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I am trying to understand the absorption of different energies/frequencies of light by electrons, and I have a few questions (I am relatively new to quantum mechanics, and may have some fundamental misconceptions).

As I understand it, photons can exist at a continuum of frequencies and therefore energies. For an electron to absorb a photon and jump to a higher energy level, it needs to absorb a photon whose energy is the difference in energy between the 2 energy levels. Does the photon need to be that exact frequency/energy to be absorbed, or is there a narrow range of frequencies/energies that will be absorbed?

If the photon needs to be the exact frequency, it seems to me that the chance of any photon would have a minimal probability of having the specific frequency required, due to the energy/frequency of photons ranging over a continuum. My instinct for this case would be that due to the uncertainty principle, photons would not have exactly defined energies/frequencies, and photons with frequencies around the correct value would have a probability of being absorbed. Is this idea correct?

If the photons can be in a range of frequencies around the correct energy, this would explain why photons have a non-0 chance of being absorbed, and also why the bands on absorption/emission spectra have varying thicknesses, but if so, what happens to the minute difference in energy between the energy of the photon and the actual value?

Which, if any of these thoughts are correct? if none, what is the correct case?

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  • $\begingroup$ Note that electron need not have exact energy, it may be in superposition state consisting of 2 or more energy levels. $\endgroup$
    – kludg
    Feb 5, 2023 at 17:14
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    $\begingroup$ Note also that an electron cannot absorb a photon. An atom, consisting of a nucleus interacting with one or more electrons, can absorb a photon. In the approximation where the nucleus is "infinitely" heavy, we can pretend like only the electrons rearrange; for some transitions it is even productive to pretend that "only one" of the electrons bound to the nucleus changes to a different state. $\endgroup$
    – rob
    Feb 5, 2023 at 21:27
  • $\begingroup$ @Rob strictly speaking electron can absorb a photon in the simple model used in the Compton scattering analysis where it absorbs incoming photon and emits a new one, with modified direction, momentum and energy. This is usually simplified as instantaneous process, but it may take some time. Then the overall process is similar to photon and atom, atom also radiates away from the excited state later. However, I think the point here is that with free electrons, there is no resonance behaviour like with atoms or molecules, the scattering is almost "featureless" as function of frequency. $\endgroup$ Feb 6, 2023 at 1:25
  • $\begingroup$ @JánLalinský Interesting. How is the four-momentum preserved in this model? Is the virtual state off-shell? $\endgroup$
    – rob
    Feb 8, 2023 at 11:42

3 Answers 3

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Does the photon need to be that exact frequency/energy to be absorbed, or is there a narrow range of frequencies/energies that will be absorbed?

"That exact frequency" is an exact frequency only for an isolated atom. In a solid made up of say $N$ atoms, each (at least) $N$-fold degenerate atomic energy level "splits" or "broadens" into solid state energy bands.

The deep core electronic states will not broaden much, but will broaden a little. Photoelectron final states will broaden a lot into "valence" and "conduction" bands. So the energy difference between initial and final states will not have to be the exact atomic energy level difference.

If you are interested in absorption due to a "deep core" electron, then the initial state can usually be well-approximated as an atomic electron state, but not the final states.

The rate for absorption due to exciting the "deep core" electron ($c$) can be calculated via Fermi's Golden Rule as something like: $$ R \sim \sum_f |\langle \psi_f|V|\psi_c\rangle|^2\delta(E_c + \hbar\omega - E_f)\;, $$ where $V$ represents the interaction after getting rid of the photon operator part. (E.g., $V\sim \epsilon\cdot\vec r$, where $\vec r$ is the electronic position operator.)

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If the photon needs to be the exact frequency, it seems to me that the chance of any photon would have a minimal probability of having the specific frequency required, due to the energy/frequency of photons ranging over a continuum.

Correct. But the transition frequencies are just centers of non-zero-width peaks in the response functions. Sharp transition frequency coming out of calculation of eigenvalues does not mean other close frequencies won't react with the system.

So "the photon" (really the external EM field) needs not have the exact frequency corresponding to two eigenstates to make the system manifest absorption. When you analyze interaction of harmonically oscillating field with a simple atom in quantum theory, the atom gets excited even for non-resonant frequencies. It may even get ionized by non-resonant frequencies, if the field is strong enough. It's just that for resonant frequencies, all this is usually much quicker and much more effective.

My instinct for this case would be that due to the uncertainty principle, photons would not have exactly defined energies/frequencies, and photons with frequencies around the correct value would have a probability of being absorbed. Is this idea correct?

More yes than no. When talking about photons in light-matter interaction, we almost always assume the ideal case of photons of definite energy and frequency. EM field in quantum theory in general however is not a nice set of sharp frequency photons. Instead, in general, the sharp photon states are "eigenstates" of the EM field and the real state is some nasty superposition of them. For example, laser light description in quantum theory is far from a definite sharp photon state, instead there is a superposition of all possible number of photons that makes the quantum EM field mimic properties of classical electric field with definite phase and intensity.

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The most bare bones explanation I can give is the following. The dynamics of a quantum system are encoded in its Hamiltonian. I suppose you agree that if the Hamiltonian contains a cross term between two electronic states, a transition between these states is possible.

Now, do you recall the potential energy due to the intercation of an electric dipole $\bf{d}$ with an applied electric field $\bf{E}$? Indeed, an interaction term appears is the Hamiltonian describing the system composed of an electromagnetic field interacting with an atom:

$$H_{int} = -\bf{d\cdot E}$$

If you decompose this interaction part on the basis of the atomic states $\{|\psi_i\rangle\}$ you obtain non-zero terms of the form

$$-\langle\psi_i|\mathbf{d\cdot E}|\psi_j\rangle|\psi_i\rangle\langle \psi_j|$$

These, as you can see, induce a possible transition between the $i$th and the $j$th atomic state. It is after quantization of the electric field that we see that $\bf E$, as an operator, is a linear combination of annihilation and creation operators for the field, which formalizes the notion of destruction and production of photons with definite frecuency and polarisation. In fact, the electric field operator gets out of the bracket we applied to decompose the Hamiltonian on the atomic states basis (since it acts on a different Hilbert space than that where the atomic states live):

$$-\langle\psi_i|\mathbf{d}|\psi_j\rangle|\psi_i\rangle\langle \psi_j|\cdot\bf E$$

Consequently the intercation term exists and the transition between electronic states is possible as long as the dipole operator has non diagonal terms in this basis, which it does. This means that a photon can be absorbed and subsequently the atom get exited, even if the photon's energy is not identically equal to that of the electronic transition.

As an example of a very simple model showing this, I suggest you read about the Jaynes-Cummings Hamiltonian and about Rabi oscillations.

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