This discussion centers on the behavior of photons in quantum mechanics, specifically regarding their states before and after measurement. It is established that photons exist in an indeterminate state prior to measurement, and upon measurement, they acquire definite properties. The conversation explores whether a measured photon remains deterministic indefinitely and how it can revert to a random state. Key experiments mentioned include the use of polarizers and the implications of wavefunction collapse in determining photon states.
PREREQUISITESPhysicists, quantum mechanics students, and anyone interested in the principles of photon behavior and measurement in quantum theory.
San K said:you can rotate but not predict the outcome?
DrChinese said:True. If the input is unknown, the output is the same but rotated, even if unknown still.
If the input is indeterminate, i.e. in a superposition of states, then after passing through the waveplate the output is also indeterminate, although in a different superposition of states.San K said:is it unknown but not indeterminate?
lugita15 said:If the input is indeterminate, i.e. in a superposition of states, then after passing through the waveplate the output is also indeterminate, although in a different superposition of states.
If it is in a superposition of polarization states to start with, then as soon as it passes through the first polarizer the wave function will collapse and it take on a state of definite polarization, either parallel to the orientation of the first polarizer or perpendicular to the orientation of the first polarizer. Then, after it passes through the second polarizer, it will have a new state of definite polarization, either parallel to the orientation of the second polarizer or perpendicular to the orientation of the second polarizer, with the probability of becoming polarized in the direction of the second polarizer being equal to the cosine squared of the difference in angle between the two polarizers.San K said:well clarified lugita, how about after passing through a first polarizer (assume it was indeterminate to start with) and then a second polarizer?
lugita15 said:If it is in a superposition of polarization states to start with, then as soon as it passes through the first polarizer the wave function will collapse and it take on a state of definite polarization, either parallel to the orientation of the first polarizer or perpendicular to the orientation of the first polarizer.
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Then, after it passes through the second polarizer, it will have a new state of definite polarization, either parallel to the orientation of the second polarizer or perpendicular to the orientation of the second polarizer, with the probability of becoming polarized in the direction of the second polarizer being equal to the cosine squared of the difference in angle between the two polarizers.
Sorry, by polarizer I didn't mean a polarized filter, which destroys photons which are polarized perpendicular, but rather a polariscope like the one discussed http://quantumtantra.com/bell2.html.mathal said:*** The problem with the logic that there are two lines of probability flowing from any polarized filter is that a polarized filter precludes this possibility.l
Thanks for that clarification. The link you provided had a neat shot of the two polarized images of a page of print seen through a calcite crystal.lugita15 said:Sorry, by polarizer I didn't mean a polarized filter, which destroys photons which are polarized perpendicular, but rather a polariscope like the one discussed http://quantumtantra.com/bell2.html.