Angular momentum state of a system

[yamata1]

Hello,
Suppose that the angular momentum of a system can take the values 0, 1, 2. One carries out a measurement of $J_z$ on this system.
What can be said about the state of the system after the measurement? To what extent can it be perfectly certain if $J_y$ and $J_x$ do not commutate with $J_z$?Is there a value which eliminates all uncertainty ? Thank you.

 

[Nugatory]

When we say something like "the angular momentum of the system can take on the values 0, 1, 2" we're generally speaking about $J^2$, not $\vec{J}$. Its eigenvalues are of the form $j(j+1)$, and the "0, 1, 2" refers to the possible values of $j$.

Although $J_x$, $J_y$, and $J_z$ do not commute with one another, all three commute with $J^2=\vec{J}\cdot\vec{J}$, the squared magnitude of the angular momentum. Thus we can find states in which there is no uncertainty about the value of $J^2$ and anyone of the three $J_i$.

 

[yamata1]

When we say something like "the angular momentum of the system can take on the values 0, 1, 2" we're generally speaking about $J^2$, not $\vec{J}$. Its eigenvalues are of the form $j(j+1)$, and the "0, 1, 2" refers to the possible values of $j$.

Although $J_x$, $J_y$, and $J_z$ do not commute with one another, all three commute with $J^2=\vec{J}\cdot\vec{J}$, the squared magnitude of the angular momentum. Thus we can find states in which there is no uncertainty about the value of $J^2$ and anyone of the three $J_i$.

Since $J_z$ and $J^2=\vec{J}\cdot\vec{J}$ form a CSCO we can say their eigenvalues completely specify the state of a system and ignore the results for $J_x$ and $J_y$?

 

[Nugatory]

Since $J_z$ and $J^2=\vec{J}\cdot\vec{J}$ form a CSCO we can say their eigenvalues completely specify the state of a system

Only if the system state happens to be an eigenfunction of those two observables. In general it's not, so we'll write the state as a superposition of the various eigenfunctions of $J_z$ and $J^2$. There's nothing special about $J_z$ here of course; we could have chosen to write the state as a superposition of eigenfunctions of $J^2$ and $J_x$ or $J_y$ instead.

and ignore the results for $J_x$ and $J_y$?

What results? We don't have a "result" for either of these unless and until we measure them. No matter what the system state is, and no matter whether we've chosen to write it as a superposition of eigenfunctions of $J_z$ and $J^2$ or something else, we can measure $J_x$ or $J_y$ and get some result.

 

[yamata1]

we can measure $J_x$ or $J_y$

But not simultaneously with $J_z$.Can it ever be said that the state of the system is perfectly known after measuring a certain value of $J_z$ or measuring both $J_z$ and $J^2$ ?

 

[mikeyork]

But not simultaneously with $J_z$.Can it ever be said that the state of the system is perfectly known after measuring a certain value of $J_z$ or measuring both $J_z$ and $J^2$ ?

The state may be perfectly known within the limits of the observer's context (choice of observables). What other meaning you might intend by ``perfectly known'' is not clear.

 

[kith]

But not simultaneously with $J_z$.Can it ever be said that the state of the system is perfectly known after measuring a certain value of $J_z$ or measuring both $J_z$ and $J^2$ ?

After a (standard textbook) measurement, the state of the system is an eigenstate of the measurement operator. So after a measurement, the state of the system is always "perfectly known" no matter what observable you are measuring.

It's just that the Heisenberg uncertainty principle doesn't allow all observables to have sharp values in a quantum state simultaneously. Finding a common eigenstate of a CSCO is the best you can do.

So yes, the state of the system is "perfectly known" if you measure $J_z$ and no, this doesn't include a sharp value for $J_x$.