Quantum Teleportation

This applies entanglement. Let there be Alice and Bob who are separated. Alice has Qubit A and B. Bob has qubit C. B and C are entangled.

Alice wants to send Bob qubit A in unknown state

∣υ(A)⟩=α∣0(A)⟩+β∣1(A)⟩\ket{\upsilon^{(A)}} = \alpha\ket{0^{(A)}}+\beta\ket{1^{(A)}}

Teleportation occurs as we can copy the state of B to C. The state of B is destroyed, so it doesn't violate No-Cloning Theorem.

Note α\alpha and β\beta are probability amplitudes.

Alice and Bob pre-share the Bell state

∣Φ+(BC)⟩=12(∣00⟩+∣11⟩)\ket{\Phi_+^{(BC)}} = \tfrac{1}{\sqrt{2}}\big(\ket{00} + \ket{11}\big)

so now we combine this

∣u(A)⟩∣Φ+(BC)⟩⟶∣trash(AB)⟩∣u(C)⟩\ket{u^{(A)}} \ket{\Phi_+^{(BC)}} \quad \longrightarrow \quad \ket{\text{trash}^{(AB)}} \ket{u^{(C)}}

Before observing it, Alice's qubit A holds the unknown state ∣u⟩\ket{u}. B,CB, C are entangled but unrelated to you. After observing it, Bob's qubit now holds state ∣u⟩\ket{u}. Alice's qubit A,BA,B turn into trash. Trash is because the two qubits collapse into one of four random bell measurements.

Example We know

∣υ(A)⟩=α∣0(A)⟩+β∣1(A)⟩\ket{\upsilon^{(A)}} = \alpha\ket{0^{(A)}}+\beta\ket{1^{(A)}}

So

∣u(A)⟩∣Φ+(BC)⟩=(α∣0(A)⟩+β∣1(A)⟩)⊗12(∣0(B)0(C)⟩+∣1(B)1(C)⟩)\ket{u^{(A)}} \ket{\Phi_+^{(BC)}} = \big(\alpha\ket{0^{(A)}} + \beta\ket{1^{(A)}}\big) \otimes \tfrac{1}{\sqrt{2}}\big(\ket{0^{(B)} 0^{(C)}} + \ket{1^{(B)} 1^{(C)}}\big) =α2(∣000⟩+∣011⟩)+β2(∣100⟩+∣111⟩)=\frac{\alpha}{\sqrt{2}}(\ket{000}+\ket{011})+\frac{\beta}{\sqrt{2}}(\ket{100}+\ket{111})

Gives

=12(∣Φ+(AB)⟩+∣Φ−(AB)⟩)⊗α∣0(C)⟩= \tfrac{1}{2}\big(\ket{\Phi_+^{(AB)}} + \ket{\Phi_-^{(AB)}}\big) \otimes \alpha\ket{0^{(C)}} +12(∣Ψ+(AB)⟩+∣Ψ−(AB)⟩)⊗α∣1(C)⟩+ \tfrac{1}{2}\big(\ket{\Psi_+^{(AB)}} + \ket{\Psi_-^{(AB)}}\big) \otimes \alpha\ket{1^{(C)}} +12(∣Ψ+(AB)⟩−∣Ψ−(AB)⟩)⊗β∣0(C)⟩+ \tfrac{1}{2}\big(\ket{\Psi_+^{(AB)}} - \ket{\Psi_-^{(AB)}}\big) \otimes \beta\ket{0^{(C)}} +12(∣Φ+(AB)⟩−∣Φ−(AB)⟩)⊗β∣1(C)⟩+\tfrac{1}{2}\big(\ket{\Phi_+^{(AB)}} - \ket{\Phi_-^{(AB)}}\big) \otimes \beta\ket{1^{(C)}}

Gives

=12∣Φ+(AB)⟩(α∣0(C)⟩+β∣1(C)⟩)⟶ ∣u(C)⟩= \tfrac{1}{2}\ket{\Phi_+^{(AB)}}\big(\alpha\ket{0^{(C)}} + \beta\ket{1^{(C)}}\big) \quad \longrightarrow\ \ket{u^{(C)}} +12∣Φ−(AB)⟩(α∣0(C)⟩−β∣1(C)⟩)⟶ σZ(C)∣u(C)⟩+ \tfrac{1}{2}\ket{\Phi_-^{(AB)}}\big(\alpha\ket{0^{(C)}} - \beta\ket{1^{(C)}}\big) \quad \longrightarrow\ \sigma^{Z(C)}\ket{u^{(C)}} +12∣Ψ+(AB)⟩(α∣1(C)⟩+β∣0(C)⟩)⟶ σX(C)∣u(C)⟩+ \tfrac{1}{2}\ket{\Psi_+^{(AB)}}\big(\alpha\ket{1^{(C)}} + \beta\ket{0^{(C)}}\big) \quad \longrightarrow\ \sigma^{X(C)}\ket{u^{(C)}} +12∣Ψ−(AB)⟩(α∣1(C)⟩−β∣0(C)⟩)⟶ iσY(C)∣u(C)⟩+ \tfrac{1}{2}\ket{\Psi_-^{(AB)}}\big(\alpha\ket{1^{(C)}} - \beta\ket{0^{(C)}}\big) \quad \longrightarrow\ i\sigma^{Y(C)}\ket{u^{(C)}}

Alice measures in Bell basis and gets four possible outcomes

1⇒Φ+2⇒Φ−3⇒Ψ+4⇒Ψ−1\Rightarrow\Phi_+\quad 2\Rightarrow \Phi_- \quad 3\Rightarrow \Psi_+\quad 4\Rightarrow \Psi_-

Example: If Alice got outcome three (i.e., Ψ+\Psi_+) then

∣ϕ3(C)⟩=1P3⟨Ψ+(AB)∣υ(A)Φ+(BC)⟩=α∣1(C)⟩+β∣0(C)⟩\ket{\phi_3^{(C)}}=\frac{1}{\sqrt{P_3}}\braket{\Psi_+^{(AB)}|\upsilon^{(A)}\Phi_+^{(BC)}}=\alpha\ket{1^{(C)}}+\beta\ket{0^{(C)}}

Alice tells Bob she got 3 Bob's operation conditional on Alice's outcome

1=I2=σZ3=σX4=σY1=I\quad 2=\sigma^Z\quad 3=\sigma^X\quad 4=\sigma^Y

Bob choose σX\sigma^X as the base so

σX∣ϕ3(C)⟩=α∣0(C)⟩+β∣1(C)⟩=∣υ(C)⟩\sigma^X\ket{\phi_3^{(C)}}=\alpha\ket{0^{(C)}}+\beta\ket{1^{(C)}}=\ket{\upsilon^{(C)}}

Bob now has reproduced ∣υ⟩\ket{\upsilon}