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Can you ever have an odd number of entangled particles?

  1. Sep 18, 2013 #1
    This should be a very easy question to answer, but oddly enough I'm not certain what that answer is. So I thought that someone here could help me.

    It's easy enough to visualize and understand a pair of entangled particles, and how measuring the state of one instantly effects the state of the other. If the first is measured to be spin up for example, then the other will be spin down. But what if you have three entangled particles? How does measuring the first one effect the other two? Would measuring one collapse the superposition of all three? Or would the first one simply fall out of entanglement with the other two? Or is it as I suspect, that you can simply never have an odd number of entangled particles?

    Every bit of logic and intuition that I have says that entanglement always results in an even number of particles. It has to. But logic and intuition can be wrong so I thought that I would ask someone who knows.

    Can you ever have an odd number of entangled particles? If so, what happens when you measure one of an entangled triplet?

  2. jcsd
  3. Sep 18, 2013 #2
    i am not sure if it's possible to have more than two entangled particle in one quantum state
    but i think if this is possible , looking at one of them would be useless , because apparently two of them are either up or down and one is the opposite , so if we look at one and you find it spin up for instance , you still have 50 % chance of finding one of the remaining two either up or down
  4. Sep 18, 2013 #3

    Doc Al

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    You might want to look into the GHZ experiment, which uses three entangled particles: GHZ experiment
  5. Sep 18, 2013 #4


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    You can have any number (greater than 1) of entangled particles, including odd numbers.

    What is unusual sounding at first - spin is most obvious - is really not hard to explain. The sum of the spin will be a conserved quantity. So a+b+c=1 or similar. Then if you measure a=1, then b+c=0, etc.

    The 3 quarks in a proton are entangled. Entangled particles are not required to be like particles, either.
  6. Sep 18, 2013 #5
    Both! There are two inequivalent ways to entangle three particles, known as the GHZ and W states (this was shown in a famous paper in 2000 with the masterfully descriptive title Three qubits can be entangled in two inequivalent ways).

    For the GHZ-type states, if you trace out ('throw away') one of the qubits, you break the entanglement of all of them, while for the W-type ones, the remaining two will be entangled.

    This has a nice topological illustration in the two different ways you can link three rings together: in analogy to the W-state, you can link them by forming three Hopf links; if you now break one ring, the other two remain linked. For the GHZ-case, you can link them in the manner of the Borromean rings, which are completely separated once you cut one of the rings.

    The picture gets more complicated once you go to four qubits; there, you have nine different ways of entangling the qubits (proven in the paper Four qubits can be entangled in nine different ways). However, those are really infinite families of states, each depending on some continuous parameter.
  7. Sep 18, 2013 #6

    Ken G

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    What a wonderful answer, thank you.
  8. Sep 19, 2013 #7
    Could you please explain what the monogamy of entanglement means in that context?
  9. Sep 19, 2013 #8
    This is something that I hadn't thought about before, that there are different ways of entangling particles. Made me take a second look at just how I visualize the whole process of entanglement. And I like it! But it does make me wonder.....just two ways, why not more? And what about more particles? And are there some ways of entangling particles that are naturally more stable than other ways? If so, do these lead to patterns that we see in other things, like atoms? This new idea is gonna keep me busy for awhile.

    This has given my active mind something to think about. Thank you very much!
  10. Sep 19, 2013 #9
    It is highly unlikely that you can find a triplets of entangled particles as far as I am concerned, the main reason for that, is that the entanglement of particules is a result/consequence of pauli's exclusion principle: say for example at the fondamental level of an atom you have only 2 electrons, if you change anything in 1of them the other will certainly change (the rest of layers contain even numbers given by 2n^2). If you can find a stable and fully saturated atom that has an odd number of electron in any of its layers then you are right. However these kinds of atoms does not exist and are a direct violation of pauli's exclusion principle. So I guess the answer is no.
  11. Sep 19, 2013 #10


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    Your opinion doesn't matter. You can entangle 3 or more particles. As far as I know, the current record in an experiment is 14, see http://link.aps.org/doi/10.1103/PhysRevLett.106.130506

    It is not. I don't think you understand what entanglement is.
  12. Sep 19, 2013 #11
    I will check this experience nonetheless to see in what conditions it has been realised.
    Last edited: Sep 19, 2013
  13. Sep 19, 2013 #12


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    As mentioned, the 3 quarks in a proton are entangled - not that you could ever see those individually. Ditto for many scenarios involving electrons in shells.

    However, there are a lot of objects which can be entangled in lab conditions. And 2 is not the limit, there is no upper limit. For example: PDC is used to create entangled pairs of photons, but the same process occasionally produces 4 or more entangled photons.

    Naturally if you entangle a lot of particles, and then measure one, the remaining group will continue to be entangled - generally with a reduced set of possible states (reduced density matrix). That process continues if you make additional observations until there is but one remaining. Now you (potentially) know its state in that basis.
  14. Sep 19, 2013 #13
    Agreed, I made a mistake when refering to the atom.
    Apologies for my mistake and thanks for adding more info to my knowledge Dr Chinese
    Last edited: Sep 19, 2013
  15. Sep 19, 2013 #14
    My pleasure!

    The same thing as it always means: you can't entangle three maximally entangled particles with a fourth one. Generally, if you have some entanglement measure [itex]E(A|B)[/itex], which measures the entanglement of the system across the partition [itex]A|B[/itex], then, for [itex]n+1[/itex] parts [itex]A,B_1,B_2,\ldots,B_n[/itex], you have the monogamy relation
    [tex]E(A|B_1) + E(A|B_2) +\ldots+E(A|B_n)\leq E(A|B_1B_2\ldots B_n),[/tex]
    that is, the total entanglement of [itex]A[/itex] with each of the [itex]B_i[/itex] can't exceed its entanglement with the rest of the system. This introduces a trade-off: the more you entangle [itex]A[/itex] with, say, [itex]B_1[/itex], the less entangled it will be with all the other subsystems.

    The way entanglement classes are defined uses a paradigm known as LOCC: local operations and classical correlation (sometimes SLOCC, with the S standing for stochastic, meaning that you don't require your operations to lead the desired result with absolute certainty). This classifies roughly the kinds of things you can do to a quantum system by manipulating its local parts, and telling other people about it. Some quantum states are equivalent under these operations; some aren't. It just so happens that for (genuine) three qubit entanglement, all entangled states are SLOCC equivalent to either the GHZ or the W state. Of course, there are other entanglement classes for three qubits, the biseparable states (in which two of the three qubits are entangled, while the third one is separable), and the fully separable ones (in which, you guessed it, no qubits are entangled). Neither of these can be transformed into one another via SLOCC (in particular, you can't generate entanglement locally---that is, if you have a qubit, and I have a qubit, there's nothing we could do to them such that they become spontaneously entangled; we'd either have to let them interact somehow, or share a 'pre-entangled' state as a resource).

    And yes, different states will in general react differently to various types of noise, some being more robust than others (to certain kinds of noise, at least).
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