# Teleportation followed by measurement: Lowering communication cost

Suppose Alice and Bob have access to shared entanglement and a classical channel and wish to simulate the following quantum protocol. Alice sends over to Bob an $$n$$-qubit state which is not known to him. Bob then does a projective measurement with $$2$$ outcomes. The measurement is known to both parties before the protocol starts. I am interested in the communication cost of simulating this protocol.

Option 1: Alice teleports her state to Bob with the cost of $$2n$$ classical bits of communication. Bob applies the unitary correction required and then measures the state.

Option 2: Alice does the projective measurement herself and simply communicates $$1$$ bit of information to Bob.

Question: If Alice cannot do the projective measurement herself, can she and Bob do better than Option 1? Can they still successfully execute this protocol but with only $$1$$-bit (or at least fewer than $$2n$$ bits) of communication?

• It takes $2n$ bits to teleport $n$ qubits, not $2\log(n)$. As for your question, it depends on which projective measurement they want to simulate; if you want to simulate a complete general one I don't think it is possible to do better than teleportation. Nov 9, 2020 at 12:13
• Thanks, will edit the incorrect log(n) claim Nov 9, 2020 at 13:33

Yes, Alice and Bob can do better than option 1. However, Alice needs to send 2 classical bits to Bob. This is more than in option 2, but is still independent of $$n$$.

The idea is that instead of performing a measurement Alice applies a unitary to rotate the measured subspace onto an auxiliary qubit and then teleports the auxiliary qubit to Bob who performs the measurement.

Note that option 2 only succeeds in transmitting measurement outcome and not the post-measurement state, so we assume that this is the goal of the protocol.

Alice's unitary

Suppose Alice has an $$n$$-qubit state $$|\psi_R\rangle$$ and let $$P_0$$ and $$P_1$$ be the measurement projectors. By definition, the projectors satisfy the completeness and orthogonality relations

$$P_0 + P_1 = I\\ P_0P_1 = 0.\tag1$$

Define the $$(n+1)$$-qubit operator $$U$$ acting on the input register $$R$$ and an auxiliary qubit $$M$$ as

$$U |\psi_R\rangle|0_M\rangle = P_0|\psi_R\rangle|0_M\rangle + P_1|\psi_R\rangle|1_M\rangle \\ U |\psi_R\rangle|1_M\rangle = P_0|\psi_R\rangle|1_M\rangle + P_1|\psi_R\rangle|0_M\rangle$$

or in block matrix form

$$U = \begin{pmatrix} P_0 & P_1\\ P_1 & P_0 \end{pmatrix}.$$

Using $$(1)$$ it is easily checked that $$U$$ is unitary.

Protocol

Alice prepares an auxiliary qubit in the $$|0_M\rangle$$ state and applies the $$U$$ to $$|\psi_R\rangle|0_M\rangle$$. Next, she teleports the state of the auxiliary qubit $$M$$ to Bob at the cost of transmitting two classical bits and consuming a Bell pair $$(|0_A0_B\rangle + |1_A1_B\rangle)/\sqrt{2}$$ where $$A$$ is the Alice's half of the pair and $$B$$ is the Bob's half. Note that teleportation preserves entanglement, so after teleportation the joint state of $$R$$ and $$B$$ is

$$|\gamma_{RB}\rangle = P_0|\psi_R\rangle|0_B\rangle + P_1|\psi_R\rangle|1_B\rangle.$$

Finally, Bob measures $$B$$ in computational basis.

Correctness

We need to prove that measuring $$P_0$$ and $$P_1$$ on the input state $$|\psi_R\rangle$$ has the same output statistics as measuring the post-teleportation state of $$B$$ in computational basis. The probability of the $$0$$ outcome in the former case is

$$p(0|\psi_R) = \langle\psi_R|P_0|\psi_R\rangle$$

and in the latter case it is

$$p(0|\rho_B) = \langle 0|\rho_B|0\rangle = \langle\psi_R|P_0|\psi_R\rangle$$

where

$$\rho_B = \mathrm{tr}_R(|\gamma_{RB}\rangle\langle\gamma_{RB}|) = \langle\psi_R|P_0|\psi_R\rangle|0_B\rangle\langle 0_B| + \langle\psi_R|P_1|\psi_R\rangle|1_B\rangle\langle 1_B|$$

is the post-teleportation state of $$B$$.

Teleportation preserves entanglement

Teleportation is often described as a protocol on three qubits $$B$$, $$C$$ and $$D$$ with $$B$$ in an arbitrary single-qubit state $$|\psi_B\rangle = \alpha|0_B\rangle + \beta|1_B\rangle$$ and $$CD$$ in the Bell state $$(|0_C0_D\rangle + |1_C1_D\rangle)/\sqrt{2}$$ at the end of which the qubit $$D$$ is in the state $$|\psi_D\rangle = \alpha|0_D\rangle + \beta|1_D\rangle$$.

Suppose that the input qubit $$B$$ is entangled with a qubit $$A$$ in a joint state

$$|\phi_{AB}\rangle = \alpha|0_A0_B\rangle + \beta|0_A1_B\rangle + \gamma|1_A0_B\rangle + \delta|1_A1_B\rangle.$$

We will show that at the end of the standard teleportation protocol the qubits $$AD$$ are in the state

$$|\phi_{AD}\rangle = \alpha|0_A0_D\rangle + \beta|0_A1_D\rangle + \gamma|1_A0_D\rangle + \delta|1_A1_D\rangle$$

thus proving that any entanglement the input qubit $$B$$ has with another system $$A$$ is preserved by the teleportation channel.

The initial state of the composite system $$ABCD$$ is

\begin{align} \frac{1}{\sqrt{2}} (&\alpha|00\rangle(|00\rangle +|11\rangle)\, + \\ & \beta|01\rangle(|00\rangle +|11\rangle)\, +\\ & \gamma|10\rangle(|00\rangle +|11\rangle)\, +\\ & \delta|11\rangle(|00\rangle +|11\rangle)) \end{align}

where we have elided subsystem labels for legibility. After the CNOT gate with $$B$$ as the control and $$C$$ as the target and the Hadamard gate on $$B$$, the state of $$ABCD$$ becomes

\begin{align} & \frac{1}{2}(\alpha|0000\rangle + \beta|0001\rangle + \gamma|1000\rangle + \delta|1001\rangle)\, + \\ & \frac{1}{2}(\alpha|0011\rangle + \beta|0010\rangle + \gamma|1011\rangle + \delta|1010\rangle)\, + \\ & \frac{1}{2}(\alpha|0100\rangle - \beta|0101\rangle + \gamma|1100\rangle - \delta|1101\rangle)\, + \\ & \frac{1}{2}(\alpha|0111\rangle - \beta|0110\rangle + \gamma|1111\rangle - \delta|1110\rangle) \end{align}

where we have grouped the terms according to the state of the $$BC$$ subsystem. It is immediate that if measurement of $$BC$$ yields $$|00\rangle$$ then $$AD$$ collapses to $$|\phi_{AD}\rangle$$ as promised. Similarly, we see that the usual $$X$$ and $$Z$$ corrections on $$D$$ conditioned on the measurement results transform each of the other three groups into $$|\phi_{AD}\rangle$$ as well. $$\square$$