1. Jan 2, 2016

T S Bailey

Suppose we have a particle pair (A, B) which are maximally entangled. Then A goes cavorting with some macroscopic system in a thermodynamically irreversible way, and becomes highly entangled with it. Does this entanglement with the new system reduce the entanglement between A and B? At first I want to say no, because if we were to measure a maximally entangled quanta and found it spinning one way (for example) we can immediately deduce the direction of spin for its partner. But since a measurement is just a process of sufficient interaction to produce entanglement with the measured quanta, and because only two systems may be maximally entangled with one another, I imagine that the degree of entanglement between A and B must now be something less than maximal. What am I missing?

2. Jan 3, 2016

Staff: Mentor

That's a mistake right off the bat.

For entangled particles A and B do not have separate existence.

Most certainly you can break the entanglement by observing just A or B - but prior to observation you cant say A or B have a separate existence.

Thanks
Bill

3. Jan 3, 2016

atyy

Yes. That is what monogamy of entanglement means.

If you measure just one of an entangled pair without knowing that it was one of an entangled pair, measuring one will not tell you anything about the other particle. You will just think the particle you measured is in a mixed state.

Also, a measurement will break the entanglement between A and B.

4. Jan 3, 2016

T S Bailey

So if we were to perform an experiment like that described in the EPR paradox we could never infer the spin of the entangled partner? If that's the case then why is there confidence that entangled states exist at all?

5. Jan 3, 2016

Staff: Mentor

You have to know in some other way (usually because you know how it was produced) that you have an entangled pair. If you know this, then when you measure the spin of one particle on a given axis you will know what the result would be if a measurement along the same axis is performed on the other particle.
Conversely, if you have a black box that is producing particle pairs and you want to know whether they are entangled, you have to perform measurements on both members of many pairs and then compare results. A single pair won't tell us anything - we'll get opposite results on the same axis 50% of the time just by random chance. But when we do many pairs and get correlations much higher than 50% we know that there is some sort of relationship between the two particles.

6. Jan 3, 2016

Staff: Mentor

We start with a quantum system consisting of the environment around A, the particle A, and the particle B. The initial state of this system is a superposition of "The environment is what it is; A is spin-up and B is spin-down" and "The environment is what it is; A is spin-down and B is spin-up"; in this state A and B are entangled. When A interacts with the environment around it, the system transitions (collapses? evolves? MWI branches? The interpretation doesn't matter) into one of two possible result states:
1) "A is spin-up and the environment is consistent with A being spin-up; B is spin-down"
2) "A is spin-down and the environment is consistent with A being spin-down; B is spin-up"
These are states in which A and the environment are entangled while A and B are not. If the environment happens to include some sort of readout (for example a dial with a needle that points up or down according to A's spin) then we will be able to know A's spin and what the result of a measurementof B's spin on that axis would be by examining that part of the environment and we'll call the interaction a measurement.

Thus, monogamy is respected and we get to know B's spin as well.

Last edited: Jan 3, 2016
7. Jan 3, 2016

Staff: Mentor

Of course you can. What you cant do is say it has the property of spin, or even a separate existence, until you observe it.

Remember QM is a theory about observations. What's going on, and that includes if objects have an existence independent of other objects, when not observed, its silent about.

Thanks
Bill