Why does a BBO crystal put after 2 slits make the interference pattern disappear?

Question

In this "Delayed choice eraser" experiment:

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.

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.

Answer 1

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.