New answers tagged quantum-optics
Let us take a pi0 decay. The Feynman graph gives the wave function of the two gammas. As the pi0 is spinless, if we measure the spin of one gamma, we know that the spin of the other gamma will be the opposite. Any entanglement comes from the wave function which mathematically describes what the two photons do. Thus if you have a setup where you ask : ...
If 2 particles such as 2 photons are in a maximally entangled state, then this means that the state of both particles is in a specific state, but the local state of only one of the 2 is perfectly mixed. So for example if 2 particles are entangled such that their polarization states are opposite than a measurement of one of the photons will yield a random ...
Instead of "destruction" think of "measurement". When one photon is measured, the other keeps its state.
I'm not an expert, but is this not simply the single-mode attenuation in the fibre? As far as I can see, the absorbed photon will either a) be re-emitted into a different mode (contributes to attenuation.) b) turn into phonons (contributes to attenuation.) c) be re-emitted into the same mode by stimulated emission (probably very rare, and would you care ...
A very interesting question. It looks like you are right, with a light enough beam splitter and a photon with enough momentum the interference effects will vanish as the beam splitter "measures" the photons. Source: Quantum Processes Systems, and Information by B. Schumacher, et al. Section 10.4.
There can be a pulse made up of many photons whose temporal length is a femtosecond, or there could be a single photo - one energy quantum, with a single wavelength. So, if you know the spectral bandwidth of your short-pulse signal, compare that with the spectral transmittance of your beam splitter. I'd suggest that unless you're right on the cutoff edge ...
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