Can we keep a photon bouncing of mirrors for a long time (eg days)?

Question

If a photon is sent into a box with perfect mirrors aligned in a way that it won’t ever be reflected out, is it possible to hold the photon for extended lengths of time with its quantum properties intact, and measure it when we want by opening a path into the detector?

Answer 1

One way to store photons is with a resonance cavity. The quality factor $Q$ is the inverse fraction of photons lost per radian of a resonance cycle. It takes $Q$ radians or $Q/2\pi$ cycles for a photon to be absorbed. $Q$ can reach values of up to 50 billion for superconducting RF resonant cavities. This a time of $5\times 10^{10} / 1.3 \times 10^9 \approx 40$ s.

https://en.m.wikipedia.org/wiki/Superconducting_radio_frequency

Answer 2

The answer for light, by Lubos Motl here gives an estimate of the time it takes for electromagnetic waves to be absorbed, $1/1,000$ of a second

When we are talking of photons, we are in the quantum regime, and in the quantum regime there may be surprises not existing in the classical solutions.

A photon hitting a perfect mirror will undergo elastic scattering with the quantum mechanical lattice of the molecules of the mirror. In the center of mass "mirror photon" the photon does not lose energy. But the laboratory is only approximately at that center of mass, because of the very large mass of the mirror lattice the lab almost coincides with the center of mass. That "almost" will give the same time results as the classical light for eventually the complete loss of energy of the photon, which will be falling towards the infrared frequencies and will be absorbed by the lattice.

There is a research direction called "quantum optics" that is growing that may have possibilities to keep a photon reflected for much longer times, and maybe an expert can answer.

Answer 3

The "storage time" of a photon in the cavity is given by the cavity round trip time, $\tau_{RT} = \frac{2L}{c}$, multiplied by the cavity Finesse $\mathcal{F}$. $c$ is the speed of light and $L$ is the length of the cavity. The cavity Finesse is essentially the inverse of the photon survival probability on a single round traip. That is, what is the probability, on a single round trip, that a photon is lost out of the cavity. The main reason for a photon to be lost is due to transmission in the mirrors, absorption in the mirrors or the intervening medium between the mirrors, or scatter off of mirror roughness.

State of the art mirror coatings can acheive transmissions at the level of few parts per million (ppm). I will assume transmission probability for each mirror of $T = 1$ ppm. Absorption in the mirrors can be less than $1$ ppm. It is very difficult to get scatter losses on the mirror surfaces to be less than $1$ ppm but I will assume for purposes of this calculation that it is possible.

The Finesse is calculated as

$$ \mathcal{F} = \frac{2\pi}{T_1+T_2+L_{tot}} \approx 300000 $$

$T_1 = T_2 = 2 \text{ ppm}$ and $L_{tot}\approx 0$. The longest moderate finesse optical cavity I know is LIGO with an arm length of 4 km. The round trip time for LIGO is then

$$ \tau_{RT} = \frac{2\times 4 \text{ km}}{3\times 10^8 \text{ m/s}} \approx 25 \mu s $$

The cavity lifetime is that

$$ \tau_{lifetime} = \tau_{RT} \mathcal{F} = 750 \text{ ms} $$

Note that this is probably one to two orders of magnitude more generous than what could actually be realized with current technology. So the answer is NO, with modern technology we could not store an optical photon in a high finesse cavity for days.

In principle, with better mirrors coatings and mirror surface we could hold a photon for longer, but it is hard to imagine storage on the timescale of days.

Answer 4

It was many times discussed here that the interaction between a photon and a mirror always is accompanied by the pressure of the photons momentum to the mirror. Or the mirror moves back, or some amount of the energy gets dissipated into inner vibrations (this is a local movement of a part of the mirror with fooling vibrations).

The re-emission (as Bill states) of the photon happens with a lower among of energy. The photon gets redshifted and that is why the EM radiation goes very fast redshifted and “heats” at the end the mirrors.

All predictions or statements about perfect mirrors will violate the second law of thermodynamics.

Answer 5

In principle if you can arrange a medium geometry such that your mirrors are in a perfect internal reflection geometry path, the only source of losses will be the internal scattering and absorption from the medium

if you want the propagating medium to be the vacuum, you are forced to rely on reflective mirror surfaces. The LIGO interferometers rely on extremely precise (and expensive) mirrors that can bounce photons a mean number of reflections of around 400 times before being scattered. This is the current state-of-the-art in reflection quality, at least that I am aware

Then there are so called "perfect dielectric mirrors" which I'm not acquainted with, so I can't tell about their shortcomings

Answer 6

Photons don’t bounce. They can be transmitted, scattered or absorbed. What looks like bouncing is a photon being absorbed and a new one being emitted. The new photon is emitted in a random direction and the odds of it coming directly back are very low. So even if both mirrors were perfectly aligned you would never get an infinite bounce. You’ll be lucky to get one so called bounce.