I'm trying to understand the following phenomenon quantum mechanically:
If I'm shooting an electron beam (using a vacuum tube for example) and I have a measuring device at a certain distance. Then I expect that the odds that I measure the electron before a time $t$ is small because (classicaly) the electron has a certain speed so it would take $\Delta t = \Delta x /\Delta v$ to measure it.
If I assume (which is probably wrong?) that the hamiltonian is time independent (assuming no external influence on the electron) then it follows that the time dependent wave function $\psi (q,t)$ can be written as follows:
$\psi (q,t) = \sum c_i(0)\exp(-i E_n t) \psi_n(q)$
- $E_n$ and $\psi_n$ energy eigenvalues and functions resp.
- $c_i(0)$: coefficient at time $t=0$$
But this implies that the square amplitude of the coefficients is constant in time. This means that the spatial distribution (energy eigenfunctions) also doesn't change. So the electron does not propagate?
- Where is the flaw in my argument.
- How to model this classical effect (shooting an electron and then measuring it) in a quantum mechanical way?
The hamiltonian is not time independent because the electron interacts with the vacuum tube and the measuring device.
I'm not really specifying the specific measuring device. Is this important?
I'm quite sure I'm missing some crucial insight here..