Why do position operators in orthogonal directions commute? In three dimensions, we have $\hat x$, $\hat y$, $\hat z$ as the position operators in the three orthogonal directions. If the components of angular momentum don't commute, why must these all commute? I can't seem to find an answer elsewhere. For example, this states that the "coordinate operators clearly commute" without explanation. Is this an experimental fact or a postulate or something else?
 A: This is an assumption that seems to be born out by the experimental evidence thus far. 
Non-commutative quantum mechanics is a speculative theory that introduces a degree of non-commutativity between the components of the position operator. This introduces a sort of minimum length scale into the theory, beyond which it is not possible to localize a particle. This is a generic prediction of a certain perspective on quantum theories of gravity. In these theories it is expected that probing particles at high enough energies will lead to the postulated non-commutativity.
A: In position representation these operators are simply equivalent to multiplying the wave function by the corresponding variable, so it is easy to see that they commute. For example,
$$
\langle \hat{x}\rangle=\int dxdy dz \psi(x,y,z)^*\hat{x}\psi(x,y,z)=
\int dxdy dz \psi(x,y,z)^*x\psi(x,y,z)=\int dxdy dz x\left|\psi(x,y,z)\right|^2,\\
\langle \hat{y}\rangle=\int dxdy dz \psi(x,y,z)^*\hat{y}\psi(x,y,z)
=\int dxdy dz \psi(x,y,z)^*y\psi(x,y,z)=\int dxdy dz y\left|\psi(x,y,z)\right|^2
$$
Thus
$$\langle \hat{x}\hat{y}\rangle=\int dxdy dz \psi(x,y,z)^*\hat{x}\hat{y}\psi(x,y,z)
=\int dxdy dz \psi(x,y,z)^*xy\psi(x,y,z)=\\\int dxdy dz xy\left|\psi(x,y,z)\right|^2=
\langle \hat{y}\hat{x}\rangle
$$
Same can be shown for the momentum operators (which are derivatives in the position representation.)

If the components of angular momentum don't commute, why must these all commute?

It is easy to see that, if $\hat{x},\hat{y},\hat{z}$ commute and $\hat{p}_x,\hat{p}_y,\hat{p}_z$ commute, then the angular momentum operators do not commute just because $\hat{x}$ does not commute with $\hat{p}_x$ (and same for $y,z$). Indeed, angular momentum operators are obtained by definition as
$$
\hat{\mathbf{L}}=\hat{x}\times\hat{p}\rightarrow
\hat{L}_x=\hat{y}\hat{p}_z-\hat{z}\hat{p}_y,
\hat{L}_y=\hat{z}\hat{p}_x-\hat{x}\hat{p}_z,
\hat{L}_z=\hat{x}\hat{p}_y-\hat{y}\hat{p}_x
$$
I leave calculating their commutation as a homework exercise (or lookup in the QM book.)
A: $\hat{x}$ operator takes a quantum state and multiplies by a number $x$ and $\hat{y}$ operator takes a quantum state multiplies by a number $y$. So since $x$ and $y$ are now numbers after the operators hit the quantum states, then $[\hat{x},\hat{y}] = [\hat{y},\hat{x}]$.
