Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 53 / 1

Take the vectors in the quantum state ensemble to be normalized, with $p_i\ge 0$ and ${\sum }_i{p}_i=1$. The corresponding density matrix isEach rank-one projector is a self-adjoint operator, so ${\rho }^{†}=\rho$. For every $|v⟩$ in the Hilbert space,proving that $\rho$ is positive semidefinite. The trace is $\mathrm{Tr}\rho ={\sum }_i{p}_i⟨{\psi }_i|{\psi }_i⟩=1$.

Conversely, diagonalize a self-adjoint operator $\rho$ in an orthonormal eigenbasis, writing $\rho ={\sum }_j{\lambda }_j|e_j⟩⟨e_j|$. Its positive semidefiniteness implies ${\lambda }_j=⟨e_j|\rho |e_j⟩\ge 0$, and its unit trace gives ${\sum }_j{\lambda }_j=1$. Thus $\{{\lambda }_j,|e_j⟩\}$ is a quantum state ensemble with precisely this density matrix. Terms with zero eigenvalue may be omitted. This argument includes rank-one pure states; an ensemble representation need not describe a genuinely mixed state.

The stated condition is the definition of an extreme point in the convex set of density matrices. Suppose first that $\rho =|\psi ⟩⟨\psi |$ with $\Vert \psi \Vert =1$, and consider any convex combination $\rho ={\sum }_i{a}_i{\rho }_i$ with all $a_i>0$. For every $|w⟩$ orthogonal to $|\psi ⟩$,Every term is nonnegative, so $⟨w|{\rho }_i|w⟩=0$ for each $i$. To see that this implies ${\rho }_{i}w=0$, expand ${\rho }_i={\sum }_j{\mu }_j|u_j⟩⟨u_j|$, ${\mu }_j\ge 0$. The equality ${\sum }_j{\mu }_j|⟨u_j|w⟩{|}^2=0$ implies ${\mu }_j⟨u_j|w⟩=0$ for every $j$, which kills ${\rho }_{i}w$. Thus each density matrix ${\rho }_i$ vanishes on ${\psi }^{\perp }$. By self-adjointness its range lies in the line $\mathbb{C}\psi$, so ${\rho }_i=c_i|\psi ⟩⟨\psi |$. Its unit trace forces $c_i=1$, establishing the required purity.

Conversely, suppose $\rho$ is not rank one. In its spectral decomposition, at least two eigenvalues are positive. Select a normalized eigenvector $|e_1⟩$ with positive eigenvalue ${\lambda }_1$; then $0<{\lambda }_1<1$. DefineThe remaining eigenvalues show that $\sigma$ is a density matrix. The decomposition $\rho ={\lambda }_1|e_1⟩⟨e_1|+(1-{\lambda }_1)\sigma$ is a convex combination of two different density matrices, neither equal to $\rho$. This violates the given purity condition. Hence the extreme points of the density-operator state space obey

Let $R=|\psi ⟩⟨\psi |$ and choose an orthonormal basis $\{|e_{\mu }⟩\}$ of ${\mathcal{H}}_2$. The reduced density matrix on $S_1$ is the partial traceHere $I_1$ is the identity on ${\mathcal{H}}_1$, and the bras and kets in the sum contract only the second tensor factor. Equivalently, if $|\psi ⟩={\sum }_{a,\mu }{c}_{a\mu }|f_a⟩\otimes |e_{\mu }⟩$, then $({\rho }_1{)}_{ab}={\sum }_{\mu }{c}_{a\mu }{c}_{b\mu }^{\ast }$. The definition is independent of the chosen basis: it is characterized by $\mathrm{Tr}(M{\rho }_1)=\mathrm{Tr}[(M\otimes I_2)R]$ for every operator $M$ on ${\mathcal{H}}_1$.

The generalized measurement postulate takes $A_i$ to be operators on ${\mathcal{H}}_2$ satisfying ${\sum }_i{A}_i^{†}{A}_i=I_2$. The Born rule and the conditional state update giveIf the outcome is not supplied to $S_1$, the appropriate post-measurement density matrix is $R^{\prime }={\sum }_i(I_1\otimes A_i)R(I_1\otimes A_i^{†})$. For any local operator $M$, cyclicity of the full trace givesSince this holds for every $M$, $\boxed{{\mathrm{Tr}}_2{R}^{\prime }={\rho }_1}$. This is the no-communication theorem.

The invariance concerns the nonselective measurement in quantum mechanics. A conditional reduced density matrix ${\rho }_{1|i}={\mathrm{Tr}}_2|{\psi }_i⟩⟨{\psi }_i|$ can change, although ${\sum }_i{p}_i{\rho }_{1|i}={\rho }_1$. For example, measuring one half of a Bell pair in the computational basis changes the other half conditionally to $|0⟩$ or $|1⟩$; before learning the outcome, its density matrix remains $I/2$. Thus distant measurement in quantum measurements cannot transmit a controllable signal without classical communication. This quantum no-signalling is compatible with special relativity, even though some entangled states have correlations incompatible with a local hidden-variable theory.