Past exam of the theoretical physics II course of the University of Cambridge / 2019 / 4

A quantum state ensemble of normalized states $|{\psi }_i⟩$, with probabilities $p_i\ge 0$ and ${\sum }_i{p}_i=1$, has density operatorIt is a positive semidefinite Hermitian operator with trace one, and the Born rule assigns a measurement outcome with projector $\Pi$ the probability $\mathrm{Tr}(\rho \Pi )$.

A pure state has $\rho =|\psi ⟩⟨\psi |$, hence ${\rho }^2=\rho$ and purity of a density operator $\mathrm{Tr}({\rho }^2)=1$. Its only nonzero eigenvalue is one. A mixed state has at least two nonzero eigenvalues and $\mathrm{Tr}({\rho }^2)<1$. The spectral theorem proves the converse criteria, since the nonnegative eigenvalues sum to one. Different quantum state ensembles may give the same density operator; the mere presence of several terms in a preparation list does not make the state mixed if all positive-weight terms represent the same quantum state ray.

Taking the outer product for A and averaging the two orthogonal projections for B givesThe off-diagonal entries in ${\rho }_A$ retain quantum coherence between the $z$-basis states. They are absent from B's classical quantum state ensemble. In particular, $\mathrm{Tr}{\rho }_A^2=1$ and $\mathrm{Tr}{\rho }_B^2=1/2$: A is a pure state, B a mixed state.

The projective measurement probabilities are $P({\sigma }_j=\pm 1)=\mathrm{Tr}[\rho (I\pm {\sigma }_j)/2]$. Both density operators have zero $z$-component of their Bloch vector, soTheir full ${\sigma }_z$ outcome distributions, not merely their mean values, coincide. Along $x$, however,Thus repeated spin measurement along $x$ distinguishes the preparations statistically. A single negative outcome excludes A, while a single positive outcome is compatible with either; under B the probability of $N$ independent positive outcomes is $2^{-N}$.

The opposite $x$-axis pure states have projectors $(I\pm {\sigma }_x)/2$. Averaging them givesTherefore C and B cannot be distinguished by any measurement on the atoms alone. Every POVM outcome has the same probability $\mathrm{Tr}(E\rho )$ for the two preparations; this is the operational equivalence of different quantum state ensembles. Accessible preparation labels or additional correlations would be extra information. C can be distinguished from A by the $x$-axis spin measurement in part (c).

The normalized singlet state has column vector $(0,1,-1,0{)}^{\mathsf{T}}/\sqrt{2}$ in the specified product basis. Its outer product isIt obeys ${\rho }_D^2={\rho }_D$ and has eigenvalues $(1,0,0,0)$, hence purity of a density operator one. The joint two-atom system is therefore a pure state, even though the quantum entanglement prevents assigning a separate pure spin state to each atom.

Taking the partial trace over the left atom in the singlet state gives the right-hand reduced density matrixThus every measurement on the right stream alone has the same statistics as B or C. The joint pure state is compatible with locally mixed states; the correlations reside in the full density operator.

The reduced density matrix has two eigenvalues equal to $1/2$. Its Von Neumann entropy, which is the entanglement entropy of the bipartite pure state, isThis is one bit when base-two logarithms are used. It is the maximum entanglement entropy for a pair of two-dimensional systems, while the joint Von Neumann entropy of the singlet state remains zero.

Take the two magnetic moments along $z$ to be $\pm \mu$, so $H=-\mu B_z{\sigma }_z$ and the two energies are $\mp \mu B_z$. The quantum Gibbs state at inverse temperature $\beta =1/(k_{B}T)$ isIts Bloch vector is $(0,0,\tanh \beta \mu B_z)$. At $B_z=0$, or in the high-temperature limit, it approaches $I/2$; at $T\to 0^{+}$ with nonzero $\mu B_z$, it approaches the lower-energy pure state. Here $\mu$ is the magnitude or signed coefficient multiplying ${\sigma }_z$ in the magnetic Hamiltonian, rather than a moment with an additional unstated factor of two.