1
$\begingroup$

I'm trying to understand the following phenomenon quantum mechanically:

Problem statement

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:

$\psi (q,t) = \sum c_i(0)\exp(-i E_n t) \psi_n(q)$

with

  • $E_n$ 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?


My Question

  1. Where is the flaw in my argument.
  2. How to model this classical effect (shooting an electron and then measuring it) in a quantum mechanical way?

Ideas

  • 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..

$\endgroup$
2
$\begingroup$

Some problems. First, in practice in this situation you're not going to want to use a discrete model, but rather use the continuum approximation. Second, your coefficients don't depend on $n$. Third, you totally ignore the momentum distribution of the electron, which is what relates the position and the time.

So what would the continuum approximation look like? Suppose that the electron gun produces a pulse of electrons with frequency distribution $f(k)$ and consider the wave function far away from both the detector and the gun, where you can approximate it as a free particle: this covers most of the time of flight. The frequency distribution implicitly tells you everything about the emission process that affects the electron's behaviour far away from the detector and the gun. Depending on what you're doing you might need to know what is happening at the detector and gun in detail, but to roughly understand the time of flight this treatment should be okay. You will want to integrate over phase changes that depend on both space and time and will be something like $\exp(ikx-E(k) t/\hbar)$. The wave function there will be: $$ \eta(x,t)\approx\frac{1}{\Delta\sqrt{2\pi}}\int_{-\infty}^\infty f(k)\exp i\left(k(x+x_0)-\frac{E(k)t}{\hbar}\right)dk. $$

If you have a wave function that is a Gaussian wrt frequency, then $$ f(k) = \exp\left(-\frac{(k-k_0)^2}{\Delta^2}\right) $$ and $$\Theta(x,t)=\sqrt{\frac{\pi}{i(\hbar/2m)t+(1/\Delta^2)}}\exp(i(k_0x-\omega_0t))\exp\left(-\frac{(x-x_0-v_gt)^2}{4(i(\hbar/2m)t+(1/\Delta^2))}\right).$$ The probability density for the wave packet is then $$ |\Theta(x,t)|^2=\frac{\pi}{(\hbar/2m)^2t^2+(1/\Delta^2)^2}\exp\left(-\frac{(1/\Delta^2)(x-x_0-v_gt)^2}{2(\hbar/2m)^2t^2+(1/\Delta^2)^2}\right), $$ which is time dependent. In many situations the frequency distribution won't be a Gaussian, but you get the idea and can adapt it to suit your purposes.

I adapted this sort of treatment from T.E. Hartman (1962). "Tunneling of a wave packet". Journal of Applied Physics 33 (12): 3427, who supposedly followed chapters 3 and 11 of Bohm's book "Quantum Theory", which I haven't read. You can probably find notes online or a book that will cover this material.

$\endgroup$
0
$\begingroup$

Yes, the modulus of the coefficients of the state in the energy basis are constant. But their phase changes, and that gives you changing observables.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.