Past exam of the introduction to quantum information course of the University of Oxford / 2017 (A14687W1)

Source: released examination PDF, bundle pages 103–106.

Solutions to all surviving material, Question 3. TODO: Questions 1 and 2 cannot be answered from this release: printed pages 2–3 are blank. No missing question text or subpart structure is inferred.

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The Pauli matrices are Hermitian matrices with zero trace, so a real Bloch vector makes $\varrho$ Hermitian and gives $\mathrm{tr}\varrho =1$. Their Pauli multiplication identity implies $(\mathbf{s}\cdot \mathbf{\sigma }{)}^2=|\mathbf{s}{|}^{2}I$. Consequently the two eigenvalues of $\mathbf{s}\cdot \mathbf{\sigma }$ are $\pm |\mathbf{s}|$, including the zero-vector case, andA density matrix must also be positive semidefinite. The displayed eigenvalues are nonnegative exactly when $\boxed{|\mathbf{s}|\le 1}$. Thus real coordinates alone are insufficient: the allowed vectors fill the Bloch ball. Its boundary consists of pure states, with eigenvalues one and zero; its center is the maximally mixed state $I/2$.

For any column vector $v$ and positive semidefinite input $\varrho$, the Kraus representation givesThus the output is positive semidefinite. The cyclic property of the trace and completeness of the Kraus operators giveMore strongly, tensoring with an arbitrary ancilla replaces $A_k$ by $A_k\otimes I$, so the same argument proves complete positivity rather than only positivity on single-system density matrices.

If $\mathcal{F}$ has Kraus operators $B_j$, the composition of quantum channels hasThis is again a trace-preserving Kraus representation. The composition of completely positive maps is completely positive even without the trace-preserving assumption; in the general case, the product Kraus operators still establish complete positivity.

Conjugation by one Pauli matrix fixes that same matrix and reverses the other two. For example, $XXX=X$, $YXY=-X$, and $ZXZ=-X$. Applying this to each component of the Bloch vector givesHence the quantum depolarizing channel isThe Kraus operators $\sqrt{1-p}I$ and $\sqrt{p/3}{\sigma }_j$ establish complete positivity for $0\le p\le 1$. At $p=3/4$ every state becomes maximally mixed. For $p>3/4$ the factor is negative: it reverses the vector as well as reducing its length. Its most negative value is $-1/3$, consistent with the complete-positivity range of the isotropic qubit map.

The universal NOT gate is the isotropic qubit map with factor $-1$. It sendsFor $|\mathbf{s}|\le 1$, the output has the same two nonnegative eigenvalues as the input, exchanged, and still has trace one. Thus it maps every single-qubit density matrix to a density matrix. Its linear extension to all operators is $\mathcal{N}(A)=\mathrm{tr}(A)I-A$, not $I-A$ for arbitrary unnormalized $A$. Positivity on one qubit does not establish complete positivity when an ancilla is present.

The input is the Bell state $\rho =|{\Phi }^{+}⟩⟨{\Phi }^{+}|$, where $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. Acting on the first qubit reverses the signs of all three terms containing its Pauli matrices, so in computational basis order $00,01,10,11$ the output isIts eigenvalues are $\boxed{-1/2,1/2,1/2,1/2}$: in particular, its expectation value in $|{\Phi }^{+}⟩$ is $-1/2$. It therefore fails to be positive semidefinite, although its trace remains one. This single entangled state proves that the universal NOT gate is positive but not completely positive, so it cannot be a deterministic physical quantum channel on an arbitrary input.