 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-3/8b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

For square-integrable wavefunctions with the stated decay, multiplication by the real coordinate

x

is Hermitian because

⟨ ψ, x χ ⟩ = ⟨ x ψ, χ ⟩ .

For

p_{x} = - i ℏ \partial_{x}

⟨ ψ, p_{x} χ ⟩ = ⟨ p_{x} ψ, χ ⟩ - i ℏ \int_{- \infty}^{\infty} [\overline{ψ} χ]_{x = - \infty}^{x = \infty} d y,

and the boundary term vanishes. The same arguments apply to

y

and

p_{y}

F

and

G

are Hermitian on a common suitable domain, then

(FG)^{†} = GF

, so

[\frac{1}{2} (FG + GF)]^{†} = \frac{1}{2} (FG + GF) .

Because

x

commutes with

p_{y}

and

y

with

p_{x}

L = x p_{y} - y p_{x}

is Hermitian. Using

p_{x} x = x p_{x} - i ℏ

and its

y

analogue,

D = \frac{1}{2} (x p_{x} + p_{x} x + y p_{y} + p_{y} y) = - i ℏ (x \partial_{x} + y \partial_{y} + 1),

D

is Hermitian as well.

Finally set

R = x \partial_{y} - y \partial_{x}

and

S = x \partial_{x} + y \partial_{y}

. Direct evaluation on a smooth test function gives

[R, S] = 0

: rotations commute with radial dilations. The scalar term in

D

also commutes with everything, and therefore

[L, D] = 0 .

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