Teleportation

Transmitting a quantum state from one place to another without that state traversing the space in between.

• Assume, that Alice wishes to send state $\ket{\chi}$ to Bob, where

  $\ket{\chi} = \alpha\ket{0} + \beta\ket{1}$

STEPS TO FOLLOW

1. Alice & Bob share an Entangled Pair of Particles:

   Alice & Bob create an entangled state, where the first member of the pair belongs to Alice & the second member of the pair belongs to Bob. The entangled state at hand being:

   $\ket{\beta_{00}} = \large\frac{\ket{0_A0_B} + \ket{1_A1_B}}{\sqrt{2}}$

   Note, that the qubits owned by Alice have been marked with a subscript A (similarly B for Bob).

   Following this, Alice Makes the state $\ket{\psi}$ interact with her member of the Entangled Qubit Pair. Giving us:

   $\ket{\psi} = \ket{\chi} \otimes \ket{\beta_{00}} = (\alpha\ket{0} + \beta\ket{1})\otimes\large\frac{\ket{0_A0_B} + \ket{1_A1_B}}{\sqrt{2}}$

   $= \large\frac{\alpha(\ket{0_A0_A0_B} + \ket{0_A1_A1_B}) + \beta(\ket{1_A0_A0_B} + \ket{1_A1_A1_B})}{\sqrt{2}}$

2. Alice Applies a CNOT Gate to her Two Qubits

   Applying CNOT (on the first two qubits, which are owned by Alice), on our new state, we get:

   $\ket{\psi’} = \large\frac{\alpha(\ket{0_A0_A0_B} + \ket{0_A1_A1_B}) + \beta(\ket{1_A1_A0_B} + \ket{1_A0_A1_B})}{\sqrt{2}}$

3. Alice Applies a Hadamard Gate to her First Qubit

   $\ket{\psi’} = \large\frac{\alpha\ket{0_A}(\ket{0_A0_B} + \ket{1_A1_B}) + \beta\ket{1_A}(\ket{1_A0_B} + \ket{0_A1_B})}{\sqrt{2}}$

   applying the Hadmard Gate to the first Qubit owned by Alice, we get:

   $\ket{\psi"} = \normalsize\alpha\large\left(\frac{\ket{0_A} + \ket{1_A}}{\sqrt{2}}\right)\frac{(\ket{0_A0_B} + \ket{1_A1_B})}{{\sqrt{2}}} \normalsize+ \beta\large\left(\frac{\ket{0_A} - \ket{1_A}}{\sqrt{2}}\right)\frac{(\ket{1_A0_B} + \ket{0_A1_B})}{\sqrt{2}}$

   

4. Alice Measures her Pair

   • Case 1: Alice Measures $\ket{00}$

     Bob has the unknown state that Alice wanted to send Bob.

   • Case 2: Alice Measures $\ket{01}$

     Bob can achieve the unknown state, by applying the $X$ gate to his qubit:

     $X(\alpha\ket{1} + \beta\ket{0}) = \alpha\ket{0} + \beta\ket{1}$

   • Case 3: Alice Measures $\ket{10}$

     Bob can achieve the unknown state, by applying the $Z$ gate to his qubit:

     $Z(\alpha\ket{0} - \beta\ket{1}) = \alpha\ket{0} + \beta\ket{1}$

   • Case 4: Alice Measures $\ket{11}$

     Bob can achieve the unknown state, by applying the $X$ gate followed by the $Z$ gate to his qubit:

     $ZX(\alpha\ket{1} - \beta\ket{0}) = \alpha Z\ket{0} - \beta Z\ket{1} = \alpha\ket{0} + \beta\ket{1}$