Quantum Mechanics – Is Delayed Choice Quantum Eraser Possible Without Retrocausality?

Question

After reviewing the more than half dozen questions on the DCQE here, this doesn't seem to have been asked or answered directly – if I've missed this piece in one of the other questions or answers please point me in the right direction and we can close this.

Using the detector notation from the DCQE Wikipedia article for brevity:

Let $D_0$ = idler photon collector plate

Let $D_1$ = no-path information detector 1

Let $D_2$ = no-path information detector 2

Let $D_1$ and $D_2$ be time-like separated from $D_0$ such that $D_0$ will register photon arrival hits some finite, measurable time prior to $D_1$ or $D_2$.

For the purposes of this question, we can ignore the cases where which-path info is collected. Frankly, we can ignore the "choice" to insert a beam splitter or not entirely. For the purposes of this question, a half-silvered mirror will remain in place (no which-path information is available).

After many runs, when I go back to correlate the arrival times of clicks at $D_0$ with clicks at $D_1$ and $D_2$, I note that 2 interference patterns are recoverable, with the location of $D_1$ idler entangled photons aligning with troughs of $D_2$, just as expected.

Since I now know the locations of the respective peaks and troughs on $D_0$ correlated with $D_1$ and $D_2$ detection, I wipe my detection plates clean and start again. I note that my first "click" at $D_0$ is in a location that corresponds to a peak in the $D_1$ fringes.

My question is simple:

Based on this knowledge, is the signal photon more likely to be detected at $D_1$? (my $D_1$ detector hasn't clicked yet due to the time-like separation).

It seems the answer must be yes.

If it IS yes, then why is there any discussion at all about backwards causation? If I can note the location of photon recording on $D_0$ and predict a higher or lower probability of recording an entangled sister photon on $D_1$ or $D_2$, then that's the ballgame. Where's the spookiness that has this experiment come up again and again?

If the answer is no in this setup, I struggle to see why, and so any clarification would be greatly appreciated.

Answer 1

The short answer is that you're right. The delayed choice experiment doesn't require backwards-in-time shenanigans, and all the pop-science articles implying this to be the case are basically garbage.

For example, here is the example delayed choice eraser circuit from my quantum circuit simulator Quirk:

The green boxes are state displays that show the probability of each possible measurement result, optionally conditioned on some other qubits' possible measurement results.

The top wire is the "choice" qubit. The second wire is the "which slit" qubit. The rest of the wires are the "where did it hit on the screen" qureg. The first two operations setup some entanglement between the which-slit qubit and the screen qureg.

The four displays on the right show that, if you group the screen measurements by the choice qubit and the which-slit qubit, then within the groupings you will see an interference pattern if and only if the choice qubit is ON.

But correlation goes both ways. Instead of thinking about how the choice qubit and which-slit qubits predict the screen measurement, we can think about how the screen measurement predicts the which-slit qubit's state.

It would be tedious to set up $2^7$ bloch sphere displays, each conditioned on a different screen measurement. Instead, let's use a single condition but cycle through offsets to the screen measurement. This circuit makes it very clear that the landing position on the screen is correlated with different states of the which-slit qubit:

This diagram also answers your main question:

[given a landing position] is the signal photon more likely to be detected at D1 [than D2]?

Yes. The value analogous to the D1-vs-D2-given-position likelihood is shown in that green box in the top right of the diagram. The chance is changing as we focus on different positions.

Notice that the qubit is spinning like crazy regardless of whether we applied the $X^{1/2}$ rotation that will be controlled by the choice qubit. All the $X^{1/2}$ rotation does is switch which axis the qubit is spinning around as I vary the screen measurement position being focused on. If there is no $X^{1/2}$ rotation, the spinning goes around the measurement axis, and so doesn't affect the probability of measuring ON-vs-OFF (i.e. we've picked a rather poor measurement axis). But if the $X^{1/2}$ is applied, then the spinning is around the Y axis instead of the Z axis, and so does translate into changes in computational-basis measurement probability.

So here is a forward-in-time collapse interpretation of the delayed choice experiment:

1. We create a situation where the which-slit qubit is entangled with a screen-landing position.
2. We measure the screen-landing position. This collapses the system into a state where the which-slit qubit is pure, but the specific direction it points is determined by the landing-position measurement.
3. We choose to measure along an axis that picks up these variations-from-landing-measurement, or to measure along a totally uninformative axis.
4. We act surprised that our choice determined whether we see correlations between the which-slit qubit measurement and the screen landing-position measurements.
5. We switch to thinking about the correlations in the opposite direction and start talking about time travel, because that's what brings in the sweet sweet clickbait ad revenue.