 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2b/section-b/35/b/ii/solution

The azimuthal derivative is

\partial_{ϕ} ψ = - i ψ

, so

\hat{J}_{z} ψ = (- i ℏ) (- i) ψ = - ℏ ψ

. For the polar part of the angular Laplacian,

\frac{1}{s i n θ} \partial_{θ} (sin θ \partial_{θ} ψ) = \frac{c o s ^{2} θ - s i n ^{2} θ}{s i n θ} e^{- i ϕ}, \frac{1}{s i n ^{2} θ} \partial_{ϕ}^{2} ψ = - \frac{1}{s i n θ} e^{- i ϕ} .

Their sum is

- 2 sin θ e^{- i ϕ} = - 2 ψ

, hence

\hat{J}^{2} ψ = 2 ℏ^{2} ψ

. Comparing these two eigenvalues with

M ℏ

and

J (J + 1) ℏ^{2}

gives $J = 1$ , $M = - 1$ . The sign of

M

follows from both minus signs in the

\hat{J}_{z}

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