Measuring J^2 & Angular Momentum in QM

[KFC]

Hi there,
I just wonder how to measure the square of angular momentum and the z componet of angular momentum in QM system? If I have two spin (marked as 1 and 2), why J^2 (total) doesn't commute with J_{1z} or J_{2z} ? or why they cannot be measured at the same time? (Well if I calculate the commutation relation, I could find it is ZERO, but what's the physical reason for that?)

 

[Manilzin]

I'm not sure how to actually measure J^2, but for your second comment: It is possible to measure both J^2 and the x or y (or z) component of J. It's not possible to measure all of them at once, though, since the x,y,z-components don't commute. So you can measure J^2 plus one of either Jx, Jy or Jz. By convention, however, we normally choose the z-component, but this is an arbitrary choice of axis, and our results don't depend on our particular choice of axis.

 

[weejee]

In a system where its energy is dependent on J^2, one can measure J^2 by measuring the energy. Otherwise, I'm not sure. Maybe it is impossible?

 

[KFC]

I'm not sure how to actually measure J^2, but for your second comment: It is possible to measure both J^2 and the x or y (or z) component of J. It's not possible to measure all of them at once, though, since the x,y,z-components don't commute. So you can measure J^2 plus one of either Jx, Jy or Jz. By convention, however, we normally choose the z-component, but this is an arbitrary choice of axis, and our results don't depend on our particular choice of axis.

Thanks for reply. But I am talking about a system of two particles, and I am measuring J^2 (total) and J_{1z} (z-component of the first particle), they cannot be measured simutneously.

 

[KFC]

In a system where its energy is dependent on J^2, one can measure J^2 by measuring the energy. Otherwise, I'm not sure. Maybe it is impossible?

Thanks. What I am wondering is: J^2 is not a classical concept. Classically, if we know the moment of inertia, we could measure the energy and then find the J^2. But in QM, there is no more "orbital", once you get the energy, how do you find J^2?

 

[jambaugh]

Hi there,
I just wonder how to measure the square of angular momentum and the z componet of angular momentum in QM system? If I have two spin (marked as 1 and 2), why J^2 (total) doesn't commute with J_{1z} or J_{2z} ? or why they cannot be measured at the same time? (Well if I calculate the commutation relation, I could find it is ZERO, but what's the physical reason for that?)

You measure the $$J_z$$ using a Stern-Gerlach magnet. A beam of particles passing through will be deflected in proportion to $$J_z$$. Reorient the S-G magnet (and turn the momentum of the particle without affecting its spin) and you can measure any of the other $$J_x$$ and $$J_y$$ components. Of course these don't commute so you can't get all three simultaneously.

However the value is not observed until the amount of deflection is measured. If you fail to do this but screen out (or selectively deflect) only those component beams which correspond to a distinct set of values and then recombine those beams then you've only effected a projection onto that subspace and not resolved each component in that space.

In principle you should be able to combine a series of such magnets so that the beam of particles splits and then recombine until the output is a set of beams indexed by $$J^2$$.

In short by not observing until the end you can execute value dependent actions on the system and just so long as the information about the values themselves are not retained in the final configuration you can measure whatever the math says you can construct from the basic operations. In this case since mathematically you can build $$J^2$$ from $$J_x$$, $$J_y$$, and $$J_z$$, so too you can mechanically create a device which will measure $$J^2$$ using component devices which act depending on these three component observables.

Now I say this but I'm trying to think of a simple such arrangement. I don't have time right now to work it out rigorously and until I or others do my argument may be suspect.