Measuring different observables simultaneously with arbitrary precision

Question

I am trying to understand how different observables can be measured at the same time with arbitrary precision.

To check if I understand it I am using an example. Let's say we have as our first observable the Hamiltonian which has the potential $V(x, y, z) = z^2$ and we want to measure it simultaneously with another observable; the angular momentum on the z-axis: $L_z$.

For this to be possible, I am aware of the fact that both operators must share the same eigenstate and commute with each other as $[A,B]:=AB-BA=0$:

$$ [\hat{L_z}, \hat{H}]| \psi \rangle = \hat{L_z} \hat{H}|\psi \rangle - \hat{H} \hat{L_z}|\psi \rangle = 0$$

Would this be enough to assert that both energy and angular momentum can be measured at the same time with arbitrary precision?

Answer 1

These are the commutation equations between two or more operators:

$$[\hat{A},\hat{B}] = \hat{A} \hat{B}-\hat{B} \hat{A} = 0$$

$$[a\hat{A},\hat{B}] = a[\hat{A},\hat{B}]$$

$$[\hat{A}-\hat{B}, \hat{C}] = [\hat{A},\hat{C}] - [\hat{B},\hat{C}]$$

$$[\hat{A} \hat{B}, \hat{C}] = [\hat{A},\hat{C}] \hat{B} + \hat{A} [\hat{B},\hat{C}]$$

Where $\hat{A},\hat{B}, \hat{C}$ are operators and $a$ is a constant.

The canonical commutation relations are:

$$[r_i, p_j] = -[p_i, r_j] = i \hbar \delta_{ij}$$

$$[r_i, r_j] = [p_i, p_j] = 0$$

Although the following commutation relation is correct:

$$[\hat{L_z}, \hat{H}]| \psi \rangle = \hat{L_z} \hat{H}|\psi \rangle - \hat{H} \hat{L_z}|\psi \rangle = 0$$

Let's delve into it. The angular momentum of a particle (with respect to the origin) is given by:

$$L = r \times p$$

From which we can know the three components of it. In this case we are interested in the z component:

$$L_z = xp_y -yp_x$$

Then:

$$[L_z, H] = [xp_y -yp_x, H]$$

Knowing these properties:

$$[\hat{A} \hat{B}, \hat{C}] = [\hat{A},\hat{C}] \hat{B} + \hat{A} [\hat{B},\hat{C}]$$

$$[r_i, p_j] = -[p_i, r_j] = i \hbar \delta_{ij}$$

We can say:

$$[L_z, H] = [xp_y -yp_x, H] = x [p_y,H] + [x,H] p_y - y [p_x,H] - [y,H] p_x = 0 $$

Finally, as $[L_z, H] = 0$ (i.e $L_z$ and $H$ commute), we can assert that both observables can be simultaneously measured with arbitrary precision.

Source: Introduction to Quantum Mechanics, David Griffiths.