Rydberg/ejected wave packets

The movies on this page show a slightly more complicated situation than the movies for H and Li electron wave packets. In H and Li, the only possible type of motion is the radial motion of the electron. In the movies on this page, there is radial motion for one highly excited electron and also core electrons that can be excited. The movie is for the case where the core electrons are not highly excited and thus only extend over a small region near the nucleus. The wave function can be written as the sum of two terms: psi(r,t)=phi_1 y_1(r,t) + phi_2y_2(r,t) where phi_1 is the ground state of the atomic ion and phi_2 is the first excited state of the ion. |y_1(r,t)|^2 is the probability for finding the excited electron at the distance r and the core electrons in state 1 at time t. |y_2(r,t)|^2 is the probability for finding the excited electron at the distance r and the core electrons in state 2. (As an example, you can think of the 1s1s2pnp "autoionizing states" in Be. These states are at the same energy as having a Be+ ion in the ground state, 1s1s2s, and a free electron.)
The first movie shows the behavior of the radial motion of one highly excited electron when the core can also be excited. This movie simulates the excitation of an atom by a pulsed laser to a high enough total energy for one electron to be ejected from the atom only if the atomic ion is left in its ground state. However, if the atomic ion is excited, the highly excited electron does not have energy to leave the atom. One frame from the movie is given below.


 

1.4 Mb MPEG movie of two channel radial wave packets

The lowest box gives the radial probability distribution for the highly excited electron when the core of the atomic ion is in its ground state. The upper box gives the radial probability distribution for the highly excited electron when the core of the atomic ion is in its lowest excited state. The probability for the highly excited electron to be in the closed channel; this is the radial integral of the probability distribution when the atomic ion is excited. The arrow is a clock that gives the time in units of the period of motion of a classical electron with the same energy as the highly excited electron when the atomic ion is excited.
Here is an interpretation of what you see in the movie. The pulsed laser has an amplitude to excite the electron into either the open or closed channel; since the initial state is compact you see one initial peak in the probability distribution of each channel move out from the origin. After one period, the probability in the closed channel returns to the region near the nucleus. When the electron is near the nucleus, it can strongly interact with the core electrons and cause a de-excitation of the atomic ion. If the de-excitation occurs, the highly excited electron gains energy and is ejected from the atom. I've chosen parameters such that the probability for causing a de-excitation during one period is roughly 1/2. Thus, a large fraction of the probability distribution will elastically scatter and remain in the closed channel. This collision is repeated until the probability to be in the closed channel decays to zero. Finally, as in the wave packets for H and Li, the probability distribution in the closed channel will disperse because the higher energy components have a longer period than the lower energy components.
The snapshot above shows the time roughly 3/2 period after the laser excitation. The two features in the probability distribution in the open channel can be interpreted as: (1) the pulse of electron probability that was directly ejected from the atom (r=1.2E5 atomic units) and (2) the pulse of electron probability that arises from the laser exciting the atom such that the core is also excited so that the highly excited electron can not escape but after one period it returns to the nucleus and scatters from the core and then is ejected from the atom.
The important lesson to learn from this movie is that the highly excited electron and the core electrons can only exchange energy when the highly excited electron is near the nucleus. This has the important consequence that if you can measure the ejected electron flux in a time dependent manner you will have measured the probability for the highly excited in the closed channel to be near the nucleus.

There is one last interesting feature that I would like to investigate. How does the electron probability move from the nucleus out to a detector that is a macroscopic distance from the atom? From simple classical mechanics you might expect that the ejected electron flux will be a series of concentric and expanding radial shells. You might also expect that the peaks seen in the movie above will disperse due to the familiar quantum dispersion of free particle wave packets. To see how the ejected electron probability evolves as it moves out from the atom, I plotted the radial probability in the open channel as a function of  r-vt. One frame from the movie is given below.

1.4 Mb MPEG movie of the ejected probability traveling away from the atom

The two bits of information above the plot gives the time in picoseconds where t=0 is the peak of the laser pulse and the leftmost point of the graph in micrometers at this time. There are a number of peaks which can be interpretted as follows. The rightmost peak at r-vt roughly equal 0 microns is from the electron that is directly ejected from the atom after excitation by the laser pulse. All other features are from an indirect ejection. The peak near r-vt equal -5 microns is from the electron moving out from the nucleus in the closed channel, not having enough energy to escape, and then returning to the nucleus where it de-excites the atomic ion and gains enough energy to escape the atom. The peaks near r-vt equal -9, -13, -17 microns are from the electron performing 2, 3,4 radial oscillations in the closed channel before de-exciting the core and escaping.
When viewing the movie, you will probably notice that only the peak near r-vt equal to 0 microns behaves as expected. From the moment it is ejected from the atom the probability in this peak spreads due to dispersion. The other peaks behave in a more complicated manner. For short times they become narrower for a time. After reaching a minimum width they start becoming broader. Even more fascinating than this peculiar behavior is the fact that at small times the peaks between -20 and -10 microns don't even exist. The radial wave packet in the closed channel has dispersed to such a large extent that after the probability distribution in the closed channel almost covers all of the classically allowed range after 2 radial oscilations.
What is the explanation for the peculiar behavior of the indirectly ejected electron probability? The answer is quite surprising: dispersion. Yes, it is dispersion that is causing the indirect peaks to initially become narrower before getting broader. The point to remember is that dispersion has a sign associated with it (do the high energy part of the packet move to the front or the back of the packet). While the electron is bound to the atom, the radial wave packet disperses by having the low energy parts of the packet move ahead of the high energy part (remember the period of motion increases with energy for a Coulomb potential). Once the electron is ejected from the atom, the packets disperse by having the high energy part of the packet move to the front of the packet. These two dispersions can cancel each other.
The amount of dispersion increases linearly with time which is why the minimum dispersion for the -9 micron peaks is reached at a later time than the -5 micron peak.