Delayed Choice Quantum Eraser: Am I missing something here?

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1. Oct 26, 2015

SpaceKidd_N7

Hello everyone,

I actually have three questions:

1. Am I missing an important detail in my understanding of how the delayed choice quantum eraser experiment is done?
2. How does one account for what takes place in the experiment without using the concept of "retrocausality" (effect before cause)?
3. 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.

2. Oct 26, 2015

Strilanc

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 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.

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 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.

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 information is permanently inaccessible, unrecoverable due to thermodynamics stopping you from reverting the measurement.

So which is it? Were they entangled or not? Well, it's complicated. 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 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.

The exact same logic applies to the delayed choose 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".

3. Oct 26, 2015

Staff: Mentor

I will leave the actual experimental setup to those into that sort of thing. I will confine myself to the conciousness comment.

That conciousness is somehow involved is one of a myriad interpretations of QM and a very very fringe one these days. The modern view of QM is that its simply a reasonable variant on ordinary probability theory:
http://arxiv.org/pdf/quant-ph/0101012.pdf

The vast majority of interpretations is simply an argument over the meaning of probability:
http://math.ucr.edu/home/baez/bayes.html

There are a ton to choose from so take your pick:
https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics

I hold to the Ensemble interpretation with a slight twist that there is no need to go into here which is a frequentest view of probability. But regardless of what you choose conciousness doesn't have to be involved any more that it needs to be involved when using probability to examine games of chance or Actuaries use it to manage financial risk.

All interpretations of QM have exactly the same underlying formalism and that formalism explains the eraser experiment - and indeed every single quantum experiment ever done. If it didn't that would be big news and mean the awarding of an instant Nobel prize because it would prove QM wrong.

As I said at the start I will leave to others to explain exactly how QM explains your particular set-up, but overall they are explained by the fact decoherence can be undone in simple cases. If you don't know about decoherence check out:
http://www.physics.drexel.edu/~tim/open/main/node2.html

Thanks
Bill

Last edited: Oct 27, 2015
4. Oct 27, 2015

zonde

You didn't mention coincidence counter in your description so it is not clear if you understand it's role. Just to be on the safe side, you don't see interference or no interference in detectors but rather in coincidences of D0 and say D1.
Instead of "photons" one talks about "photon probability amplitudes". Say $1/\sqrt{2}$ part of photon probability amplitude goes through one slit and $1/\sqrt{2}$ part of photon probability amplitude goes through other slit. In D1 and D2 these two probability amplitudes interfere while in D3 and D4 they don't.
Photon (a la particle) can't pass trough both slits. This is sort of self contradictory statement.

5. Oct 27, 2015

stevendaryl

Staff Emeritus
I think that delayed choice experiments do call into question what we mean by a "measurement" or "decoherence". Normal measurements involve amplifying some microscopic state variable (such as particle position, or spin, or the presence or absence of a photon) into a macroscopic state variable that can be the basis for a permanent record. The "collapse" interpretation of QM assumes that the wave function is put into a definite state of whatever was measured at the moment the measurement took place. The theory of decoherence replaces the need for an actual measurement by any interaction in which the state of the system being studied affects the environment in an irreversible way.

In the case of a "delayed choice" experiment, though, the "measurement" is not a macroscopic event, but a microscopic event, which is itself subject to quantum mechanics. Depending on how this microscopic evidence is handled long after the "measurement" takes place can affect whether or not the measurement "collapsed the wave function" (that is--destroyed interference effects between different alternatives). To me, this shows that the collapse interpretation is not really consistent; you can't say that measurement causes an instantaneous wave function collapse, if it's not determined until long afterwards that a measurement even took place. I suppose it's still possible to hold onto the idea that macroscopic, irreversible measurements collapse the wave function, but that just seems like clinging to ignorance to me. For practical reasons, we can't do a "delayed choice" type experiment if the measurement is macroscopic, but I don't see any reason to think that there is anything standing in the way other than enormous calculational (and measurement precision) difficulties.

Last edited: Oct 27, 2015
6. Oct 28, 2015

zonde

Do you mean that BBO crystal is measuring photon beam? What justification you have for such interpretation of experiment?

Measurement is produced by coincidence counter. It gives out numbers that represent relative frequencies of different outcomes.
And this experiment (as other entanglement experiments) is unusual because macroscopic amplification (in photodiodes) happens before measurement (in coincidence counter) not at the same time as usual. And because of this delayed choice experiments indeed calls into question what we mean by a "measurement" and "decoherence".