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Categories: Physics, Quantum mechanics.

WKB approximation

In quantum mechanics, the Wentzel-Kramers-Brillouin or simply the WKB approximation is a technique to approximate the wavefunction $\psi(x)$ of the 1D 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:

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

The left-hand side of the rearranged version is simply the momentum squared, so we know that the magnitude of the momentum $p(x)$ is:

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

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:

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

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 $2 \pi / k(x)$, 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:

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

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:

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

The heart of the WKB approximation is its assumption that quantum effects are sufficiently weak that we only need to consider the first two terms of this expansion, i.e. $\hbar^2$ is so small that it is negligible. Therefore, our approximated wavefunction $\psi(x)$ now looks like this:

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

Inserting the expansion’s first two terms into our equation for $\chi(x)$ gives:

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

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

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

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

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

We can use this to simplify the latter exponential in $\psi(x)$ using integration by substitution:

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

In the WKB approximation for $E > V$, the solution $\psi(x)$ is therefore 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 or over 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 wavefunction 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.

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