General commutation question

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?

Answer 1

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.

Answer 2

This commutation relation does make sense. I guess you are inspired by the commutation relation of the angular momentum, and taking it makes a clear example:

The components of the angular momentum follow the commutation relation

$$[L_x, L_y] = i \hbar L_z $$

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.