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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 33B / b / Solution

For

∣ J, J ⟩

, tracing out subsystem 2 gives the reduced density matrix

ρ_{1} = ∣ j_{1}, j_{1} ⟩ ⟨ j_{1}, j_{1} ∣,

which is pure, so its entanglement entropy is zero.

For

∣ J, J - 1 ⟩

, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and gives

ρ_{1} = \frac{j _{1}}{J} ∣ j_{1}, j_{1} - 1 ⟩ ⟨ j_{1}, j_{1} - 1∣ + \frac{j _{2}}{J} ∣ j_{1}, j_{1} ⟩ ⟨ j_{1}, j_{1} ∣.

Its nonzero eigenvalues are

j_{1} / J

and

j_{2} / J

, so the Von Neumann entropy is

S = - \frac{j _{1}}{J} lo g \frac{j _{1}}{J} - \frac{j _{2}}{J} lo g \frac{j _{2}}{J} .

When

j_{1} = j_{2}

, both eigenvalues are

1/2

and

S = lo g 2

, the maximal entropy for a rank-two reduced state.

Solved by gpt-5.6-sol high.

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