# [Physics] Why doesn’t a typical beam splitter cause a photon to decohere

decoherenceopticsquantum-optics

In many experiments in quantum mechanics, a single photon is sent to a mirror which it passes through or bounces off with 50% probability, then the same for some more similar mirrors, and at the end we get interference between the various paths. This is fairly easy to observe in the laboratory.

The interference means there is no which-path information stored anywhere in the mirrors. The mirrors are made of 10^20-something atoms, they aren't necessarily ultra-pure crystals, and they're at room temperature. Nonetheless, they act on the photons as very simple unitary operators. Why is it that the mirrors retain no or very little trace of the photon's path, so that very little decoherence occurs?

In general, how do I look at a physical situation and predict when there will be enough noisy interaction with the environment for a quantum state to decohere?

However, you didn't do that. You started the mirror off in a thermal state at room temperature. This state can be considered as a superposition of different momentum states of the mirror1, with a phase associated to each one. If you change the momentum by a small amount, the phase associated to this state in the superposition only changes by a small amount. Now, let $p_\gamma$ and $p_m$ be the original momenta of the photon and the mirror, and let $\Delta p_\gamma$ be the change in the momentum when the photon bounces off the mirror. When you send the photon towards the mirror, the original state $p_m$ (photon passes through) will end up in the same configuration as the original state $p'_m = p_m - \Delta p_\gamma$ (photon bounces off). These two states $p_m$ and $p'_m$ had nearly the same phase before you aimed the photon at the mirror, so they will interfere, and if the phase on these two states are really close, the interference will be nearly perfect.