I’m going to try and keep this question as simple as possible in the hopes that I get a simple answer.

Recently, I’ve been reading more about Quantum Erasers and experiments with them, particularly this one. In the experiment, a photon is split into two entagled photons (A and B) after passing through a “dual slit”. Photon A immediately gets its position detected by a detector. Photon B meanders around a little, then has a 50/50 chance of getting its quantum state (and information about which slit it went through) erased or measured. If it gets erased, we see an interference pattern in A’s detection, if it gets measured we see no pattern. Very weird, but no paradox yet.

My problem is, what happens if instead of erasing B’s state (or not) by chance right away, what if we put B in a holding cell for a bit. Then, we shoot a lot more photons through, putting all the B’s in holding cells and measuring the positions of all the A’s. Then, we look at the positions at which the A’s were detected. If we see an interference pattern, we immediately measure the states of all of the B photons to see which slit they went through. If we see no interference pattern, we erase the information in all the B photons. (We’re crazy scientists intent on causing a paradox).

What’s wrong with my logic?

How are you going to “put B in a holding cell for a bit”?

We have many options, we could use mirrors, or light slowing materials, or just send it to jupiter. I like the Jupiter idea best, actually.

My friend on Jupiter wants to send me a message. In this version of the experiment, I send all my B photons to this friend on Jupiter. Now, when he gets them (a few minutes or an hour later, I forget how far it is), he erases all the states of Photon B if he wants to sent me a zero, and measures them if he wants to send me a 1. I then look at my A photons to see if there’s an interference pattern or not. If there is a pattern, I write down a zero, and if there’s no pattern I write down a 1. In this way, he communicates with me at faster than the speed of light, something that’s not possibly even using quantum teleportation.

I’ve done some more reading on the topic, though, and I think the answer is that I never see any interference pattern, since the states of the B photons are still “Available”. I guess the first photon hitting the plate would have an interference pattern, but subsequent ones “know” that the other photons haven’t had their states… arg my head hurts again.

We’ve done this before (here’s one thread). Your source of confusion is in what is meant by “seeing an interference pattern.” The detector for photon A is a single-photon position detector, and will record a single point position for each entangled A-B pair you create. If you look at all of these positions, *you will see no interference pattern*; this is true regardless of what you do with the B photon. You only see an interference pattern when you consider the conditional statistics (that is, the measurements of A conditioned on a particular measurement outcome for B). Schematically, you can imagine making one of two types of measurements on B: one which measures “which-path” information about A and B, which I will call an X (position) measurement, and one which does not provide which-path information, which I will call P (for momentum). Each has two possible results, which I will call “+1” and “-1”. The scatter plots for A detections, conditioned on these different measurement types and results, will look something like this (warning, horrid ASCII graphics ahead):

```
measurement density of A detections vs. position
X measured ::::::++++++++####################++++++++::::::
X=+1 ::::::++++++++####################++++++++::::::
X=-1 ::::::++++++++####################++++++++::::::
P measured ::::::++++++++####################++++++++::::::
P=+1 ::: +++ ++# ### ### ### +++ :::
P=-1 ::: +++ ### ### ### #++ +++ :::
```

Here :, +, and # represent increasing densities of detections (space represents no detections).

Notice that whether you choose to measure X or P, you see no interference fringes at A; you can’t alter the measurement results at A by what you choose to do with B. If you measure X, and look at the conditional pattern for X=+1 or X=-1, you still see no interference. However, if you measure P, and then look at the conditional pattern for P=+1 or P=-1, you *do* see interference. The fringes are shifted depending on the measurement result, in such a way that the sum of the two patterns looks just like the no-interference pattern. (The “P measured” plot is just the sum of the “P=+1” and “P=-1” plots below it; and similarly for the “X measured” plot.)

These results, of course, require that you know the results of measurements on both A and B; so you can’t use it for superluminal tricksiness. When you save all of your B photons, all you do is make it impossible to see the coincident interference until you finally get around to making the measurements.

From what I understand of the article, the issue is how we choose which photons to erase. If they are erased by a quantum-mechanically random process, then there will be a correlation between whether or not they are erased and the probability pattern at D0. If, on the other hand, you have a friend on Jupiter deciding which photons to erase, that would not be quantum-mechanically random, so you would not see such a correlation for the patten at D0.

Thanks for that great explanation, **Omphaloskeptic**. Now that I know there’s no actual interference pattern on the plate everything makes perfect sense. I’m chagrined to admit that although I did a search for past threads, I didn’t look past the first 3 or 4 results. :smack: