Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 66 (Quantum Information Theory)

In the state-picture convention, a Stinespring representation of a completely positive map $\mathcal{N}:\mathcal{L}({\mathcal{H}}_Q)\to \mathcal{L}({\mathcal{H}}_{Q^{\prime }})$ consists of an auxiliary Hilbert space ${\mathcal{H}}_E$ and a linear map $V:{\mathcal{H}}_Q\to {\mathcal{H}}_{Q^{\prime }}\otimes {\mathcal{H}}_E$ such thatThe partial trace discards the environment. For example, from a Kraus representation $\mathcal{N}(X)={\sum }_j{K}_{j}X{K}_j^{†}$, take $V={\sum }_j{K}_j\otimes |j{⟩}_E$. The adjoint, or observable-picture, form is ${\mathcal{N}}^{†}(Y)=V^{†}(Y\otimes I_E)V$.

A general completely positive map does not require $V$ to be an linear isometry of Hilbert spaces. If $\mathcal{N}$ is trace preserving, then $V^{†}V={\sum }_j{K}_j^{†}{K}_j=I$, so $V$ is an linear isometry of Hilbert spaces. This is the Stinespring dilation of a quantum channel, which will be used in the data-processing proof.

By spectral decomposition, the Von Neumann entropy is the Shannon entropy of the eigenvalues:Throughout, logarithms have base two, so the Von Neumann entropy and Shannon entropy are measured in bits. The continuous convention $0{\log }_{2}0=0$ includes zero eigenvalues.

For a bipartite density operator ${\rho }_{AB}$, let $S(A{)}_{\rho }=S({\rho }_A)$, with ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$, and similarly for other systems. The quantum conditional entropy and coherent information areUnlike classical conditional entropy, quantum conditional entropy can be negative. For a Bell state, $S(AB)=0$ and $S(B)=1$, giving $H(A|B)=-1$ and coherent information equal to one bit.

Use the Schmidt decomposition of the bipartite pure state:The two reduced density matrices, obtained by partial trace, areTheir nonzero eigenvalues are identical, even if the two ambient dimensions differ. Zero eigenvalues contribute no Von Neumann entropy, soThis common Von Neumann entropy is the entanglement entropy of the bipartite pure state.

Use two results: the isometric Stinespring dilation of a quantum channel, and Strong subadditivity of Von Neumann entropy. Let $V:B\to B^{\prime }E$ dilate the given operation, and defineAn linear isometry of Hilbert spaces preserves the nonzero eigenvalues, so $S(B^{\prime }E{)}_{\omega }=S(B{)}_{\rho }$ and $S(A{B}^{\prime }E{)}_{\omega }=S(AB{)}_{\rho }$. Subtracting the two coherent information expressions givesThe last inequality is Strong subadditivity of Von Neumann entropy, in the form $S(A{B}^{\prime })+S(B^{\prime }E)\ge S(B^{\prime })+S(A{B}^{\prime }E)$. Thus the data-processing inequality for coherent information isThe lost coherent information is precisely the quantum conditional mutual information between the reference $A$ and discarded environment $E$, conditional on the retained output $B^{\prime }$.

Take a purification of a density operator $|\Psi {⟩}_{ABCD}$ of ${\rho }_{ABC}$. Apply Strong subadditivity of Von Neumann entropy to the reduced state on $ACD$, with $A$ the conditioning system:Complementary subsystems of a pure state have the same Von Neumann entropy, by the Schmidt decomposition. Therefore $S(AD)=S(BC)$ and $S(ACD)=S(B)$. Substitution givesThis is weak monotonicity of quantum entropy. The purification of a density operator is only a proof device; no purity assumption is imposed on ${\rho }_{ABC}$.

Use superdense coding. For the two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies $Z^a{X}^b$ to her half of the shared Bell state, where $X,Z$ are the Pauli X gate and Pauli Z gate. The encoded states areThese four Bell states are orthonormal. Alice then transmits her one qubit to Bob, who now holds both qubits. A Bell-basis measurement identifies the message with probability one.

More explicitly, Bob can apply a CNOT gate with the received qubit as control and his original qubit as target, followed by a Hadamard gate on the control. The CNOT gate sends the encoded state toand the Hadamard gate sends this to $|a⟩\otimes |b⟩$. Computational-basis projective measurements therefore read out $a,b$ exactly. One transmitted qubit, assisted by one previously shared Bell pair, communicates two classical bits.

Use one-bit remote preparation of real qubit states. Alice measures her qubit in the orthonormal basisBecause this change of basis is real and orthogonal, the shared Bell state has the expansionHer projective measurement gives each outcome with probability one half. She sends $m=0$ for the first outcome and $m=1$ for the second. Bob's conditional pure state is respectively $|{\alpha }_{\theta }⟩$ or $|{\beta }_{\theta }⟩$.

For $m=0$ Bob does nothing. For $m=1$ he applies the fixed unitary operatorwhere $Y$ is the Pauli Y gate. Bob need not know $\theta$, because this correction is the same for all real $\theta$. Both outcomes produce the exact density operator $|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|$, using one classical bit and only local operations.

This remote state preparation consumes the shared entangled state. Before receiving the bit, Bob has the mixture $(|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|+|{\beta }_{\theta }⟩⟨{\beta }_{\theta }|)/2=I/2$, independent of $\theta$, as required by quantum no-signalling. The real-state restriction is what permits a fixed unitary correction of the orthogonal outcome.

The positive part of a Hermitian operator and negative part of a Hermitian operator are the positive operatorsThus $X=X_{+}-X_{-}$ and $X_{+}{X}_{-}=0$. In this convention the negative part itself is nonnegative. This is the positive-negative decomposition of the Hermitian operator.

The operator absolute value is defined by the unique positive operator square rootThe scalar identity $|\lambda |=\max (\lambda ,0)+\max (-\lambda ,0)$ applied in the spectral decomposition proves

For any linear operator, the trace norm is $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}$, the sum of its singular values. For a Hermitian operator,The Holevo–Helstrom theorem concerns minimum-error discrimination of two density operators $\rho ,\sigma$. If their prior probabilities are $p$ and $1-p$, the maximum success probability over all quantum measurements isAn optimal binary POVM declares $\rho$ on the positive spectral subspace of $p\rho -(1-p)\sigma$ and declares $\sigma$ on its negative spectral subspace; zero-eigenvalue vectors can be assigned either way. For equal priors the formula becomes $P_{\mathrm{succ}}^{\ast }=\frac{1}{2}+\frac{1}{4}\Vert \rho -\sigma {\Vert }_1$.

The constraints $-I\le T\le I$ are in the order of Hermitian operators. They imply $-1\le ⟨{\alpha }_i|T|{\alpha }_i⟩\le 1$. Using the spectral decomposition of $X$,Choose $T=\mathrm{sgn}(X)$, with eigenvalues $+1$ on the positive spectral subspace, $-1$ on the negative spectral subspace, and zero on the kernel. It satisfies the constraints and attains equality. The trace-norm variational principle for Hermitian operators is therefore

To prove the Holevo–Helstrom theorem, write a binary POVM as $\{E,I-E\}$, where $0\le E\le I$, and associate the first outcome with $\rho$. Any larger outcome set followed by a binary decision can be grouped into this form. Put $\Delta =p\rho -(1-p)\sigma$. Its success probability isThe substitution $T=2E-I$ bijects the allowed effects with the interval $-I\le T\le I$. Since $\mathrm{Tr}\Delta =2p-1$,Choosing $E$ as the positive spectral projection of $\Delta$ attains this value, with an arbitrary decision on its kernel. This proves the optimal success probability and constructs an optimal measurement.

For two density operators, the trace distance and the unsquared quantum fidelity areThe unsquared convention matters: if $P=|\psi ⟩⟨\psi |$ is a pure state, then $\sqrt{\rho }P\sqrt{\rho }$ is a rank-one positive operator with its sole nonzero eigenvalue equal to $⟨\psi |\rho |\psi ⟩$. Hence

Write $x_i=⟨{\psi }_i|X|{\psi }_i⟩$, which is real because $X$ is a Hermitian operator. Setwith $\mathrm{sgn}(0)=0$. This Hermitian operator satisfies $-I\le T\le I$. The trace-norm variational principle for Hermitian operators gives the diagonal absolute-sum bound for the trace norm:

For $X=\rho -P$, with $P=|\psi ⟩⟨\psi |$, extend $|\psi ⟩$ to an orthonormal basis $|{\psi }_1⟩=|\psi ⟩,|{\psi }_2⟩,\dots$. Put $r=⟨\psi |\rho |\psi ⟩$. The first diagonal entry is $r-1\le 0$, and all the others are nonnegative. Their sum is $1-r$, because $\mathrm{Tr}\rho =1$. The bound therefore yields $\Vert \rho -P{\Vert }_1\ge 2(1-r)$. Using the definitions of trace distance and quantum fidelity,This pure-target lower bound on trace distance is attained whenever $\rho$ has no coherence between $|\psi ⟩$ and its orthogonal complement. The proof used the trace-norm variational principle for Hermitian operators, together with positivity and normalization of a density operator.

The Shannon entropy of a finite-valued random variable isAn epsilon-sufficient set is a subset $S\subseteq {\mathcal{A}}_X$ retaining all but at most $ϵ$ of the probability:Usually $0\le ϵ\le 1$. Sufficiency here concerns retained probability mass, rather than a sufficient statistic.

Use the weak notion of a typical set. For a word $z^n=(z_1,\dots ,z_n)$ write $P_Z^n(z^n)={\prod }_{i=1}^n{P}_Z(z_i)$. The definition isA weakly typical sequence therefore has probability between $2^{-n(H(Z)+\delta )}$ and $2^{-n(H(Z)-\delta )}$. Words containing zero-probability symbols are excluded. These two bounds and the weak law of large numbers provide the three conclusions below.

Let $Y_i=-{\log }_2{P}_Z(Z_i)$. Since the alphabet is finite, these independent and identically distributed random variables are finite almost surely, have mean $H(Z)$, and have finite variance $v$. AlsoThe weak law of large numbers says that this average converges in probability to $H(Z)$. ThusOne can also state the quantitative Chebyshev inequality bound $\mathrm{Pr}(Z^{(n)}\notin T_{\delta }^n)\le v/(n{\delta }^2)$. If $v=0$, every word of positive probability is in the typical set. This is the finite-alphabet asymptotic equipartition property.

Every word in the typical set has probability at least $2^{-n(H(Z)+\delta )}$. ConsequentlyRearranging proves the upper estimate in the typical-set cardinality bounds:

Fix $0\le ϵ<1$ and $\delta >0$. By the asymptotic equipartition property, all sufficiently large $n$ satisfy $\mathrm{Pr}(Z^{(n)}\notin T_{\delta }^n)\le (1-ϵ)/2$. For any epsilon-sufficient set $S_n$,Each word in this intersection has probability at most $2^{-n(H(Z)-\delta )}$. Hencewhich gives the minimum size of a sufficient set for an IID source:The Chebyshev inequality used above supplies the sufficient quantitative condition $n\ge 2v/[{\delta }^2(1-ϵ)]$. No assumption that $S_n$ itself is a typical set is needed; its intersection with one provides the bound.

Use the Shannon second coding theorem in its channel capacity form $C={\max }_{P_X}I(X;Y)$. The finite-alphabet erasure channel erases independently of the input. On an unerased output Bob knows $X$ exactly; on the erasure output his posterior distribution is still $P_X$. Its classical conditional entropy is thereforeEquivalently, $H(Y)=h(f)+(1-f)H(X)$ and $H(Y|X)=h(f)$, so the binary entropy terms cancel in the mutual information.

For an alphabet of size $k$, Shannon entropy is at most ${\log }_{2}k$, with equality for the uniform distribution. Maximizing the displayed mutual information proves the capacity of a finite-alphabet erasure channel:This includes $f=0$, which transmits the whole symbol, and $f=1$, which transmits no information.

For $t<1$, normalize the remaining probabilities by $q_j=P_X(j)/(1-t)$, $j=2,\dots ,k$. Directly splitting the Shannon entropy sum givesHere $h$ is the binary entropy. The Shannon entropy of a distribution on $k-1$ points is at most ${\log }_2(k-1)$. For example, nonnegativity of its Kullback-Leibler divergence from the uniform distribution gives ${\log }_2(k-1)-H(q)\ge 0$. Thus the entropy bound with one prescribed probability isFor $t<1$, equality holds precisely when the remaining probabilities are all $(1-t)/(k-1)$. For $t=1$, the distribution is deterministic and both sides are zero. The displayed formula is for $k\ge 2$; a one-point alphabet simply has zero Shannon entropy.

Put $p_i=⟨{\psi }_i|\rho |{\psi }_i⟩$. The rank-one dephasing is ${\rho }^{\prime }={\sum }_i{p}_i|{\psi }_i⟩⟨{\psi }_i|$. If $p_i=0$, positivity givesso $\rho |{\psi }_i⟩=0$. The kernel of ${\rho }^{\prime }$ is exactly the span of these zero-probability basis vectors and is therefore contained in the kernel of $\rho$. Taking orthogonal complements proves support inclusion under rank-one dephasing:Here the support of a positive operator is the orthogonal complement of its kernel.

On that support, ${\log }_2{\rho }^{\prime }$ is diagonal in the measurement basis, givingConsequently the relative-entropy identity for rank-one dephasing isKlein's inequality gives $D_{\mathrm{rel}}(\rho \Vert {\rho }^{\prime })\ge 0$, since both density operators have trace one. For singular $\rho$, first restrict to $\mathrm{supp}{\rho }^{\prime }$ where ${\rho }^{\prime }$ is positive definite, replace $\rho$ by $(\rho +\eta I)/(1+\eta \dim \mathrm{supp}{\rho }^{\prime })$, and let $\eta \downarrow 0$. The support inclusion ensures that the limit is finite. ThusThis proves entropy increase under nonselective projective measurement using Klein's inequality. Equality holds exactly when $\rho ={\rho }^{\prime }$, so the original density operator was already diagonal in the chosen basis.

Choose an orthonormal basis of the joint $d^2$-dimensional Hilbert space whose first vector is the given purification of a density operator $|\rho {⟩}_{QR}$. Apply rank-one dephasing to ${\sigma }_{QR}$ in this basis, and denote the resulting diagonal probabilities by $p_1,\dots ,p_{d^2}$. Their first entry isThe preceding entropy increase under nonselective projective measurement and entropy bound with one prescribed probability yieldTherefore the quantum Fano inequality isThe quantity $f^2$ is the entanglement fidelity of $\mathcal{N}$ on ${\rho }_Q$; it is already a squared overlap, so it is $f^2$, rather than $f$, that enters the binary entropy. The argument is an instance of the entropy bound from overlap with a pure state in dimension $d^2$. At $f^2=1$ the output is the original pure state and its Von Neumann entropy is zero. For $d=1$ the system is trivial and the same zero-entropy conclusion holds without evaluating ${\log }_2(d^2-1)$.