region I towards the boundary (x=0 to x=L) at E<U.

Solution

Given:

$T = \frac{\psi^*(L) \psi(L)}{\psi^*(0)\psi(0)}$ where, $\psi(x)=\psi(0)e^{i\pm(\int k(x)dx+C_1)} \nonumber$

For tunneling to occur, E < U. So,

$$k(x)=-i\sqrt{\frac{2m(U(x)-E)}{\hbar^2}}$$ for E < U(x), assuming U(x) = U.

Plugging in $$k(x)$$ to solve for the wavefunction,

\begin{align*} \psi(x) &=\psi(0)e^{i\left(\pm\int_{0}^{x}-i\sqrt{\frac{2m(U(x)-E)}{\hbar^2}}dx\right)} \\[4pt] &=\psi(0)e^{\left(-\int_{0}^{x}\sqrt{\frac{2m(U-E)}{\hbar^2}}dx\right)} \\[4pt] &=\psi(0)e^{\left(-\left(\sqrt{\frac{2m(U-E)}{\hbar^2}}\right)x\right)} \end{align*}

Thus solving for the the tunneling probability $$T$$,

\begin{align*} T &= \frac{\psi^*(L) \psi(L)}{\psi^*(0)\psi(0)} \\[4pt] &= \frac{\psi(0)e^{\left( - \left( \sqrt{\frac{2m(U-E)}{\hbar^2}}\right) L \right)} \psi(0)e^{\left( - \left( \sqrt{\frac{2m(U-E)}{\hbar^2}}\right) L \right)}}{\psi(0)e^{\left( - \left( \sqrt{\frac{2m(U-E)}{\hbar^2}}\right) 0 \right)} \psi(0)e^{\left( - \left( \sqrt{\frac{2m(U-E)}{\hbar^2}}\right) 0 \right)}} \\[4pt] &= e^{\left( -2 \left( \sqrt{\frac{2m(U-E)}{\hbar^2}}\right) L \right)} \end{align*}

## Contributors

• Vy Pham (UC Davis, Chemistry 110A, Fall 2016)