Forgive me for lack of formality and possibly incorrect understanding, but hopefully someone can both help to explain the intuition and also add mathematical formalism.

In classical electrodynamics, absorption of electromagnetic radiation by a solid, for example a collection of classical dipoles, may be thought of as a destructive interference effect. Oscillating light that is incident on a dielectric material will cause out of phase oscillation of the charges in the material, which will in turn radiate more electromagnetic fields. The incident and radiated fields conspire at lowest order to provide the exponential decay of the Beer-Lambert law. Note that this is an interference effect that relies on the addition of fields.

Now in quantum electrodynamics, where we consider a quantum system interacting with the quantized electromagnetic field, absorption can be considered rather to be a multiplicative effect, whereby the quantum state is multiplied by an annihilation operator which removes one quantum from the electromagnetic field.

Can someone help me rationalize the difference? I’m puzzled that quantum absorption seems to rely on “multiplicative interference” rather than “additive interference?” Thanks!


It's still an additive and not a multiplicative effect.

To model absorption for a quantum state, you look at an output of a quantum beam splitter.

A pure single photon will experience "absorption" by losing some probability amplitude. If your electric field amplitude would be reduced by half by this process, that same reduction will occur but to the probabilities instead.

This could be written as $|1\rangle \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$

Here you can see that it is a additive interference, not a multiplicative interference. The multiplicative confusion I think stems from the use of a^dagger to represent states. I would recommend that you just work it bra's and ket's until your more comfortable with the more advanced methods of representing states.

  • $\begingroup$ For wave addition in the classical case, in the case of complete absorption we have a positive wave plus a negative wave canceling out. However, as you have put it in the quantum case we simply have the 0 photon ket, which is not related to a linear combination of two 1 photon let's since that's not how quantum mechanics works. So there is not an additive destructive interference effect in the quantum theory here. I'm wondering how we can see an additive interference giving us the 0 ket in the quantum theory. $\endgroup$ – Ian Sep 10 at 14:57
  • $\begingroup$ So if your question is: "how can additive interference cause photons to 'vanish'", then you can see what from a mach-zender interferemeter in the link I sent you. If you connect the two outputs of one quantum beam splitter to the other, then you will see this linear interference. It's rare people work with interference that is not linear. $\endgroup$ – Steven Sagona Sep 10 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.