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PressXAI-UniversityTM - Quantum Entanglement
Adrian W. Pasieka Ikane

Socrates "I know that I know nothing."
Qubit "I don't show , but I know."

'(2021 03 21)

The traditional way of looking into a quantum entanglement, as some mysterious connection between the quantum states, doesn't necessarily have to be right.

A brief summary for a superdense code algorithm is presented below, using the nomenclature in accordance with the ANSI Common Lisp, allowing an easy verification in the PressXAI-Quantum-SimulatorTM.


All operations made on the vectors and matrices are also presented in form of the Lists, which is very useful when analyzing and coding the whole system.

1) The qubit A is in a state (A 0 1).
2) The qubit B is in a state (B 1 0).

Please note that in a traditional approach the qubit A and B are in the same state (A 1 0) and (B 1 0), but there isn't any theoretical requirement for it, and there is a reason for starting from a different states, explained later on.

Definitions:

(defun QUANTUM-HADAMARD (target reg)
"First order gate"
(let ((m (quantum-new-matrix 2 2)))
(if (quantum-objcode-put +HADAMARD+ target)
(return-from quantum-hadamard))
(setf (aref (quantum-matrix-t1 m) 0) (sqrt (/ 1.0 2.0)))
(setf (aref (quantum-matrix-t1 m) 1) (sqrt (/ 1.0 2.0)))
(setf (aref (quantum-matrix-t1 m) 2) (sqrt (/ 1.0 2.0)))
(setf (aref (quantum-matrix-t1 m) 3) (- (sqrt (/ 1.0 2.0))))
(quantum-gate1 target m reg)))

(defun QUANTUM-CNOT-FT (control target reg)
"Fault-tolerant version of Controlled NOT gate"
(let ((tmp 0)
(lambda 0.0))
(setq tmp qec-type)
(setq qec-type 0)
(setq lambda (quantum-get-decoherence))
(quantum-set-decoherence 0)
(quantum-cnot control target reg)
(quantum-cnot (+ control qec-width) (+ target qec-width) reg)
(quantum-set-decoherence lambda)
(quantum-cnot (+ control (* 2 qec-width)) (+ target (* 2 qec-width)) reg)
(quantum-qec-counter 1 0 reg)
(setq qec-type tmp) ) )

(defun QUANTUM-KRONECKER (reg1 reg2)
"Compute the Kronecker product of two quantum registers"
(let* ((reg-width (+ (quantum-reg-width reg1) (quantum-reg-width reg2)))
(reg-size (* (quantum-reg-size reg1) (quantum-reg-size reg2)))
(reg-hashw (+ 2 reg-width))
(l1 0)
(reg (make-quantum-reg
:width reg-width
:size reg-size
:hashw reg-hashw
:state (make-array reg-size :element-type 'integer :adjustable 't)
:amplitude (make-array reg-size
:element-type 'complex
:adjustable 't
:initial-element #c(0.0 0.0))
:hash (make-array (ash 1 reg-hashw)
:adjustable 't
:element-type 'integer))))
(loop for i from 0 below (quantum-reg-size reg1) do
(loop for j from 0 below (quantum-reg-size reg2) do
(setq l1 (+ j (* i (quantum-reg-size reg2))))
(setf (aref (quantum-reg-state reg) l1)
(logior (ash (aref (quantum-reg-state reg1) i) (quantum-reg-width reg2))
(aref (quantum-reg-state reg2) j)))
(setf (aref (quantum-reg-amplitude reg) l1)
(* (aref (quantum-reg-amplitude reg1) i) (aref (quantum-reg-amplitude reg2) j)))))
reg))

3) The qubits are prepared, and entangled through the QUANTUM-HADAMARD transform from (A 0 1) to (A 1/sqrt2 -1/sqrt2), and the QUANTUM-CNOT-FT on their superposition, which yields the result (AB 1/sqrt2 0 0 -1/2sqrt2), as described below.

(QUANTUM-CNOT-FT

(QUANTUM-KRONECKER

(QUANTUM-HADAMARD (A 0 1)

(A 1/sqrt2 -1/sqrt2) ) (B 1 0)

(AB 1/sqrt2 0 -1/sqrt2 0) )

(AB 1/sqrt2 0 0 -1/sqrt2) )

Next, as the popular superdense code algorithm indicates, the qubit A is been sent to Alice, and the qubit B to Bob.

Here is the first question - what in reality is the state of the qubit A received by Alice?.

If this is a A qubit in its initial state , before a QUANTUM-HADAMARD, it means that this is not the entangled qubit. If it is after the QUANTUM-HADAMARD, and the QUANTUM-CNOT-FT , it means that its states can't be modified by Alice, without collapsing the entangled state, which creates the basis of the theory.

So, the assumption taken here is that the A qubit sent to Alice is in the state after the QUANTUM-HADAMARD, and before the QUANTUM-CNOT-FT, which creates (A1/sqrt2 - 1/sqrt2).

4) The are four possible states and three modifications of the (A 1/sqrt2 - 1/sqrt2) by Alice, which work as some key code describing the classical bits a and b, which we want to send the information about to Bob.

(a0b0 transformed by nothing)
(= after transformation (A 1/sqrt2 - 1/sqrt2))

(a0b1 ((1 0) (0 -1)))
(= after transformation (A 1/sqrt2 1/sqrt2))

(a1b0 (( 0 1) (1 0)))
(= after transformation (A -1/sqrt2 1/sqrt2))

(a1b1 (( 0 1) (-1 0)))
(= after transformation (A -1/sqrt2 -1/sqrt2))

Now Alice sends the A qubit to Bob, and there are two transforms performed, which reverse the entanglement process, the first one is made on the pairs (AB) by the QUANTUM-CNOT-FT, bringing the pair back to the state (AB 1/sqrt2 0 -1/sqrt2 0).

The second one is the QUANTUM-HADAMARD on the qubit A, which creates the following states below.

The rotation properties are not obvious, because if a quantum state has some distance to pass, to turn into another state, it should take a different time. This property could be possibly measured.

Also, applying any transform, on any state, changes its physical superposition, not only the amplitude value, and it requires the other states to change their superpositions, too. It resembles a similarity to a continuous entanglement.


(A 1/sqrt2 -1/sqrt2) into (A 0 1) = 135 degrees left rotation up

(A 1/sqrt2 1/sqrt2) into (A 1 0) = 45 degrees right rotation down

(A -1/sqrt2 1/sqrt2) into (A 0 -1) = 135 degrees left rotation down

(A -1/sqrt2 -1/sqrt2) into (A -1 0) = 45 degrees right rotation up

The above results show that there is no necessity for talking about a mysterious entanglement state, or a spooky action at a distance.

This state is absolutely necessary to bind the qubit A and B, to keep the qubit B in an unequivocal state, otherwise the output could not have been measured properly, as the B qubit wouid rotate, sustaining a superposition, without and eigenvalue.

There is no such a thing like a superposition OR an entanglement, the entanglement is a special case of the superposition, where a KRONECKER-PRODUCT can't be derived.

Both Einstein and Bohr could be right, and this opens up a new field for research and experiments, with the results to be used in a quantum information and the other sciences.


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