# Measuring different observables simultaneously with arbitrary precision

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?

• It looks like you're trying to take scalar products of operators. Please add state vectors to your scalar products or make a distinction between states and the angular momentum operator. To say that two operators commute you don't need scalar products. Operators A, B commute if $AB=BA$ or equivalently $[A,B]:=AB-BA=0$. A state in which both observables can be measured with certain result is any common eigenvector of the two operators. – Adomas Baliuka Oct 11 '18 at 9:12
• @AdomasBaliuka So using scalar products to prove both operators commute is not the right method? Then how can I do so? – JD_PM Oct 11 '18 at 9:37
• You seem to be confused about the notation and distinction between state vectors and operators. Read about the axiomatic formulation of quantum mechanics from textbooks such as Shankar, Ballentine or Nakahara (this one is on quantum computing). Operators act on states $\hat L_z |\phi \rangle$. You can multiply two operators and have them act on states $\hat L_y \hat L_z |\phi \rangle$. The multiplication of two operators is generally a new operator $\hat A = \hat L_y \hat L_z \neq \hat L_z \hat L_y = \hat B$. – GodotMisogi Oct 11 '18 at 11:51

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.