![12. Which of the following operators commute? 𝐴 (𝑓 (𝑥) = [𝑓 (𝑥)]2 𝐵(𝑓 (𝑥)) = 𝑥𝑓(𝑥) 𝐶 (𝑓 (𝑥)) = 𝑑𝑓/𝑑𝑥 a. A commutes with B b. B commutes with C c. A commutes with C d. No pair of these commute](data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTkyMCAxMDgwIiB3aWR0aD0iMTkyMCIgaGVpZ2h0PSIxMDgwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==)
12. Which of the following operators commute?
𝐴 (𝑓 (𝑥) = [𝑓 (𝑥)]2 𝐵(𝑓 (𝑥)) = 𝑥𝑓(𝑥) 𝐶 (𝑓 (𝑥)) = 𝑑𝑓/𝑑𝑥
a. A commutes with B
b. B commutes with C
c. A commutes with C
d. No pair of these commute
Commutator Definition
Two operators P and Q commute when PQ = QP, or [P,Q] = 0. Apply both orders to a test function and check equality. Multiplication and differentiation often fail to commute due to product rule effects.
Operator Definitions
- A(f) = [f(x)]² – Squaring operator
- B(f) = x f(x) – Multiplication by x operator
- C(f) = df/dx – Differentiation operator
Option Analysis
A with B (Squaring vs Multiplication by x)
Compute A(Bf) = [x f(x)]² = x² f(x)² and B(Af) = x [f(x)]². Both yield x² f² symbolically, but detailed check shows non-zero difference like x(x-1)f² generally. They do not commute universally.
B with C (Multiplication vs Differentiation)
B(Cf) = x f'(x), C(Bf) = d/dx(xf) = f + x f’. Commutator equals -f(x) ≠ 0.
A with C (Squaring vs Differentiation)
A(Cf) = [f'(x)]², C(Af) = d/dx[f²] = 2 f f’. Commutator is (-2f + f’) f’ ≠ 0 generally.
Complete Commutator Table
| Pair | ABf or BCf etc. | BAf or CBf etc. | Commutator | Commutes? |
|---|---|---|---|---|
| A-B | x² f² | x f² | Non-zero | No |
| B-C | x f’ | x f’ + f | -f | No |
| A-C | (f’)² | 2 f f’ | Non-zero | No |
CSIR NET Solution
Correct answer: d. No pair of these commute. Verify with smooth functions like f(x) = sin x; numerical checks confirm differences. Master this for quantum operator questions in exams.
