Fiziqs said:
But therein lies my problem, for this only seems to work if there are just two entangled particles. For any number more than that it doesn't seem to work. It does hold true for momentum as DrChinese indicated, but it doesn't hold true for something like spin, in which there are only two distinct outcomes, like up or down. in such a case, measuring the state of one particle within any set of particles greater than two, would seem to have no effect. The remaining particles will still be both up and down.
Part of the problem here is that you're you are using the handwaving English language description of quantum states and entanglement instead of looking at the math (it is well to remember that entanglement appeared in the math before it was recognized in experiments).
In the English language description, we talk about the individual particles and their state - "This one is up, this one is down, this one changed from up to down". However, a quantum system is a single system with a single wave function even if we're tempted to think of it as containing multiple particles. So let's consider a quantum system on which we can perform either of two measurements, which I'll call A and B. If A and B commute, then the system has states that are eigenstates of both operators and it makes sense to say that in some states "A has the value x and B has the value y"; we will write that particular eigenstate state as ##|xy\rangle##. Another possible eigenstate state is the one in A has the value p and B has the value q, which we would write as ##|pq\rangle##. (If A and B did not commute, they would not have a complete set of common eigenstates and we'd find that the uncertainty principle prevented us from making constructing states in which A and B both have definite values).
OK, now let's say our quantum system is prepared in a superposition of those two states: ##\frac{sqrt{2}}{2}|\psi\rangle=|xy\rangle+|pq\rangle##. If we perform a measurement of A on this state, half the time the wave function will collapse to ##|xy\rangle## and we'll get the result x; the other half of the time it will collapse to ##|pq\rangle## and we will get the result p. Either way the final state of the system will be one in which we know what the result of a measurement of B would be, when and if we make such a measurement. That's entanglement; and we say the initial state ##\frac{sqrt{2}}{2}|\psi/rangle=|xy\rangle+|pq\rangle## was an entangled state of the two observables A and B. Note that A and B can be any two commuting observables of the quantum system - they don't have to refer to the same or different particles, or any particles at all for that matter, one of them could be spin and the other could be energy, the quantum system might be one that we don't naturally think of as containing two particles... All that's necessary is that we be able to put the quantum system into a state that can be written in the entangled form for some two observables and we have successfully entangled those tw observables.
If you go back and substitute "spin of first particle" for "A", "spin of second particle" for "B", "+" and "-" for "x", "y", "p", and "q", you will have the mathematical description of the two-particle entangled spin experiment.
In this framework it's a lot easier to see how three-particle entanglement might work. If we have three observables commuting A, B, and C each taking on either of two possible values, then an eigenstate will be written as something like ##|+--\rangle## where the three signs represent the values of A, B, C in that order. A state like ##\frac{\sqrt{2}}{2}(|+--\rangle+|-++\rangle## or ##\frac{\sqrt{2}}{2}(|+--\rangle+|-+-\rangle## is entangled; a measurement of A will cause the wave function to collapse to either ##|+--\rangle## or ##|-++\rangle##, states in which we know more about what will happen when and if we subsequently measure B or C. (Be warned that this is a highly contrived example; in any realistic situation the wave function would be such that even after collapse B and C would remain entanged).
