WKB approximation

The WKB approximation allows us to approximately solve the Schrödinger equation for arbitrary slowly-varying potentials. It is also called the semiclassical approximation.

We begin by writing the classical momentum at some position $x$ as $p(x) = \sqrt{2m (E-V(x))}$ where $E$ is the total energy. The Schrödinger equation then becomes

\begin{align*} -\hbar^2 \frac{d^2\psi}{d x^2} = p^2(x) \psi(x). \end{align*}

Or, in terms of the momentum operator, $\hat p^2 \psi(x) = p^2(x) \psi(x)$ .

Now, we can write our solution wavefunction as a complex exponential. The real component of $S$ corresponds to a phase, and the imaginary component corresponds to the probability density.

\begin{align*} \psi(x) = e^{i S(x)/\hbar}. \end{align*}

When the potential is constant, we expect to recover the plane wave solution, so it is logical to write the wavefunction in this form. Substituting into the Schrödinger equation we find

\begin{align*} -\hbar^2 \frac{d ^2 }{d x^2} \left( e^{i S(x)/\hbar} \right) &= -\hbar^2 \frac{d }{d x} \left( \frac{i}{\hbar} S'(x) e^{iS(x)/\hbar} \right) \\ &= -\hbar^2 \left( \frac{iS''}{\hbar} - \frac{(S')^2}{\hbar^2} \right) e^{iS(x)/\hbar}\\ p^2(x) &= (S'(x))^2 - i\hbar S''(x). \end{align*}

Now begins the approximation. In the semiclassical case we are considering, de Broglie wavelength of the particle is very small compared to the variation of the potential. We observe that for a constant potential, the differential equation is solved by $S' = p_0$ , such that $\hbar S'' = 0$ . Thus we use $\hbar$ as the expansion parameter and expand $S$ .

\begin{align*} S(x) = S_0(x) + \hbar S_1(x) + \hbar^2 S_2(x) + \cdots \end{align*}

We follow a similar approach to what we did with our derivation of perturbation theory and substitute in the expansion, arguing that the equation must hold for all small $\hbar$ .

\begin{align*} (S_0' + \hbar S_1' + \hbar^2 S_2'' + \cdots)^2 - i\hbar (S_0'' + \hbar S_1'' + \hbar^2 S_2'' + \cdots) - p^2(x) = 0. \end{align*}

Collecting terms

\begin{align*} (S_0')^2 - p^2(x) + \hbar(2S_0' S_1' - iS_0'') + \cdots = 0. \end{align*} \\ \begin{align*} S_0' &= p(x) \\ 2S_0'S_1' &= i S_0'' \end{align*}

The equations can be easily solved

\begin{align*} S_0(x) &= \pm \int_{x_0}^x p(x') \d x' \\ S_1(x) &= \frac{iS_0''}{2S_0'} = \frac{ip'}{2p} \\ iS_1(x) &= - \frac{1}{2} \ln p(x). \end{align*}

We see that $S_0$ is real and $S_1$ is imaginary, corresponding to the phase and probability density of the state. Plugging these into $\psi$ we find the general WKB solution

\begin{align*} \psi(x) = \frac{A}{\sqrt{p(x)}} \exp \left( \pm \frac{i}{\hbar} \int_{x_0}^x p(x') \d x' \right). \end{align*}

It is more useful to consider the classically allowed and forbidden region of the solution separately. In the allowed region, where $E > V(x)$ , we expect to find a sinusoid. In the forbidden region we expect a decaying exponential. The two solutions must agree at the boundary, but as we see from the $1/\sqrt{p(x)}$ factor, the WKB solution will diverge.

We define a wavenumber in the allowed region such that $\psi'' = -k^2 \psi$ , or $k(x) = p(x) / \hbar$ . Then the solution takes the form of two waves, one propagating left and one right.

\begin{align*} \psi_\text{allowed}(x) = \frac{A}{\sqrt{k(x)}} \exp \left( -i \int_{x_0}^x k(x') \d x' \right) + \frac{B}{\sqrt{k(x)}} \exp \left( i \int_{x_0}^x k(x') \d x' \right). \end{align*}

In the forbidden region where $E < V(x)$ , we define an analogous constant $\kappa(x) = ip(x)/\hbar = \sqrt{2m(V(x)-E)}/\hbar$ and write the wavefunction as a decaying and growing exponential.

\begin{align*} \psi_\text{forbidden} = \frac{C}{\sqrt{\kappa(x)}} \exp \left( - \int_{x_0}^x \kappa(x') \dx' \right) + \frac{D}{\sqrt{\kappa(x)}} \exp \left( \int_{x_0}^x \kappa(x') \dx' \right). \end{align*}

To connect the solutions in at the boundary $x_0$ we use the WKB connection coefficients. These can be derived either by solving exactly the differential equation $\frac{d ^2\psi}{d x^2} = x\psi$ or by doing some complex integration tricks. They yield

\begin{align*} \frac{2}{\sqrt{k(x)}} \cos \left( \int_x^{x_0} k(x') \dx' - \frac{\pi}{4} \right) &\impliedby \frac{1}{\sqrt{\kappa(x)}} \exp \left( -\int_{x_0}^x \kappa(x') \dx' \right) \\ -\frac{1}{\sqrt{k(x')}} \sin \left( \int_{x}^{x_0} k(x') dx'- \frac{\pi}{4} \right) &\implies \frac{1}{\sqrt{\kappa(x)}} \exp \left( \int_{x_0}^x \kappa(x') dx' \right). \end{align*}

The arrows are important: these are not strict equalities. Rather, to avoid errors from exponentially larger solutions dominating, we must start by considering the region where the solution is the smallest and then apply the coefficients to find larger solutions.