How do you compute Recombination Probability ?

[Widdekind]

When a free electron is incident upon a bare proton, how does one compute the "recombination" probability, for the formation of a neutral atom?

 

If the electron "wave packet" is moving at even thermal velocities, 100-1000 m/s, the initial interaction, between the two particles, is surely fairly fleeting -- perhaps a trillionth of a second (??). When the electron "wave packet" is far away from the proton, the Overlap Integral (and, hence, Transition Matrix Element) is ~0. Then, a trillionth of a second later, the scattered "wave packet" ricochets away, and the Overlap Integral returns to ~0. In between, briefly, the Overlap Integral increases, to some maximum value, and then declines, again, back down to negligibility. Would the effective "recombination" probability, be computed, from the maximum Overlap Integral ??

[swansont]

Why would the interaction time be a trillionth of a second? The electrostatic force has an infinite range.

[Widdekind]

In practice, wouldn't the electrostatic potential be "strong" only w/in a few nm of the nucleus?

 

 

 

 

 

Atomic electron orbitals can be decomposed, as the sum, of inflowing & outgoing spherical waves (e^ikr/r), which, taken together in opposing pairs, produce Standing Waves (M.Wolff. Schrodinger's Universe). For the Hydrogen 1S state, ~e^-ar/r, the overlap integral:

 

[math]< k_r | \Phi_{1S} > \; \propto \int_0^{\infty} 4 \pi r^2 \frac{e^{i k r}}{r} \frac{e^{-\alpha r}}{r} dr \propto \frac{\alpha + i k}{\alpha^2 + k^2}[/math]

where, in the last step, the imaginary exponential has been decomposed, into sines & cosines, and those integrals computed "by parts". This makes the fractional contribution:

 

[math]\left| < k_r | \Phi_{1S} > \right|^2 \propto \frac{1}{\alpha^2 + k^2}[/math]

Since [math]\alpha \approx a_B^{-1}[/math], we see that the Heisenberg Uncertainty Principle is obeyed. Since plane waves, comprising (electron) wave packets, naturally scatter, off of nuclei, into spherical waves, one can start to see, how a simple scattering interaction, could "generate" or "set up" such "spherical standing waves" (which would then, presumably, be "augmented" & "exaggerated" during the "quantum jump", from free particle, to bound orbital, even as the "tails" fizzled away).

Edited September 7, 2010 by Widdekind

[Widdekind]

Why would the interaction time be a trillionth of a second? The electrostatic force has an infinite range.

 

Insofar, as any interaction, can be described, as "beginning" ([math]P \approx 0[/math]), "peaking" ([math]P = P_{max}[/math]), and "ending" ([math]P \approx 0[/math]), then isn't the "peak probability" (P_max) what quantum theory calculates, for the transition probability? If so, could [math]\partial P / \partial t = 0[/math]) "trigger" the "dice roll" of w.f. 'collapse' ??

[swansont]

I'm guessing that the electron recombination with a bare proton could be estimated classically. To first order, it's probably just a matter of whether the electron can shed enough energy to become bound (KE < |PE|)

[Widdekind]

I'm guessing that the electron recombination with a bare proton could be estimated classically. To first order, it's probably just a matter of whether the electron can shed enough energy to become bound (KE < |PE|)

 

That seems completely consistent with quantum theory. The de Broglie wavelength, of a 14 eV electron, is roughly:

 

[math]\frac{p^2}{2 m_e} = 14 \, eV[/math]

 

[math]p = \frac{h}{\lambda}[/math]

 

[math]\therefore \lambda = \sqrt{\frac{h^2}{2 m_e E_0}} = 3.3 \AA[/math]

That's comparable, to the atomic distance scales, of the available electron orbitals. Thus, electrons with really high incident kinetic energies, would start having de Broglie wavelengths smaller than the size of atoms. If the incident electron's wave function oscillated quickly, across the nuclear positive potential well, the overlap integral would start to self-cancel, leading to lower & lower transition probabilities. (An electron, whose de Broglie wavelength was a_B, would have a KE of ~535 eV (1/2 keV).)

 

 

EDIT: Technically, wouldn't the overlap integral involve the time-frequency phase factors, [math]e^{i \omega t}[/math]? Would beat frequencies, [math]e^{i (\omega - \omega_0) t}[/math], play any important part in the process ?? Or, would those phase factors cancel out, when you took the square modulus, in going from the overlap integral, [math]< \Psi_e | \Phi_{1S} >[/math], to the transition probability, [math]| < \Psi_e | \Phi_{1S} > |^2[/math] ?

 

Edit 2: How would you compute the overlap integral, of an incident plane wave, [math]e^{i \vec{k} \vec{r}}[/math], with the ground state, [math]\Phi_{1S} \propto e^{- \alpha r} / r[/math] ? With only 1 power of r in the denominator, does the integral still converge ?

 

Edit to Edit 2: On second thought, I calculate that the overlap integral, between an incident plane wave, and the 1S ground state, is [math]\propto \frac{1}{k^2 + \alpha^2}[/math], so that the overlap dramatically declines when [math]k > \alpha[/math], to wit, when the de Broglie wavelength is less than a_B.

Edited September 13, 2010 by Widdekind