Does entanglement interfere with causality?

[student34]

Suppose someone entangled 2 particles many years ago and kept particle E here on Earth and sent particle S a light year away from Earth. So the observation of particle E on Earth would fix the state of particle S a light year away. But they did it in such a way that observing particle E would in turn cause a supernova a light year away from Earth.

Now general relativity says that if I am moving with a certain velocity, the future for particle S will already have been determined in my frame of reference.

But how can this be if the supernova has not happened yet?

 

[PeterDonis]

they did it in such a way that observing particle E would in turn cause a supernova a light year away from Earth.

It can't cause the supernova instantaneously; whatever is causing the supernova has to go from Earth to one light-year away no faster than the speed of light.

the observation of particle E on Earth would fix the state of particle S a light year away.

This is interpretation dependent; not all interpretations of QM say that anything "real" happens when particle E is observed.

Now general relativity says that if I am moving with a certain velocity, the future for particle S will already have been determined in my frame of reference.

No, that's not what relativity says. (Actually special relativity is sufficient for this, you don't need general relativity.) A particular interpretation of relativity, the "block universe" interpretation, says it, but that's not the only possible interpretation of relativity.

See this Insights article and comments thread:

https://www.physicsforums.com/threads/the-block-universe-refuting-a-common-argument-comments.843000/

Note, though, that regardless of what position you take on the "block universe" interpretation, since, as above, whatever causes the supernova can't travel any faster than light, it is impossible for the events "measurement of E on Earth" and "supernova one light-year away from Earth" to be spacelike separated, so their time ordering is the same in every frame and the issue you are raising here doesn't apply anyway.

Note also that you really have two separate questions here: one is a QM question (what, if anything, happens to particle S when particle E is measured), and the other is a relativity question (the one about the "block universe"). These really belong in two separate threads in two different forums (this forum for the QM one and the relativity forum for the relativity one, since it has nothing whatever to do with QM).

 

[Dale]

But they did it in such a way that observing particle E would in turn cause a supernova a light year away from Earth.

What is this “such a way”? I don’t think something like that is possible.

 

[student34]

What is this “such a way”? I don’t think something like that is possible.

We can imagine something like the Schrödinger's cat thought experiment but on a much larger scale. So maybe instead of a supernova it would just be a very powerful bomb.

 

[student34]

It can't cause the supernova instantaneously; whatever is causing the supernova has to go from Earth to one light-year away no faster than the speed of light.

That is fine. A spaceship flies the contraption out there. 100 years later, or whatever, someone has particle E and observes going at a certain velocity as to make particle S be in their past.

 

[PeterDonis]

That is fine. A spaceship flies the contraption out there. 100 years later, or whatever, someone has particle E and observes going at a certain velocity as to make particle S be in their past.

I don't think you fully grasped what I said. Let me try again.

You have two events: "measurement of E" and "supernova". Those two events must be timelike or null separated: that means their time ordering is the same in all frames. There is no way to make the supernova be to the past of the measurement of E in any frame; it's in the future of the measurement E in every frame.

Now, pick any event on particle S's worldline that is spacelike separated from the "measurement of E" (which is what's required for the time ordering of the events to be frame dependent, i.e., you could "change" it by changing your state of motion). The supernova is in S's future at every single one of those events. That includes events that are "simultaneous" with the measurement of E in some frame, events that are "to the past" of the measurement of E in some frame, and events that are "to the future" of the measurement of E in some frame.

So if it's a problem that the supernova in S's future is "determined" in some frame, it's a problem in every frame. There's no need to even invoke relativity of simultaneity. Nor is there any need to invoke quantum entanglement between particles E and S. You basically have a problem with determinism, which is a completely separate issue from any of those.

 

[DrChinese]

Suppose someone entangled 2 particles many years ago and kept particle E here on Earth and sent particle S a light year away from Earth. So the observation of particle E on Earth would fix the state of particle S a light year away. But they did it in such a way that observing particle E would in turn cause a supernova a light year away from Earth.

Now general relativity says that if I am moving with a certain velocity, the future for particle S will already have been determined in my frame of reference.

As far as anyone knows (and as far as experiments demonstrate, and quantum mechanics predicts): observing E does not change S any more or less than observing S changes E. The result is order/sequence independent, reference frame does not matter. If you think this violates causality, that's one way to look at things - but certainly not the only way. The subject of QM and causality/determinism/etc. is one that is debated endlessly.

You might try checking out the Quantum Interpretations and Foundations subforum to learn more.

 

[Dale]

We can imagine something like the Schrödinger's cat thought experiment but on a much larger scale. So maybe instead of a supernova it would just be a very powerful bomb.

Schrodingers cat didn’t involve entanglement. It was just a single radioactive particle that decayed or not. I don’t see the relevance.

 

[DaTario]

Schrodingers cat didn’t involve entanglement. It was just a single radioactive particle that decayed or not. I don’t see the relevance.

Dear Dale, Schroedinger's cat involves coherent superposition of states, but in some sense we can say that it also involves entanglement, since after the photon has crossed the beam splitter, its state may be written as
$$
\vert \psi \rangle = \vert 1 0 \rangle + \vert 0 1 \rangle
$$
where $\vert 1 0 \rangle$ means one photon in the vertical channel of the BS and no photon in the horizontal channel of the BS. Thus, the states of these two channels seem to be in some sense entangled.

Furthermore, after the possible click in the detector, the system can be described by the following state:
$$
\vert \psi \rangle = \vert 1 0\;\; LC\rangle + \vert 0 1 \;\; DC\rangle
$$
where LC and DC means living cat and dead cat respectively. It seems to me that an entangled state may be recognized here.

Best wishes.

* states are not normalized here.

 

[PeterDonis]

in some sense we can say that it also involves entanglement

The state you wrote down here is not an entangled state. There's only one degree of freedom (one photon); you can't have an entangled state with only one degree of freedom.

the states of these two channels seem to be in some sense entangled

The channels themselves don't have states. Only the photon does.

after the possible click in the detector, the system can be described by the following state

That state is entangled, yes (though it's not an entangled state of the photon by itself, but of the photon plus the cat), but its relevance is interpretation dependent. Discussions about interpretations of QM are off topic here; they belong in the interpretatations subforum.

 

[Dale]

since after the photon has crossed the beam splitter

What photon and what beamsplitter?

https://en.m.wikipedia.org/wiki/Schrödinger's_cat

Are you talking about some other cat? There is just an unstable atom and a Geiger counter with the poison and the cat. No beamsplitters etc.

 

[StevieTNZ]

What photon and what beamsplitter?

https://en.m.wikipedia.org/wiki/Schrödinger's_cat

Are you talking about some other cat? There is just an unstable atom and a Geiger counter with the poison and the cat. No beamsplitters etc.

I think the point was well emphasised that there is entanglement in the Schrödinger cat thought experiment.

 

[vanhees71]

Schrodingers cat didn’t involve entanglement. It was just a single radioactive particle that decayed or not. I don’t see the relevance.

Schrödinger's cat example is precisely about entanglement. The state of the cat is entangled with the state of the particle ("decayed" or "not decayed"). Of course, it's oversimplifying to describe the cat in terms of "dead" and "alive" as if these were pure quantum states of the cat, but it was about the fact that in principle macroscopic objects can be entangled with "quantum objects" (like a single radioactive nucleus). Of course, decoherence makes the possible entanglement between the cat and the radioactive nucleus practically unobservable, and the states "dead" and "alive" are pretty much coarse grained macroscopic states, which also behave macroscopically.

 

[Dale]

The state of the cat is entangled with the state of the particle ("decayed" or "not decayed")

Ah, ok.

There is still no beamsplitter, and I still don’t see the connection with the OP. @student34 please describe explicitly the experiment you have in mind

 

[vanhees71]

What the OP most probably has in mind is the Bohmian version of the EPR paradoxon, which is the most simple version of it using a "two-state observable". As an example take a neutral pion at rest decaying to two photons. Since the pion is a pseudoscalar particle the total angular momentum of the two photons is 0, and they fly back to back in, say, direction $\pm \vec{k}=\pm k \vec{e}_z$. This prepares an entangled two-photon state of the type
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (\hat{a}^{\dagger}(k,x) \hat{a}^{\dagger}(-k,y) - \hat{a}^{\dagger}(k,y) \hat{a}^{\dagger}(k,x)).$$
You have to consider the $\hat{a}^{\dagger}(k,j)$ as the creation operators of photonic wave packets with polarization $j \in \{x,y \}$.

Now in the OP's scenario you put detectors at $z_{1,2}=\pm 1 \text{Ly}$ and measure the polarization (say you put a polarizing beam splitter like a birefringent crystal at the places).

It's clear that the polarization states of each photon is completely indetermined, i.e., they are exactly unpolarized, but if the observer at $z_1=1 \text{Ly}$ finds his photon to be $x$-polarized he instantaneously knows that the observer at $z_2=-1 \text{Ly}$ must find her photon to be $y$-polarized and vice versa.

Now if you take the "collapse hypothesis" a la some flavors of Copenhagen as a physical process indeed, the polarization of the photon instantaneously changes by the measurement of the other photon's polarization. This seems to be violating Einstein causality since it should take 2 Ly for the signal of the measurement at the one place can influence anything at the other place.

The resolution of the paradox is, as is also well known in this forum, still subject to debates and in this sense it's a matter of opinion. For me personally there is only one satisfactory explanation, i.e., only one choice for the physical resolution of the paradox, and this is choice of the minimal statistical interpretation of the quantum state, which takes the meaning of the quantum state as only that of what's observable, and that are probabilities for the outcome of measurements of previously indetermined observables, here the polarization of the single photons.

It is also clear that the description of the measurements of the photons at the far distant places is done with "local equipment" and the only satisfactory description of the interaction of the single photons with the polarizing beam splitter and the detector at the corresponding place, here is relativistic local quantum field and thus by construction there cannot be any faster-than-light influences of one measurement on the other measurement.

So what the state of the two photons before the measurement simply says is that both observers measure completely unpolarized single photons, but there is the 100% correlation between the measured polarizations, because this correlation is already due to the preparation in this polarization-entangled state, and the reason for being in this state is simply the conservation of the total angular momentum. I.e., the quantum state describes the preparation of the two-photon system before any measurements were done, and this preparation procedure implies both, (a) the total indeterminism of the polarization of the single photon and (b) the 100% correlation between the outcome of the polarization measurements at far distant places.

No FTL communication is thus needed to explain this 100% correlation. However, if Alice and Bob want to verify that these correlations are really there, they have to exchange the outcome of measurements. E.g., for each photon Alice measures she sends a message to Bob about her outcome, and he can compare it to what he measured on his photon. However, of course, he has to wait for 2 years before Alice's messages reaches him. So there's no faster-than-light communication possible with this entanglement, because neither can Alice predetermine in any way the outcome of her measurement, because what she observes is just an ideally unpolarized photon, i.e., she cannot provide Bob with a predetermined plan for communication via the outcome of her photon-polarization measurement and in this way send instantaneous messages. All Alice and Bob can do is to "a posteriori" confirm the predicted 100% correlation between the outcome of their polarization measurement, for which they need at least two years to communicate about.

With the minimal statistical interpretation there are never any contradictions between relativistic (Einstein) causality and relativistic quantum field theory, because relativistic quantum field theory is constructed in such a way that it can never violate this causality, because it's assuming that there are only local interactions (which is the very reason already for classical physics to introduce the concept of fields instead of actions at a distance, here partiularly the electromagnetic field for electromagnetic interactions) and that the microcausality constraint holds, according to which any local observable operator commutes with the Hamilton density operator for space-like separated arguments.

With that statement you can move the entire thread to the quantum interpretation section and debate over the philosophical implications ;-)).

 

[student34]

I found this https://phys.org/news/2017-04-physicists-violate-local-causality.html . I guess this answers my question.

 

[PeterDonis]

I found this https://phys.org/news/2017-04-physicists-violate-local-causality.html . I guess this answers my question.

What the article means by "violate local causality" is "violate the Bell inequalities". But you can't use violations of the Bell inequalities to send signals or make information travel to distant locations faster than light.

 

[student34]

What the article means by "violate local causality" is "violate the Bell inequalities". But you can't use violations of the Bell inequalities to send signals or make information travel to distant locations faster than light.

But then it also says to violate the Bell inequalities is to violate local realism, "If a Bell inequality is violated, then either locality or realism (or simply "local realism") has also been violated."

 

[PeterDonis]

But then it also says to violate the Bell inequalities is to violate local realism, "If a Bell inequality is violated, then either locality or realism (or simply "local realism") has also been violated."

That's because "local realism" means "not violating the Bell inequalities". There is no additional content to "local realism", so the article is just restating the same thing in different words.

As a general rule, phys.org articles are not good sources; they are notorious for misrepresenting what the actual physics says.

 

[student34]

That's because "local realism" means "not violating the Bell inequalities". There is no additional content to "local realism", so the article is just restating the same thing in different words.

As a general rule, phys.org articles are not good sources; they are notorious for misrepresenting what the actual physics says.

I found the paper, it was made by people from legitimate universities. The abstract says, "Our results provide a proof-of-principle experiment of generalizations of Bell's theorem for networks, which could represent a potential resource for quantum communication protocols." from http://europepmc.org/article/MED/28300068 .

 

[PeterDonis]

I found the paper, it was made by people from legitimate universities.

Yes, the paper was. That doesn't mean the phys.org article was. Note that the paper does not say the same thing the phys.org article says. What the paper says is much more careful.

 

[PeroK]

I found the paper, it was made by people from legitimate universities. The abstract says, "Our results provide a proof-of-principle experiment of generalizations of Bell's theorem for networks, which could represent a potential resource for quantum communication protocols." from http://europepmc.org/article/MED/28300068 .

I fail to see what "a potential resource for quantum communication protocols" has to do with FTL signalling. Which, reading between the lines, appears to be the issue here.

 

[student34]

Yes, the paper was. That doesn't mean the phys.org article was. Note that the paper does not say the same thing the phys.org article says. What the paper says is much more careful.

But if it is evidence for Bell's theorem for networks, isn't that evidence that events can be connected faster than the speed of light?

 

[PeroK]

But if it is evidence for Bell's theorem for networks, isn't that evidence that events can be connected faster than the speed of light?

I knew it!

 

[student34]

I fail to see what "a potential resource for quantum communication protocols" has to do with FTL signalling. Which, reading between the lines, appears to be the issue here.

Yes, the part that you quoted is the part of my quote that is not relevant.

 

[Dale]

What the OP most probably has in mind

Wouldnt it be best to let the OP clarify what they have in mind?

 

[student34]

Wouldnt it be best to let the OP clarify what they have in mind?

Yeah, I forgot that there is no entanglement in Schrödinger's cat in the same way it would have to be for my thought experiment.

So imagine an apparatus set up in such a way that if it detects some physical change, then the supernova is triggered.

I am not sure about this, but wouldn't fixing an entangled electron on Earth potentially cause something physical to happen with the other electron in some very delicate contraption?

For example, if we observe an electron on Earth that is entangled with an electron one light year from here, then doesn't that enable the electron to interact with objects, such as emit photons or the ability to interact with nearby electrons?

 

[f95toli]

But if it is evidence for Bell's theorem for networks, isn't that evidence that events can be connected faster than the speed of light?

No, quantum communication in this context (mainly*) means "quantum key distribution"; i.e. distributing keys for cryptography in a secure way. QKD DOES rely entanglement and versions of the experiments that were originally designed to test Bell inequalities are now used for QKD.
Again, It has nothing to do with FTL communication.

(*)There are a few other areas/applications that fall under the "quantum communication" umbrella; e.g. networked quantum computing; but 95% of the time quantum communication is synonymous with QKD

 

[PeroK]

Yeah, I forgot that there is no entanglement in Schrödinger's cat in the same way it would have to be for my thought experiment.

So imagine an apparatus set up in such a way that if it detects some physical change, then the supernova is triggered.

I am not sure about this, but wouldn't fixing an entangled electron on Earth potentially cause something physical to happen with the other electron in some very delicate contraption?

For example, if we observe an electron on Earth that is entangled with an electron one light year from here, then doesn't that enable the electron to interact with objects, such as emit photons or the ability to interact with nearby electrons?

It works like this. Suppose you and I are a long way apart and we each have one of a pair of entangled particles. You want to send me a message. Our code is spin-up = yes and spin-down = no.

You measure your particle and if you get spin-up, then you know I get spin down and hence the message is "no". And if you get spin-down, then I get spin-up and the message is "yes".

Now, you want to to send me the message "yes". So, you measure your particle and if it's spin-down, then bingo I get the message "yes". But, if you get spin-up, then I get the message "no". Which is not what you intended.

And, in fact, even if you forget to measure your particle (or you don't want to send a message that day), when I measure mine I still get one of spin-up or spin-down and have no way to know that you didn't actually want to send a message that day.

As you have no way to control the result of your measurement and therefore no way to control the result of mine, you cannot influence what result I get or what message I receive. That means it's not a message from you at all, it's just some random result that is independent of what you choose to do.

There are literally dozens of threads on here with the same question: why can't I use quantum entanglement to send a message FTL?

 

[Dale]

So imagine an apparatus set up in such a way that if it detects some physical change, then the supernova is triggered.

I am not sure about this, but wouldn't fixing an entangled electron on Earth potentially cause something physical to happen with the other electron in some very delicate contraption?

No. Measuring an entangled electron does not produce any measurable change on the other particle.