Conservation of information in double-slit experiment

peterfreed

I am a novice and am aware of many possible mistakes in my reasoning below; I am looking for guidance as to what they are.

1. For sake of argument, assume that a photon carries a single bit of information.

2. If a single photon is shot towards a classic double-slit apparatus, it is registered on the back wall and therefore generates one bit (for sake of argument) of information for an observer, who would have been unable to predict the photon's ultimate destination prior to it hitting the back wall. When billions of such photons form an interference pattern on the back wall, the location of each can be statistically predicted to the point where the interference pattern is assured, but on a case-by-case basis they cannot be. Thus each generates a bit of information.

3. If the same single photon is shot through a double-slit apparatus with a detector on each slit, it passes through one or the other slit, and when the detector beeps a single bit of information is generated. When the photon hits the back wall, a second bit of information is NOT generated, as its location can be predicted based on the first bit of information. Thus information is conserved.

4. If - in obvious violation of quantum mechanics - the photon were to pass through both slits simultaneously as a wave, cause two beeps, and then hit the back wall at some unpredictable location, three bits of information would be generated. Indeed, it is not possible to imagine a scenario in which the photon could be recorded passing through one or both slits and recorded on the back wall in an unpredictable location without generating > 1 bit of information.

Based on the above, can it be said that detectors at the slits cause the wavefunction to collapse at the slit because not to do so would violate the law of conservation of information?