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)}} ∣ υ ( A ) ⟩ = α ∣ 0 ( A ) ⟩ + β ∣ 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 α and β \beta β are probability amplitudes .
Alice and Bob pre-share the Bell state
∣ Φ + ( B C ) ⟩ = 1 2 ( ∣ 00 ⟩ + ∣ 11 ⟩ ) \ket{\Phi_+^{(BC)}} = \tfrac{1}{\sqrt{2}}\big(\ket{00} + \ket{11}\big) ∣ Φ + ( B C ) ⟩ = 2 1 ( ∣ 00 ⟩ + ∣ 11 ⟩ )
so now we combine this
∣ u ( A ) ⟩ ∣ Φ + ( B C ) ⟩ ⟶ ∣ trash ( A B ) ⟩ ∣ u ( C ) ⟩ \ket{u^{(A)}} \ket{\Phi_+^{(BC)}} \quad \longrightarrow \quad \ket{\text{trash}^{(AB)}} \ket{u^{(C)}} ∣ u ( A ) ⟩ ∣ Φ + ( B C ) ⟩ ⟶ ∣ trash ( A B ) ⟩ ∣ u ( C ) ⟩
Before observing it, Alice's qubit A holds the unknown state ∣ u ⟩ \ket{u} ∣ u ⟩ . B , C B, C B , C are entangled but unrelated to you.
After observing it, Bob's qubit now holds state ∣ u ⟩ \ket{u} ∣ u ⟩ . Alice's qubit A , B A,B A , 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)}} ∣ υ ( A ) ⟩ = α ∣ 0 ( A ) ⟩ + β ∣ 1 ( A ) ⟩
So
∣ u ( A ) ⟩ ∣ Φ + ( B C ) ⟩ = ( α ∣ 0 ( A ) ⟩ + β ∣ 1 ( A ) ⟩ ) ⊗ 1 2 ( ∣ 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) ∣ u ( A ) ⟩ ∣ Φ + ( B C ) ⟩ = ( α ∣ 0 ( A ) ⟩ + β ∣ 1 ( A ) ⟩ ) ⊗ 2 1 ( ∣ 0 ( B ) 0 ( C ) ⟩ + ∣ 1 ( B ) 1 ( C ) ⟩ )
= α 2 ( ∣ 000 ⟩ + ∣ 011 ⟩ ) + β 2 ( ∣ 100 ⟩ + ∣ 111 ⟩ ) =\frac{\alpha}{\sqrt{2}}(\ket{000}+\ket{011})+\frac{\beta}{\sqrt{2}}(\ket{100}+\ket{111}) = 2 α ( ∣ 000 ⟩ + ∣ 011 ⟩ ) + 2 β ( ∣ 100 ⟩ + ∣ 111 ⟩ )
Gives
= 1 2 ( ∣ Φ + ( A B ) ⟩ + ∣ Φ − ( A B ) ⟩ ) ⊗ α ∣ 0 ( C ) ⟩ = \tfrac{1}{2}\big(\ket{\Phi_+^{(AB)}} + \ket{\Phi_-^{(AB)}}\big) \otimes \alpha\ket{0^{(C)}} = 2 1 ( ∣ Φ + ( A B ) ⟩ + ∣ Φ − ( A B ) ⟩ ) ⊗ α ∣ 0 ( C ) ⟩
+ 1 2 ( ∣ Ψ + ( A B ) ⟩ + ∣ Ψ − ( A B ) ⟩ ) ⊗ α ∣ 1 ( C ) ⟩ + \tfrac{1}{2}\big(\ket{\Psi_+^{(AB)}} + \ket{\Psi_-^{(AB)}}\big) \otimes \alpha\ket{1^{(C)}} + 2 1 ( ∣ Ψ + ( A B ) ⟩ + ∣ Ψ − ( A B ) ⟩ ) ⊗ α ∣ 1 ( C ) ⟩
+ 1 2 ( ∣ Ψ + ( A B ) ⟩ − ∣ Ψ − ( A B ) ⟩ ) ⊗ β ∣ 0 ( C ) ⟩ + \tfrac{1}{2}\big(\ket{\Psi_+^{(AB)}} - \ket{\Psi_-^{(AB)}}\big) \otimes \beta\ket{0^{(C)}} + 2 1 ( ∣ Ψ + ( A B ) ⟩ − ∣ Ψ − ( A B ) ⟩ ) ⊗ β ∣ 0 ( C ) ⟩
+ 1 2 ( ∣ Φ + ( A B ) ⟩ − ∣ Φ − ( A B ) ⟩ ) ⊗ β ∣ 1 ( C ) ⟩ +\tfrac{1}{2}\big(\ket{\Phi_+^{(AB)}} - \ket{\Phi_-^{(AB)}}\big) \otimes \beta\ket{1^{(C)}} + 2 1 ( ∣ Φ + ( A B ) ⟩ − ∣ Φ − ( A B ) ⟩ ) ⊗ β ∣ 1 ( C ) ⟩
Gives
= 1 2 ∣ Φ + ( A B ) ⟩ ( 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)}} = 2 1 ∣ Φ + ( A B ) ⟩ ( α ∣ 0 ( C ) ⟩ + β ∣ 1 ( C ) ⟩ ) ⟶ ∣ u ( C ) ⟩
+ 1 2 ∣ Φ − ( A B ) ⟩ ( α ∣ 0 ( C ) ⟩ − β ∣ 1 ( C ) ⟩ ) ⟶ σ Z ( 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)}} + 2 1 ∣ Φ − ( A B ) ⟩ ( α ∣ 0 ( C ) ⟩ − β ∣ 1 ( C ) ⟩ ) ⟶ σ Z ( C ) ∣ u ( C ) ⟩
+ 1 2 ∣ Ψ + ( A B ) ⟩ ( α ∣ 1 ( C ) ⟩ + β ∣ 0 ( C ) ⟩ ) ⟶ σ X ( 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)}} + 2 1 ∣ Ψ + ( A B ) ⟩ ( α ∣ 1 ( C ) ⟩ + β ∣ 0 ( C ) ⟩ ) ⟶ σ X ( C ) ∣ u ( C ) ⟩
+ 1 2 ∣ Ψ − ( A B ) ⟩ ( α ∣ 1 ( C ) ⟩ − β ∣ 0 ( C ) ⟩ ) ⟶ i σ Y ( 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)}} + 2 1 ∣ Ψ − ( A B ) ⟩ ( α ∣ 1 ( C ) ⟩ − β ∣ 0 ( C ) ⟩ ) ⟶ i σ Y ( C ) ∣ 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_- 1 ⇒ Φ + 2 ⇒ Φ − 3 ⇒ Ψ + 4 ⇒ Ψ −
Example:
If Alice got outcome three (i.e., Ψ + \Psi_+ Ψ + ) then
∣ ϕ 3 ( C ) ⟩ = 1 P 3 ⟨ Ψ + ( A B ) ∣ υ ( A ) Φ + ( B C ) ⟩ = 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)}} ∣ ϕ 3 ( C ) ⟩ = P 3 1 ⟨ Ψ + ( A B ) ∣ υ ( A ) Φ + ( B C ) ⟩ = α ∣ 1 ( C ) ⟩ + β ∣ 0 ( C ) ⟩
Alice tells Bob she got 3
Bob's operation conditional on Alice's outcome
1 = I 2 = σ Z 3 = σ X 4 = σ Y 1=I\quad 2=\sigma^Z\quad 3=\sigma^X\quad 4=\sigma^Y 1 = I 2 = σ Z 3 = σ X 4 = σ Y
Bob choose σ X \sigma^X σ X as the base so
σ X ∣ ϕ 3 ( 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)}} σ X ∣ ϕ 3 ( C ) ⟩ = α ∣ 0 ( C ) ⟩ + β ∣ 1 ( C ) ⟩ = ∣ υ ( C ) ⟩
Bob now has reproduced ∣ υ ⟩ \ket{\upsilon} ∣ υ ⟩
Continuous Identity