The uncertainty principle for angular momentum and angular position

Question

For $$L_{z}=xp_{y}-yp_{x}$$ we see that angular position in the $x-y$ plane is canonically conjugate, $$\theta_{x-y}=\mathrm{tan}^{-1}\left(\frac{y}{x}\right)$$ that is, $$\{\theta_{x-y},L_{z}\}=1$$ where $\{\}$ represent Poisson brackets. Applying canonical quantisation, we see that $$[\hat{\theta}_{x-y},\hat{L}_{z}]=i\hbar$$ therefore, applying the generalised uncertainty principle, we should have $$\Delta\hat{\theta}_{x-y}\Delta\hat{L}_{z}\geq\frac{\hbar}{2}$$ This is where I have some trouble though. We know that angular momentum in QM is quantised in units of $\hbar$, and hence measurements of $L_{z}$ lead to exact measurements. Hence some angular state of $\hat{L_{z}}$, $$\hat{L}_{z}|m\rangle=m\hbar|m\rangle$$ has an uncertainty of angular momentum of 0, $\Delta \hat{L}_{z}=0$. This doesn't make sense, as it seems to suggest that $\Delta\hat{\theta}_{x-y}=\infty$, i.e eigenstate of angular momentum correspond to states where we have complete negligence to the angle of the state in the $x-y$ plane in position space. So I guess my question is, do we have to treat the generalised uncertainty principle differently for angular momentum since it is a discrete observable?

Answer 1

You don't even need the uncertainty principle for a contradiction here: $\widehat{L}_z$ is hermitian ($\widehat{L}_z^\dagger=\widehat{L}_z$) and therefore by applying $\cdot^\dagger$ to your last equation, we get $\langle m|\widehat{L}_z=m\hbar\langle m|$, which directly results in the contradiction: $$i\hbar =i\hbar\langle m|m\rangle =\langle m|[\widehat\theta_{x-y},\widehat{L}_z]|m\rangle =\langle m|\widehat\theta_{x-y}\widehat{L}_z|m\rangle -\langle m|\widehat{L}_z\widehat\theta_{x-y}|m\rangle =m\left(\langle m|\widehat\theta_{x-y}|m\rangle -\langle m|\widehat\theta_{x-y}|m\rangle\right)=0$$ I think the problem here is the canonical quantization, which simply might not preserve the Poisson bracket. An important result of canonical quantization is that there is no quantization map $Q$ from the maps on a classical phase space to a quantum Hilbert space, so that: $$[Q(f),Q(g)]=i\hbar Q(\{f,g\})$$ is always fulfilled.