- #176
AnssiH
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Yes of course; polarization is usually associated with wave behavior - it refers to the direction of the oscillation. A wave can be a pulse that is oscillating horizontally, vertically, in any diagonal angle, circularly and so on.zonde said:So you say there is continuous wave. Fine. But please explain, does this continuous wave have a property "polarization", or no? When it passes polarizer, is the amplitude of the wave changing as we rotate the polarizer around the clock?
So yes linear polarizers would eat away a direction component from that polarization - some of the amplitude of the wave is eaten away as per cos^2(relative angle). This appears to be what polarization filters are doing (as per experiments).
Note that if a beam splitter outputs two beams at orthogonal polarization, it would yield 50% of the randomly polarized input energy into each path (minus some energy loss from imperfect materials).
But this is an excellent question because it's important to keep in mind what exactly do we suppose happens upon an electron - a wave system in itself - absorbing some energy.
The absorption thresholds are related to the frequency of the wave - makes sense if we are looking at harmonic wave systems (as per Max Planck's theory). It's not possible to achieve the next possible harmonic mode by simply increasing amplitude, you actually have to incite a higher frequency to a harmonic oscillator. But a continuous amplitude can still become reduced to the point where there is no oscillation occurring anymore! At some point before it becomes exactly zero, the implied velocity of the oscillation of the signal starts to approach 0 smoothly.
Usual QM wave function perspective of this is mathematically identical, the only difference is that we would say "the probability of detection starts to approach 0". If you assume Schrödinger Equation represents also an actual wave, the final expectations are the same because even when the actual wave energy starts to approach 0, you will still have noise from the rest of the universe making it possible for a n absorption to occur. So in all practicality, your wave model only gives you the probability of a detection - unless you somehow manage to take the entire universe into account.
The noise being all the excess energies left over from absorption events all across the universe. Which btw is an interesting question I don't here too many people asking themselves - in a photon picture, what happens to the excess energy of a photon that an electron is not able to absorb?
Cheers,
-Anssi