From 922a0bbeb81f9a0297c6a728d243cbec75cf9c3b Mon Sep 17 00:00:00 2001 From: Prefetch Date: Mon, 29 Mar 2021 09:15:42 +0200 Subject: Expand knowledge base, move WKB approximation --- .../index.pdc | 207 --------------------- 1 file changed, 207 deletions(-) delete mode 100644 content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc (limited to 'content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc') diff --git a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc b/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc deleted file mode 100644 index cf44fc8..0000000 --- a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc +++ /dev/null @@ -1,207 +0,0 @@ ---- -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 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 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. - - -## 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. -- cgit v1.2.3