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I am doing a scattering simulation of a Gaussian wave packet on a finite square well. I have solved numerically the Schroedinger equation and I know the values of the wave function after the scattering process before and after the square well.

To be more clear, this is the situation after the scattering process: (the wave function is not normalized, the biggest packet is moving towards left, the smallest towards right)enter image description here

I know that the tasmission coefficient is defined as:

$T=\dfrac{\vec{j}_{trasmitted}}{\vec{j}_{incident}}$

where the flux $\vec{j}$ is defined as:

$\vec{j}(x)=\psi^*(x)\frac{\hat{p}}{2m}\psi(x)-\psi(x)\frac{\hat{p}}{2m}\psi(x)^*$

Now, for an unnormalizable function like $Ce^{ikx}$ I am able to determine $T$, it is easy because the flux is constant. But in this case the flux is not constant before and after the square well.

Maybe the probability to find the particle after the square well (and so the area under the part of the normalized $|\psi|^2$ that crossed the barrier) is the correct value of the transmission coefficient?

If my hypotesis is correct, could you give me some references where it is explained?

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The square modulus of the transmission coefficient ($|T|^2$) is the transmission probability, and you have the data to calculate that for your wave-packets (i.e. it's just $|\psi|^2$ as you surmised, if you normalized your wavefunction). You can compare that to the expected transmission coefficient for plane waves with the same average energy as your wave-packet. I don't know how well the comparison will be since you have uncertain energy, but you can use wider wave-packets and at least see if it seems to be converging to the plane wave solution.

Incidentally it is a good idea to check the norm of your wave function, it can help determine if anything is going wrong with your numerical simulation.

PS: This http://arxiv.org/abs/quant-ph/0301114 might be of some value. It is an explicit solution for a wave-packet on a square barrier (i.e. like your problem, but only one 'edge'.)

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