3
$\begingroup$

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?

$\endgroup$
2
  • 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. $\endgroup$ Nov 9 '20 at 12:13
  • $\begingroup$ Thanks, will edit the incorrect log(n) claim $\endgroup$ Nov 9 '20 at 13:33
1
$\begingroup$

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. $\square$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.