# 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?

• It's a postulate. Mar 29, 2018 at 9:33
• Also I really don't recommend using any resource that feels the need to say "obviously" and "clearly" 4 times in one page. Mar 29, 2018 at 9:34

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.)
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.)
$$\hat{x}$$ 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 $$[\hat{x},\hat{y}] = [\hat{y},\hat{x}]$$.