Commit eed1e421 by Juan Torres

### Update wkb_wf.md

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 ... ... @@ -49,24 +49,31 @@ display(HTML(common.to_svg_jshtml(anim))) ## Heuristic derivation The WKB wavefunction can be derived as follows by considering the following assumptions: * A smooth potential can be decomposed into piecewise constant pads. * There is no back reflection in a smooth potential. The later can be justified by considering sufficiently small pads. In general, a wavefunction has an amplitude and a phase, that is, \$\$ \psi(x) = A(x)e^{i\phi(x)}. \$\$ First, let us focus on the phase. By moving along a single pad, the phase evolve as \$\Delta \phi = \phi(x_{i+1})-\phi(x_{i})\$. Over a constant potential, the acquired phase is \$\Delta \phi = p(x_i) \Delta x / \hbar\$, where \$p(x)=\sqrt{2m(E-V(x))}\$. Therefore, the total phase can be obtaining by summing the contributions of all the pads, that is, \$\$ \psi(x) \sim \exp\left( \frac{i}{\hbar} \sum_{j=0}^N p(x_j) \Delta x\right) = \exp\left( \frac{\pm i}{\hbar} \int_{x_0}^x p(x') d x'\right). \$\$ Note that, for the moment, it is implicitly assumed that \$E>V(x)\$. In the last equality, the sum is taken to the continuum limit. Second, consider the amplitude. The current is given by \$j(x) = |\psi(x)^2| v(x)\$ where the velocity is \$v(x) = p(x)/m\$. Since there is no back reflection, the current is constant. Therefore, \$|\psi(x)^2| \sim 1/p(x)\$. From here, we find that the amplitude of the wavefunction goes as \$A(x) \sim 1/\sqrt{p(x)}\$. Then, the WKB wavefunction will be given by, \$\$ \psi_{WKB}(x) = \frac{1}{\sqrt{p(x)}} \exp\left( \frac{i}{\hbar} \int_{x_0}^x p(x') d x'\right). \$\$ ### WKB for evanescent waves We assumed that \$E>V(x)\$, but this heuristic derivation holds for \$E < V(x)\$ as well. In this case, the wavefunction does not accumulate a phase, but accumulates a decaying amplitude. On the formal level, \$p(x)=\sqrt{2m(E-V(x))}=i\sqrt{2m(V(x)-E)}=i|p(x)|\$. Therefore, the WKB function in a region where \$E < V(x)\$ is, \$\$ \psi_{WKB}(x) = \frac{1}{\sqrt{p(x)}} \exp\left( \frac{\pm 1}{\hbar} \int_{x_0}^x p(x') d x'\right). \$\$ ## Formal derivation \ No newline at end of file ## Formal derivation
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