Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 50 / 5

Label the atomic ground state $|0⟩$ and excited state $|1⟩$, with $0\le p\le 1$. The process is the amplitude damping channel, with Kraus operatorsThey satisfy $A_0^{†}{A}_0+A_1^{†}{A}_1=I$, so the resulting Kraus representation is trace preserving.

The operator $A_1$ represents emission: it maps the excited state to $\sqrt{p}$ times the ground state and annihilates the ground state. Its probability for an initially excited atom is $p$. The operator $A_0$ represents no emission: it leaves the ground state unchanged and retains the excited amplitude with factor $\sqrt{1-p}$. This branch is not generally the identity operation; the absence of a photon itself changes the conditional relative amplitudes. For an arbitrary atomic density operator, the emission probability is $\mathrm{Tr}(A_1\rho A_1^{†})=p{\rho }_{11}$ and the no-emission probability is $1-p{\rho }_{11}$.

A unital quantum channel preserves the identity operator: $\Phi (I)=I$. For equal input and output dimensions, this is equivalent to preserving the maximally mixed density operator.

For the amplitude damping channel,Thus the channel is not unital for $p>0$; it is unital only at $p=0$, when it is the identity channel. In particular, it biases the maximally mixed state toward the atomic ground state. The trace-preservation condition involves ${\sum }_j{A}_j^{†}{A}_j$, whereas unitality involves the opposite products ${\sum }_j{A}_j{A}_j^{†}$.

Take an environment initially in its vacuum state $|0{⟩}_E$, with $|1{⟩}_E$ recording an emitted photon. In atom-environment order, put $q=\sqrt{1-p}$ and $u=\sqrt{p}$. A unitary dilation of amplitude damping is defined byThe middle two images are orthonormal, and they are orthogonal to the two fixed images. Thus this completion really is a unitary operator. For the occupied initial environmental subspace, it leaves a ground-state atom unchanged and splits an excited-state atom into no-emission and emission alternatives with probabilities $1-p$ and $p$.

Tracing out the environment after this unitary evolution givesThe environment matrix elements are $⟨0{|}_{E}U|0{⟩}_E=A_0$ and $⟨1{|}_{E}U|0{⟩}_E=A_1$, recovering the Kraus operators in part (a). More explicitly, the basis operators transform asApplying linearity to the atomic density operator therefore yieldsThe lost excited population goes to the ground state, the trace stays one, and coherences decrease by $\sqrt{1-p}$. Discarding the distinguishable environmental photon record is precisely what turns the joint unitary evolution into a noisy atomic quantum channel.

Let the quantum channel send $B$ to $B^{\prime }$. From its Kraus representation, choose a Stinespring representation $V:B\to B^{\prime }E$ and formThe channel output is ${\sigma }_{A{B}^{\prime }}={\mathrm{Tr}}_E{\sigma }_{A{B}^{\prime }E}$. Isometric invariance of Von Neumann entropy givesTherefore the quantum mutual information satisfies $I(A:B{)}_{\rho }=I(A:B^{\prime }E{)}_{\sigma }$. Expanding its loss after discarding the environment givesThe final expression is the quantum conditional mutual information $I(A:E\mid B^{\prime })$. The Strong subadditivity of Von Neumann entropy states exactly that it is nonnegative. HenceThis proves data processing for quantum mutual information for any mixed input and any local quantum channel, not just the atomic channel in earlier parts. The correlations lost are exactly the conditional correlations with the discarded environment.