---
title: "WKB approximation"
sort_title: "Wkb approximation" # sic
date: 2021-02-22
categories:
- Quantum mechanics
- Physics
layout: "concept"
---

In quantum mechanics, the **Wentzel-Kramers-Brillouin** or simply the **WKB
approximation** is a technique 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 rearranged version 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 true in classical mechanics,
but not necessarily in quantum mechanics.
We rewrite the Schrödinger equation:

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

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 just not allowed; a ball
cannot simply go through a potential bump without the necessary energy.
On the other hand, in quantum physics, particles can **tunnel** through barriers.

Luckily, the only thing 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-physical region ($$E < V$$) it is exponential.
Note that for $$E \approx V$$ the approximation breaks down,
because of the appearance of $$p(x)$$ in the denominator.


## References
1.  D.J. Griffiths, D.F. Schroeter,
    *Introduction to quantum mechanics*, 3rd edition,
    Cambridge.
2.  R. Shankar,
    *Principles of quantum mechanics*, 2nd edition,
    Springer.