Hello everyone,
I actually have three questions:
- Am I missing an important detail in my understanding of how the delayed choice quantum eraser experiment is done?
- How does one account for what takes place in the experiment without using the concept of "retrocausality" (effect before cause)?
- If the photon passes through both slits, wouldn't the BBO crystal produce 4 photons? If it does, what happens in that case, and if it doesn't, then why?
Okay so here is my understanding of the experiment. A laser fires a photon at a double slit. It can either go through slit A (red), slit B (blue), or both. After the double slit, there's a nonlinear optical crystal (BBO) that converts the photon into two entangled photons. A Glan-Thompson prism diverges these two entangled photons. One of them (called the signal photon) goes towards the detector D0 while the other (idler photon) goes towards a prism PS and is deflected depending on whether it follows path A or path B. An idler photon following path A passes through a beam splitter BSb where it can either reflect and go to D4 or transmit, reflect off of mirror Mb and then either reflect off of BSc and enter D2 or transmit and enter D1 (sorry about the run-on sentences, trying to keep this short). An idler photon following path B will either reflect off of BSa or transmit and reflect off Ma and arrive at BSc where it will either go to D2 or D1. Detectors D1 and D2 always give interference patterns, while D3 and D4 only show diffraction without interference. If the idler photon enters D4, then we know that it passed through slit A, if D3, then slit B. What ends up happening though is that whether or not the signal photon displays interference at D0 depends on whether the idler photon enters D1/D2 or D3/D4. If the idler photon enters D1/D2, there will be an interference pattern at D0. If the idler photon enters D3/D4, there will not be an interference pattern at D0.
I'm not studying this for a class or anything, I've just been having a discussion with someone about the role of consciousness within these double-slit experiments. They used this as an example of how consciousness can effect matter. I, however, have a very hard time accepting this. There just has to be another explanation that does not involve retrocausality. If there isn't, then my friend would have to be right; somehow the signal photon knows whether or not we will have the path information (it is "erased" at D1/D2). I know some people believe consciousness plays a role in the original double-slit experiment, but I know that it doesn't. In that experiment, the reason why the photon acts like a particle is not because it knows a physicist is attempting to measure it, but because of the way it physically interacts with the detector. The delayed time version can't be explained this way. I'm not really too familiar with entangled particles, I only understand the main concept. All 5 detectors are the same kind of detector correct? The only difference between the last 4 is that we the observers know that D3/D4 will let us know the path information, while D1/D2 will not. How in the world would a photon "know" this??? There just has to be something I'm missing here. I would really like to know what the explanations that don't involve retrocausality (Ex. of explanation using retrocausality: if the idler photon arrives at say D3/D4, then it will "go back in time" and make sure that the original photon only passes through one slit, even if it originally passed through both) are. Wikipedia says that this paper provides such an explanation, however, I'm having some trouble understanding it http://arxiv.org/abs/1103.0117.
Thank you for taking the time to read all of this.
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These comments are for @Timaeus. I have so many, that if I try posting them in the comments section of their answer, I'll end up having to post nearly a dozen of them and I've already got that warning that comments should not involve discussions yesterday. I don't know if posting too many comments might redflag me as a spammer or someone who's not following the rules, but I don't want to take that chance.
Thank you very much for taking the time to help me with this, I appreciate it. I’d just like to comment on a few of your responses.
“This question doesn't make sense”
Why do you think my second question doesn’t make sense? Dont you see how someone who does not have an established background in QM can understand why the results of this experiment happened the way it did by considering retrocausality? I am by no means saying that I believe retrocausality is possible. All I’m saying is that without the necessary knowledge, someone may not be able to conceptualize why this experiment had these results without considering retrocausality.
“Passing through both slits is one of your misunderstandings.”
In your response to my 3rd question, are you saying that a single photon cannot pass through two slits? What are the “sectors” you mention? By “single state”, are you’re implying that the wave function of the photon has collapsed before going through the double slit right? Doesn’t the photon need to interact with something, like a some kind of measuring device, in order for that to happen? This is why I asked about what you meant by sectors. From the original paper that published the results of this experiment: “This reflects the wave property (both-path) of photon 1.” “The which-path or both-path information of a quantum can be erased or marked by its entangled twin even after the registration of the quantum.” If the photons never pass through both slits, then what did the authors mean by “both-path”. Here is the link to the paper: http://arxiv.org/abs/quant-ph/9903047
“so if you had a state that only made a detector outside one slit fire then this crystal would make two photons.”
You also kept mentioning detectors outside each slit. In the descriptions of the setup of this experiment and even in the original paper, though, detectors placed before the slits are never mentioned. This is what I meant when I said “the delayed time version can't be explained this way.” In the original double slit experiment, there is a detector placed before the slits and that’s what causes the photon’s wave function to collapse into a single state. In the delayed time version though, there is no mention of anything being placed between the laser and the slits that the light can interact with and take on a single state.
“And thinking it is following. a blue path or a red path is as wrong as thinking a vector in the 2d plane has to be on the x axis or the y axis. It's that totally wrong.”
I’m also confused about why you were saying that we cannot get the path (you call it red/blue state) information from D3 or D4. If you’re right, then why does everyone, even the original paper that published the results of this experiment say that we can? I thought that was one of the most important details of this experiment. From the original paper: “The registration of D 3 or D 4 provides which-path information (path A or path B) of photon 2 and in turn provides which-path information of photon 1 because of the entanglement nature of the two-photon state of atomic cascade decay.” http://arxiv.org/abs/quant-ph/9903047
“Totally totally totally wrong.”
Why was I that wrong about saying “the only difference between the last 4 is that we the observers know that D3/D4 will let us know the path information,…”, this is supposed to be the whole point of this experiment. D1 and D2 “erases” the path information. The quote I previously mentioned: “The registration of D 3 or D 4 provides which-path information (path A or path B) of photon 2…”. Here’s another: “The triggering of detectors D 1 or D 2 erases the which-path information.” http://arxiv.org/abs/quant-ph/9903047
“Yes there is. But its absolutely everything.”
How can I have “absolutely everything” wrong? Honestly, all I did was summarize Wikipedia and the original paper. If I explained absolutely everything wrong, then the paper is absolutely wrong as well.
“Its like you learned quantum mechanics from someone that wanted to make it seem mysterious.”
Why do you say that? I never said that I believe retrocausality is taking place. In fact I explicitly stated that I don’t. And actually, since when is QM not mysterious? You make it seem as if the results of this experiment are intuitive. After enough time passes, we notice that an interference pattern forms at D0 only when the idler photons entangled with the signal photons, that generate that pattern, hit D1 or D2. We also notice that an interference pattern is not formed at D0 only when the idler photons entangled with the signal photons, that do not form the pattern, hit D3 or D4. This is why some people resort to retrocauslity. Original paper: “It was predicted that the “joint detection” counting rate R01 (joint detection rate between D0 and D1) and R02 will show interference pattern when detector D0 is scanned along its x-axis. This reflects the wave property (both-path) of photon 1. However, no interference will be observed in the “joint detection” counting rate R03 and R04 when detector D0 is scanned along its x-axis. This is clearly expected because we now have indicated the particle property (which-path) of photon 1.” To make the results of this experiment even less intuitive, the signal photon reaches D0 before the idler photon reaches any of the other detectors. “The experiment is designed in such a way that L0, the optical distance between atoms A, B and detector D0, is much shorter than Li , which is the optical distance between atoms A, B and detectors D1, D2, D3, and D4, respectively. So that D0 will be triggered much earlier by photon 1.” This is the precise reason why people think that the idler photon “travels back in time” to make sure the signal photon demonstrates wave properties if it hits D1/D2 or particle properties if it hits D3/4. http://arxiv.org/abs/quant-ph/9903047
“and in your entire post I never say evidence that you know a single tiniest bit of quantum mechanics (you might, bit you didn't show it).”
I disagree. I’d say that a statement like this, “in that experiment, the reason why the photon acts like a particle is not because it knows a physicist is attempting to measure it, but because of the way it physically interacts with the detector”, would indicate that someone does know at least a tiny bit about QM. I don’t see the point of making statements like that or the one about who’s been teaching me quantum mechanics. How are those statements relevant? All I was doing here was trying to gather information and terms that I am not yet aware of so that I can use them as guidelines in knowing what to look up in order for me to gain a better understanding of this experiment. You can’t expect someone learning QM to understand everything perfectly the first time around. It may not have been your intention, but I got the impression that you were criticizing me. Nonetheless, thanks again.
Best Answer
This question was cross-posted to physics forums word for word. I'll give the same basic answer I gave there.
Consciousness is never part of any quantum mechanical explanation. Every experiment runs the same whether or not a person is in the room.
Retrocausality is also not required here. For example, the Copenhagen interpretation explains the delayed choice eraser with instantaneous non-local partial collapse and the many worlds interpretation explains it with worlds staying coherent and interfering. Those are the two most popular interpretations.
Thinking of the delayed choice eraser in terms of an optical experiment muddles the issue, in my opinion. We can create the same basic effect with a much simpler system, involving three qubits.
Analogous Simpler Situation
Suppose you have the state $\psi = \frac{1}{2} \left|000\right\rangle + \frac{1}{2} \left|110\right\rangle + \frac{1}{2} \left|011\right\rangle + \frac{1}{2} \left|101\right\rangle$. That is to say: you have three qubits, the first two qubits are each initialized into the half-and-half state $\frac{1}{\sqrt{2}} \left|0\right\rangle + \frac{1}{\sqrt{2}} \left|1\right\rangle$, and then the third qubit is conditionally toggled so that its value tells you whether the first two qubits differ or not.
Now, run some bell tests with the first two qubits. You'll find that they don't violate any bell inequalities, and fail any other test of entanglement. They aren't entangled.
But, if you later measure the third qubit, and split the tests you did on the other two qubits into a "third qubit was 0" group and a "third qubit was 1" group, you'll see that within each group there are bell inequalities being violated! So the first two qubits were entangled all along.
BUT, if you measure the third qubit along the X axis instead of the Z axis we've been working with, then you'll never be able to split the two groups apart and see the entangled sub-cases. The distinguishing information becomes permanently inaccessible, unrecoverable due to thermodynamics stopping you from reverting the measurement.
So which is it? Were they entangled? Not entangled? Only entangled when we made the right measurement? I would say that they are entangled, but in an unusual way that's harder to detect. The third qubit tells you what type of entanglement exists between the first two qubits (entangled to agree, or entangled to disagree). Each subcase is entangled, but the cases are complementary in a way that hides any signal of entanglement if you count them together instead of individually.
Whether or not we choose to measure the correct axis of the third qubit doesn't determine whether the original two qubits are entangled or not, it determines whether we have the information needed to split the results into the two complementary sub-cases. If you try to simplify the situation into just "two-particles-maximally-entangled" vs "not-entangled", or into "is-just-a-particle" vs "is-just-a-wave", you're throwing away the context needed to understand what's going on.
Mapping Back
The exact same logic applies to the delayed choice quantum eraser experiment, except there's an extra value involved and you're looking for interference patterns instead of passing bell tests. No consciousness. No retrocausality. Just "did we get and use the distinguishing information needed to group the lack-of-interference pattern into two complementary interference patterns"