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Suppose I set up the double-slit experiment using photons as my particle. Behind the left slit I place a beam splitter that points some of the light off in the direction of a camera (represented as C in the diagram below). We'll say that the beam splitter splits a light beam passing through it into two equal beams. The light that passes through the beam splitter unaffected and the light that passes through the right slit continue on to another camera (represented as CCC in the diagram below). We'll say that camera C is closer to the beam splitter than camera CCC is.

     | |
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   C-\ |
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If I fire a single photon through this apparatus, there's about a $1/4$ chance that it will be detected by camera C and about a $3/4$ chance that it will be detected by camera CCC. When the wavefunction reaches camera C, the photon is forced to either be detected by camera C or have the part of the wavefunction leading to camera C disappear.

Question: In the case that the photon is not detected by camera C, how is the $1/4$ probability from that branch of the wavefunction redistributed to the remaining two branches of the wavefunction, the branch went through the left slit and straight through the beam splitter and the branch that went through the right slit? The first starts with about $1/4$ probability and the second starts with about $1/2$ probability. Does the first get a new probability of $1/3$ and the second a new probability of $2/3$? Does the first get a new probability of $1/2$ while the second keeps its old probability of $1/2$?

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Probabilities in quantum mechanics have the same meaning as probabilities in classical systems. When you throw a dice and it comes out 6, has the probability distribution changed? It is still 1 in 6 flat. Wavefunctions though change before and after an experimental setup, but a new wave function has to be computed taking into account the new boundary conditions, not the old wavefunction which generates a probability distribution for different boundary conditions. In the dice example, it would be as if the rule was that throwing six takes you to the rulette table and its probability functions – anna v Dec 8 '12 at 10:39
up vote 1 down vote accepted

Starting without the beam splitter, just to establish notation, label the slits 1 and 2, the laser L and a point on the screen x. Then $\langle 1| L \rangle$ is the (complex) amplitude to go from the laser to slit 1. $\langle x| 1 \rangle$ is the amplitude to go from slit 1 to a point on the x on the screen. The amplitude to do that path combination is $\langle x| 1 \rangle\langle 1| L \rangle$. Given that it must land somewhere on the screen we know $$ \sum_x(|\langle x| 1 \rangle\langle 1| L \rangle + \langle x| 2 \rangle\langle 2| L \rangle |^2)=1$$ (or integral...)

If we now place a beamsplitter B right after slit 2, then instead we have$$|\langle c| B \rangle \langle B| 2 \rangle \langle 2| L \rangle|^2+\sum_x(|\langle x| 1 \rangle\langle 1| L \rangle + \langle x| B \rangle \langle B| 2 \rangle\langle 2| L \rangle |^2) =1$$ Now we know a few things about the constituents: $$|\langle 1|L \rangle |^2=0.5 $$ $$|\langle 2|L \rangle |^2=0.5 $$ $$|\langle B|2 \rangle |^2=1 $$ $$|\langle c|B \rangle |^2=0.5 $$

So the first term is $\frac{1}{4}$ (propagation into the camera) and the sum of the other terms must be $\frac{3}{4}$ (propagation to somewhere on the screen).

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Does this mean that the pattern observed at CCC would not be affected by whether C is present or not? – yakiv Dec 11 '12 at 4:10
I guess you mean the beamsplitter is still present, but you remove the camera. If you imagine a single photon at a time (OK not a laser source), then if it arrives on the screen, you still don't know if it's come through slit 1 or 2, so an interference pattern would still be present. The pattern might be shifted with respect to the pattern obtained with no beamsplitter if the beamsplitter applies a phase shift. – twistor59 Dec 11 '12 at 7:54
Yes, I meant that the beamsplitter would still be present but without the camera off to the left. Just to be absolutely clear, would having no camera off to the left give the same interference pattern that having a camera off to the left would give (assuming the presence of the beamsplitter in both cases)? Thanks for spending time on this question. – yakiv Dec 11 '12 at 18:53
My understanding is that whether or not camera c is present to capture the reflected photons, or whether those photons are free to travel indefinitely will make no difference to the observed pattern. The 50% of the energy that's transmitted by the beamsplitter contributes to the interference regardless. – twistor59 Dec 11 '12 at 19:49

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