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Past exam of the mathematics course of the University of Cambridge

2024

Paper 2

A <density operator> satisfies
$$
\rho=\rho^\dagger,\qquad \rho\succeq0,\qquad\operatorname{tr}\rho=1.
$$
An $N\times N$ Hermitian <matrix> has $N^2$ real parameters, and the trace condition removes one, leaving
$$
\boxed{N^2-1}
$$
real free parameters.

If $A_H(t)=U(t)^\dagger A_SU(t)$, equality of <expectation values> requires
$$
\operatorname{tr}(\rho_HU^\dagger A_SU)
=\operatorname{tr}(U\rho_HU^\dagger A_S).
$$
Therefore
$$
\boxed{\rho_S(t)=U(t)\rho_HU(t)^\dagger}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 35B / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 35B / a / Solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 35B / a / Solution

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