Commit fff56d8a by Michael Wimmer

### text edits for section on WKB wave function

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 ... ... @@ -15,9 +15,8 @@ common.configure_plotting() Before the start of this lecture, you should be able to: - Write down the Schrodinger equation. - Write down the Schrödinger equation of a particle in a one-dimensional potential. - Solve first order ordinary differential equations. - Write down a one dimensional Hamiltonian. !!! summary "Learning goals" ... ... @@ -25,22 +24,34 @@ common.configure_plotting() After this lecture you will be able to: - Derive the WKB wavefunction for the propagating and evanescent regimes from a intuitive and formal approach. - Write down the general solution of Schrödinger equation of a particle in a one-dimensional potential in WKB approximation. - Explain and discuss the limits of the assumptions used to derive the WKB wavefunction. ## Heuristic derivation ## Premise The WKB wavefunction can be derived following physical intuition. We will follow this approach before doing a formal derivation. Let us recall that: In the chapter on the WKB approximation we will focus on a specific problem: namely solving the quantum problem of one particle in a one-dimensional potential $V(x)$, where $x$ is the spatial coordinate in 1D. The stationary Schrödinger equation for the wave function $\psi(x)$ reads in this case $$-\frac{\hbar^2}{2m} \frac{d}{dx} \psi(x) + V(x) \psi(x) = E \psi(x)\,. \tag{1}\label{eq:schrodinger}$$ Our goal now will be to derive an approximate result for the wave function $\psi(x)$ if the potential $V(x)$ is smooth (where smooth will still have to be properly defined). In the previous section, we learned that smooth potentials prevent backscattering. Using this intuition we will first derive an approximation for the wave function making use of some heuristic arguments. Afterwards we will justify this rather intuitive derivation by showing that we can get the same result using a formal approach, too. The approximation for the wave function we are deriving here is typically referred to as Wentzel-Kramer-Brillouin (WKB) approximation, named after the physicists Gregor Wentzel, Hendrik Anthony Kramers und Léon Brillouin. ## Intuitive derivation To derive the WKB approximation in an intuitive fashion we will use two main assumptions: * A smooth potential can be decomposed into piecewise constant pads. * There is no back reflection in a smooth potential. * A smooth potential can be decomposed into piecewise constant pads. A general ansatz for a wavefunction has an amplitude and a phase, that is, We can without restriction rewrite the wavefunction as $$\psi(x) = A(x)e^{i\phi(x)}. \psi(x) = A(x)e^{i\varphi(x)}. \tag{2}\label{eq:ansatz}$$ ### Propagating waves: $E > V(x)$ where $A(x)$ is the amplitude and $\varphi(x)$ the phase (with both being real). python inline import matplotlib.pyplot as plt ... ... @@ -74,9 +85,9 @@ ax[1].plot(x2, y2, label=r'$\psi(x)$', color='red') ax[1].plot(x1, y3, color='red', linestyle='dotted', linewidth=0.5) ax[1].set_ylim(-0.1, 8.5); ax[1].vlines(x=[1, 2], ymin=1, ymax=7, linestyle='dashed', color='red') ax[1].text(0.7, 3.1, r'$\phi_i$', fontsize=18, color='red') ax[1].text(2.1, 3.1, r'$\phi_{i+1}$', fontsize=18, color='red') ax[1].text(1.4, 6.5, r'$\Delta\phi$', fontsize=18, color='red') ax[1].text(0.7, 3.1, r'$\varphi_i$', fontsize=18, color='red') ax[1].text(2.1, 3.1, r'$\varphi_{i+1}$', fontsize=18, color='red') ax[1].text(1.4, 6.5, r'$\Delta\varphi$', fontsize=18, color='red') for axes in ax: axes.legend(fontsize=12) ... ... @@ -86,92 +97,127 @@ for axes in ax: fig.show()  First, let us focus on the phase as can be shown in the right plot above. When the wavefunction moves along a single potential step, the phase acquires a constant shift proportional to the wavevector and the travelled distance, First, let us focus on the phase $\varphi(x)$ shown in the right plot above. When moving along a single potential step, the phase acquires a constant shift proportional to the wavevector and the travelled distance, $$\phi(x_{i+1}) = \phi(x_{i})+ \Delta \phi. \varphi(x_{i+1}) = \varphi(x_{i})+ \Delta \varphi.$$ Over a constant potential, the acquired phase is $\Delta \phi = p(x_i) \Delta x / \hbar$, where $$p(x)=\sqrt{2m(E-V(x))}. Here we make use of the fact that there is no backreflection, so that it is sufficient to consider the contribution of a single wave. Over a constant potential, the acquired phase is \Delta \varphi = p(x_i) \Delta x / \hbar, where we define the momentum at position x as$$ The phase can be obtaining by summing the contributions of all the steps, that is, p(x)=\sqrt{2m(E-V(x))} \tag{3}\label{eq:momentum}. $$\psi(x) \sim \exp\left( \frac{i}{\hbar} \sum_{j=0}^N p(x_j) \Delta x\right) =\exp\left( \frac{\pm i}{\hbar} \int_{x_0}^x p(x') d x'\right). The total phase of the wave function is then obtaining by summing the contributions of all the steps, that is,$$ In the last equality, we take the limit $\Delta x \rightarrow 0$. Second, let us consider the amplitude. In quantum mechanics, probability currents are conserved. The current is given by, \psi(x) \sim \exp\left(\pm \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). $$j(x) = |\psi(x)^2| v(x) = |A(x)|^2 \frac{p(x)}{m},$$ where the velocity is $v(x) = p(x)/m$. Since the current is constant, we find that the amplitude of the wavefunction goes as In the last equality, we take the limit $\Delta x \rightarrow 0$ to convert the summation to an integral. Having found an expression for $\varphi(x)$, let us consider the amplitude $A(x)$. To this end we note that the solution we have been constructing so far is a travelling wave - remember, we assumed that there is no backreflection! Such a travelling wave has an associated probability current $$j(x) = \frac{\hbar}{2 m i} \left(\psi(x) \frac{d}{dx} \psi^*(x) - \psi^*(x) \frac{d}{dx} \psi(x)\right)\,. \tag{4}$$ Plugging the ansatz \eqref{eq:ansatz} for the wave function into this expression for the current, we find $$A(x) \sim 1/\sqrt{p(x)}. j(x) = \frac{\hbar}{m} A^2{x} \frac{d \varphi}{dx} = A^2(x) \frac{p(x)}{m}\,.$$ Finally, the WKB wavefunction will be given by, From quantum mechanics we know that the probability current is conserved, i.e. $j(x) = j_0 = \text{const}$. We thus solve for $A(x)$ to find $$\psi_{WKB}(x) = \frac{1}{\sqrt{p(x)}} \exp\left( \pm \frac{i}{\hbar} \int_{x_0}^x p(x') d x'\right). A^2(x) = \frac{m j_0}{p(x)}\quad \Rightarrow \quad A(x) = \sqrt{\frac{m j_0}{p(x)}}.$$ ### Evanescent waves: $E < V(x)$ 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, Typically, the proportionality constants $m$ and $j_0$ are left out and we write $$p(x)=\sqrt{2m(E-V(x))}=i\sqrt{2m(V(x)-E)}=i|p(x)|. A(x) \sim \frac{1}{\sqrt{p(x)}}.$$ Therefore, the WKB function in a region where $E < V(x)$ is, At first glance this seems weird - now the amplitude doesn't even have the correct units! The reason for doing this is that in most cases it will be sufficient to derive expressions of the wave function up to a global, constant factor as we will see in the explicit examples derived in the next sections. If you however need a properly normalized wave function in your application, remember to normalize it properly at the end. Combining our results for $\varphi(x)$ and $A(x)$ we then arrive at the wave function in WKB approximation: $$\psi_{WKB}(x) = \frac{1}{\sqrt{|p(x)|}} \exp\left( \frac{\pm 1}{\hbar} \int_{x_0}^x p(x') d x'\right). \psi_{WKB}(x) \sim \frac{1}{\sqrt{p(x)}} \exp\left( \pm \frac{i}{\hbar} \int_{x_0}^x p(x') d x'\right).$$ !!! summary "Energy dependence and turning points" - Observe that the energy enter as a parameter instead of being an outcome of calculations. - When $V(x_0)=E$, the amplitude $A$ of $\psi$ diverges. This means that the WKB approximation is no longer valid. Observe that these points are a function of energy, i.e. $x_0 = x_0(E)$. - When $V(x_0)=E$, the amplitude $A$ of $\psi$ diverges. This means that the WKB approximation is no longer valid, i.e. we do need to restrict ourselves to strictly $E>V(x)$ at this point. Observe that these points are a function of energy, i.e. $x_0 = x_0(E)$. ## Formal derivation Starting from the Schrodinger equation, We start again from the Schrödinger equation, and rewrite it using the definition \eqref{eq:momentum} of the momentum $p(x)$: \begin{equation}\label{eq:schrod} -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi(x) + V(x) \psi(x) = E\psi(x) \longrightarrow \psi^{''}(x) + \frac{p^2(x)}{\hbar^2} \psi(x) = 0. \end{equation} \begin{align}\label{eq:schrod} & -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi(x) + V(x) \psi(x) = E\psi(x)\\ \Rightarrow \quad & \psi''(x) + \frac{p^2(x)}{\hbar^2} \psi(x) = 0. \end{align} Consider $E>V(x)$. Then, $p(x)$ is real. Consider a general wavefunction as ansatz, i.e. $\psi(x) = A(x) \exp(i\varphi(x))$, where $A$ and $\varphi$ are real. Then, the second spatial derivative of $\psi$ is, We consider $E>V(x)$, so $p(x)$ is real. We use again the general wave function ansatz \eqref{eq:ansatz}, i.e. $\psi(x) = A(x) \exp(i\varphi(x))$, where $A$ and $\varphi$ are real. In the following we will suppress the $x$-dependence in the notation and simply write $\psi=A e^{i\varphi}$. The second spatial derivative of $\psi$ is $$\psi^{''} = A^{''}e^{i\varphi}+2i A^{'} \varphi^{'} e^{i\varphi}- A (\varphi^{'})^2 e^{i\varphi} + i A \varphi^{''} e^{i\varphi}. \psi'' = A''e^{i\varphi}+2i A' \varphi' e^{i\varphi}- A (\varphi')^2 e^{i\varphi} + i A \varphi'' e^{i\varphi}.$$ By replacing this into the original equation. Then, one finds, Inserting this expression in the Schrödinger equation and dividing by $e^{i \varphi}$ gives $$A^{''} + 2i A^{'} \varphi^{'} - A ( \varphi^{'} )^2 + i A \varphi^{''} + \frac{p^2}{\hbar^2} A = 0. A'' + 2i A' \varphi' - A ( \varphi' )^2 + i A \varphi'' + \frac{p^2}{\hbar^2} A = 0,$$ We have obtained two equations: one for the real and one for the imaginary part. That is, Both the real and imaginary part of this equation need to be zero individually. Since both $A$ and $\varphi$ are real, we obtained two simpler equations: A \varphi^{''} + 2 A^{'} \varphi^{'} = 0, \qquad A^{''} - A ( \varphi^{'} )^2 + \frac{p^2}{\hbar^2} A = 0. \begin{align} A \varphi'' + 2 A' \varphi' &= 0 \tag{5}\label{eq:imagpart}\\ A'' - A ( \varphi' )^2 + \frac{p^2}{\hbar^2} A &= 0 \tag{6}\label{eq:realpart}. \end{align} The solution for the imaginary part can be obtained by multiplying the equation by $\varphi^{'}$. That is, The solution to \eqref{eq:imagpart}can be obtained by multiplying the equation by $\varphi'$. That is, \varphi^{'}\left(A \varphi^{''} + 2 A^{'} \varphi^{'}\right) = (A^2 \varphi^{'})^{'} = 0 \rightarrow A = \frac{c}{\sqrt{\varphi^{'}}}. \begin{align} & \varphi'\left(A \varphi'' + 2 A' \varphi'\right) = (A^2 \varphi')' = 0 \\ \Rightarrow \quad & A = \frac{c}{\sqrt{\varphi'}}. \end{align} To solve for the real part, let us assume that $A^{''} << A(\varphi^{'})^2$. Then, we find that the equation becomes, To solve \eqref{eq:realpart}, let us assume that $A'' \ll A(\varphi')^2$, i.e. we drop $A''$. Then, we find that the equation can be solved as A(\varphi^{'})^2 + \frac{p^2}{\hbar^2} A = 0 \rightarrow \varphi^{'} = \mp \frac{p}{\hbar} \rightarrow \varphi(x) = \mp \frac{1}{\hbar}\int_{x_0}^x p('x) dx'. \begin{align} & -A (\varphi')^2 + \frac{p^2}{\hbar^2} A = 0\\ \Rightarrow \quad & (\varphi')^2 = \frac{p^2}{\hbar^2}\\ \Rightarrow \quad & \varphi' = \pm \frac{p}{\hbar}\\ \Rightarrow \quad & \varphi(x) = \pm \frac{1}{\hbar}\int_{x_0}^x p('x) dx'. \end{align} Then, we can conclude that the WKB wavefunction is given as, $$\psi(x) = \frac{1}{\sqrt{p(x)}} \exp\left( \mp \frac{i}{\hbar} \int_{x_0}^x p(x') d x'\right). \psi(x) = \frac{1}{\sqrt{p(x)}} \exp\left( \pm \frac{i}{\hbar} \int_{x_0}^x p(x') d x'\right).$$ For the case $E < V(x)$, we use the same argument as before to find exponential decaying and exponential growing solutions. !!! summary "Smooth potential approximation" We have assumed that $A^{''} << A(\varphi^{'})^2$. This implies that the length scale over wich $V(x)$ changes, $\Delta L$, is much larger than the wavefunction's wavelength, $\lambda$. Note that We have assumed that $A'' \ll A(\varphi')^2$. What does this approximation mean physically? We can expect that the amplitude $A$ changes over the length scale over wich $V(x)$ changes. If we define this length scale as $\Delta L$, we can estimate that $A' \sim \frac{1}{\Delta L}$ and $A'' \sim \frac{1}{\Delta L^2}$. On the other hand, $A(\varphi')^2$ is dominated by the change in $\varphi$ that is given by the oscillatory nature of the wave function, and characterized by the wave length $\lambda$. We can thus estimate $A (\varphi')^2 \sim \frac{1}{\lambda^2}$. with this, we find the condition A^{''} \sim \frac{1}{\Delta L} << A(\varphi^{'})^2 \sim \frac{1}{\lambda^2} \longrightarrow \lambda << \Delta L. \begin{align} & A'' \ll A(\varphi')^2\\ \Rightarrow \quad & \frac{1}{\Delta L^2} \ll \frac{1}{\lambda^2}\\ \Rightarrow \quad & \lambda \ll \Delta L. \end{align} This indeed implies that the potential $V(x)$ should change slowly compared to the wave length $\lambda$. ### Evanescent waves: $E < V(x)$ We assumed in our derivation that $E>V(x)$, but it is straightforwardly extended to $E < V(x)$ as well. In this case, the wavefunction does not accumulate a phase, but accumulates a decaying amplitude. On the formal level, when $E< V(x)$ it is useful to write $$p(x)=\sqrt{2m(E-V(x))}=i\sqrt{2m(V(x)-E)}=i|p(x)|.$$ where $|p(x)|$ is now a positive and real function. The WKB function then is given by $$\psi_{WKB}(x) = \frac{1}{\sqrt{|p(x)|}} \exp\left( \frac{\pm 1}{\hbar} \int_{x_0}^x |p(x')| d x'\right).$$ ## Summary The WKB wavefunction can be derived from two assumptions: a smooth potential, and a slowly varying wavefunction. The general solution is, We have derived the wave functions in WKB approximation under the assumption of a slowly changing potential - slow compared to the wave length. Since the Schrödinger equation \eqref{eq:schrodinger} is a second-order differential equation, the general solution will be a linear superposition of the two fundamental solutions ($\pm$) we found above: $$\begin{split} \psi(x)_{E > V(x)} &= \frac{A}{\sqrt{p(x)}}e^{\frac{i}{\hbar} \int_x^{x_1} p(x') dx'} + \frac{B}{\sqrt{p(x)}}e^{-\frac{i}{\hbar} \int_x^{x_1} p(x') dx'},\\ \psi(x)_{E < V(x)} &= \frac{C}{\sqrt{|p(x)}|}e^{\frac{1}{\hbar} \int_x^{x_1} |p(x')| dx'} + \frac{D}{\sqrt{|p(x)|}}e^{-\frac{1}{\hbar} \int_x^{x_1} |p(x')|dx'}. \psi(x)_{E > V(x)} &= \frac{A}{\sqrt{p(x)}}e^{\frac{i}{\hbar} \int_{x_0}^{x} p(x') dx'} + \frac{B}{\sqrt{p(x)}}e^{-\frac{i}{\hbar} \int_{x_0}^{x} p(x') dx'},\\ \psi(x)_{E < V(x)} &= \frac{C}{\sqrt{|p(x)}|}e^{\frac{1}{\hbar} \int_{x_0}^{x} |p(x')| dx'} + \frac{D}{\sqrt{|p(x)|}}e^{-\frac{1}{\hbar} \int_{x_0}^{x} |p(x')|dx'}. \end{split}$$
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