Understanding - how universe/reality plays dice

[DrChinese]

you can rotate but not predict the outcome?

True. If the input is unknown, the output is the same but rotated, even if unknown still.

 

[San K]

True. If the input is unknown, the output is the same but rotated, even if unknown still.

is it unknown but not indeterminate?

 

[lugita15]

is it unknown but not indeterminate?

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]

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.

well clarified lugita, how about after passing through a first polarizer (assume it was indeterminate to start with) and then a second polarizer?

 

[lugita15]

well clarified lugita, how about after passing through a first polarizer (assume it was indeterminate to start with) and then a second polarizer?

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.

 

[mathal]

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.

*** 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. Delta Kilo pointed this out very lucidly to me.
Tests have shown that my example of horizontal-diagonal-vertical polarized filters results in <1/8 the output of the original signal which precludes any signal from an orthogonal signal out of any polarized filter getting through by any means. This notion (of double exit from polarized filters) may be my fault, and for that, I apologize. mathal

 

[lugita15]

*** 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

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]

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.

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.
mathal