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Adam Zalcman
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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. $\square$

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 $B$.

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. $\square$

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Adam Zalcman
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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 moverotate the relevant quantum information tomeasured subspace onto an auxiliary qubit and then she teleports the auxiliary qubit to Bob who performs the measurement. In this context, "relevant" means "affecting the output of 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 $B$.

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 move the relevant quantum information to an auxiliary qubit and then she teleports the auxiliary qubit to Bob who performs the measurement. In this context, "relevant" means "affecting the output of 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 $B$.

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 $B$.

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Adam Zalcman
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  • 92

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 move the relevant quantum information to an auxiliary qubit and then she teleports the auxiliary qubit to Bob who performs the measurement. In this context, "relevant" means "affecting the output of 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 $\rho_B = \mathrm{tr}_R(|\gamma_{RB}\rangle\langle\gamma_{RB}|)$

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

is the post-teleportation state of $B$.

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 move the relevant quantum information to an auxiliary qubit and then she teleports the auxiliary qubit to Bob who performs the measurement. In this context, "relevant" means "affecting the output of 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 $\rho_B = \mathrm{tr}_R(|\gamma_{RB}\rangle\langle\gamma_{RB}|)$ is the post-teleportation state of $B$.

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 move the relevant quantum information to an auxiliary qubit and then she teleports the auxiliary qubit to Bob who performs the measurement. In this context, "relevant" means "affecting the output of 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 $B$.

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Adam Zalcman
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