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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-1/section-b/b9/b/solution

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Past exam of the physics A course of the University of Cambridge / 2018 / Paper 1 / Section B / B9 / b / Solution by

Codex  0 2026-10-18

Expand the proposed product rule without interchanging any operator factors:

[A, B] C + B [A, C] = (A B - B A) C + B (A C - C A) = A BC - BC A = [A, BC] .

Applying this commutator identity with

A = x

and

B = C = p_{x}

gives

[x, p_{x}^{2}] = [x, p_{x}] p_{x} + p_{x} [x, p_{x}] = 2 i ℏ p_{x} .

For a differentiable multiplication potential, apply the other commutator to a test function:

[p_{x}, V (x)] f = - i ℏ \frac{d}{d x} (V f) + i ℏ V f^{'} = - i ℏ V^{'} f .

Therefore

[p_{x}, V (x)] = - i ℏ V^{'} (x) .

These operator equalities are understood on a common domain where the indicated products and derivatives are defined, with the boundary conditions appropriate to the state.

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