Quantum Commutator: Angular and Linear Momentum

This solution calculates the commutator $[L_x, p_y]$ for a particle, involving the angular momentum operator component ($L_x$) and the linear momentum operator component ($p_y$).

Commutator Calculation Steps

1. Define Angular Momentum Component: The x-component of the angular momentum operator $\vec{L}$ is defined as $\vec{L} = \vec{r} \times \vec{p}$. Its x-component is:

   $ L_x = y p_z - z p_y $

2. Set up the Commutator: We need to compute the commutator $[L_x, p_y]$. Substitute the definition of $L_x$:

   $ [L_x, p_y] = [y p_z - z p_y, p_y] $

3. Apply Linearity: Use the property $[A - B, C] = [A, C] - [B, C]$ to separate the terms:

   $ [L_x, p_y] = [y p_z, p_y] - [z p_y, p_y] $

4. Evaluate Individual Commutators: Use the standard quantum mechanical commutation relations: $[p_i, p_j] = 0$, $[x_i, p_j] = i\hbar \delta_{ij}$, and the product rule $[AB, C] = A[B, C] + [A, C]B$. Here, indices correspond to coordinates (1=x, 2=y, 3=z).

   • For the first term, $[y p_z, p_y]$:

     $ [y p_z, p_y] = y [p_z, p_y] + [y, p_y] p_z $

     We know $[p_z, p_y] = 0$ (commutator of different momentum components) and $[y, p_y] = [x_2, p_2] = i\hbar \delta_{22} = i\hbar$.

     $ [y p_z, p_y] = y(0) + (i\hbar) p_z = i\hbar p_z $

   • For the second term, $[z p_y, p_y]$:

     $ [z p_y, p_y] = z [p_y, p_y] + [z, p_y] p_y $

     We know $[p_y, p_y] = 0$ (commutator of the same momentum component with itself) and $[z, p_y] = [x_3, p_2] = i\hbar \delta_{32} = 0$ (different coordinate and momentum components).

     $ [z p_y, p_y] = z(0) + (0) p_y = 0 $

5. Combine Results: Substitute the evaluated individual commutators back into the equation from Step 3:

   $ [L_x, p_y] = (i\hbar p_z) - (0) = i\hbar p_z $

The commutator $[L_x, p_y]$ is equal to $i\hbar p_z$.