0
$\begingroup$

I know about Inverse Compton Scattering, but is it theoretically possible to take photons with frequencies so low they cannot be detected, increase their energy and hence make them detectable?

$\endgroup$
6
  • 2
    $\begingroup$ Sure. Just bounce them off of a mirror moving at a very high speed toward the photon source. $\endgroup$ – S. McGrew Jan 17 at 4:31
  • 1
    $\begingroup$ Something along these lines does happen to the cosmic microwave background. The hot electrons in a galaxy cluster can knock the photons of the CMB to higher energy, making the CMB look hotter there. en.wikipedia.org/wiki/Sunyaev%E2%80%93Zeldovich_effect $\endgroup$ – kaylimekay Jan 17 at 5:42
  • $\begingroup$ @S.McGrew Do you know of anyone who has conducted experiments to do this? I would love to read about that. $\endgroup$ – Juan Jimenez Jan 17 at 5:49
  • $\begingroup$ Or you can drop the photon into a gravitational well. $\endgroup$ – Roger Wood Jan 17 at 6:01
  • $\begingroup$ If you can do any of these, the photons can be detected. If you realy mean that they "cannot be detected", by definition there is no way to change anything about them. $\endgroup$ – nasu Jan 17 at 6:13
3
$\begingroup$

Here is the Feynman diagram for Compton scattering:

Compt

A Feynman diagram is the pictorial representation of of the integral needed to calculate an interaction crossection, and depending on how you read the axis it describes the corresponding interaction. If the x axis is taken as time, it describes the scattering of an electron on a photon, and a new photon appearing.

Because of lepton number conservation the outgoing electron is not called "new" , but because the photon's only identification is its $energy=hν$ there is no handle to call it the same photon, it is a photon of different energy, and thus a new photon.

You ask

is it theoretically possible to take photons with frequencies so low they cannot be detected

Theoreticaly there is no bound on the energy of the photon so yes, an electron could scatter off one and a higher energy photon appear , the electron losing energy.

Here is a copy of the low frequency part of the electromagnetic spectrum

enter image description here

( the ultra high frequency working labels are with respect to radio waves, which are lower)

Experimentally, though single photon detection seems to have gone down to the infrared frequencies, the detectors are complicated systems, not single electrons, example here.

$\endgroup$
0
2
$\begingroup$

This is a partial answer because I am not going to discuss the many different ways it is possible to change photons' frequency.

I want to emphasize an issue related to the way the question has been formulated. I think it may be useful to appreciate the meaning of changing the frequency of a photon, whatever method is used.

Photons are definitely neither like small billiard balls traveling at the speed of the light nor like classical wavepackets. Their behavior is expressed within Quantum Electrodynamics (QED) in terms of the presence of one or more excited states of the quantum description of the electromagnetic field.

The kind of statements QED allows is about the presence or not of one or more excited states carrying well-defined energy and momentum.

Every change described by QED is always expressed in terms of the probability that, given the state of the system at time $t_0$, there is another state at time $t$.

In practice, a meaningful description of every process involving a single photon will always be stated as given a state with one photon of wavevector ${\bf k}$ and frequency $\omega$ at the time $t_0$, there is a certain probability that the state of the system at a later time $t$ will consist of a photon of wavevector ${\bf k'}$ and frequency $\omega'$. There is nothing in the formalism that could say that we deal with the same photon. Actually, the QED formalism describes the process as the annihilation of the original state with one photon and creating a new state again with one photon.

I apologize if I wrote about things well-known to the author of this question but I feel useful to stress them for the general reader.

$\endgroup$

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.