Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 65 / 5

The Bell basis consists ofLet $Z={\sigma }_z$, $X={\sigma }_x$. The Pauli operators $Z\otimes Z$ and $X\otimes X$ commute: each of the two local anticommutations contributes a minus sign, so they cancel. Their eigenvalue table isTo make the bit values literally zero or one rather than signed eigenvalues, the commuting Bell parity and phase observables areThey distinguish all four states jointly. The parity bit records whether the computational bits agree; the phase bit records the relative sign.

Apply a Hadamard gate to qubit $A$, followed by a controlled-NOT gate with $A$ controlling $B$. The resulting Bell-basis conversion circuit is $U=\mathrm{CNOT}(H\otimes I)$ and givesThus $00,01,10,11$ map respectively to ${\Phi }^{+},{\Psi }^{+},{\Phi }^{-},{\Psi }^{-}$. The first input bit becomes the phase bit and the second becomes the parity bit.

Both gates are self-inverse, but inversion of their product reverses the order:So the same two gates convert the Bell basis back to the computational basis when applied in reverse order: first the controlled-NOT, then the Hadamard. Using the forward order twice does not generally work, since these gates do not commute. For example, $U|{\Phi }^{+}⟩=(|00⟩+|01⟩-|10⟩+|11⟩)/2$, which is not a computational-basis state.

Write a bipartite pure state in fixed product bases as $|\psi ⟩={\sum }_{ij}{M}_{ij}|i⟩|j⟩$. The Schmidt decomposition theorem, equivalently the singular value decomposition of $M$, says that its Schmidt rank equals $\mathrm{rank}M$. A branch of local operations is a product operator $A\otimes B$, so its coefficient matrix isThe elementary rank inequality $\mathrm{rank}(AM{B}^{\mathsf{T}})\le \mathrm{rank}M$ proves Schmidt-rank contraction under product operators. Normalizing a nonzero branch does not change matrix rank.

For an adaptive LOCC protocol, a complete classical transcript selects one local Kraus operator at each stage. Multiplying the local operators along that transcript still gives one product $A_t\otimes B_t$. ConsequentlyClassical communication changes which product operator is chosen, not this rank bound. If the overall output is a pure state, every nonzero branch must be proportional to that same vector, so its Schmidt rank also cannot increase. If the outcome is discarded and the output is mixed, pure-state Schmidt rank is not defined; the induced decomposition instead shows that its Schmidt number is at most the original rank. This includes probabilistic filtering as well as deterministic pure-state conversion.

The final printed direction is reversed under the majorization convention stated in the same part. A completely depolarizing unital quantum channel sends a pure qubit state to $I/2$, giving spectra $r=(1,0)$ and $s=(1/2,1/2)$. The proposed $r\prec s$ fails already at the first partial sum, since $1>1/2$. The valid conclusion isHere is its general proof. Choose eigenbases $\rho ={\sum }_i{r}_i|u_i⟩⟨u_i|$ and $\sigma ={\sum }_j{s}_j|v_j⟩⟨v_j|$, and define the eigenvalue mixing matrix of a unital quantum channelIts entries are nonnegative by positivity. Trace preservation gives ${\sum }_j{A}_{ji}=1$ for every column. Unitality gives ${\sum }_i{A}_{ji}=⟨v_j|\Lambda (I)|v_j⟩=1$ for every row. Thus $A$ is a doubly stochastic matrix, and linearity givesThe supplied doubly stochastic characterization of majorization then yields $s\prec r$, completing the intended corrected result. This proves spectral majorization under a unital quantum channel. A unital quantum channel can make a spectrum more uniform; it cannot generally make the output spectrum majorize the input. The counterexample resolves the false printed assertion rather than treating the reversed inequality as a convention change.