3 ms·
You can only "film" something when the photons hit the "film" (a.k.a the image sensor). There's no way to record photons flying by.
by pathseeker 7y ago
You can only "film" something when the photons hit the "film" (a.k.a the image sensor). There's no way to record photons flying by.
- fastball 7y agoIs it not possible for the energy of a photon to to disrupt EM fields enough to be detectable, without the photon actually being incident on a sensor?
- pbhjpbhj 7y agoKinda, no. By observing a photon you perturb the system, you wouldn't get a result for what that photon would have done had it not been perturbed. AIUI if you detect the photon then it interacted with some part of your sensing apparatus, so it didn't then go on to do what it should have. Thus we're left to look at the aftermath and impute the situation that produced it.
- fsh 7y agoYes, this is called a "quantum non-demolition measurement" and has actually been done. One technique is to use highly excited atoms (Rydberg atoms) which are extremely sensitive to external electric fields. In this way, the field caused by a single photon can be detected without absorbing the photon. There is a famous paper from Serge Haroche's group where they used this method to detect the presence of single microwave photons in a superconducting cavity (https://www.nature.com/articles/nature05589 https://www.nature.com/articles/nature05589). However, the measurement still collapses the wavefunction of the system. Performing a QND measurement to determine which slit the photon takes in a double-slit experiment simply destroys the interference pattern.
- xhrpost 7y agoCurious, what is "the field generated by a single photon" made of? Off hand, I think the force carrier for EM fields is the photon no? So, the photon is emitting other photons (virtual I think?) then?