From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Thu, 20 Oct 2022 18:25:31 +0200 Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex' --- source/know/concept/wkb-approximation/index.md | 62 +++++++++++++------------- 1 file changed, 31 insertions(+), 31 deletions(-) (limited to 'source/know/concept/wkb-approximation') diff --git a/source/know/concept/wkb-approximation/index.md b/source/know/concept/wkb-approximation/index.md index efeeca6..ef57a3b 100644 --- a/source/know/concept/wkb-approximation/index.md +++ b/source/know/concept/wkb-approximation/index.md @@ -9,13 +9,13 @@ 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 +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 +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: $$\begin{aligned} @@ -25,13 +25,13 @@ $$\begin{aligned} \end{aligned}$$ The left-hand side of the rearranged version is simply the momentum squared, -so we define the magnitude of the momentum $p(x)$ accordingly: +so we define the magnitude of the momentum $$p(x)$$ accordingly: $$\begin{aligned} p(x) = \sqrt{2 m (E - V(x))} \end{aligned}$$ -Note that this is under the assumption that $E > V$, +Note that this is under the assumption that $$E > V$$, which is always true in classical mechanics, but not necessarily in quantum mechanics. We rewrite the Schrödinger equation: @@ -42,7 +42,7 @@ $$\begin{aligned} = \dvn{2}{\psi}{x} + \frac{p^2}{\hbar^2} \psi \end{aligned}$$ -If $V(x)$ were constant, and by extension $p(x)$ too, then the solution +If $$V(x)$$ were constant, and by extension $$p(x)$$ too, then the solution is easy: $$\begin{aligned} @@ -51,9 +51,9 @@ $$\begin{aligned} \end{aligned}$$ 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 +$$V(x)$$ 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: $$\begin{aligned} @@ -61,9 +61,9 @@ $$\begin{aligned} = \psi(0) \exp\!\Big(\!\pm\! \frac{i}{\hbar} \int_0^x \chi(\xi) \dd{\xi} \Big) \end{aligned}$$ -$\chi(\xi)$ is an unknown function, which intuitively should be related -to $p(x)$. The purpose of the integral is to accumulate the change of -$\chi$ from the initial point $0$ to the current position $x$. +$$\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} @@ -71,8 +71,8 @@ $$\begin{aligned} = \psi(0) \exp\!\bigg( \!\pm\! \frac{i}{\hbar} \Big( \int \chi(x) \dd{x} - C \Big) \bigg) \end{aligned}$$ -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 $\psi(0)$. +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: $$\begin{aligned} @@ -81,7 +81,7 @@ $$\begin{aligned} \chi(x) = \pm \frac{\hbar}{i} \frac{\psi'(x)}{\psi(x)} \end{aligned}$$ -Next, we insert this ansatz for $\psi(x)$ into the Schrödinger equation +Next, we insert this ansatz for $$\psi(x)$$ into the Schrödinger equation to get: $$\begin{aligned} @@ -91,7 +91,7 @@ $$\begin{aligned} = \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi \end{aligned}$$ -Dividing out $\psi$ and rearranging gives us the following, which is +Dividing out $$\psi$$ and rearranging gives us the following, which is still exact: $$\begin{aligned} @@ -99,9 +99,9 @@ $$\begin{aligned} = p^2 - \chi^2 \end{aligned}$$ -Next, we expand this as a power series of $\hbar$. This is why it is +Next, we expand this as a power series of $$\hbar$$. This is why it is called *semiclassical*: so far we have been using full quantum mechanics, -but now we are treating $\hbar$ as a parameter which controls the +but now we are treating $$\hbar$$ as a parameter which controls the strength of quantum effects: $$\begin{aligned} @@ -109,9 +109,9 @@ $$\begin{aligned} \end{aligned}$$ 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 +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 +$$\hbar$$, not $$\hbar^2$$ or higher. Inserting the first two terms of this expansion into the equation: $$\begin{aligned} @@ -119,8 +119,8 @@ $$\begin{aligned} &= p^2 - \chi_0^2 - 2 \frac{\hbar}{i} \chi_0 \chi_1 \end{aligned}$$ -Where we have discarded all terms containing $\hbar^2$. At order -$\hbar^0$, we then get the expected classical result for $\chi_0(x)$: +Where we have discarded all terms containing $$\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 @@ -128,7 +128,7 @@ $$\begin{aligned} \chi_0(x) = p(x) \end{aligned}$$ -While at order $\hbar$, we get the following quantum-mechanical +While at order $$\hbar$$, we get the following quantum-mechanical correction: $$\begin{aligned} @@ -138,7 +138,7 @@ $$\begin{aligned} \chi_1(x) = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)} \end{aligned}$$ -Therefore, our approximated wave function $\psi(x)$ currently looks like +Therefore, our approximated wave function $$\psi(x)$$ currently looks like this: $$\begin{aligned} @@ -158,7 +158,7 @@ $$\begin{aligned} = \frac{1}{\sqrt{p(x)}} \end{aligned}$$ -In the WKB approximation for $E > V$, the solution $\psi(x)$ is thus +In the WKB approximation for $$E > V$$, the solution $$\psi(x)$$ is thus given by: $$\begin{aligned} @@ -167,7 +167,7 @@ $$\begin{aligned} } \end{aligned}$$ -What if $E < V$? In classical mechanics, this is just not allowed; a ball +What if $$E < V$$? In classical mechanics, this is just not allowed; a ball cannot simply go through a potential bump without the necessary energy. On the other hand, in quantum physics, particles can **tunnel** through barriers. @@ -178,7 +178,7 @@ $$\begin{aligned} p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)} \end{aligned}$$ -And then take the absolute value in the appropriate place in front of $\psi(x)$: +And then take the absolute value in the appropriate place in front of $$\psi(x)$$: $$\begin{aligned} \boxed{ @@ -186,10 +186,10 @@ $$\begin{aligned} } \end{aligned}$$ -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. +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 -- cgit v1.3