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?

  • 2
    $\begingroup$ 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. $\endgroup$ Commented Nov 9, 2020 at 12:13
  • $\begingroup$ Thanks, will edit the incorrect log(n) claim $\endgroup$ Commented Nov 9, 2020 at 13:33

1 Answer 1


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.


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.


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

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


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

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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