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In Townsend's Quantum Mechanics textbook, he shows that $\hat{J}^2$, the squared magnitude of the angular momentum, and $\hat{J}_z$, the generator of rotations about the $z$-axis should commute. I understand the proof, but I would like some help understanding why, from an an intuitive point of view, these operators should commute.

At the end of the chapter, Townsend writes that it makes sense that they commute because the magnitude squared of the angular momentum vector (i.e. $\hat{J}^2$) is not affected by rotations. Could you elaborate on this? Why does this mean that we can simultaneously measure the angular momentum squared and one component of angular momentum?

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The physical intuition (of why the commutator $$[\hat{J}_z,\hat{J}^2]~=~0\tag{1}$$ vanish) is (as Townsend writes) that $\hat{J}_z$ is the generator of rotations around the $z$-axis. But the square of the angular momentum $$\hat{J}^2~=~e^{i\phi \hat{J}_z}\hat{J}^2e^{-i\phi \hat{J}_z}\tag{2}$$ is not changed by a rotation $\phi$. Taylor expansion of eq. (2) around $\phi=0$ leads to eq. (1).

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Rotations are a symmetry transformation in group theory. Scalars are any quantity that is invariant under rotations. In math notation that's $$\phi \rightarrow \phi$$ Vectors are quantities that transform using one rotation matrix. So under a rotation matrix, a vector transforms like $$x_i \rightarrow \sum_{j=1}^N R_{ij}\, x_j.$$ Both scalars and vectors are examples of objects called tensors. We categorize tensors by what we call their "rank", where the rank is the number of rotation matrices it takes to transform them. So, the inertia tensor, which is rank two, takes two rotation matrices $$I_{ij} \rightarrow \sum_{n,m=1}^N R_{in} R_{jm} I_{nm}.$$

It's important to understand that the definition of rotations is that they preserve the length of vectors. So the length of any vector quantity is, by definition, a scalar one. Combine that with the fact that you can add or multiply scalars to get new scalars, and $\vec{J}^2$ becomes a scalar.

These definitions are also true in quantum mechanics, when physical quantities are promoted to operators. So, $\hat{x}^2 + \hat{y}^2 + \hat{z}^2$ will also be rotation invariant, as will $\hat{p}_x^2 + \hat{p}_y^2 + \hat{p}_z^2$. No individual vector component will be invariant under all possible rotations, though.

Interestingly, though I can't think of an example where it's used, $\hat{J}_z$ will always commute with the $z$-components of any vector operator - because rotations around the $z$-axis leave the $z$-components of vectors invariant.

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Some aspects of the theory of quantum angular momentum are closely related to the theory of classical angular momentum.

To see this and connect with the language of your specific example, imagine we make a rotation about $\hat z$. This rotation is done through the matrix \begin{align} R_z(\alpha)&=\left(\begin{array}{ccc} \cos(\alpha)&-\sin(\alpha)&0\\ \sin(\alpha)&\cos(\alpha)&0\\ 0&0&1\end{array}\right)\, ,\\ &=e^{-i\alpha J_z}=\sum_{k} \frac{(-i \alpha)^k}{k!} J_z^k\, , \qquad J_z=\left(\begin{array}{ccc} 0&-1&0\\ 1&0&0\\ 0&0&0\end{array}\right)\, . \end{align} You can do the same to obtain $J_y$ and $J_x$. If you do so you will find that $J_z^2+J_y^2+J_x^2$ is actually proportional to the unit matrix $\hat 1$, which commutes with any component of angular momentum. This simply reflect that fact that the length of a vector does not change under rotation. Note also that the matrices for $J_z,J_x$ and $J_y$ do not commute as matrices; of course the corresponding operators do not commute as operators either.

We therefore broadly expect from classical physics arguments that, if $\hat J^2$ is to be connected to the length squared of the angular momentum vector $J^2$, it will commute with any generator of rotation $\hat J_k$: this is just the quantum version of the classical result.

Two operators can be measured “simultaneously” when they have simultaneous eigenvectors. Quoting from “The interpretation of Quantum Mechanics” by Roland Omnes:

Mathematics tells us that, when two self-adjoint operators $A$ and $B$ commute, they can be simultaneously diagonalised. This means there exists an orthogonal basis (at least in the sense of Dirac) consisting of vectors $\vert a,b,r\rangle$ where $a$ is in the spectrum of $A$, $b$ in the spectrum of $B$, and $r$ a degeneracy index. The probability if finding the values $(a,b)$ when simultaneously measuring $A$ and $B$ is then written as $$ p(a,b)=\sum_r\vert \langle a,b,r\vert\psi\rangle\vert^2$$

[Here, “$a$ is in the spectrum of $A$” means $a$ is an eigenvalue of $A$].

Moreover, one easily shows that both $\Delta A$ and $\Delta B$ are simultaneously $0$ for the common eigenvectors. Thus, in your specific example, we can ask if the state $\vert jm\rangle$ has a well-defined total angular momentum $\hbar^2 j(j+1)$ simultaneously with a well defined projection $\hbar m$.

Note that “simultaneous” here does not refer the measurement to two quantities done at the same time in a lab. The latter is called a joint measurement. To emphasize this, you can have a look at this nice but somewhat intense paper on joint measurements which in particular quotes early in the paper work by Margenau and Park:

[Margenau and Park] notice that (a) the uncertainty relations have nothing to do with simutaneous measurements, since these relations regard standard deviations of measurement results which are obtained by measuring the observables separately.

There is also a short and pedagogical paper by Raymer [Uncertainty principle for joint measurement of noncommuting variables. American Journal of Physics 62.11 (1994): 986-993] which gives some details on joint measurements for position and momentum.

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