![12. Which of the following operators commute? π΄ (π (π₯) = [π (π₯)]2Β Β π΅(π (π₯)) = π₯π(π₯)Β Β πΆ (π (π₯)) = ππ/ππ₯ a. A commutes with B b. B commutes with C c. A commutes with C d. No pair of these commute](https://www.letstalkacademy.com/wp-content/uploads/2025/12/slide_27.jpg)
12. Which of the following operators commute?
π΄ (π (π₯) = [π (π₯)]2Β Β π΅(π (π₯)) = π₯π(π₯)Β Β πΆ (π (π₯)) = ππ/ππ₯
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)]Β² β Squaring operator
- B(f) = x f(x) β Multiplication by x operator
- C(f) = df/dx β Differentiation operator
Option Analysis
A with B (Squaring vs Multiplication by x)
Compute A(Bf) = [x f(x)]Β² = xΒ² f(x)Β² and B(Af) = x [f(x)]Β². Both yield xΒ² fΒ² symbolically, but detailed check shows non-zero difference like x(x-1)fΒ² generally. They do not commute universally.
B with C (Multiplication vs Differentiation)
B(Cf) = x f'(x), C(Bf) = d/dx(xf) = f + x fβ. Commutator equals -f(x) β 0.
A with C (Squaring vs Differentiation)
A(Cf) = [f'(x)]Β², C(Af) = d/dx[fΒ²] = 2 f fβ. Commutator is (-2f + fβ) fβ β 0 generally.
Complete Commutator Table
| Pair | ABf or BCf etc. | BAf or CBf etc. | Commutator | Commutes? |
|---|---|---|---|---|
| A-B | xΒ² fΒ² | x fΒ² | Non-zero | No |
| B-C | x fβ | x fβ + f | -f | No |
| A-C | (fβ)Β² | 2 f fβ | 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.
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