Results

Each circuit has been diagnostically traced without collapsing the wavefunction i.e. it has not been measured yet and is still useable by Bob for downstream quantum gate steps.

At the end of the teleporation process Alice's and Bob's first and second qubits have become randomized |xx-> with x as 50% |0> and 50% |1>.

This destruction of original state information in the first and second qubit also makes them reusable as a randomized entangled qubits for further quantum gate stages.

For example, a CNOT gate with Bob's german qubit as control and his swahili qubit as target would have the effect of aligning the swahili qubit state with the german qubit state. In this application scenario Alice's input english qubit state would then also become the swahilis qubit state via the teleported german qubit

The probability of each state vector is shown in each three qubit circuit: mouse over bars for detail.

The |0> state of the teleported qubit is shown in the pattern of darker horizontal bars i.e. |000>, |010>, |100> and |110> each having '0' in the third qubit in the state vector |--0>, eg. |xoom, sufuri, null >

The |1> state of the teleported qubit is shown in the lighter horizontal bars i.e. |001>, |011>, |101> and |111> each having '1' in the third qubit in the state vector |--1>, eg. | plaub, nne, vier >.

Circuits 0 to 4 show a pattern of 100% darker to 100% lighter with 2 being 50% dark and light. This result is exactly as expected according to the test code shown below using i % 5, where 'M' is the number of steps:

# test |0> |1> mixtures i%M

Alice is sending the quantum state qubit of the first qubit in the circuit identified by the english language name of the circuit, eg. zero. The test's code shows that the phase of the english qubit is in five steps 0,1,2,3,4 between |0> for 'zero' in circuit 0 to |1> for 'four' in circuit 4.

M = 5  # Modulo phase input to qubits
Q = 0.25  # phase delta from |0> to |1>
R = Q / (M-1)   # phase step size

# test |0> |1> mixtures i%M
def englishPhase(i):
 y = (i % M) * R
 return y
	

Alice's second qubit was |-0-> in the circuits 0,1,2,3 and 4
versus |-1-> in the circuits 5,6,7,8 and 9. The probabilities for each state vector is the same as in in the circuits 0,1,2,3 and 4. This demostrates that the second qubit performs the function of a neutral intermediary between the top row of the 3 qubit circuit and the bottom row in the teleporation process. Alice's neutral intermediary 2nd qubit can be in either |0> or |1> input state and there will be no effect on the teleportation result.

# test |0> |1> combinations int(i/M)%2
def latinPhase(i):
 y = int(i / M) % 2 * Q
 return y
	

Alice's third qubit was |--0> in the circuits 0 through 9
versus |--1> in the circuits 10 through 9. The probabilities for each state vector is reversed as shown by the mirror image symmetry in the state probability plots in the boundary between circuits nine and ten, nineteen and twenty and each following ten circuit group.

# test |0> |1> combinations int(i/2*M)%2
def koreanPhase(i):
 y = int(i / (2*M)) % 2 * Q
 return y