---
title: "Wentzel-Kramers-Brillouin approximation"
firstLetter: "W"
publishDate: 2021-02-22
categories:
- Quantum mechanics
- Physics

date: 2021-02-22T21:38:35+01:00
draft: false
markup: pandoc
---

# Wentzel-Kramers-Brillouin approximation

In quantum mechanics, the **Wentzel-Kramers-Brillouin** or simply the **WKB
approximation** is a method to approximate the wave function $\psi(x)$ of
the one-dimensional time-independent Schrödinger equation. It is an example
of a **semiclassical approximation**, because it tries to find a
balance between classical and quantum physics.

In classical mechanics, a particle travelling in a potential $V(x)$
along a path $x(t)$ has a total energy $E$ as follows, which we
rearrange:

$$\begin{aligned}
    E = \frac{1}{2} m \dot{x}^2 + V(x)
    \quad \implies \quad
    m^2 (x')^2 = 2 m (E - V(x))
\end{aligned}$$

The left-hand side of the rearrangement is simply the momentum squared,
so we define the magnitude of the momentum $p(x)$ accordingly:

$$\begin{aligned}
    p(x) = \sqrt{2 m (E - V(x))}
\end{aligned}$$

Note that this is under the assumption that $E > V$, which is always the
case in classical mechanics, but not necessarily so in quantum
mechanics, but we stick with it for now. We rewrite the Schrödinger
equation:

$$\begin{aligned}
    0
    = \dv[2]{\psi}{x} + \frac{2 m}{\hbar^2} (E - V) \psi
    = \dv[2]{\psi}{x} + \frac{p^2}{\hbar^2} \psi
\end{aligned}$$

If $V(x)$ were constant, and by extension $p(x)$ too, then the solution
is easy:

$$\begin{aligned}
    \psi(x)
    = \psi(0) \exp(\pm i p x / \hbar)
\end{aligned}$$

This form is reminiscent of the generator of translations. In practice,
$V(x)$ and $p(x)$ vary with $x$, but we can still salvage this solution
by assuming that $V(x)$ varies slowly compared to the wavelength
$\lambda(x) = 2 \pi / k(x)$, where $k(x) = p(x) / \hbar$ is the
wavenumber. The solution then takes the following form:

$$\begin{aligned}
    \psi(x)
    = \psi(0) \exp\!\Big(\!\pm\! \frac{i}{\hbar} \int_0^x \chi(\xi) \dd{\xi} \Big)
\end{aligned}$$

$\chi(\xi)$ is an unknown function, which intuitively should be related
to $p(x)$. The purpose of the integral is to accumulate the change of
$\chi$ from the initial point $0$ to the current position $x$.
Let us write this as an indefinite integral for convenience:

$$\begin{aligned}
    \psi(x)
    = \psi(0) \exp\!\bigg( \!\pm\! \frac{i}{\hbar} \Big( \int \chi(x) \dd{x} - C \Big) \bigg)
\end{aligned}$$

Where $C = \int \chi(x) \dd{x} |_{x = 0}$ is the initial point of the definite integral.
For simplicity, we absorb the constant $C$ into $\psi(0)$.
We can now clearly see that:

$$\begin{aligned}
    \psi'(x) = \pm \frac{i}{\hbar} \chi(x) \psi(x)
    \quad \implies \quad
    \chi(x) = \pm \frac{\hbar}{i} \frac{\psi'(x)}{\psi(x)}
\end{aligned}$$

Next, we insert this ansatz for $\psi(x)$ into the Schrödinger equation
to get:

$$\begin{aligned}
    0
    &= \pm \frac{i}{\hbar} \dv{(\chi \psi)}{x} + \frac{p^2}{\hbar^2} \psi
    = \pm \frac{i}{\hbar} \chi' \psi \pm \frac{i}{\hbar} \chi \psi' + \frac{p^2}{\hbar^2} \psi
    = \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi
\end{aligned}$$

Dividing out $\psi$ and rearranging gives us the following, which is
still exact:

$$\begin{aligned}
    \pm \frac{\hbar}{i} \chi'
    = p^2 - \chi^2
\end{aligned}$$

Next, we expand this as a power series of $\hbar$. This is why it is
called *semiclassical*: so far we have been using full quantum mechanics,
but now we are treating $\hbar$ as a parameter which controls the
strength of quantum effects:

$$\begin{aligned}
    \chi(x) = \chi_0(x) + \frac{\hbar}{i} \chi_1(x) + \frac{\hbar^2}{i^2} \chi_2(x) + ...
\end{aligned}$$

The heart of the WKB approximation is its assumption that quantum effects are
sufficiently weak (i.e. $\hbar$ is small enough) that we only need to
consider the first two terms, or, more specifically, that we only go up to
$\hbar$, not $\hbar^2$ or higher. Inserting the first two terms of this
expansion into the equation:

$$\begin{aligned}
    \pm \frac{\hbar}{i} \chi_0'
    &= p^2 - \chi_0^2 - 2 \frac{\hbar}{i} \chi_0 \chi_1
\end{aligned}$$

Where we have discarded all terms containing $\hbar^2$. At order
$\hbar^0$, we then get the expected classical result for $\chi_0(x)$:

$$\begin{aligned}
    0 = p^2 - \chi_0^2
    \quad \implies \quad
    \chi_0(x) = p(x)
\end{aligned}$$

While at order $\hbar$, we get the following quantum-mechanical
correction:

$$\begin{aligned}
    \pm \frac{\hbar}{i} \chi_0'
    = - 2 \frac{\hbar}{i} \chi_0 \chi_1
    \quad \implies \quad
    \chi_1(x) = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)}
\end{aligned}$$

Therefore, our approximated wave function $\psi(x)$ currently looks like
this:

$$\begin{aligned}
    \psi(x)
    &\approx \psi(0) \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int \chi_0(x) \dd{x} \Big) \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
\end{aligned}$$

We can reduce the latter exponential using integration by substitution:

$$\begin{aligned}
    \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
    &= \exp\!\Big( \!-\! \frac{1}{2} \int \frac{\chi_0'(x)}{\chi_0(x)} \dd{x} \Big)
    = \exp\!\Big( \!-\! \frac{1}{2} \int \frac{1}{\chi_0}\:d\chi_0 \Big)
    \\
    &= \exp\!\Big( \!-\! \frac{1}{2} \ln\!\big(\chi_0(x)\big) \Big)
    = \frac{1}{\sqrt{\chi_0(x)}}
    = \frac{1}{\sqrt{p(x)}}
\end{aligned}$$

In the WKB approximation for $E > V$, the solution $\psi(x)$ is thus
given by:

$$\begin{aligned}
    \boxed{
        \psi(x) \approx \frac{A}{\sqrt{p(x)}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
    }
\end{aligned}$$

What if $E < V$? In classical mechanics, this is not allowed; a ball
cannot simply go through a potential bump without the necessary energy.
However, in quantum mechanics, particles can **tunnel** through barriers.

Conveniently, all we need to change for the WKB approximation is to let
the momentum take imaginary values:

$$\begin{aligned}
    p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)}
\end{aligned}$$

And then take the absolute value in the appropriate place in front of
$\psi(x)$:

$$\begin{aligned}
    \boxed{
        \psi(x) \approx \frac{A}{\sqrt{|p(x)|}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
    }
\end{aligned}$$

In the classical region ($E > V$), the wave function oscillates, and
in the quantum-mechanical region ($E < V$) it is exponential. Note that for
$E \approx V$ the approximation breaks down, due to the appearance of
$p(x)$ in the denominator.