Quantum Teleportation II

In the previous notebook we illustrated how Bell states arise naturally in nature. We can also
construct a quantum circuit that creates Bell states from an ordinary 2-Qubit such as |00⟩. Consider
the circuit, which we will call a Bell gate (B)

Lets suppose that the state | 0 0 ⟩ is the input state. After going through the Hadamard gate it becomes

H |0 ⟩ |0⟩  = (1/(Sqrt[2]))( |0⟩+|1⟩) |0⟩ = (1/(Sqrt[2])) ( |00⟩+|10⟩ )

the 2-Qubit then goes through a C-not gate which turns the 2- Qubit into
(1/(Sqrt[2])) ( |00⟩+|11⟩ )  which we recognize as the |(Subscript[β, 00])⟩ Bell state.

Class exercise

Evaluate

B |01⟩
B |10⟩
B |11⟩

Now that we know how to make Bell states, what do we do with them? One important application
is Quantum Teleportation. Assume Alice and Bob can share a photon or electron pair that is in a Bell
state. eg.

|(Subscript[β , 00])⟩   =    (|R R⟩ - |L L ⟩)/(Sqrt[2])

Alternatively, Alice and Bob can make a Bell state using the circuit given above. Now consider the following
circuit

The top Qubit  |ψ⟩ = α|0⟩+ γ|1⟩  belongs to Alice, she wants to send it to Bob. So Alice takes another Qubit (photon, electron etc. ) |(Subscript[q, 2])⟩, and puts it into  a Bell state with Bob’s qubit |(Subscript[q, 3])⟩.  After the three Qubits leave
the Bell gate the system state looks like


|ψ⟩ | (Subscript[β, 00])⟩ = |ψ⟩(1/(Sqrt[2])) ( |00⟩+|10⟩ ) = (α|0⟩+ γ|1⟩) (1/(Sqrt[2])) ( |00⟩+|10⟩ )

After exiting the Bell gate we hit the C-not gate connecting the first and second Qubit.
Exiting from the C-not the system is in the state.

 α|0⟩(1/(Sqrt[2])) ( |00⟩+|10⟩ ) + γ|1⟩(1/(Sqrt[2])) ( |10⟩+|00⟩ )        ( Can you see why ?)

Now the first Qubit goes through a Hadamard gate and we get

αH |0⟩(1/(Sqrt[2])) ( |00⟩+|10⟩ ) + γH |1⟩(1/(Sqrt[2])) ( |10⟩+|00⟩ ) =

(1/2) α ( |0⟩ + |1⟩ ) ( |00⟩+|10⟩ ) + (1/2) γ ( |0⟩- |1⟩)   ( |10⟩+|00⟩ )

which I am going to re-arrange as

(1/2) |00⟩(α |0⟩ + γ |1⟩ )  + (1/2) |01⟩ (α |1⟩ + γ |0 ⟩ )  + (1/2) |10⟩(α |0⟩ -  γ |1⟩ ) + (1/2) |11⟩ (α |1⟩ -  γ |0⟩ )

Remember, the first two Qubits belong to Alice while the last Qubit belongs to Bob. Now let’s
look at the output states, before measurements are made.

(1/2) |00⟩(α |0⟩ + γ |1⟩ )  + (1/2) |01⟩ (α |1⟩ + γ |0 ⟩ )  + (1/2) |10⟩(α |0⟩ -  γ |1⟩ ) + (1/2) |11⟩ (α |1⟩ -  γ |0⟩ )  
 
Alice now makes a measurement of her two Qubits. What possible results can she get from her measurements?

Answer 00, 01, 10,11

Question: what is the probability that Alice gets 00

Answer: 1/4

Suppose she does get 00, what does the system collapse into?

Answer: using the generalized Born rule

|00⟩ (α |0⟩ + γ |1⟩ ) 

After this collapse, Alice is sure that Bob has the desired state |ψ ⟩, but how does Bob know this? 
Alice calls up Bob on the telephone and tells him that she just got the result 00 for her measurement
and therefore Bob has the desired state |ψ⟩ encoded in his Qubit. Mission accomplished !

But Alice does not always get 00, she only gets that 25% of the time. She might also get the result
01. What does the state collapse into if this is the case ?

(1/2) |00⟩(α |0⟩ + γ |1⟩ )  + (1/2) |01⟩ (α |1⟩ + γ |0 ⟩ )  + (1/2) |10⟩(α |0⟩ -  γ |1⟩ ) + (1/2) |11⟩ (α |1⟩ -  γ |0⟩ ) 

if Alice measures 01 the state collapses into
|01⟩ (α |1⟩ + γ |0 ⟩ )

Alice then knows that Bob’s Qubit is given by  α |1⟩ + γ |0 ⟩ which is not the desired state (note |0⟩ and
|1⟩ ) are interchanged. Not to worry!  Alice calls Bob and tells him; Hey Bob, I just measured 01 you need
to send your Qubit through Pauli X gate in order to get |ψ⟩  (Question: Is Bob cheating here?  Is he in a sense measuring is Qubit by putting it through a Pauli gate ? )

This procedure of Alice measuring and calling Bob to tell him of her results, is shown in the diagram


What should Bob do if Alice calls him and tells him she measured 01 ?
What if she measured 11 ?