photonsreflectionspecial-relativityspeed-of-lightvisible-light
When a photon of light hits a mirror does the exact same photon of light bounce back or is it absorbed then one with the same properties emitted? If the same one is bounced back does it's velocity take all values on $[-c,c]$ or does it just jump from $c$ to $-c$ when it hits the mirror?
Or, is the phenomenon of a mirror better explained using a wave analogy? If so, what is this explanation?
I just began learning light and optics and I am just so confused about what reflection, refraction and absorption is.
Below I will assume there are two media with different indices of refraction: the first medium is the one from which the incident light arrives and the second medium is that which the incident light either reflects off of or refracts and propagates through.
I think a large part of my confusion stems from exactly what happens at an microscopic level when light hits an objects that causes it to be either reflected, refraction and absorbed?
On a classical level, some of the incident radiation is almost always reflected, transmitted, and absorbed/attenuated (e.g., wave amplitude decreases as it propagates through the medium due to absorbed photons converting to internal energy of the medium). In most cases during your first introduction to optics, you can entirely ignore absorption/attenuation. Thus, you need only calculate the reflection and transmission coefficients to determine the ratios of reflected and transmitted to the incident radiation.
Like during reflection, does the object not absorb and transmit light at all, causing it to have no choice but to bounce back?
At a quantum level, things are a little different but here is the sugar-coated version using the example of sunlight hitting a typical, green leaf. Let's ignore the sun's complex light spectrum for the moment and just assume nice black-body radiation.
When you see a green leaf on a tree/plant, you are seeing the specific range of frequencies/wavelengths reflected by the material in the leaf (i.e., chlorophyll). You see only green because the medium absorbed and/or transmitted the other frequencies/wavelengths. If you held a leaf over a white background next to a completely opaque object (e.g., steel disk), you may be able to notice that the shadow cast by the leaf is not as dark as that of the opaque object (one needs to limit light so it only hits the leaf prior to the white background otherwise scattering from adjacent locations can cause similar effects). This is due to some of the light being transmitted through the leaf. The rest is absorbed and converted to internal energy, e.g., heat (basically, the particles jostle/oscillate faster resulting in a higher mean random kinetic energy).
At an atomic level, when photons are incident on a new medium, they are absorbed by the atoms/electrons and then re-emitted or converted to internal energy.
The re-emitted photons can be at the same frequency/wavelength as when they were absorbed or at a different frequency/wavelength. In the former case, the absorbing atom/electron is effectively unchanged by the interaction. In the latter case, the absorbing electron needs to change its energy to account for the different energy of the re-emitted photon (e.g., change orbital level).
The absorption and re-emission process is stochastic, thus, the direction of the re-emitted photon can be random relative to its incident direction. Statistically, some photons are re-emitted propagating back into the medium from which they originated while others try to go further into the material. The ensemble average of all these absorptions and re-emissions leads to the macroscopic, classical approximations we call optics.
As I understand it, the question boils down to, "Why doesn't a mirror collapse the wavefunction of a photon that reflects off the mirror?". The answer is that the photon does not change the state of the mirror. After the photon has been reflected, the mirror is unchanged. There is no way to prove that the photon struck the mirror without also detecting the photon's path downstream. The only state-changing event that occurs to the photon is its detection where it strikes a screen (or camera sensor, or an observer's eye, etc.).
It is not correct that the wavefunction collapses upon interaction. It is "less incorrect" to say that the wavefunction collapses upon detection. "Less incorrect" requires a bit of explanation.
Detection is an interaction that results in an observer "knowing" that the wavefunction has collapsed. Yes, it's a vaguely circular definition. In the many worlds view of quantum mechanics, an observer "detects" the state of a particle and in doing so splits his world into as many different independent alternative worlds as there are possible values for the state of the particle.
In the case of the two-slit interferometer, the observer detects the position ("state") of each photon that strikes a screen. That "detection" (according to the MW view) is not really an observation of what the photon's state (i.e., location) is, but rather a projection of the observer's world onto one of the possible values of the photon's state. In a sense, the observer's world splits into all the possible worlds that would result immediately after the detection event, depending on the possible different state values that the photon might have.
The gedankenexperiment of Schroedinger's cat can be generalized to help explain this. Suppose we put the observer inside a box that is totally isolated from the rest of the universe, and the observer in the box detects a photon on a screen. We cannot know where on that screen the observer detected the photon, until we open the box and look at the observer's records. Moreover, according to Bell's theorem, from our perspective, the observer himself is a quantum object -- so the location of the photon's impact on the screen does not even have an actual value from our perspective until we open the box. The value is not hidden; it is indeterminate. The observer inside the box is sure to think he knows where the photon hit his screen, but from our perspective the observer is in a superposition of states until we "detect" what his state is: until we "open the box". As far as we are concerned, the observer co-exists in all possible states corresponding to different places the photon strikes his screen; and in each state he is sure he knows where the photon struck his screen -- but for each of his different states the photon landed a different place.
All that is background for saying that as long as there is no possible way to know that the photon reflected off a given mirror in an interferometer, the photon wavefunction takes all available paths including those in which it does not reflect of the mirror -- and it will thus form the interference pattern we observe.
If we were to make the mirror so tiny and thin that its recoil could in principle be detected , there would be no interference pattern. I know some folks will ponder what might happen in the gray area between using an extremely tiny & thin mirror and using a normal several-to-many grams mirror. I don't know if anyone has done the experiment, but I'll bet what happens is that the interference pattern's contrast is reduced as the mirror's recoil approaches detectability. It would be an experiment worth doing.
Best Answer
How do mirrors work? is closely related to your question, if not a precise duplicate.
We normally think of photon scattering as absorbing the original photon and emitting a new one with a different momentum, so in your example of the mirror the incoming photon interacts with the free electrons in the metal and is absorbed. The oscillations of the free electrons then emit a new photon headed out from the mirror. Unlike e.g. electrons, photon number isn't conserved and photons can be created and destroyed whenever they interact.