In this "Delayed choice eraser" experiment:

experiment schematic

if one looks at all the photons detected by $D_0$ regardless of their coincidence with the other detectors, one sees no interference pattern.

It is obvious by taking the sum of coincidence patterns $\sum_1^4<D_0,D_i>$, since the $<D_0,D_1>$ and $<D_0,D_2>$ patterns have a phase difference of $\pi$, and the $<D_0,D_3>$ and $<D_0,D_4>$ don't show any interference to begin with.

simulated results

So not looking at all at what the idler photon does, and only looking at $D_0$ we would see no interference.

But without the BBO after the 2 slits, we'd see Young's double slit fringes.

My question is: Why does the BBO destroy the interference pattern from the double slits.

  • $\begingroup$ Would it be because the atoms in the crystal are entangled with the photons and we lack the information about their state? Just guessing, but whenever an active optical element goes into a light path one might want to look at the atomic physics of that material... after all, one can't even model a laser without understanding this in detail. $\endgroup$ – CuriousOne Aug 16 '16 at 11:23

It is indeed because of the entanglement that is produced by the BBO's that the interference is lost. In quantum information technology the loss of coherence often results because the system becomes entangled with the environment.

Let's start with the following coherent superposition $$ |\psi\rangle = |a\rangle+|b\rangle . $$ The interference will show up when we do a projective measurement $$ \langle\psi|P|\psi\rangle = \langle a|P|a\rangle+\langle b|P|b\rangle + \langle b|P|a\rangle+\langle a|P|b\rangle , $$ where $P$ is a projection operator. The first two terms are just constants given by those parts of $|a\rangle$ and $|b\rangle$ that survive the projection process. The last two terms would give the interference $$ \langle b|P|a\rangle+\langle a|P|b\rangle \sim \cos(\theta) . $$

Now consider the case where the BBO (or the environment) causes entanglement (represented by an operation $E$). Then we would have $$ E|\psi\rangle = |\psi'\rangle = |a\rangle|c\rangle+|b\rangle|d\rangle . $$ In this case the parts given by $|c\rangle$ and $|d\rangle$ are not observed; they are ignored. So we need to perform a partial trace to remove them $$ \rho={\rm tr}_{cd} \{ |\psi'\rangle \langle\psi'| \} = |a\rangle\langle a|+|b\rangle\langle b| $$ The result is a density matrix that represents a completely mixed state. When we now do the projective measurement we get $$ {\rm tr} \{ P \rho \} = {\rm tr} \{ P (|a\rangle\langle a|+|b\rangle\langle b|) \} = \langle a|P|a\rangle+\langle b|P|b\rangle $$ So we see that the interference terms are missing. A mixed state does not have any coherence left and therefore cannot show interference.

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  • $\begingroup$ So the BBO produces something the looks like a statistical mixture if we ignore the entanglement, but when we cleverly measure a super-position of the entangled twin, we can filter out the interference pattern because we take the whole density matrix, and not just the trace? Can you show how taking the trace of the density matrix is equivalent to ignoring |c> and |d>? $\endgroup$ – Mr.WorshipMe Aug 22 '16 at 9:27
  • $\begingroup$ @Mr.WorshipMe: Yes, there are sometimes ways that one can perform measurements on the `ignored' parts to recover the original coherence. $\endgroup$ – flippiefanus Aug 22 '16 at 12:47
  • $\begingroup$ @Mr.WorshipMe: taking the trace is like doing a measurement where the projection operator is the identity. This is like saying that we measure it in all possible ways so that anything goes. If one thinks about it, this type of measurement really gives one no information at all. Therefore it comes down to being ignored. $\endgroup$ – flippiefanus Aug 22 '16 at 12:49

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