Past exam of the computer science course of the University of Oxford / 2023 / Quantum information / 3

A bipartite pure state is entangled exactly when it is not a product state: there are no normalized quantum states $|u{⟩}_A,|v{⟩}_B$ such thatEquivalently, its Schmidt rank exceeds one, or either reduced density matrix is a mixed state. This definition concerns quantum entanglement across the specified division into systems $A$ and $B$.

An entangling gate is a unitary transformation $U$ on two quantum systems for which at least one pure product input has an entangled output:It need not create quantum entanglement on every product state. For example, the controlled-NOT gate leaves $|00⟩$ a product state but maps $|+⟩|0⟩$ to a Bell state, so it is an entangling gate.

The sign factor separates as $(-1{)}^{x+y}=(-1{)}^x(-1{)}^y$. Consequently the quantum state factors into a tensor product of quantum systems:It is a product state, so it is not entangled.

Grouping by the first qubit givesThe vectors on both sides form orthonormal bases, so this is a Schmidt decomposition with two nonzero Schmidt coefficients $1/\sqrt{2}$. Hence the state is maximally entangled. Independently, its coefficient matrix has determinantwhich excludes a product state by the two-qubit product-state determinant criterion. Each reduced density matrix is $I/2$.

Alice holds the input qubit $Q$ in $|\psi {⟩}_Q=a|0⟩+b|1⟩$ and her half $A$ of a shared Bell pair $|{\varphi }_{+}{⟩}_{AB}$; Bob holds $B$. Alice applies a controlled-NOT gate $Q\to A$, then a Hadamard gate on $Q$. In the system order $Q,A,B$, the initial quantum state isAfter these two quantum gates, direct expansion yields the quantum teleportation identityAlice performs computational-basis measurements on $Q,A$, and sends the two classical bits $m,n$ to Bob. Each outcome has probability $1/4$, independently of the input pure state. Bob applies the indicated Pauli gates: first $X^n$, then $Z^m$, so his total correction is $Z^m{X}^n$. In every branch,This proves exact quantum teleportation using one Bell pair and two classical bits. Alice's original qubit has been measured; the protocol transfers the quantum state without leaving her another copy.

Write the arbitrary initial pure state asThe reference vectors need not be normalized or orthogonal. By linearity, the same quantum teleportation calculation, with the identity operator on $R$, gives the post-gate quantum stateThus each outcome still has probability $1/4$. Bob's Pauli gate correction leaves the normalized joint quantum stateAll operations act trivially on $R$, so the result is independent of its holder or location. This establishes teleportation as an identity channel on a reference, preserving quantum entanglement with the untouched reference. Since every corrected branch is the same, discarding the classical outcomes also leaves the required joint density matrix.

Combine the hypothetical protocol with superdense coding. Before a message is chosen, Alice and Bob share a further Bell pair $|{\varphi }_{+}{⟩}_{QR}$, with $Q$ at Alice and $R$ at Bob, in addition to any resources needed for the hypothetical quantum teleportation protocol. To encode two classical bits $u,v$, Alice applies the Pauli gates $Z_Q^u{X}_Q^v$, preparingThe four choices are the four mutually orthogonal Bell states. Alice now uses the hypothetical protocol to teleport $Q$ to Bob using only one classical bit. Its assumed preservation of the joint quantum state gives Bob both $B$ and $R$ inBob performs a Bell-basis measurement. Explicitly, a controlled-NOT gate $B\to R$, followed by a Hadamard gate on $B$, givesA computational-basis measurement therefore recovers both message bits with certainty after only one transmitted classical bit. The initial Bell pairs are shared before the message is chosen and carry no message themselves. This proves the classical communication lower bound for exact qubit teleportation: a perfect protocol preserving an external reference needs at least two transmitted classical bits.