Categories: Physics, Quantum mechanics.

# WKB approximation

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:

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

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