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Quantum entanglement

  1. Apr 5, 2004 #1
    Quantum entanglement, if I understand it correctly, specifies the phenomenon where the scattered pieces of a particle somehow know the whereabouts of its other pieces.
    Does that mean that by taking a neutron from atom A and placing it away from atom A, the atom would know where the missing neutron is?
    Does the atom have an exterior surface, like a ball (as is often depicted in science books)?
  2. jcsd
  3. Apr 5, 2004 #2


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    Particles don't "know" about each other.

    Quantum entanglement means that, in some situations when you bounce two particles off each other, the particles continue to be related after the collision. In a common example, some collisions will leave one particle in a spin up state, and the other in a spin down state.

    Until you actually measure one of the particles, however, they are both in a superposition (combination) of both spin up and spin down states. Only when you measure the spin does the particle (randomly) select its pure state.

    At the same "instant" that you measure one particle's spin, the other particle, across the room or across the galaxy, automatically enters its appropriate (opposite) spin state.

    At first glance, it may seem as though information is being transmitted instantaneously -- the measurement of one particle can instantaneously affect the state of a particle all the way across the galaxy! However, you're really not transmitting information. When you measure the spin of a particle, you cannot select whether you would like it to be spin up or spin down -- it enters one state or the other, at random. The other particle picks the opposite (yet still random) state.

    You can transmit information using this mechanism, but you need some way to tell the person at the other end the results of your own measurements. You still need a "classical" channel, like a light beam or a radio wave, to communicate. Nothing is being transmitted faster than light.

    And no, the atom does not really have a "surface" as in the macroscopic sense.

    - Warren
  4. Apr 7, 2004 #3
    Are you familiar with quantum information science?
  5. May 28, 2004 #4
    Sorry to bump this but... chroot:

    The idea behind quantum computing is to manipulate the particles, which act as qubits, so that when the superposition is collapsed the probability of the desired answer emerging will be highest. I don't know how this is actually done, but couldn't it be used to operate one of the entangled so that you can choose what it will 'read' when measured?

    Well, you are transmitting information... it is just meaningless. :smile:
  6. May 28, 2004 #5
    Quantum Information

    I thought I would post this as I am interested in the possibilties and from what I read from a conversation that is not apparent to me anywhere, that this is not possible? This needs to be done for LiGO, and if we can do it for numerical relativity why can we not retranslated LIGO information?

    One context in which entanglement has recently played an important role is the emerging field of "quantum information". In a classical two-state system, a single binary digit can completely specify the state (e.g. 1 for "up", 0 for "down"); the system thus carries one classical "bit" of information, and a set of L such classical systems carries L bits. A quantum two-state system, such as a single spin-½ particle, is already richer because two real numbers (e.g. the polar and azimuthal components of the unit vector n) are needed to specify the state. Moreover, these numbers can vary continuously between certain limits. The system is said to carry one quantum bit or "qubit" of information, and a set of L two-state quantum systems can carry L qubits. However, only one of the two binary values, 0 or 1, will be detected when the qubit is measured. A classical 5-bit register can store exactly one of 32 different numbers: i.e. the register can be in one of 32 possible configurations 00000, 00001, 00010, ... , 11111 representing the numbers 0 to 31. But a quantum register composed of 5 qubits can simultaneously store up to 32 numbers in a quantum superposition. Once the register is prepared in a superposition of many different numbers, we can perform mathematical operations on all of them at once. The operations are unitary transformations that entangle the qubits. During such an evolution each number in the superposition is affected, so we are performing a massive parallel computation.

    This means that a quantum computer operating on L qubits can, in a single computational step, perform the same mathematical operation on 2L different input numbers, and the result will be a superposition of all the corresponding outputs. In order to accomplish the same task, any classical computer has to repeat the computation 2L times, or has to use 2L different processors working in parallel. In this way a quantum computer offers an enormous gain in the use of computational resources, such as time and memory, although only in certain types of computation.

    Whether such a "quantum computer" can realistically be built with a value of L that is large enough to be of practical use is a topic of much debate. However, the mere possibility has led to an explosive renaissance of interest in the host of curious and classically counterintuitive properties associated with entangled states. Other phenomena that rely on nonlocal entanglement, such as quantum teleportation and various forms of quantum cryptography, have also been demonstrated in the laboratory


    See Introduction to Cryptology

    I'll butt out now.....
  7. May 28, 2004 #6
    It's my research area, so let me know if you have any questions about it.

    According to the mathematics of information theory, you are not transmitting information. The transmitted information is measured by something called the mutual information, which is always zero for measurements on two halves of an entangled state.

    However, from a foundational perspective, you can ask whether you really are transmitting information at the "subquantum" level, but it is just washed out by the statistics of quantum theory. This can be the case in hidden variable theories, like in Bohmian mechanics.

    There are other interpretations of QM where absolutely nothing is transmitted.
  8. May 28, 2004 #7
    May I ask how the other particle (call it particle B) "picks" the opposite state? Is it just a matter of us assigning the opposite state to it after we measure the spin of particle A?

    Sorry, I'm being ignorant here.
  9. May 28, 2004 #8
    Can you give some further elaboration on what this means?
  10. May 29, 2004 #9
    It's mostly due to conservation of momentum:

    We are not the ones "assigning" it its state. Before we measure the spin of the first particle, the two of them together are in a superposition. Particle A is both up and down and so is particle B. But by measuring any of them, the superposition collapses and a spin is assigned to both particles. In this case each particle has a 50-50 chance to be either up or down, but we know for certain that if one of them is up the other will be down.

    That said, sometimes I don't understand what the "big deal" about this experiment is. Let's say you put two pieces of paper with different numbers in a big black hat. You let two people choose a piece of paper, without checking what their number is. They could be light years away, but whenever you check which number one of them picked, you instantly know the number of the other one. I think this is also what Einstein wanted to say in this:

    At the moment of their creation, both particles 'choose' a spin - one will be up and the other will be down. Just because we don't know which spin each of them has, does not mean it's indeterminate. And the fact that when we measure the spin of one particle we can know the spin of the other isn't that startling, because after all there are only two possibilities.

    That is not to say I don't believe in quantum mechanics... I admit I don't know enough about it to decide that yet. And now someone will jump and say "there's nothing to believe about it - just accept it." :wink:
    Last edited: May 29, 2004
  11. May 29, 2004 #10
    It's a rather long story, but I'll try to give some basic insight into what it means.

    The classical measure of information is the Shannon entropy. Suppose, a random variable [tex]X[/tex] has a probability distribution given by [tex]p_X(j)[/tex], where [tex]j[/tex] runs over the possible values that [tex]X[/tex] can take. The Shannon entropy is given by

    [tex]H(X) = - \sum_j p_X(j) \log_2 p_X(j) [/tex]

    It measures the amount of information or "degree of surprise" that you will have when you do an experiment with outcomes distributed according to [tex]p_X[/tex]. For example, suppose you toss a fair coin with the probabilities of getting heads and tails both equal to 1/2. Then, you know nothing about what the outcome of the coin toss will be, so you will be equally surprised if it is heads or tails. The entropy is 1 in this case. On the other hand, if you have a completely biased coin that only ever lands on heads, then you already know the outcome of every future coin toss, so you gain no information by tossing the coin. The entropy is 0 in this case.

    Shannon entropy has an operational interpretation. If you are given N instances of a random variable [tex]X[/tex], then you can store the data in [tex]N H(X)[/tex] bits and recover it perfectly without any loss (although strictly this is only true in the limit [tex]N \rightarrow \infty[/tex]). This is what is going on in compression programs like WinZip or gzip, that you might use on your computer.

    When you have two random variables, [tex]X, Y[/tex] it is possible that they might be correlated, e.g. they might actually be two copies of the same variable, or [tex]Y[/tex] might be obtained by subjecting [tex]X[/tex] to some noise process. Generally, they will have a joint probability distribution [tex]p_{X,Y}(j,k)[/tex] and a joint entropy

    [tex]H(X,Y) = - \sum_{j,k}p_{X,Y}(j,k) \log_2 p_{X,Y}(j,k)[/tex]

    The marginal probability distributions are given by

    [tex]p_X (j) = \sum_k p_{X,Y} (j,k)[/tex]
    [tex]p_Y (k) = \sum_j p_{X,Y} (j,k)[/tex]

    from which we can define the entropies of the individual random variables. The mutual information is given by

    [tex]I(X:Y) = H(X) + H(Y) - H(X,Y)[/tex]

    and it can be interpreted as the amount of information we learn about Y by looking at the outcomes of X. This is the same as the amount of information we learn about X by looking at the outcomes of Y.

    For further details and justification of these definitions, look at any textbook on Information Theory (Cover and Thomas is my favourite).

    Now, it can be shown that for any pair of measurements done on an entangled state, the resulting random variables have mutual information equal to zero. Thus, no information is transmitted in such an experiment.

    Note that, information theory works at an entirely operational level. All these quantities are related to the amount of data (bits) that have to be transmitted to perform some communication task. Thus, it says nothing about whether the underlying physics of a process is local or non-local. However, note that quantum mechanics (at least the standard formalism) is operational too, in that it only tells us how to calculate probabilities of measurement outcomes, and says nothing about what happens between measurements. Therefore, we can't really conclude anything about locality from experiments on entangled particles. To do so, requires the extra assumption of realism, i.e. there are some physical parameters that really exist that determine the outcomes of mesurements and they exist whether or not we actually make the measurement. Bell's theorem tells us that we have to throw out either locality or realism.

    This is actually a very controversial question in the foundations of quantum theory. In some interpretations of the theory, the wavefunction only represents the knowledge of the experimenter about the quantum system. In those interpretations, it is just a matter of assigning a state. The problem is that the rules for state assignment are very different from those of classical statistics and proponents of these interpretations have to give some justification of why this is the case. They also have to explain how classical probability theory emerges on the scale of everyday objects. Both of these things are very problematic and there is no universal agreement on the best way to approach them.

    On the other hand, some interpretations treat the wavefunction as an element of reality, i.e. it is not just a description, but really exists in the physical world. Such interpretations often run into problems when we try to combine them with relativity theory because of the nonlocality.
  12. May 29, 2004 #11
    there's no a dictionary of quantum entanglement, so I will appreciate a definiton of these words:
    -entanglement swapping
    -entanglement flow
    -entanglement entropy

    And is my information that the maximum number of entangled particles achieved to date is 4. No one has achieved to entangle 5 particles. I wonder, can you entangle one particle with more than other one? FOr example, given 3 particles entangled:

    can you achieve that A is entangled with B and C art the same time, or you can only achieve that A is entangled only with B?
    So in general, the question is: can a particle be entangled with an arbitrary number of other particles at the same time, or only with other one?

    PS: Oh wait, there's a nice explanation of entanglement swapping in this page
    Last edited: May 29, 2004
  13. May 30, 2004 #12
    Entanglement can occur between many particles at the same time if set up correctly. A recent paper details the entanglement of several thousand atoms, and another experiment may have entangled even more.

  14. May 30, 2004 #13
    Since you found an article on entanglement swapping, I will try to explain the other two:

    -entanglement flow: I don't think this is a rigorous concept, with a well established definition, but it refers to how the entanglement properties of a quantum state change during dynamical evolution.

    -entanglement entropy: This is one of the most commonly used measures of entanglement. Given N copies of an entangled state, you would like to convert them into maximally entangled states (often called Bell states, singlet states or EPR pairs) so that you can use them to do things like teleportation, quantum cryptography and entanglement swapping. The maximum number of such states that you can make from your N states is N x (entropy of entanglement) if you are not allowed to have any direct interaction between the two havles of the entangled states during the conversion.

    OK, stricly speaking, this is only true for pure states, as N tends to infinity and by "not allowed to have any direct interaction" I mean that the conversion must be done by local operations and classical communication (LOCC). However, these comments can be ignored unless you are interested in learning the details.

    Mathematically, a two-party pure state can always be written as:

    [tex]\sum_j \sqrt{p_j} \Ket{\psi_j} \otimes \Ket{\phi_j}[/tex]

    where [tex]\Ket{\psi_j}[/tex] is an orthonormal basis for the first system and [tex]\Ket{\phi_j}[/tex] is a (generally different) orthonormal basis for the second system. This is called the Schmidt decomposition. The [tex]p_j[/tex]'s form a probability distribution and the entropy of entanglement is the Shannon entropy of this distribution (see my earlier post for a definition of Shannon entropy).

    Generally, you can have entanglement between any number of systems. However, multi-party entanglement is much more complicated than the two-party case, and much less is known about it.

    One interesting thing is a property known as the "monogamy" of entanglement. This says that the more A is entangled with B, the less A can be entangled with C. In the extreme case, if A is maximally entangled with B, then A cannot be entangled with C at all.

    However, there is also "genuine" multi-party entanglement, which is not reducible to entanglement among pairs. For example, you can have a state where on throwing away C, A is not entangled with B; on throwing away A, B is not entangled with C; on throwing away B, C is not entangled with A; but, taken together, A B and C are entangled.

    This has applications in cryptography in a protocol known as secret sharing. Suppose you own a very profitable company and you know that you are going to die soon. You would like the members of your board of directors to choose someone to sell the company to after you are gone. You trust them to make a good decision collectively, but they are prone to disagreement and you do not trust their judgement on an individual level.

    To sell the company they need a security code. By encoding the security code in a multiparty entangled state, you can ensure that it can only be decoded by the board if they all decide to cooperate. Even more than that, if only one board member does not cooperate, then the rest of the board cannot get any information about the code at all, not even a single digit of it.
  15. May 30, 2004 #14
    Great post slyboy. Baez gives also a succint definition of entanglement entropy in this page
    "...this being standard jargon for the way that two parts of a quantum system can each have entropy due to correlations, even though the whole system has zero entropy"
    I went to a favorite of mine, Arxiv, and searched for the last entry with the word entanglement. It appeared me this
    "Realistic operation of an entangler : a density matrix approach"
    WIll some day be possible to construct entanglers that entangle not only particles, but macroscopic objects, say animals?. Don't will be great to entangle your cat with your dog?
  16. May 30, 2004 #15
    No, No great sin:) We don't want to repeat past mistakes do we:) Remember the FLY

    I enjoyed slyboys information as well. GHZ entanglement issue for consideration?

    I look forward to reading more.
  17. May 31, 2004 #16
    The GHZ state is precisely the sort of state that is entangled between all the parties, but there is no entanglement between any pair.

    There are usually several papers on entanglement every day. I don't know so much about the paractical side of things, but here is a selection of classic papers on the theoretical side:

    The basics of two-party entanglement

    Basics of multiparty entanglement

    Basics of secret sharing

    I have chosen early papers on most topics, so they don't require as much mathematical sophistication as some of the later ones. To appreciate the significance of this stuff it is a good idea to have a basic familiarity with quantum information theory. Reading the introduction to Nielsen and Chuang's book would be a good way to do this. There are also more recent topics, such as the application of entanglement measures to other areas of physics like condensed matter, which I haven't included.
  18. May 31, 2004 #17
    Appreciate it.

    What do you think of this?

    Are you working out of the PI Institute?
  19. Jun 1, 2004 #18
    It's a speculative idea that has the advantage of being experimentally testible. I don't think it is any better or worse than other speculative ideas in theoretical physics, of which there are many.

  20. Jun 1, 2004 #19
    What are your feelings on Anton Zeilinger and the ideas around quantum teleportation?
  21. Jun 1, 2004 #20
    He is one of the leading experimentalists in the foundations of quantum theory and quantum information. I have a lot of respect for him.

    Quantum teleportation is a well established idea and it has been verified experimentally. It is a key building block of many ideas in quantum information theory.

    The interpretation of what is going on in teleportation is slightly more problematic and depends on your interpretation of quantum theory. I prefer interpretations in which the wavefunction represents our knowledge or information about a system, so that there is no need for an instantaneous transfer of information in teleportation. I think that Zeilinger subscribes to a similar view (although don't quote me on that).
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