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diff --git a/source/know/concept/wkb-approximation/index.md b/source/know/concept/wkb-approximation/index.md index ef57a3b..fb04414 100644 --- a/source/know/concept/wkb-approximation/index.md +++ b/source/know/concept/wkb-approximation/index.md @@ -8,24 +8,25 @@ categories: 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 -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 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: $$\begin{aligned} - E = \frac{1}{2} m \dot{x}^2 + V(x) - \quad \implies \quad + E = \frac{1}{2} m (x')^2 + V(x) + \qquad \implies \qquad m^2 (x')^2 = 2 m (E - V(x)) \end{aligned}$$ The left-hand side of the rearranged version is simply the momentum squared, -so we define the magnitude of the momentum $$p(x)$$ accordingly: +so we know that the magnitude of the momentum $$p(x)$$ is: $$\begin{aligned} p(x) = \sqrt{2 m (E - V(x))} @@ -38,8 +39,9 @@ We rewrite the Schrödinger equation: $$\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 + &= \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)$$ too, then the solution @@ -50,20 +52,20 @@ $$\begin{aligned} = \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$$, but we can still salvage this solution +In practice, $$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$$ is the -wavenumber. The solution then takes the following form: +$$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$$. +$$\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} @@ -71,111 +73,118 @@ $$\begin{aligned} = \psi(0) \exp\!\bigg( \!\pm\! \frac{i}{\hbar} \Big( \int \chi(x) \dd{x} - C \Big) \bigg) \end{aligned}$$ -Where $$C = \int \chi(x) \dd{x} |_{x = 0}$$ is the initial point of the definite integral. +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: $$\begin{aligned} - \psi'(x) = \pm \frac{i}{\hbar} \chi(x) \psi(x) - \quad \implies \quad - \chi(x) = \pm \frac{\hbar}{i} \frac{\psi'(x)}{\psi(x)} + \psi'(x) + = \pm \frac{i}{\hbar} \chi(x) \psi(x) \end{aligned}$$ -Next, we insert this ansatz for $$\psi(x)$$ into the Schrödinger equation -to get: +We insert this ansatz for $$\psi(x)$$ into the Schrödinger equation to get: $$\begin{aligned} 0 &= \pm \frac{i}{\hbar} \dv{(\chi \psi)}{x} + \frac{p^2}{\hbar^2} \psi - = \pm \frac{i}{\hbar} \chi' \psi \pm \frac{i}{\hbar} \chi \psi' + \frac{p^2}{\hbar^2} \psi - = \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi + \\ + &= \pm \frac{i}{\hbar} \chi' \psi \pm \frac{i}{\hbar} \chi \psi' + \frac{p^2}{\hbar^2} \psi + \\ + &= \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi \end{aligned}$$ -Dividing out $$\psi$$ and rearranging gives us the following, which is -still exact: +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: +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: $$\begin{aligned} - \chi(x) = \chi_0(x) + \frac{\hbar}{i} \chi_1(x) + \frac{\hbar^2}{i^2} \chi_2(x) + ... + \chi(x) + = \chi_0(x) + \frac{\hbar}{i} \chi_1(x) + \frac{\hbar^2}{i^2} \chi_2(x) + \cdots +\end{aligned}$$ + +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: + +$$\begin{aligned} + \psi(x) + &\approx \psi(0) \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int \chi_0(x) \dd{x} \Big) \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big) \end{aligned}$$ -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: +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 $$\chi_0(x)$$: +Where we have discarded all terms containing $$\hbar^2$$. +At order $$\hbar^0$$, we then get the expected classical result for $$\chi_0(x)$$: $$\begin{aligned} 0 = p^2 - \chi_0^2 - \quad \implies \quad - \chi_0(x) = p(x) + \qquad \implies \qquad + \chi_0(x) + = p(x) \end{aligned}$$ -While at order $$\hbar$$, we get the following quantum-mechanical -correction: +While at order $$\hbar$$, we get the following quantum-mechanical correction: $$\begin{aligned} \pm \frac{\hbar}{i} \chi_0' = - 2 \frac{\hbar}{i} \chi_0 \chi_1 - \quad \implies \quad - \chi_1(x) = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)} + \qquad \implies \qquad + \chi_1(x) + = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)} \end{aligned}$$ -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: +We can use this to simplify the latter exponential in $$\psi(x)$$ +using integration by substitution: $$\begin{aligned} \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big) &= \exp\!\Big( \!-\! \frac{1}{2} \int \frac{\chi_0'(x)}{\chi_0(x)} \dd{x} \Big) - = \exp\!\Big( \!-\! \frac{1}{2} \int \frac{1}{\chi_0}\:d\chi_0 \Big) + \\ + &= \exp\!\Big( \!-\! \frac{1}{2} \int \frac{1}{\chi_0}\:d\chi_0 \Big) \\ &= \exp\!\Big( \!-\! \frac{1}{2} \ln\!\big(\chi_0(x)\big) \Big) - = \frac{1}{\sqrt{\chi_0(x)}} - = \frac{1}{\sqrt{p(x)}} + \\ + &= \frac{1}{\sqrt{\chi_0(x)}} \end{aligned}$$ -In the WKB approximation for $$E > V$$, the solution $$\psi(x)$$ is thus -given by: +In the WKB approximation for $$E > V$$, +the solution $$\psi(x)$$ is therefore given by: $$\begin{aligned} \boxed{ - \psi(x) \approx \frac{A}{\sqrt{p(x)}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big) + \psi(x) + \approx \frac{A}{\sqrt{p(x)}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big) } \end{aligned}$$ -What if $$E < V$$? In classical mechanics, this is just not allowed; a ball -cannot simply go through a potential bump without the necessary energy. +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: $$\begin{aligned} - p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)} + p(x) + = \sqrt{2 m (E - V(x))} + = i \sqrt{2 m (V(x) - E)} \end{aligned}$$ And then take the absolute value in the appropriate place in front of $$\psi(x)$$: @@ -186,12 +195,13 @@ $$\begin{aligned} } \end{aligned}$$ -In the classical region ($$E > V$$), the wave function oscillates, and -in the quantum-physical region ($$E < V$$) it is exponential. +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, |