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

• I
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?

## Answers and Replies

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J is not a scalar.

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

Mentor
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##).

vanhees71 and annaphys
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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)##.

annaphys and vanhees71