# Quantum Tunnelling with Delta Potential

I'm trying to create an animation of Quantum Tunnelling like this one.

I've been learning some QM on my own, so please forgive and correct any mistakes.

I considered the potential barrier $\alpha \delta(x)$ where $\alpha$ is a real constant and $\delta$ is Dirac's.

I assumed a wave coming in from the left (travelling to the right), that either reflects off the barrier or tunnels through the barrier. Solving the time independent Schrodinger equation gave me $$\psi(x) = \left\{ \begin{array}{ccc} 1\mathrm e^{\mathrm ikx} + R\mathrm e^{-\mathrm ikx} & : & x<0 \\ T\mathrm e^{\mathrm i kx} & : & x > 0 \end{array}\right.$$ where $k = \frac{\sqrt{2mE}}{\hbar}$. Here $|R|^2$ gives the probability of the wave reflecting and $|T|^2$ the probability of the wave tunnelling through the barrier.

We want $\psi$ to be continuous and we want, as $\varepsilon \to 0^+$, $$-\frac{\hbar^2}{2m}\int_{-\varepsilon}^{\varepsilon}\frac{\mathrm d^2\psi}{\mathrm dx^2}~\mathrm dx+\alpha\int_{-\varepsilon}^{\varepsilon}\delta(x)\psi(x)~\mathrm dx= E\int_{-\varepsilon}^{\varepsilon}\psi(x)~\mathrm dx$$ $$\lim_{\varepsilon \to 0}\left[\frac{\mathrm d\psi}{\mathrm dx}\right]_{-\varepsilon}^{\varepsilon} = \frac{2m\alpha}{\hbar^2}\psi(0)$$ Applying these conditions gives me $$R=\frac{\alpha}{2\mathrm ik-\alpha} \ \ \ \mbox{ and } \ \ \ T=\frac{2\mathrm ik}{2\mathrm ik-\alpha}$$

$$\psi(x) = \left\{ \begin{array}{ccc} \mathrm e^{\mathrm ikx} + \left(\frac{\alpha}{2\mathrm ik-\alpha}\right)\mathrm e^{-\mathrm ikx} & : & x<0 \\ \left(\frac{2\mathrm ik}{2\mathrm ik-\alpha}\right)\mathrm e^{\mathrm i kx} & : & x > 0 \end{array}\right.$$

Including the time-dependent term $\varphi(t)=\mathrm e^{-\mathrm iEt/\hbar} = \mathrm e^{-\mathrm ik^2\hbar t/2m}$ gives

$$\psi(x)\varphi(t) = \left\{ \begin{array}{ccc} \mathrm e^{\mathrm ik(x+k\hbar t/2m)} + \left(\frac{\alpha}{2\mathrm ik-\alpha}\right)\mathrm e^{-\mathrm ik(x-k\hbar t/2m)} & : & x<0 \\ \left(\frac{2\mathrm ik}{2\mathrm ik-\alpha}\right)\mathrm e^{\mathrm ik(x+k\hbar t/2m)} & : & x > 0 \end{array}\right.$$

I've looked at $|\psi(x)\varphi(t)|^2$ and this is independent of $t$.

Griffiths mentions taking a linear combination of the $\psi(x)\varphi(t)$, but does not give any details.

Any ideas?

• Solving the time-independent Schrodinger equation gives you eigenfunctions, which by definition don't change with time (except up to a phase). The correct thing to do is either start with an initial wavepacket and solve the time-dependent Schrodinger equation numerically, or Fourier expand the wavepacket in terms of your eigenfunctions. – Jahan Claes Jul 27 '17 at 21:02
• As an aside, using the notation $\theta(t)$ in this context for anything other than the Heaviside step function can be a bit confusing - it's best avoided unless there's a good reason. – Emilio Pisanty Jul 27 '17 at 21:26
• @EmilioPisanty Thank you - I've changed $\theta$ to $\varphi$. – Fly by Night Jul 27 '17 at 22:00
• @JahanClaes I'm not really sure what you mean by a wave packet. Is it what we get by summing $\psi_k(x)\varphi_k(t)$ for $k$ in an open interval? $$\int_{k \in I} f(k)\psi_k(x)\varphi_k(t)~\mathrm dk$$ You mention the Fourier expansion. I don't know if this is related, but if I have $\Psi(x,0)$ and I want to find $\Psi(x,t)$ then I use the Fourier Transform like Griffiths on pages 61 and 62. – Fly by Night Jul 27 '17 at 22:04

Luckily, you've already done most of the required work, in building out the relevant continuum eigenstates $\psi_k(x)$ (and therefore their associated TDSE solutions, $e^{-i\hbar k^2 t/2m}\psi_k(x)$), and all you need is to assemble those into a wavepacket. The way that's normally done is by starting with a gaussian on the left and with momentum to the right, $$\psi_0(x,t_0) = N \exp\left(-\frac{1}{2\sigma^2}(x-x_0)^2+ip_0x\right),$$ decompose that via a Fourier transform into a sum of plane waves, extend those plane waves into the barrier eigenstates you've found, add the time-dependent phase, and then do the Fourier transform back into position space.
• Thanks for your reply. As you say, my eigenfunctions are not physical states - they're not normalisable. Griffiths (on page 75) makes that point, and goes on to say that we need to make a linear combinations of these states. I thought that maybe it would be an integral over a short range of $k$ values, say $$\int_{K-\varepsilon}^{K+\varepsilon} f(k)\psi_k(x)\varphi_k(t)\mathrm e^{\mathrm ixk}~\mathrm dk$$ I am probably totally wrong though! – Fly by Night Jul 27 '17 at 22:11