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Atoms can absorb photons, and when they do, an electron "orbiting" that atom moves up an energy level. I was wondering what the dynamics of this interaction was? Precisely how is the photon absorbed, from a wave function perspective?

My mental picture is that the wave function for the photon would have to go to zero at the end of the interaction (since it has been absorbed by the atom), and the wavefunction for the atom (lets say hydrogen for simplicity) would have to change to the wavefunction for excited hydrogen.

This is just a vague picture though, and I was wondering if anyone had done the mathematics of this interaction, and whether anyone had made a physically correct animation of it. What does the dynamics of an atom absorbing a photon look like?

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  • $\begingroup$ quantummechanics.ucsd.edu/ph130a/130_notes/node465.html $\endgroup$
    – G. Smith
    Dec 11, 2020 at 5:46
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    $\begingroup$ Your mental picture is overly classical. An incoming photon only has a certain probability of being absorbed. The ground state only has a certain probability of transitioning to an excited state. $\endgroup$
    – G. Smith
    Dec 11, 2020 at 5:49
  • $\begingroup$ @G.Smith I was considering the specific case where a photon is absorbed. Obviously there is a probability that it wont be, but this is not in the interest of the question. $\endgroup$
    – user400188
    Dec 11, 2020 at 5:58
  • $\begingroup$ What I meant is that photon’s field does not “go to zero” because there is a probability that it is not absorbed. When a quantum process has various outcomes, you can’t make animations of evolving fields for one particular outcome. It’s like asking for an animation of how the cat gets from alive+dead to alive or dead. The mathematics is clear but the “mental picture” is not, because our brains don’t perceive the world as quantum mechanical. $\endgroup$
    – G. Smith
    Dec 11, 2020 at 6:31
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    $\begingroup$ This applet shows an animation of the electron wavefunction when irradiated by a resonant electromagnetic wave of constant amplitude. If the field contains only 1 photon the effect on the atom is of course very small, but the resulting superposition of orbitals contains a small contribution of the excited state. $\endgroup$
    – A. P.
    Dec 12, 2020 at 21:08

2 Answers 2

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Let's start by looking at the behavior of an atom irradiated by a resonant electromagnetic wave of constant amplitude and frequency $\omega$. For simplicity, let's use hydrogen.

Classical electric field

The electric field polarizes the atom by moving the electron along the axis of polarization. To appropriately account for the discrete energy eigenstates of the electron ($|1s\rangle$, $|2s\rangle$, $|2p\rangle$, etc.) we need to treat it quantum mechanically. So instead of a point particle we need to describe it by its wavefunction. In the ground state it occupies a $1s$ orbital.
If you, for example, excite it at the frequency resonant with the $1s \to 2p_x$ transistion, the wavefunction describing the electron is displaced by the electic field along the $x$ axis. It can then be described by a superposition of the wavefunctions associated with the $1s$ and $2p_x$ orbitals. $$\left| \Psi(t) \right\rangle = \alpha(t) \left| 1s \right\rangle + \beta(t) e^{-i \omega t} \left| 2p_x \right\rangle$$ After some time the electron will be fully in the $\left| 2p_x \right\rangle$ state. Driving it further at the same frequency $\omega$ will not inrease its energy (as expected from a harmonic oscillator), but transition it back to the ground state. This is in line with how Einstein described stimulated emission in his famous 1917 paper Zur Quantentheorie der Strahlung, Phys. Z. 18 121-128 (1917). In §2b) he wrote:
If a Planck resonator [the atom] is in a radiation field, the energy of the resonator can be changed by the transfer of energy from the electromagnetic field to the resonator; this energy can be positive or negative depending on the phases of the resonator and of the oscillating field.
Better than any wordy description is a visualization of the wavefunctions, like this applet. I got a pretty good intuition for this process by just watching the electron undergoing many Rabi cycles.

The movement of the electron creates an electric field by itself, which interferes with the incident beam. On the transmission side this interference is destructive, so that a weak, strongly focused beam can be completely blocked.

Single photon

Since a single photon holds only a finite amount of energy you need to care about its spatial extend. Typically, single photons are created by other atoms, which in free space emit in all directions. For simplicity, let's not care about this and assume that we managed to transform its emission pattern into a Gaussian beam focused on our observed atom.
The temporal envelope of a photon created by spontaneous emission is a decaying exponential, but the excitation probability is highest, when it's a rising exponential. In fact, if the photon temporally and spatially matches a time-reversed spontaneously emitted photon it would be absorbed with a probability of $100\,\%$. In that case the photon is completely gone at the moment the absorption is finished. $$\left| 1s \right\rangle_{\text{electron}} \left| 1 \right\rangle_{\text{photon}} \rightarrow \left| 2p_x \right\rangle_{\text{electron}} \left| 0 \right\rangle_{\text{photon}}$$ The more realistic case is that the photon does not fulfill the previous conditions and therefore is only partially absorbed, so that the final state is a superposition of the photon being absorbed and having passed the atom: $$\alpha \, \left| 1s \right\rangle_{\text{electron}} \left| 1 \right\rangle_{\text{photon}} + \beta \, \left| 2p_x \right\rangle_{\text{electron}} \left| 0 \right\rangle_{\text{photon}}$$ One can increase the interaction efficiency $\beta$ for non-ideal photon shapes by embedding the atom in a suitable environment, for example in a cavity.
The latter superposition can be collapsed by a measurement. For example if you place a detector after the atom and you detect a photon you know that it must have passed the atom. Since atom and photon are entangled it also affects the state of the atom. It will be completely in the $| 1s \rangle$ state then. Else, if you detected no photon (and unrealistically made sure that you would inevitably detect one if it was there) the atom would end up completely in the excited state $| 2p_x \rangle$.
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You want the scattering experiment "γ+Hydrogen", a neutral target.

The available energy levels of hydrogen have a width, thus if one could do the experiment the way $e^+e^-$ experiments are done, the energy levels will appear as resonances to this scattering.

The problem is the low energies , ev, needed to scan the region ,I do not think this can be done experimentally.

Theoreticaly a Feynman diagram field theoretic method should be developed, to predict the data given the energy levels available. I found this article, so people have been working along these lines.

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