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| -rw-r--r-- | content/know/category/quantum-mechanics.md | 3 | ||||
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| -rw-r--r-- | latex/know/concept/pauli-exclusion-principle/source.md | 2 | ||||
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diff --git a/content/know/category/quantum-mechanics.md b/content/know/category/quantum-mechanics.md index 12c62d6..18f0fdb 100644 --- a/content/know/category/quantum-mechanics.md +++ b/content/know/category/quantum-mechanics.md @@ -18,3 +18,6 @@ Alphabetical list of concepts in this category. ## T * [Time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/) + +## W +* [Wentzel-Kramers-Brillouin approximation](/know/concept/wentzel-kramers-brillouin-approximation/) diff --git a/content/know/concept/index.md b/content/know/concept/index.md index 4171da7..372a6c6 100644 --- a/content/know/concept/index.md +++ b/content/know/concept/index.md @@ -18,3 +18,6 @@ Alphabetical list of concepts in this knowledge base. ## T * [Time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/) + +## W +* [Wentzel-Kramers-Brillouin approximation](/know/concept/wentzel-kramers-brillouin-approximation/) diff --git a/latex/know/concept/pauli-exclusion-principle/source.md b/latex/know/concept/pauli-exclusion-principle/source.md index e9e4d42..8870c0c 100644 --- a/latex/know/concept/pauli-exclusion-principle/source.md +++ b/latex/know/concept/pauli-exclusion-principle/source.md @@ -87,7 +87,7 @@ $$\begin{aligned} \ket{\Psi(a, b)}_f = C \big( \ket{a}\ket{b} - \ket{b}\ket{a} \big) \end{aligned}$$ -This state called *antisymmetric* under exchange: switching $a$ and $b$ +This state is called *antisymmetric* under exchange: switching $a$ and $b$ causes a sign change, as we would expect for fermions. Now, what if the particles $x_1$ and $x_2$ are in the same state $a$? diff --git a/latex/know/concept/wentzel-kramers-brillouin-approximation/source.md b/latex/know/concept/wentzel-kramers-brillouin-approximation/source.md new file mode 100644 index 0000000..f862004 --- /dev/null +++ b/latex/know/concept/wentzel-kramers-brillouin-approximation/source.md @@ -0,0 +1,188 @@ +% 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. |