Quantum eraser experiment in enclosure if data is not recorded, destroyed, or encrypted

Question

I'm trying to understand something precisely and finding it difficult to trust videos on YouTube, conversations on Reddit, etc.

This existing question actually comes very close to mine: Quantum eraser without the quantum eraser. There, the asker asks whether removing detection mechanisms, i.e. letting particles "fly off into space, undisturbed and undetected," would restore the interference pattern since the which-way information is not observed. To which user @EmilioPisanty replies that the which-way information is still available in principle, and so the interference pattern would not come back. This user also says,

The only way to restore the interference pattern is to completely erase the which-way information, in a way that makes the reconstruction of which slit the photons went through impossible even in principle.

My question is, what counts as making it impossible? In contrast to the other asker's question where particles fly off into space, what if the experiment was performed in an enclosure where no particles (at least no photons) can escape, and the only thing that makes it out of the enclosure is electrical data from the screen. What if the screen's data is fed to a faulty circuit or computer that isn't recording the data? What if the data is one-way encrypted and the key is destroyed in memory?

Maybe what I'm asking is, "making it impossible" to reconstruct for whom?

Is the answer that, at the time it arrived at the screen, it was in principle possible to reconstruct the which-way information, and therefore what happens afterward to the data is of no import?

Answer 1

When information is copied out of a system that copying suppresses interference: this process is called decoherence

https://arxiv.org/abs/quant-ph/0105127

What matters has nothing do with whether somebody has observed the result or could observe the result. If the information is copied out and dumped in a black hole or travels away at the speed of light so nobody can catch up to it and observe it interference will still be suppressed. What matters is that the information was copied out and nothing else.

Answer 2

So I do not think this will answer your question perfectly but I would like to give a try to explain how another (simpler) quantum earaser works and maybe the answer to your question can be constructed from it.

So the textbook way to understand the quantum earaser is when we have an initial photon beam through a non-linear crystal (type I SPDC and picking out the entangled photons (where the ordinary and extra-ordinary wave cones overlap).

This creates a two photons (signal and idler) with the same polarization lets define it as horizontally polarized i.e. $|H\rangle_s$ and $|H\rangle_i$. The photons are sent through a set-up as shown in the picture:

Here the basis is the Hong-Ou-Mandel experiment but we add for the idler beam a rotator (usually a half wave plate) which results in the idler beam to be rotated by an angle $\theta$ to $$|H\rangle_i \rightarrow |\theta\rangle_i = \cos\theta |H\rangle_i + \sin\theta |V\rangle_i$$. After that the photons are not identical anymore and "which-way" information is created. After the beam splitter the photon wavefunction is

$$|\Psi_{out}\rangle =\cos\theta \frac{i}{2}(|2H\rangle_1 |0\rangle_2+|0\rangle_1|2H\rangle_2)+\sin\theta \frac{1}{2}(i|H,V\rangle_1|0\rangle_2+i|0\rangle_1|H,V\rangle_2+|H\rangle_1|V\rangle_2-|V\rangle_1|H\rangle_2)$$

So here we see that if $\theta = 0$ we get our Hong-Ou-Mandel experiment back where either two photons enter detector 1 or detector 2 but not one at detector 1 and another one at detector 2. This is a result of the interference of the wavefunctions canceling the one in each detector possibility. So here we have interference and the coincidence probability (the probability to similtaneously find a photon in detector and 2) is thus $P_c=0$. However, when $\theta \neq 0$ we created the "which-way" information leading to distinguishable photons resulting in coicidences (i.e. photons reaching detector 1 and 2 at the same time) depending on the rotation angle $\theta$ as $P_c = \frac{1}{2}\sin^2\theta$ which equals 1/2 for a half wave plate ($\theta=\pi/2$).

Now we come to the earaser part: instead of just counting the coincidence irrespective of the polarization of the photons we now place polarizers in front of the detectors projecting the photon wavefunction into the basis $$|\theta_\text{i}\rangle =|H\rangle_i \cos\theta_i+|V\rangle_i \sin\theta_i$$ where $i=1,2$ for polarizer 1 and 2 respectively. Calculating the the coincidence porbability now we obtain $$P_c = |\langle\theta_1|\langle\theta_2|\Psi_{\text{out}} (\pi/2)\rangle|^2=\frac{1}{2}\sin^2({\theta_2-\theta_1})$$ Note that I inserted a half wave plate for the rotator to simplify the expression i.e. $\theta=\pi/2$. Moreover, when $\theta_1=\theta_2$ the coincidence probability vanishes and we have our indistinguishability back. Hence the "which-way information is earased" by the polarizers and we

Now let us come back to your question: what counts as earasing the "which way" information? I think it is interesting to see what would happen when we would measure the the photons in detector 1 under an angle $\theta_1$ and consider the faulty circuit idea you mentioned. What I think this means quantum mechanicaly is that you only measure the photons detected on detector 1 with a polarization $\theta_1$ and photons having any polarization on detector 2 (we loose this information by the faulty circuitery encryption etc.). This would lead to a coincidence probability of: $$P_c = \frac{1}{2\pi}\int_0^{2\pi} \text{d} \theta_2\ \frac{1}{2}\sin^2(\theta_1-\theta_2)=\frac{1}{4}>0.$$ Hence we still have distinguishability. So even if we lose information on one detector (by any means) we cannot recover the interference by indistinguishable particles leading to a coincidence probability>0.

I get this example is not the same as with a double slit experiment and detecting which port the photon went. But here one can clearly see that only under special circumstances you can make the indistinguishable photons distinguishable by the half wave plate and indistinguishable again by the earaser.