Question Statement:
From the pairs of operators given below, identify the ones which commute. Here l
and j correspond to the orbital angular momentum and the total angular
momentum, respectively.
Options:
-
(A) l2, j2
-
(B) j2, j
-
(C) j2, l
-
(D) l, j
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.21)**
**1. Quantum Angular Momentum Algebra**:
- Orbital angular momentum $\vec{l}$ and total angular momentum $\vec{j} = \vec{l} + \vec{s}$.
- Fundamental commutation relations:
$[l_i, l_j] = i\hbar \epsilon_{ijk} l_k, \quad [j_i, j_j] = i\hbar \epsilon_{ijk} j_k, \quad [l_i, s_j] = 0$
**2. Step-by-Step Verification of Operator Pairs**:
- **Pair (A) $[l^2, j^2]$**:
$j^2 = l^2 + s^2 + 2\vec{l}\cdot\vec{s}$
Since $[l^2, l_i] = 0$ and $[l^2, s_i] = 0$, $l^2$ commutes with $\vec{l}\cdot\vec{s}$, so $[l^2, j^2] = 0$. **(A) COMMUTES**.
- **Pair (B) $[j^2, j_z]$**:
By canonical angular momentum algebra, total magnitude operator $j^2$ commutes with any of its components $j_z$: $[j^2, j_z] = 0$. **(B) COMMUTES**.
- **Pair (C) $[j^2, l_z]$**:
$[j^2, l_z] = [l^2 + s^2 + 2(l_x s_x + l_y s_y + l_z s_z), l_z]$
Since $[l_x, l_z] = -i\hbar l_y$ and $[l_y, l_z] = i\hbar l_x$, $[j^2, l_z] = 2(-i\hbar l_y s_x + i\hbar l_x s_y) \ne 0$. (C) DOES NOT COMMUTE.
- **Pair (D) $[l_z, j_z]$**:
$j_z = l_z + s_z \implies [l_z, j_z] = [l_z, l_z + s_z] = [l_z, l_z] + [l_z, s_z] = 0 + 0 = 0$
**(D) COMMUTES**.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, B, D)**.