Two Spin 1/2 Particles: Stot=0 Outcome?

[K8181]

I have a hypothetical situation that I am trying to work through...

Say there are two spin 1/2 particles, and the system is known to be in a total spin state of Stot=0. An observer comes along and determines the first particle to have a spin component (S1z) of hbar/2. Now say another observer comes along and measures the z component for the second particle (S2z). What is the outcome?

My solution is this: I believe the Stot information is lost when the first observer measures S1z, so that then there is a 50% chance of the second observer measuring hbar/2, and a 50% chance for -hbar/2.

Is this right?

Thanks very much.

 

[so-crates]

Well, the problem is, how do you distinguish between the two particles ?

 

[K8181]

I found the problem in a book, and it says that the first observer is trained to measure particle one, and the second is trained to measure particle two. I don't think distinguishing between them is the crux of the problem. I am more concerned about eigenstates of angular momentum and that sort of thing. The experiment is some fictitious scenario invented to illustrate a point.

 

[Galileo]

If the total spin is zero, then there's only one spin-state the system can be in: the singlet |0 0\rangle configuration. What are the possible outcomes when you measure S_{1_z}?

 

[Doc Al]

My solution is this: I believe the Stot information is lost when the first observer measures S1z, so that then there is a 50% chance of the second observer measuring hbar/2, and a 50% chance for -hbar/2.

Is this right?

No. The spin measurements of the two particles are correlated. Can you write the overall singlet state in terms of combinations of the individual S1z and S2z states? (I'm making the same point made by Galileo.)

 

[K8181]

I guess my confusion is...why is the particle still in the singlet state once the measurement of S1z has been carried out?

 

[reilly]

K8181 -- You are somewhat correct. The measurement of one of the particles can influence the other if they are in a bound state in which case the interaction between measurement and particle 1 also can perturb particle 2. For example, consider a helium atom with the electrons in a total spin 0, S-state. If we use a spin 0 probe, then the matrix elements for any spin flip will be zero. On the other hand, if we use a spin 1 probe, spin flip is possible, and the initial spin correlation will not survive the interaction. (I'm assuming the the probe comes in as an S wave, so the total probe angular momentum is = probe spin.)

If, as in EPR, the two electrons travel well away from their source, then their correlation will be unchanged by measurement.

So, it all depends on whether angular momentum is conserved by the measurement, and/or whether the two are sufficiently distant to preclude radiation, or whatever, from measurement 1 going to measurement 2 prior to the second measurement.

Regards,
Reilly Atkinson

 

[kleinwolf]

Local measurement of singlet state

I guess my confusion is...why is the particle still in the singlet state once the measurement of S1z has been carried out?

The answer is clearly given by quantum mechanics for this measurement :

Operators :

-measurement operator in A is S1z
-measurement operator in B is 1 (identity operator, since you don't measure in B)

Final state=Eigenstates

-final state in A is |+> or |->
-final state in B is |\phi>= \left(\begin{array}{c} cos(\phi)\\ sin(\phi)\end{array}\right)

Hence the final state after a measurement of S1 without measuring in B is either |+>|\phi> or |->|\phi> with some (by quantum mechanics) given probablitities...

However if you measure S1z and S2z, then the final state is |+-> or |-+>...so measuring or not S2z influence the result, a usual fact in quantum mechanics

I don't know if this is clear...?