When quantum systems are entangled, they have a "grade of entanglement" which can be quantified e.g. as the entropy of entanglement. There also are states of "maximum entanglement", e.g. the Bell states.

But what does it mean practically if quantum systems are less than maximum entangled, e.g. with half of the maximum entropy? Does it mean that there is a probability of 50% that the systems are (maximum) entangled? Or that they will decide in the moment of measurement with a 50:50 chance to behave like maximum entangled or non-entangled? Or does it mean something else?

Or in other words: If an observable is measured for two pairs of identical particles under the same conditions, except that the first pair is maximum entangled and the second pair less the maximum entangled, what will be the difference in the measurement results?

(A similar question was already asked in a comment here by Passiday on May 24 '13 6:18, but not really answered.)

[Edit: bolded the second, better version of the question.]

  • $\begingroup$ how can the two pairs be "under the same conditions" if they are in different states? $\endgroup$
    – fqq
    Jun 18, 2016 at 19:27
  • $\begingroup$ The conditions are not exactly the same, see "except ...". The pairs of particles are prepared in basically the same way, but with different grades of entangelement (see e.g. arXiv:1511.03445 on how to do so). Then they are measured in the same way. $\endgroup$
    – following
    Jun 21, 2016 at 9:43
  • $\begingroup$ What is your question? There are four sentences with question mark, and they are not asking the same. $\endgroup$ Jun 22, 2016 at 1:53
  • $\begingroup$ The bold question. $\endgroup$
    – following
    Jun 22, 2016 at 11:08

1 Answer 1


For an illustrative example, suppose the state space of a single particle is two-dimensional, say with orthogonal basis $X,Y$.

Now consider a two-particle system in the pure state $$\alpha X\otimes X+\beta X\otimes Y +\gamma Y\otimes X+\delta Y\otimes Y$$ with $\alpha^2+\beta^2+\gamma^2+\delta^2==1$.

This state is called maximally entangled if $\alpha^2+\gamma^2=\beta^2+\delta^2=1/2$ and $\alpha\beta+\gamma\delta=0$. (Exercise: This condition is independent of the chosen basis.)

This defines maximal entanglement for a pure state. One can extend the definition to mixed states, but there's already enough here to answer your question. A pure state can be non-maximally entangled, therefore a non-maximally entangled state need not be non-trivially mixed, therefore no, a non-maximally entangled state need not have any non-zero probability of being maximally entangled.

As far as the rest, if your model calls for a collapse at the moment of measurement, then if you measure the first particle and find that it's in an eigenstate $Z=pX+qY$ of your measurement, then you can see that the two-pair system collapses into a state of the form $Z\otimes W$, which is not entangled at all (regardless of whether it was maximally or non-maximally entangled to begin with).

  • $\begingroup$ Thanks for the clarifications about pure and mixed states and about the result of a collapse. I am confused now by the paradox statement "a non-maximally entangled state need not have any non-zero probability of being maximally entangled"; and I still don't understand what is the difference between measurement results when measuring either maximally entangled or non-maximally entangled systems. $\endgroup$
    – following
    Jun 18, 2016 at 10:11
  • $\begingroup$ 1) There are pure states that are entangled but not maximally entangled. 2) Because these states are pure, there is no sense in which they have any probability of being anything else. Therefore they have zero probability of being maximally entangled. 3) Therefore, when you ask whether a non-maximally entangled state must have some positive probability of being maximally entangled, the answer is no. $\endgroup$
    – WillO
    Jun 18, 2016 at 14:40
  • 1
    $\begingroup$ Ok, now this point of your explanation is clear. However, my original question is still unanswered. $\endgroup$
    – following
    Jun 18, 2016 at 19:13
  • $\begingroup$ @following: I do not understand what part of your original question you think is still unanswered. With maximal entanglement, the outcome of any observation has a 50/50 chance of coming out either way. With less-than-maximal entanglement, this is not true. What more do you want to know? $\endgroup$
    – WillO
    Jun 19, 2016 at 2:21
  • $\begingroup$ I will try to reword the question, so that it's clearer: How can we empirically measure the entropy of entanglement? How to set up such an experiment, and what will be the results? How does it work (e.g. in Copenhagen terms)? $\endgroup$
    – following
    Jun 19, 2016 at 11:23

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