What's the importance of the squared of the angular momentum?

[annaphys]

In quantum mechanics one sees what J^2 can offer but why do we even consider looking at the eigenstates and eigenvalues of J^2 and a component of J, say J_z? Why don't we just use J?

 

[Vanadium 50]

J is not a scalar.

 

[annaphys]

So the only reason we use J^2 is because it's a scalar?

 

[PeterDonis]

why do we even consider looking at the eigenstates and eigenvalues of J^2 and a component of J, say J_z? Why don't we just use J?

Because the different components of $J$ don't commute. So there is no set of states that is a complete set of eigenstates for both $J^2$ and $J$. The best we can do is to find a complete set of eigenstates for both $J^2$ and one of the three components of $J$, usually defined to be $J_z$ (i.e., we define our $z$ axis to point along this component of $J$).

 

[PeroK]

So the only reason we use J^2 is because it's a scalar?

$J^2$ is easier to work with because it's defined as $J^2 = J_x^2 + J_y^2 + J_z^2$. You could try to work with $J = \sqrt{J^2}$ as the magnitude of total AM, but I suspect it would be more awkward to work with.

As $J^2$ can be diagonalised, then $J$ would have the square roots of all the diagonal entries. I think everything would work out, except that the eigenvalues of $J$ would be the square roots of the ones you get using $J^2$. I'd say it's more convenience and convention.

Note that $J$ is not to be confused with $\vec J = (J_x, J_y, J_z)$.