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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 4

Contract the two Levi-Civita symbols:
$$
\begin{aligned}
L_iL_i
&=\varepsilon_{ijk}\varepsilon_{i\ell m}
X_jP_kX_\ell P_m\\
&=(\delta_{j\ell}\delta_{km}-\delta_{jm}\delta_{k\ell})
X_jP_kX_\ell P_m.
\end{aligned}
$$
Moving momenta past positions with
$P_kX_\ell=X_\ell P_k-i\hbar\delta_{k\ell}$ gives the <squared orbital angular momentum in position and momentum operators>:
$$
\boxed{
L^2=X^2P^2-(X\cdot P)^2+i\hbar X\cdot P}.
$$
Therefore
$$
\boxed{c_1=i\hbar,\qquad c_2=1,\qquad c_3=-1}.
$$

In position representation,
$$
P=-i\hbar\nabla,
\qquad
X\cdot P=-i\hbar r\partial_r,
$$
and
$$
\nabla^2=\partial_r^2+\frac2r\partial_r+\frac1{r^2}\nabla_{S^2}^2.
$$
Also $(r\partial_r)^2=r^2\partial_r^2+r\partial_r$. Substitution into the Cartesian identity gives
$$
\begin{aligned}
L^2
&=-\hbar^2r^2\nabla^2
+\hbar^2(r\partial_r)^2
+\hbar^2r\partial_r\\
&=-\hbar^2\nabla_{S^2}^2,
\end{aligned}
$$
because every radial derivative cancels. Hence
$$
\boxed{L^2=-\hbar^2\nabla_{S^2}^2},
$$
the <spherical Laplacian from orbital angular momentum>.

Solution

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

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

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

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