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Is the particle entanglement a boolean property? That is, when we consider two given particles, is the answer to the question "are they entangled" always either "yes" or "no" (or, of course, "we are not sure if it's yes or no")? Is there such thing as partial entanglement?

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No, it's not a Boolean property. Entanglement between two quantum systems (could be particles, or anything else) could be partial, and can be quantified using different measures. In the specific example of Bell states, the two systems (each of them with 2 states $|0\rangle$ and $|1\rangle$) are said to be maximally entangled with the entanglement entropy being 1 qubit.

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But without involving multi-particle systems, ie can two photons be partially entangled? Any reference to an experiment where this is achieved? –  Passiday May 23 '13 at 22:10
@Passiday, two photons are a mutli-particle system. More generally, if two particles are entangled, then you have to describe them as a two-particle system. –  Colin McFaul May 23 '13 at 22:22
I think @Passiday means to ask if each system needs to be multiparticle. If that's the question: There is no such requirement. Any quantum system will do and two or more of them can be entangled. Recall that in quantum mechanics, even a single particle system has multiple states (2 possible photon polarizations, Spin up/down, the many levels of a harmonic oscillator or particle in a box, etc.) –  Siva May 23 '13 at 22:51
Yes, I meant the simplest possible particle interaction. Ok, so, let's say, two photon entanglement entropy can be 0.5. Doest that mean that in 50% of measurements, the two such entangled photons behave as fully (=1) entangled? –  Passiday May 24 '13 at 6:18
@Passiday : No, photons are, or are not, entangled. This has nothing to do with measurements. If you want to have a measure of entanglement for a 2-photon state like $a|0>|0> + b|0>|1> + c|1>|0> + d|1>|1> $, you could use $E = 2* \frac{|ad - bc|}{(a^2 +b^2 + c^2 + d^2)} $. For instance, the state $$|0>|0> + |1>|1> $$ is maximally entangled and has $E = 1$ –  Trimok May 24 '13 at 9:23

Yes, the answer is always either yes or no. Being entangled means that they are not in a separable state. A state is called separable if and only if it can be written in the form $\rho = \sum_i p_i \rho_i^A \otimes \rho_i^B$, where $\rho_i^A$ refer to the state of particle $A$ and $\rho_i^B$ to $B$. This is always either the case or not the case, regardless of how much entanglement precisely we assign to the two particles. All standard entanglement measures satisfy the property that $\rho$ is entangled if and only if $E(\rho) > 0$ (ie. $E(\rho)=0$ for all separable states).

To find out whether given two particles are entangled, people have developed procedures known as entanglement witnesses, and research into the entanglement witnesses has formed a sub-discipline in quantum information theory. Bell inequalities can also be thought of as entanglement witnesses since their violation implies the presence of entanglement.

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Ok, but once the E(ro) is greater than zero, it can be of any value up to 1. Essentially, it is a continuous value characterizing state of entanglement between A and B. The entanglement-ness becomes a yes/no property only because we define it with boolean expression (E(ro) > 0). –  Passiday May 24 '13 at 6:12
It's the other way around. Entanglement measures are defined so that they are $E(\rho)>0$ if and only if the state $\rho$ is entangled. There are also many measures of entanglement. The form a separable state takes, however, was defined decades before the measures of entanglement were developed. For a nice review I recommend reading arxiv.org/abs/quant-ph/0504163 –  SMeznaric May 24 '13 at 9:46

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