- #1
mrspeedybob
- 869
- 65
Here is a thought experiment to illustrate the problem...
Suppose you have a light source capable of emitting single photons on command from a computer. these photons travel through a barrier with two parallel slits and then strike a charge coupled device. The computer records each strike and compiles an image.
After a large number of photons have traveled through the experiment the image recorded by the computer will be the interference pattern of the photons wave functions.
Now suppose that the time of each photon emission is compared with the time of its arrival at the CCD. For all locations except for the line equidistant from both slits the travel time will be different for a photon traveling through slit A then one traveling through slit B. Making this comparison is strictly a software function, every physical characteristic of the experiment remains the same but somehow this determination collapses the wave function of each photon and the interference pattern is not produced. So far so good.
Does recording and then immediately destroying the time data without observing it collapse the wave function? If so, how is this different then not recording it?
If it doesn't then what happens if the image and time data are recorded separately and moved to distant locations. If a decision is made to observe or destroy the time data is there some sort of "action-at-a-distance" that affects what will be seen when the image is observed? What if the image is observed before a decision is made about whether to observe or destroy the time data?
Suppose you have a light source capable of emitting single photons on command from a computer. these photons travel through a barrier with two parallel slits and then strike a charge coupled device. The computer records each strike and compiles an image.
After a large number of photons have traveled through the experiment the image recorded by the computer will be the interference pattern of the photons wave functions.
Now suppose that the time of each photon emission is compared with the time of its arrival at the CCD. For all locations except for the line equidistant from both slits the travel time will be different for a photon traveling through slit A then one traveling through slit B. Making this comparison is strictly a software function, every physical characteristic of the experiment remains the same but somehow this determination collapses the wave function of each photon and the interference pattern is not produced. So far so good.
Does recording and then immediately destroying the time data without observing it collapse the wave function? If so, how is this different then not recording it?
If it doesn't then what happens if the image and time data are recorded separately and moved to distant locations. If a decision is made to observe or destroy the time data is there some sort of "action-at-a-distance" that affects what will be seen when the image is observed? What if the image is observed before a decision is made about whether to observe or destroy the time data?