Transmission coefficient of a Gaussian wave packet through a potential barrier

I have simulated the scattering of a gaussian wave packet with a potential barrier (Crank-Nicolson), and through many simulations I have determined the dependence of the transmission coefficient with the height of the potential barrier. However, when trying to compare it to the theoretical function it fails.

I suppose that that is mainly because the derivation of the coefficient is done for wavefunctions of the form $$\sim e^{ikx}$$. The theoretical coefficient that I've been using is [Sakurai, Modern Quantum Mechanics, Appendix B3]:

$$\begin{cases} T=\{ 1+[V_0^2/4E(V_0-E)]\sinh ^2(2a\sqrt{2m(V_0-E)/\hbar^2}) \} ^{-1}\quad \mathrm{if} \quad EV_0\\ \end{cases}$$

And I get what you can see in the image, which is quite off.

There have been some changes of units, so that everything is easier to treat numerically, $$m=1/2, \ \hbar=1, \ V_0=\lambda k^2, \ E=\hbar k^2/2m=k^2$$ and $$\lambda$$ would be the x-axis. My parameters are: $$a=200$$ and $$k=0.4398$$.

I think that the main problem has to do with the expression being only useful for a wave that is totally determined in the momenta-space, as I said, $$\sim e^{ikx}$$, but I've been researching for an expression for gaussians and failed.

• Please label your axes. How are you defining/computing the transmission coefficient for a Gaussian wave packet? – Puk Jun 22 '20 at 1:57
• The y axis is T and as I've said, the x axis is lambda (related to the height of the potential barrier V_0). The definition for T is already in the text, just substituting the parameters after the figure, to make an easier expression. – Adrián David Jun 22 '20 at 2:06

If your incident wavefunction is $$\psi_i=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty{A(k)e^{i(kx-\omega_kt)}dk},$$ the transmitted wavefunction is $$\psi_t=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty{\alpha(k)A(k)e^{i(kx-\omega_kt)}dk}$$ where the transmission coefficient for an incident wave $$e^{ikx}$$ is $$T(k)=|\alpha(k)|^2$$. The transmission coefficient for the wave packet would be $$T'=\frac{\int\limits_{-\infty}^\infty|\psi_t|^2dx}{\int\limits_{-\infty}^\infty|\psi_i|^2dx}=\frac{\int\limits_{-\infty}^\infty T(k)|A(k)|^2dk}{\int\limits_{-\infty}^\infty|A(k)|^2dk},$$ where Parseval's relation has been used. In other words, the transmission probability is the average of $$T(k)$$ weighted by $$|A(k)|^2$$. The denominator would be just $$1$$ if your initial wavefunction is normalized.
Try expressing your wavefunction in momentum space (i.e. computing $$A(k)$$) and computing $$T'$$ as above to verify your results.
• Great, is $T(k)$ just the same expression that I've used before (but converting E to k) or is it another one that I don't know of? – Adrián David Jun 22 '20 at 2:51