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B Entangled particles

  1. Apr 26, 2016 #1
    Pardon me for asking a very simple question, but this is something that I'm confused about. If we have a pair of entangled particles, and we measure the state of one of the particles, then the state of the other particle becomes fixed instantaneously. But what if we have three entangled particles, and we measure one of them, what happens to the other two? Why? How does measuring one of them tell us the state of the other two?

    But isn't this what happens when we interact with a group of entangled particles? Interacting with one particle causes the whole system to collapse? It would almost seem as if there must be a causal order among the particles, a chain if you will. Such that interacting with one particle has an instantaneous ripple effect across all of the particles.

    Or am I simply misunderstanding entanglement?
  2. jcsd
  3. Apr 26, 2016 #2


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    If 3 particles are entangled on some basis, and you measure 1 of the 3 on that basis: the other 2 remain entangled on that basis. Their entanglement will be different than before, probably best to consider some example. My example may not be exactly correct, but I think you will get the idea by analogy.

    You have N=3 entangled particles with momenta of p1, p2, p3 such that p1+p2+p3=Ptotal (but p1, p2, p3 are not individually known, only the total momentum is known). You measure p1 to have momentum of P1. You now have p2+p3=Ptotal-P1 which is still entangled for those N-1=2 particles. This analogy would hold for greater N as well.

    Logically, if you have a larger number N of entangled particles: learning something about 1 of N does not constrain the values of the remaining N-1 particles very much. But if N=2, then it does and that is one reason (of several) why N=2 is usually discussed rather than N=3 or more.
  4. Apr 26, 2016 #3
    quantum entanglement is different than a 'classical entangled system' -about which we interact normally/see them correlated and tied by interactions ....
    so that if we disturb one of them the 'disturbance' travels to all like ripple effect. its the continuous motion of some wavelike properties.
    imagine a house of cards - one card gets displaced and the house falls in a variety of ways and these correlations could be defined and explained on the basis of basic physical laws.

    In QE(quantum entangled) particles - a measurement of one property of its state(say spin state)
    -affects the other particle's spin state wherever the particle may reside at the time of such 'measurement';
    may be they are at two ends of our universe.
    that is the 'weiredness' of the entangled state Einstein referred to as spooky action- a ghost like feature.
    and now a days people are finding the 'QE' in experiments performed with such particles.
  5. Apr 26, 2016 #4
    But therein lies my problem, for this only seems to work if there are just two entangled particles. For any number more than that it doesn't seem to work. It does hold true for momentum as DrChinese indicated, but it doesn't hold true for something like spin, in which there are only two distinct outcomes, like up or down. in such a case, measuring the state of one particle within any set of particles greater than two, would seem to have no effect. The remaining particles will still be both up and down.

    This leads me to a problem with DrChinese's use of momentum as an example, because as far as I know, momentum isn't a measurable thing. It's a mathematically derived thing. I measure something at position "A" and then measure it again at position "B", and then derive the momentum from the results. I then extrapolate that momentum to the unmeasured particles within the set. So what I'm wondering, at least in the case of DrChinese's example, does the particle really have such a thing as momentum?
  6. Apr 26, 2016 #5
    well , not an expert in 'multi particle entangled systems - you can get an idea in the following experiment where a triplet got entangled and they were more stable.....

    To make the entangled triplets, researchers from Canada, the U.S. and Sweden started with a single blue photon that was polarized both horizontally and vertically. Being able to hold two states at once is another property of quantum particles, and it's why computer scientists are interested in quantum physics. Particles that are able to hold two states at once potentially can hold more information than classical computers with machinery that can only hold one state at a time.


    Photon Detector

    The physicists used chips like this one to detect single photons of light.

    The research team sent this quantum blue photon through a crystal that turned it into two less energetic, red, entangled photons with matching polarizations -- either horizontal or vertical. Next, they sent one of those red photons through another crystal that transformed it into two less energetic, infrared, entangled photons. The infrared photons happened to still be entangled with the remaining red photon, and voilà: three entangled photons.

    Further tests demonstrated the triplets were truly entangled, and getting that to work correctly is rare. There's only a one-in-1 billion chance that the first step of the process creates two entangled photons. Then, there's a one-in-1 million chance that the second step of the process will create the entangled triplet.

    The international team http://www.nature.com/nphoton/journal/vaop/ncurrent/full/nphoton.2014.218.htm about their work this week in the journal Nature Photonics.> quote closed
    Last edited by a moderator: May 7, 2017
  7. Apr 26, 2016 #6


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    Part of the problem here is that you're you are using the handwaving English language description of quantum states and entanglement instead of looking at the math (it is well to remember that entanglement appeared in the math before it was recognized in experiments).

    In the English language description, we talk about the individual particles and their state - "This one is up, this one is down, this one changed from up to down". However, a quantum system is a single system with a single wave function even if we're tempted to think of it as containing multiple particles. So let's consider a quantum system on which we can perform either of two measurements, which I'll call A and B. If A and B commute, then the system has states that are eigenstates of both operators and it makes sense to say that in some states "A has the value x and B has the value y"; we will write that particular eigenstate state as ##|xy\rangle##. Another possible eigenstate state is the one in A has the value p and B has the value q, which we would write as ##|pq\rangle##. (If A and B did not commute, they would not have a complete set of common eigenstates and we'd find that the uncertainty principle prevented us from making constructing states in which A and B both have definite values).

    OK, now let's say our quantum system is prepared in a superposition of those two states: ##\frac{sqrt{2}}{2}|\psi\rangle=|xy\rangle+|pq\rangle##. If we perform a measurement of A on this state, half the time the wave function will collapse to ##|xy\rangle## and we'll get the result x; the other half of the time it will collapse to ##|pq\rangle## and we will get the result p. Either way the final state of the system will be one in which we know what the result of a measurement of B would be, when and if we make such a measurement. That's entanglement; and we say the initial state ##\frac{sqrt{2}}{2}|\psi/rangle=|xy\rangle+|pq\rangle## was an entangled state of the two observables A and B. Note that A and B can be any two commuting observables of the quantum system - they don't have to refer to the same or different particles, or any particles at all for that matter, one of them could be spin and the other could be energy, the quantum system might be one that we don't naturally think of as containing two particles.... All that's necessary is that we be able to put the quantum system into a state that can be written in the entangled form for some two observables and we have successfully entangled those tw observables.

    If you go back and substitute "spin of first particle" for "A", "spin of second particle" for "B", "+" and "-" for "x", "y", "p", and "q", you will have the mathematical description of the two-particle entangled spin experiment.

    In this framework it's a lot easier to see how three-particle entanglement might work. If we have three observables commuting A, B, and C each taking on either of two possible values, then an eigenstate will be written as something like ##|+--\rangle## where the three signs represent the values of A, B, C in that order. A state like ##\frac{\sqrt{2}}{2}(|+--\rangle+|-++\rangle## or ##\frac{\sqrt{2}}{2}(|+--\rangle+|-+-\rangle## is entangled; a measurement of A will cause the wave function to collapse to either ##|+--\rangle## or ##|-++\rangle##, states in which we know more about what will happen when and if we subsequently measure B or C. (Be warned that this is a highly contrived example; in any realistic situation the wave function would be such that even after collapse B and C would remain entanged).
  8. Apr 26, 2016 #7


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  9. Apr 26, 2016 #8


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    It would hold true in spin as has already been nicely mentioned by several others. It does not come out as black and white as in most 2 particle examples.

    You can entangle all kinds of systems in all kinds of ways. It is not always so easy to demonstrate that, as it can get very complicated very fast. :) Here are a couple of examples.


  10. Apr 26, 2016 #9
    Thanks everyone, it's gonna take my 9th grade educated brain a while to muddle through all of this stuff, but hopefully I'll be able to glean some sense out of it. Your help is much appreciated.

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