Is the beta barium borate crystal an observer in the delayed choice quantum eraser double split experiment?

Question

I'm a little confused about the top answer to this question: Variation of delayed choice quantum eraser

He says "if you simply detect all signal photons and make no distinction between them, there will be no interference pattern on the screen"

But in the standard vanilla double slit experiment with no observer, there is an interference pattern between all the photons.

Question. Does this mean the Beta Barium Borate Crystal is effectively an "observer" for the set of all the signal photons that hit D0? If not, then what is?

EDIT: I guess my question wasn't clear

I want to know why there is no interference pattern among all the signal photons. This is a double-slit experiment, and normally a double-slit experiment causes an interference pattern (among all the photons) unless there's an observer at the slits. So why doesn't this one?

EDIT 2: I understand that the interference pattern shown by only the D1 events and that shown by only the D2 events cancel out to produce no interference pattern. However, that is only a correlated result. It doesn't explain why there is no interference pattern. My question pertains only to the cause of the lack of interference.

Answer 1

If we look at the double slit experiment, states corresponding to the slits can be called $| A \rangle, | B \rangle$, and supposed normed. They are not orthogonal because, otherwise, there would not be interferences. One may write :

$| B \rangle = \cos k x| A \rangle + \sin k x| A_\perp \rangle$ , where $A_\perp $ is a state orthogonal to $A$.

The total state is $|\psi \rangle = | A \rangle + | B \rangle = (1 + \cos kx)| A \rangle + \sin kx | A_\perp \rangle \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad (1)$,

and the probability is proportionnal to :

$|\psi|^2 = 2 + 2 \cos kx \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad (2) $

So, we found interferences, as expected.

Now, with the delayed choice quantum eraser double split experiment, a very simple modelisation of the total state could be written :

$|\psi \rangle = (| A_1 A_2 \rangle + | A'_1 A'_2 \rangle) + (|B_1 B_2 \rangle + | B'_1 B'_2 \rangle)\quad\quad\quad\quad\quad\quad\quad\quad\quad (3)$

Here the couples $(A_1,A'_1)$, $(A_2,A'_2)$,$(B_1,B'_1)$, $(B_2,B'_2)$ are orthogonal states, because of the photons entanglement.

Now the relation between $B_1$ and $A_1$ or $B_2$ and $A_2$ is the same as the relation between $B$ and $A$ in the double slit experiment (before the BBO doubling).

$| B_1 \rangle = \cos k x| A_1 \rangle + \sin k x| A_{1\perp} \rangle\quad\quad\quad\quad\quad\quad\quad\quad\quad (4)$

$| B_2 \rangle = \cos k x| A_2 \rangle + \sin k x| A_{2\perp} \rangle\quad\quad\quad\quad\quad\quad\quad\quad\quad (5)$

And now, we define $B'_1$ and $B'_2$ such that $B_1, B'_1$ and $B_2, B'_2$ are orthogonal.

$| B'_1 \rangle = -\sin k x| A_1 \rangle + \cos k x| A_{1\perp} \rangle\quad\quad\quad\quad\quad\quad\quad\quad\quad (6)$

$| B'_2 \rangle = -\sin k x| A_2 \rangle + \cos k x| A_{2\perp} \rangle\quad\quad\quad\quad\quad\quad\quad\quad\quad (7)$

So, the final expression for $\psi$ is :

$|\psi \rangle = (| A_1 A_2 \rangle + | A'_1 A'_2 \rangle )+ (|A_1 A_2 \rangle + | A_{1\perp} A_{2\perp} \rangle)\quad\quad\quad\quad\quad\quad\quad\quad\quad (8)$

We see that the phase dependence in $x$ has disappeared, this means that the 2-qbit density matrix has no dependence in $x$.

Now, if we measure only the first qbit (the signal photon), we have to take the partial trace of the 2-qbit density matrix, to obtain the 1-qbit density matrix. But, because, in the 2-qbit density matrix, there is no phase dependence in x, it will be the same thing in the 1-qbit density matrix.

So, finally, there is no global interference pattern.

You will note, that the fact, that the signal and idler photons are entangled, is fundamental, for instance the state $| A_1 A_2 \rangle + |B_1 B_2 \rangle$ has a phase dependence in $x$

You may be interested by the original article

Answer 2

No, it has nothing to do with " Beta Barium Borate Crystal being an observer". $D_3$ and $D_4$ do not contribute to the interference pattern, and $D_1$ and $D_2$ have an inverse interference pattern. So if you detect all signal photons whose idlers are detected in ($D_1,D_2,D_3,D_4$), you have no interference pattern.

[EDIT] More precisions:

The signal photon events can be separated in idlers photon detections ($D_1,D_2,D_3,D_4$). For each detection, there is a interference pattern, which can be seen (roughly) as a probability of finding, at $D_0$ a signal photon at somme distance $x$:

$$p_{D_1}(x) = A ~cos^2(kx)$$ $$p_{D_2}(x) = A ~sin^2(kx)$$ $$p_{D_3}(x) = p_{D_4}(x) = B$$

Now if we count all the signal photon events corresponding to all idlers photons detections, we have :

$$p_{TOT}(x) \sim p_{D_1}(x) + p_{D_2}(x) + p_{D_3}(x) + p_{D_4}(x) $$

That is :

$$p_{TOT}(x) \sim A+2B$$

We see, that there is no interference pattern if we take in account all events. It will be the same, in we take in account only $D_1$ and $D_2$, in this case, we would have $p_{TOT}(x) \sim p_{D_1}(x) + p_{D_2}(x) = A$

[EDIT2]

If you keep only the $D_1$ events, or the $D_2$ events, you have a full interference pattern.

If you keep only the $D_1$ events and the $D_3$ events, you have a smaller interference pattern.

All combinations of events are possible. The pattern depends on which events one keeps in the analysis.