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  Does the interaction between a delocalized electron-proton pair tend to localize them?

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I'm sure the answer is totally obvious to QM/QFT experts, but it is currently puzzling me.

For simplicity, we could consider a thought experiment where a large cubic or spherical box is filled with electron and proton fields that have identical, centered, symmetric amplitude distributions, that is initially diffuse everywhere in the box (though not necessarily uniform).

Considering the form of the system hamiltonian, how will the ampliutde distributions change over time? In particular, will they tend to concentrate them into a localized bound state?

Thank you for your time!

asked Oct 22, 2014 in Theoretical Physics by Paolo Vicari [ no revision ]

1 Answer

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It depends on the initial states and on the box size.

Let us first consider two free charged particles - electron and positron for simplicity.

If initially their relative motion kinetic energy $E_{rel.}$ is smaller than the negative potential energy of their interaction $-e^2/|\mathbf{r}_1 -\mathbf{r}_2|$ at the distance $|\mathbf{r}_1 -\mathbf{r}_2|$ where you switch the interaction on, then the particles get coupled and form a bound state. They will move together and in this sense they become localized one with respect to the other, but the center of mass will still be free (non localized).

In the opposite case the charges fly apart from each other.

The same may happen in a box. The minimal kinetic energy in a box is proportional to $1/L^2$ and the first perturbative correction to it due to Coulomb interaction $V_{00}\propto\int \sin(\pi x_1)\sin(\pi y_1)\sin(\pi z_1)\cdot \sin(\pi x_2)\sin(\pi y_2)\sin(\pi z_2)\frac{d^3r_1d^3r_2}{|\mathbf{r}_1 -\mathbf{r}_2|}$ is proportional to $-1/L$. So for a large box the kinetic energy of their relative motion can be smaller than the negative potential energy of their attraction and they can get coupled relative to each other. The center of mass, however, will be moving "freely" in the box.

In the opposite case of small box the interaction will not be able to create a neutral system moving as a whole in this box. Similarly, if you box is large, but the initial state has high relative kinetic energy ($\propto n^2/L^2$), then no bound state creation is possible, although the total wave function will be distorted with respect to the initial $\sin(\pi n x_1)\sin(\pi n y_1)\sin(\pi n z_1)$ and $\sin(\pi n x_2)\sin(\pi n y_2)\sin(\pi n z_2)$. However, if you include radiative losses due to their mutual scattering (but not due to collisions with walls), the extra kinetic energy of their relative motion will be progressively lost until the charges get coupled and descend into the ground state of their relative motion.

answered Oct 30, 2014 by Vladimir Kalitvianski (102 points) [ revision history ]
reshown Nov 1, 2014 by Vladimir Kalitvianski

@Vladimir Kalitvianski Hi Vladimir, thank you for your kind answer.

I would like to ask for a few clarifications.

If initially their relative motion kinetic energy E rel. is smaller than the negative potential energy of their interaction −e 2 /|r 1 −r 2 | at the distance |r 1 −r 2 | where you switch the interaction on, then the particles get coupled and form a bound state.

How can we meaningfully talk of r1 and r2 position vectors when we are explicitly considering delocalized particles? Do we have to consider a specific localized realization of a two-particle system belonging to a 'delocalized ensemble' to decide whether they will form a bound state?

They will move together and in this sense they become localized one with respect to the other, but the center of mass will still be free (non localized).

This means that in case we detect both particles, we can expect them to be found close to each other in most cases, but we have no clue where their center of mass will be. In other words, we cannot expect the particle density cloud associated with each single particle to sharpen over time. Does it mean it remains the same as we started with, or that its deformations are not expected to change its qualitatively delocalized nature?

And, lastly, would the conclusions be the same if the same system were studied within a QFT paradygm?

Thank you,

@PaoloVicari: First I was speaking of classical particles (to explain physics), so their relative distance makes a perfect sense.

In QM and in a box, particles are "delocalized" indeed, that is why calculations (for example, the energy correction) is calculated as an integral over the box.

The internal interaction cannot influence the center of mass position, at least, out of box. Out of box you have a product of two plane waves $\rm{e}^{\rm{i}\mathbf{p}_1\mathbf{r}_1}\rm{e}^{\rm{i}\mathbf{p}_2\mathbf{r}_2}=\rm{e}^{\rm{i}\mathbf{P}\mathbf{R}}\rm{e}^{\rm{i}\mathbf{p}\mathbf{r}}$ where $\mathbf{P}$ is the center of mass momentum and $\mathbf{p}$ is the relative motion momentum (and $\mathbf{R}$ and $\mathbf{r}$ are the CM and relative coordinates). When you switch the interaction on, the plane wave of the relative motion will be changed into a wave function $\psi(\mathbf{r})$ of a bound state, but the CM will remain the same. Of course, I mean the case when the wave packets of particles, modeled here as plane waves, coincide while switching the interaction on. I.e., I do not consider a scattering problem.

I think that similar behavior will be in a large box where the bound states are possible.

If no particle creation/absorption is foreseen, QFT will give the same results.

@VladimirKalitvianski: Thank you for the clarifications Vladimir. In summary, the delocalization of the center of mass of the two-particle system is not influenced by the formation of a bound state between the component particles. This makes perfect sense.

@PaoloVicari: Yes, it only is the relative motion that gets "localized". The bound system as a whole can stay "delocalized". The total wave function is $\Psi(\mathbf{R},\mathbf{r})$ and it involves a delocalized center of mass.

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