# General commutation question

If I have three general observables, $$\hat{C}$$, $$\hat{H}$$, and $$\hat{L}$$, and the commutation relation between $$\hat{C}$$ and $$\hat{H}$$ is given by,

$$[\hat{C}, \hat{H}] = \hbar \hat{L}$$

At the moment, assume nothing about the observables. Since the general uncertainty principle is given (in Griffiths QM) as:

$$\sigma_A^2\sigma_B^2 \geq \left( \frac{1}{2i}\langle [\hat{A}, \hat{B}] \rangle \right)^2$$

Then,

\begin{align} \sigma_C^2\sigma_H^2 &\geq \left( \frac{1}{2i}\langle [\hat{C}, \hat{H}] \rangle \right)^2\\ \sigma_C\sigma_H &\geq \frac{1}{2i}\langle [\hat{C}, \hat{H}] \rangle = \frac{\hbar}{2i}\langle \hat{L}\rangle \end{align}

Now suppose that I make an observation of $$\hat{C}$$ with no uncertainty. Since I know nothing about $$\hat{L}$$, nor its expectation value, can I say whether I can determine $$\hat{H}$$ with no uncertainty?

Suppose $$\hat{L}$$ is location, the same as the traditional position operator. The commutation relation implies that in general $$\hat{C}$$ and $$\hat{H}$$ do not commute, yet surely $$\langle\hat{L}\rangle$$ can be zero, and so,

$$\sigma_C\sigma_H = 0$$

Which means that I can know $$\hat{H}$$ with no uncertainty. Is this a contradiction? Does the commutation relation described even make sense?

Yes the uncertainty relation does make sense and is correct in all cases. However it is of quite limited usefulness, if $$L$$ is not just a commuting number (as it is for momentum and position), because then the right hand side will depend on the state and does therefore not imply a fixed threshold. For $$\sigma_C$$ to be zero, you'll have to be in a very specific set of states and then the relation then simply implies, that for that set of states, $$\left< L \right>$$ will have to be zero as well.
Another point, where you have to be careful with your argumentation, is if you do consider operators with a continuous spectrum such as momentum and position: Those do not have proper eigenstates (that is, eigenstates that lie in the Hilbert space of normalizable states), so there is no state with $$\sigma_x = 0$$, however, if you select a series of states such that $$\sigma_x \to 0$$, then for that sequence of states you will always have $$\sigma_p \to \infty$$ in such a way, that for all points in the sequence $$\sigma_x \sigma_p \ge \frac 1 2 \hbar$$, and in this sense the inequality also holds in the limit.
$$[L_x, L_y] = i \hbar L_z$$
If we were to configure a quantum system in such a way that the expected value $$\langle L_z \rangle = 0$$, given a measure of one of the components, e.g. $$L_x$$, as $$L^2$$ can be determined in a compatible way with $$L_x$$, you could find $$L_y$$ via $$L^2 = L_x^2 +L_y^2$$, and since the system was constructed to have $$L_z = 0$$ there is no further uncertainty than that of the measurement.