 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-1/section-a/2/solution

[x, p] = i ℏ I

on a common invariant domain of the operator products. Write

A = p x

, so

x p = A + i ℏ I

. Associativity then gives

x p^{2} x = (x p) (p x) = (A + i ℏ I) A = A^{2} + i ℏ A,

whereas

p x^{2} p = (p x) (x p) = A (A + i ℏ I) = A^{2} + i ℏ A .

Therefore

x p^{2} x = p x^{2} p .

p = - i ℏ d / d x

f

gives

x (- ℏ^{2}) \frac{d ^{2} ( x f )}{d x ^{2}} = - ℏ^{2} (x^{2} f^{''} + 2 x f^{'}) = (- i ℏ) \frac{d}{d x} [x^{2} (- i ℏ f^{'})] .

This is the equality of two quadratic position-momentum orderings; it does not make arbitrary rearrangements of noncommuting linear operators interchangeable.

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