From 38fa687d6ea66e6dc7c357723798c1a8770bf00f Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 20 Feb 2021 20:56:46 +0100
Subject: Add "Wentzel-Kramers-Brillouin approximation"

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+% Wentzel-Kramers-Brillouin approximation
+
+
+# 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 also
+known as the *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 is then of the 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 quatum 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 to assume that quantum effects are
+sufficiently weak (i.e. $\hbar$ is small enough) that we only need to
+consider the first two terms, or, more generally, 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, whereas
+in the quantum 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.
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