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Teleportation

Abstract

Details of the task of quantum state Teleportation, in both the qubit and higher dimensional case. The fidelity of teleportation and teleportation from the perspective of the resource theory of entanglement is also discussed.

Keywords:TeleportationEntanglementLOCC.

Setup

Let there be two spatially separated parties, Alice (A) and Bob (B), who share a d=2d=2 dimension maximally entangled bipartite state,

Φ00+AB=0A0B+1A1B2  HA2HB2.\ket{\Phi^{+}_{00}}_{AB} = \frac{\ket{0}_{A}\ket{0}_{B}+\ket{1}_{A} \ket{1}_{B}}{\sqrt{2}}~\in~\mathcal{H}_{A}^{2} \otimes \mathcal{H}_{B}^{2}.

It is assumed they have the ability to perform any operation on a quantum system in their respective labs and that they can send any amount of classical information to each other 💭.

These operations are known as the set of local operations and classical communication (LOCC(A:B)LOCC(A:B)) 💭. They are a physically motivated set of operations as local manipulation of quantum states and classical communication between spatially separated parties are much easier to perform then the transportation of quantum states between spatially separated parties.

Alice has a single copy of the qubit,

ψA1=α0A1+β1A1  HA1,\ket{\psi}_{A_1} = \alpha \ket{0}_{A_1} + \beta \ket{1}_{A_1}~\in~\mathcal{H}_{A_{1}},

in her laboratory that she wants to send to Bob. The subscript A1A_1 shows that the state is in Alice’s lab. Given Alice and Bob can only send classical information to each other, Alice cannot send the qubit to Bob directly.

Teleportation Protocol

Alice can instead send the qubit ψA1\ket{\psi}_{A_1} to Bob via the quantum teleportation protocol.

The quantum teleportation protocol states that it is possible for Alice to exactly send the state ψA1\ket{\psi}_{A_1} to Bob using: a maximally entangled state, a measurement in a maximally entangled basis, two classical bits and one local unitary.

The Protocol:

The initial state shared between Alice and Bob is

ΨA1AB=ψA1Φ00+AB  HA1HAHB,=[α0A1+β1A1]12[0A0B1+1A1B].\begin{align*} \ket{\Psi}_{A_{1}AB} &= \ket{\psi}_{A_1} \otimes \ket{\Phi^{+}_{00}}_{AB}~\in~\mathcal{H}_{A_1} \otimes \mathcal{H}_{A} \otimes \mathcal{H}_{B}, \\ &= \big[ \alpha \ket{0}_{A_1} + \beta \ket{1}_{A_1} \big] \otimes \frac{1}{\sqrt{2}}\big[ \ket{0}_{A}\ket{0}_{B_1}+\ket{1}_{A} \ket{1}_{B} \big]. \end{align*}

This state, ΨA1AB\ket{\Psi}_{A_{1}AB}, can then be expanded and rewritten as

ΨA1AB=122[0A10A+1A11A][α0B+β1B]+122[0A10A1A11A][α0Bβ1B]+122[0A11A+1A10A][β0B+α1B]+122[0A11A1A10A][β0B+α1B].\begin{align*} \ket{\Psi}_{A_{1}AB} = \frac{1}{2 \sqrt{2}} \big[ \ket{0}_{A_1}\ket{0}_{A} + \ket{1}_{A_1}\ket{1}_{A} \big] \big[ \alpha \ket{0}_{B} + \beta \ket{1}_B \big] \\ + \frac{1}{2 \sqrt{2}} \big[ \ket{0}_{A_1}\ket{0}_{A} - \ket{1}_{A_1}\ket{1}_{A} \big] \big[ \alpha \ket{0}_{B} - \beta \ket{1}_B \big] \\ + \frac{1}{2 \sqrt{2}} \big[ \ket{0}_{A_1}\ket{1}_{A} + \ket{1}_{A_1}\ket{0}_{A} \big] \big[ \beta \ket{0}_{B} + \alpha \ket{1}_B \big] \\ + \frac{1}{2 \sqrt{2}} \big[ \ket{0}_{A_1}\ket{1}_{A} - \ket{1}_{A_1}\ket{0}_{A} \big] \big[ - \beta \ket{0}_{B} + \alpha \ket{1}_B \big] . \end{align*}

Firstly, Alice measures the operator, OO, which has the Bell states as its eigenvectors,

O=λ00Φ00+Φ00++λ01Φ01+Φ01++λ10Φ10+Φ10++λ11Φ11+Φ11+,O = \lambda_{00} \ket{\Phi^{+}_{00}}\bra{\Phi^{+}_{00}} + \lambda_{01} \ket{\Phi^{+}_{01}}\bra{\Phi^{+}_{01}} + \lambda_{10} \ket{\Phi^{+}_{10}}\bra{\Phi^{+}_{10}} + \lambda_{11} \ket{\Phi^{+}_{11}}\bra{\Phi^{+}_{11}},

where λ00λ01λ10λ11\lambda_{00} \neq \lambda_{01} \neq \lambda_{10} \neq \lambda_{11}, and

Φ00+=0A0B+1A1B2,Φ01+=0A0B1A1B2,Φ10+=0A1B+1A0B2,Φ11+=0A1B1A0B2,\begin{align*} \ket{\Phi_{00}^{+}} &= \frac{\ket{0}_{A}\ket{0}_{B}+\ket{1}_{A} \ket{1}_{B}}{\sqrt{2}}, \\ \ket{\Phi_{01}^{+}} &= \frac{\ket{0}_{A}\ket{0}_{B}-\ket{1}_{A} \ket{1}_{B}}{\sqrt{2}}, \\ \ket{\Phi_{10}^{+}} &= \frac{\ket{0}_{A}\ket{1}_{B}+\ket{1}_{A} \ket{0}_{B}}{\sqrt{2}}, \\ \ket{\Phi_{11}^{+}} &= \frac{\ket{0}_{A}\ket{1}_{B}-\ket{1}_{A} \ket{0}_{B}}{\sqrt{2}}, \\ \end{align*}

on the spaces HA1HA\mathcal{H}_{A_{1}} \otimes \mathcal{H}_{A}. Alice gets one the eigenvalues of OO as her measurement outcome.

After Alice measures OO, the state of Bob’s system is dependent on Alice’s measurement outcome:

λ00α0+β1  HB,λ01α0β1  HB,λ10β0+α1  HB,λ11β0+α1  HB.\begin{align*} \lambda_{00} &\rightarrow \alpha \ket{0} + \beta \ket{1}~\in ~ \mathcal{H}_{B}, \\ \lambda_{01} &\rightarrow \alpha \ket{0} - \beta \ket{1}~\in ~ \mathcal{H}_{B}, \\ \lambda_{10} &\rightarrow \beta \ket{0} + \alpha \ket{1}~\in ~ \mathcal{H}_{B}, \\ \lambda_{11} &\rightarrow - \beta \ket{0} + \alpha \ket{1}~\in ~ \mathcal{H}_{B}. \\ \end{align*}

Therefore, based on her measurement outcome Alice knows which state Bob has in his system. Bob, however, does not know which of the above states he has in his system.

Secondly, Alice communicates her measurement outcome to Bob, therefore informing him of the state of his system. Given there were four possible outcomes, {λ00,λ01,λ10,λ11} \{ \lambda_{00}, \lambda_{01}, \lambda_{10}, \lambda_{11} \}, she needs to send Bob two bits (22=42^2 = 4):

λ00Alice sends 00λ01Alice sends 01λ10Alice sends 10λ11Alice sends 11\begin{align*} \lambda_{00} &\rightarrow \textrm{Alice sends}~ 00 \\ \lambda_{01} &\rightarrow \textrm{Alice sends}~ 01 \\ \lambda_{10} &\rightarrow \textrm{Alice sends}~ 10 \\ \lambda_{11} &\rightarrow \textrm{Alice sends}~ 11 \\ \end{align*}

After receiving the two bits from Alice, Bob knows which of four states he has (but not α \alpha and β \beta explicitly.)

Lastly, and conditioned on the bits he receives from Alice, Bob applies one of four unitaries:

00Bob applies I,01Bob applies Z,10Bob applies X,11Bob applies Y.\begin{align*} 00 &\rightarrow \textrm{Bob applies}~ \mathbb{I}, \\ 01 &\rightarrow \textrm{Bob applies}~ Z, \\ 10 &\rightarrow \textrm{Bob applies}~ X, \\ 11 &\rightarrow \textrm{Bob applies}~ Y. \\ \end{align*}

After this, Bob is left with the state ψA1\ket{\psi}_{A_{1}} in his space.

Hence, using one maximally entangled state, one measurement in the Bell basis, two classical bits and one local unitary, the state ψA1\ket{\psi}_{A_1} has been “sent” from Alice to Bob. However, the physical state has not been sent, the information that defines the state has been.

Diagram
A graphical depiction of the quantum teleportation protocol.

A graphical depiction of the quantum teleportation protocol.

Fidelity of Teleportation

One does not need a maximally entangled state to perform teleportation. However, using a partially entangled state - a state that is entangled but not maximally entangled - will reduce the quality of the teleportation, as measured by the fidelity of teleportation. This is a measure of how similar the final quantum state in Bob’s space is to the initial state Alice is trying to teleport.

Details

The average fidelity of teleportation is given by

F(ρAB)=maxTθtrA1AT(θρAB)θdθ,s.t T  LOCC(A1A:B),\begin{align*} F(\rho_{AB}) &= \max_{\mathcal{T}} \int \bra{\theta} \textrm{tr}_{A_{1}A}\mathcal{T}(\theta \otimes \rho_{AB}) \ket{\theta} d \theta, \\ & \textrm{s.t}~\mathcal{T}~\in~LOCC(A_{1}A:B), \end{align*}

where ρAB\rho_{AB} is a shared state between Alice and Bob, θ \theta is the state being teleported, where dimθ=dθ\textrm{dim} \theta = d_{\theta}, and T\mathcal{T} is the teleportation channel (this packages up the measurement of OO, the classical communication and the local rotation). The fidelity is a measure of how well a state ρAB \rho_{AB} performs on average at teleporting all possible states in the space from Alice to Bob.

Properties

  • 0F(ρAB)10 \leq F(\rho_{AB}) \leq 1
  • F(ρAB)=1F(\rho_{AB}) = 1 iif ρAB\rho_{AB} is the maximally entangled state.
  • If ρAB\rho_{AB} is a separable state, then there exists a classical strategy such that F(ρABsep)=2/(dθ+1)F(\rho_{AB}^{\textrm{sep}}) = 2/(d_{\theta}+1).
  • The fidelity of teleportation has been related to the entanglement fraction as
F(ρAB)=f(ρAB)dθ+1dθ+1,\begin{align*} F(\rho_{AB}) &= \frac{f(\rho_{AB})d_{\theta}+1}{d_{\theta}+1}, \end{align*}

where

f(ρAB)maxEΦ00+E(ρAB)Φ00+E  LOCC(A:B).\begin{align*} f(\rho_{AB}) &\coloneqq \max_{\mathcal{E}} \bra{\Phi^{+}_{00}} \mathcal{E}(\rho_{AB}) \ket{\Phi^{+}_{00}} \\ & \mathcal{E} ~\in~LOCC(A:B). \end{align*}

The entanglement fraction maximises the overlap between the state ρAB\rho_{AB} and the maximally entangled state over all LOCC(A:B)LOCC(A:B) protocols.

Higher Dimensional Systems

The teleportation protocol can be generalised to send higher dimensional states between two spatially separated parties. A state ρA1\rho_{A_1}, where dimρA1=dA1\textrm{dim} \rho_{A_1} = d_{A_1}, can be teleported between Alice (A) and Bob (B) using a bipartie maximally entangled state of local dimension dA1d_{A_1},

Φ00+AB=12i=0dA11ii  HAdA1HBdA1.\ket{\Phi^{+}_{00}}_{AB} = \frac{1}{\sqrt{2}} \sum^{d_{A_1}-1}_{i=0} \ket{ii}~\in~\mathcal{H}_{A}^{d_{A_1}}\otimes \mathcal{H}_{B}^{d_{A_1}}.

Firstly, Alice measures the operator with a basis of maximally entangled states,

{(IWa,b)(Φ00+AB):(a,b)  {0,,dA11}},\{ (\mathbb{I} \otimes W_{a,b}) \big( \ket{\Phi^{+}_{00}}_{AB} \big) : (a,b)~\in~\{0, \ldots, d_{A_1}-1\} \},

where Wa,bW_{a,b} are the so called Heisenberg-Weyl operators. See generating a maximally entangled basis for more details on this.

Secondly, Alice needs to send the results of her measurement outcome, which can be capture by pair of values (a,b)(a,b), to Bob. There are d2d^{2} different pairs of values (a,b)(a,b), meaning Alice needs to send n=log2(d2)n= \log_{2}(d^{2}) bits to Bob.

Lastly, if Bob receives the bits from Alice associated to a pair of values (a,b)(a',b'), Bob applied Wa,bW_{a',b'} to his local system. Bob then has the dA1d_{A_1} dimensional state in his system.

Teleportation in the Resource Theory of Entanglement

In the teleportation protocol, after Alice measures OO, Alice and Bob no longer share a maximally entangled state; the maximally entangled state has been “consumed” in order to implement the teleportation protocol. Given one cannot perform teleportation without an entangled state and it is consumed in the process it is considered a resource for the task.

The teleportation protocol is an example of simulating a channel in a resource theory. Specifically, the teleportation channel allows the simulation of a noiseless quantum channel in the resource theory of entanglement.

Details

The allowed operations of the resource theory of entanglement are LOCC(A:B)LOCC(A:B) - local operations and classical communication between two spatially separated parties Alice (AA) and Bob (BB). The free states are the set of all product states, as any product state between Alice and Bob can be generated via LOCC(A:B)LOCC(A:B) and, given any product state between AA and BB, no entanglement can be generated via LOCC(A:B)LOCC(A:B).

There is no channel in the set of allowed operations to send a quantum state from Alice to Bob. However, when the two parties share an entangled state - a maximally resourceful state - there exists an allowed operation, that is an LOCC(A:B) LOCC(A:B) channel, that is able to simulate a noiseless quantum channel that sends a quantum state from Alice to Bob. Formally,

trA1AT(ρA1Φ+Φ+AB)simulatesNAB(ρA1),\textrm{tr}_{A_{1}A}\mathcal{T}(\rho_{A_1} \otimes \ket{\Phi^{+}}\bra{\Phi^{+}}_{AB}) \xrightarrow{simulates} \mathcal{N}_{A \rightarrow B}(\rho_{A_1}),

where T  LOCC(A1A:B)\mathcal{T}~\in~LOCC(A_{1}A:B) is the quantum channel for the teleportation protocol and N  LOCC(A:B)\mathcal{N}~\notin~\textrm{LOCC}(A:B) is the quantum channel that noiselessly sends the quantum state ρA1 \rho_{A_1} from Alice to Bob.

As measured by the fidelity of teleportation, Alice and Bob can perform lower quality teleportation if ρAB\rho_{AB} is entangled but not maximally entangled. In this case, the channel that is simulated is not noiseless, but instead noisy. Bob’s final state will therefore not be exactly the state Alice tried to send. This furthers the notion of entanglement being a resource for this task; the amount of entanglement Alice and Bob shared dictates how noiselessly a quantum state can be sent from Alice to Bob. This can be succinctly formalised through the following equation

qq+2[cc][qq],qq + 2[c \rightarrow c] \geq [q \rightarrow q],

meaning that one entangled state, qqqq, and 2 classical bits, 2[cc]2[c \rightarrow c] simulate a noiseless quantum channels sending a quantum state, [qq][q \rightarrow q]. The inequality represents the fact that the noiseless channel is only achievable if qqqq is the maximally entangled state.

References
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  2. Popescu, S. (1994). Bell’s inequalities versus teleportation: What is nonlocality? Physical Review Letters, 72(6), 797–799. 10.1103/physrevlett.72.797
  3. Horodecki, M., Horodecki, P., & Horodecki, R. (1999). General teleportation channel, singlet fraction, and quasidistillation. Physical Review A, 60(3), 1888–1898. 10.1103/physreva.60.1888