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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 33C / b / Solution

Contract the two Levi-Civita symbols:

L_{i} L_{i} = ε_{ijk} ε_{i ℓ m} X_{j} P_{k} X_{ℓ} P_{m} = (δ_{j ℓ} δ_{km} - δ_{jm} δ_{k ℓ}) X_{j} P_{k} X_{ℓ} P_{m} .

Moving momenta past positions with

P_{k} X_{ℓ} = X_{ℓ} P_{k} - i ℏ δ_{k ℓ}

gives the squared orbital angular momentum in position and momentum operators:

L^{2} = X^{2} P^{2} - (X \cdot P)^{2} + i ℏ X \cdot P .

Therefore

c_{1} = i ℏ, c_{2} = 1, c_{3} = - 1 .

In position representation,

P = - i ℏ\nabla, X \cdot P = - i ℏ r \partial_{r},

and

\nabla^{2} = \partial_{r}^{2} + \frac{2}{r} \partial_{r} + \frac{1}{r ^{2}} \nabla_{S^{2}}^{2} .

Also

(r \partial_{r})^{2} = r^{2} \partial_{r}^{2} + r \partial_{r}

. Substitution into the Cartesian identity gives

L^{2} = - ℏ^{2} r^{2} \nabla^{2} + ℏ^{2} (r \partial_{r})^{2} + ℏ^{2} r \partial_{r} = - ℏ^{2} \nabla_{S^{2}}^{2},

because every radial derivative cancels. Hence

L^{2} = - ℏ^{2} \nabla_{S^{2}}^{2},

the spherical Laplacian from orbital angular momentum.

Solved by gpt-5.6-sol high.

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