# How to measure observables

1. Nov 2, 2009

### clacker

Question about measuring observables. If have 2 particle system the particle seperation
Q=x1-x2 and total momentum P=p1+p2 are observables of the system as a whole and are
commuting. How do you measure these observables. It would seem the only way to
measure the seperation is to measure the individual particle positions x1,x2 and subtract
them. With the total momentum you measure the particle momentum's seperately and add.
However, since Q,P commute you should be able to measure Q,P simultaneously. But if do
it by measuring individual particle properties you have violated uncertainty principle since
you have both position and momentum for each particle. So is there some way to measure
Q or P without determining x1 and x2 or p1 and p2 or are Q and P nonmeasurable and only really mathematical fictions.

2. Nov 2, 2009

### Bob_for_short

Your P and Q belong to different, independent subsystems or quasi-particles. They are directly measurable. See Chapters 1 and 3 in http://arxiv.org/abs/0811.4416.

3. Nov 2, 2009

### clacker

My question is not the measuring of P and Q together, but how do you measure just Q say. If you measure each particles position seperately then subtract I don't think you are actually in an eigenstate of Q but rather in 2 seperate 1 particle eigenstates.

4. Nov 2, 2009

### Bob_for_short

No, not even so because the variables x1 and x2 are correlated (not independent). The relative motion is a subsystem with its own properties - proper frequencies, angular momenta, energies, etc.

5. Nov 2, 2009

### clacker

so you're saying that if I measure x1 and x2 that put's me into an eigenstate of Q.
What about the fact that Q and P commute. If I now measure P haven't I got
x1,x2,p1,p2 and now know each particles position and momentum with certainty thus
violating the uncertainty principle.

6. Nov 2, 2009

### sweet springs

Hi.

I considered it on such a case that x1 and x2 are not positions of the different particles but different directions say x and y of the same particle.

Let a particle be on the plane. Position detector of 100% sensitivity is placed on each point of the plane except on the line x - y = a. If no detector works when you make observation, you succeed to prepare eigenstate of Q of eigenvalue a.
It's the case of particle beam shoot on the screen with line slit. Momentum component along with the slit line, say px + py = P is not disturbed by the position measurement at the slit. So P and Q commute.

I hope it will make any sense.
Regards.

Last edited: Nov 2, 2009