Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 35 / 4

For a density matrix $\rho$ on a finite-dimensional Hilbert space, its Von Neumann entropy in bits iswhere ${\lambda }_j$ are its eigenvalues and $0{\log }_{2}0=0$. It is the Shannon entropy of the spectrum, is zero for a pure state, and is invariant under unitary conjugation.

The concavity inequality is true, but its printed equality condition is false. At $b=1$ equality always holds, even for distinct states; for example take ${\rho }_1=|0⟩⟨0|$ and ${\rho }_2=|1⟩⟨1|$. The correct statement on $0\le b\le 1$ isFor $0<b<1$, entropy is therefore strictly concave.

We give the matrix inequality needed for the proof. Define the quantum relative entropy $D(\tau \Vert \sigma )=\mathrm{tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ if the support of $\tau$ is contained in that of $\sigma$, and $+\infty$ otherwise. Diagonalize the two states with eigenvalues $r_i,s_j$, and put $q_{ij}=|⟨i|j⟩{|}^2$. Each row and column of $q$ sums to one. The elementary scalar inequalityfollows from $x\ln x-x+1\ge 0$, with equality exactly at $x=1$; use limits at zero. ConsequentlyEquality forces $r_i=s_j$ whenever $q_{ij}>0$. Hence every eigenvector $|i⟩$ of $\tau$ is also a $\sigma$-eigenvector with eigenvalue $r_i$, so $\tau =\sigma$. Conversely identical states give zero relative entropy. This proves nonnegativity of quantum relative entropy and its equality condition, including states with zero eigenvalues under the support convention.

Now set $\rho =b{\rho }_1+(1-b){\rho }_2$. For $0<b<1$ its support contains both component supports. Direct expansion givesSince both coefficients are positive, equality holds exactly when both component states equal $\rho$, equivalently ${\rho }_1={\rho }_2$. At either endpoint the mixture has only one component and equality is automatic. This proves the corrected strict concavity of Von Neumann entropy.

False. Take the Bell stateThe joint state is pure, so $S(\rho )=0$. Taking either partial trace kills the off-diagonal terms and gives ${\rho }_1={\rho }_2=I_2/2$, whose two eigenvalues are $1/2$. Thus $S({\rho }_1)=S({\rho }_2)=1$, contradicting the proposed inequality $1\le 0$. Quantum entanglement allows a pure joint state to have mixed subsystem states.

True: quantum entropy is subadditive. Write $\sigma ={\rho }_1\otimes {\rho }_2$. Positivity ensures the support of $\rho$ lies in that of $\sigma$. Indeed, if $|v⟩\in \ker {\rho }_1$, then ${\sum }_j⟨v,j|\rho |v,j⟩=0$; all summands are nonnegative, so ${\rho }^{1/2}|v,j⟩=0$ for every $j$. Thus $\rho$ annihilates $\ker {\rho }_1\otimes {\mathcal{K}}_2$, and similarly the kernel from the other factor.

On the resulting support, ${\log }_2({\rho }_1\otimes {\rho }_2)={\log }_2{\rho }_1\otimes I+I\otimes {\log }_2{\rho }_2$. Using the defining property of the partial trace and nonnegativity of quantum relative entropy proved in part (i),HenceEquality holds exactly for the product state $\rho ={\rho }_1\otimes {\rho }_2$. This proves Subadditivity of Von Neumann entropy directly.