Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 33B / b

The unperturbed ground state is $|0,0⟩$ with energy $\hslash \omega$. Sincethe first-order energy correction vanishes and first-order nondegenerate perturbation theory gives

Put $g=\lambda /(4{\omega }^2)$. Normalization through order $g^2$ givesThusTracing over $B$ removes the cross terms. The reduced density operator of a weakly coupled oscillator pair is thereforewhose trace is one to the required order.

If the eigenvalues of a density operator are $p_j\ge 0$ with ${\sum }_j{p}_j=1$, thenwith equality exactly when one eigenvalue is one, namely for a pure state. In dimension $d$, Cauchy--Schwarz gives $\gamma \ge 1/d$, with equality for the maximally mixed state. On an infinite-dimensional space the universal lower bound is zero as an infimum.

The eigenvalues found in part (ii) give the purity of a density operatorThe displayed truncation lies between zero and one when $|\lambda |\le 2\sqrt{2}{\omega }^2$. Perturbation theory requires the stronger condition $|\lambda |\ll {\omega }^2$; the exact coupled oscillator is stable only for $|\lambda |<{\omega }^2$. The reduction in purity records entanglement between $A$ and $B$.