/ #Teleportation 

Quantum Teleportation (Part 1): Preliminaries

Here we start exploring a core of Quantum Computing: Quantum Algorithms. Before we dive into the dark and deep pool of such algorithms, it’s helpful to get familiar with a fundamental constraint to Quantum Computing. In contrast, I’ll also introduce you how powerful qubits get in solving problems through a basic protocol: Quantum Teleportation, the “Hello, world!” of Quantum Programming.

No-cloning theorem

In terms of computer architechture, it’s convenient to copy the state of a bit in order to create a new qubit exactly the same. The protocol that seems really simple in the classical scenario is nonetheless proven impossible to perform quantumly. You should understand that while measuring a bit is extremely easy, just $0$ or $1$ with nearly perfect certainty, that process is difficult as mentioned in my previous posts. You can still measure a qubit, but what you really get is just no more than probabilities. They are not even coefficients, what are essential to fully illustrate a qubit state, and just obtained in a statistical manner rather than deterministically. For those reasons, one will never get suffcient information to recreate another version from an arbitrary qubit in general.

You may think that although we cannot recreate the qubit by knowing all of its amplitudes, we can copy its state to another without knowing the state. Unfortunatly, the cloning procedure is also unaccomplishable with the disproof as follows.

Let’s formalize the problem. Trong was given a qubit that can be in any state $|\psi\rangle$, and he want to duplicate his qubit’s state through a transformation that acts on his qubit and the second qubit initialized in $|0\rangle$. The whole process is displayed in the figure below.

Example image

We want a unitary transformation $U$ as a map that $U|\psi\rangle |0\rangle = |\psi\rangle |\psi\rangle$: $$(\alpha|0\rangle + \beta|1\rangle) \otimes |0\rangle \overset{U}{\rightarrow} (\alpha|0\rangle + \beta|1\rangle)(\alpha|0\rangle + \beta|1\rangle) \quad \forall \alpha, \beta\in \mathbb{C}$$ Second qubit replicating the first qubit means $|00\rangle \overset{U}{\rightarrow} |00\rangle$ and $|10\rangle \overset{U}{\rightarrow} |11\rangle$, which leads to the transformation of the LHS $$\begin{matrix} \alpha|00\rangle + \beta|10\rangle & \overset{U}{\rightarrow} & \alpha \cdot U|00\rangle + \beta \cdot U|11\rangle \\ & = & \alpha |00\rangle + \beta |11\rangle \end{matrix}$$ On the other hand, the expanded RHS is $$(\alpha|0\rangle + \beta|1\rangle)(\alpha|0\rangle + \beta|1\rangle) = \alpha^2 |00\rangle + \alpha\beta |01\rangle + \alpha\beta |10\rangle + \beta^2 |11\rangle$$ So we have a contradiction as $$\alpha |00\rangle + \beta |11\rangle = \alpha^2 |00\rangle + \alpha\beta |01\rangle + \alpha\beta |10\rangle + \beta^2 |11\rangle$$ only when $\alpha = 1$ or $\beta = 1$.

As you can see, it’s impossible to clone a qubit state unless the qubit is in the classical state $0$ and $1$. The no-cloning theorem introduces a fundamental barrier to the field as there isn’t a universal machine that handles the cloning problem. Imagining that you are free to clone an arbitrary qubit with no idea about its state, i.e. there exists a cloning unitary, you could overcome the post-measurement collapse of a qubit by performing as many experiments as you desire on the clones to figure out the true state of that qubit.

Bell State Circuit

We have already learned about Bell states through one of my posts on quantum entanglement. There are four Bell states totally, $|\Psi^+\rangle$, $|\Psi^-\rangle$, $|\Phi^+\rangle$, $|\Phi^-\rangle$. The Bell states are essential in several algorithms, one of which is teleportation. So, learning how to construct a Bell state is a necessity before we proceed towards the protocol.

Example image

Let’s get familiar with drawings in quantum circuits. An operator (usually called ‘gate’ in circuits) is a square that contains a letter standing for the meaning of the gate. Controlled-NOT (CNOT) gate includes a large black dot connected with $\otimes$. The black dot holds for the control qubit, and the $\oplus$ (addition in 2-modulus) is the operator that will act on the target qubit if the control qubit is $|1\rangle$. Following the circuit in the figure, we will be able to create the Bell state $\displaystyle{|\Phi^+\rangle = \frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |11\rangle}$.

First, we have two qubits both initialized in a basis state. $$|\psi\rangle = |00\rangle$$ Followed from the Hadamard gate acting on the first qubit, the composite state transforms into $$\begin{matrix} |\psi\rangle &\leftarrow& \displaystyle{\left(\frac{1}{\sqrt{2}} |0\rangle + \frac{1}{\sqrt{2}} |1\rangle\right) \otimes |0\rangle} \\ &=& \displaystyle{\frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |10\rangle} \end{matrix}$$ The CNOT gate adjusts components of the composite state $|\psi_1\rangle$ where the first qubit state is $|1\rangle$ and keeps the others the same. The performed transformation on the second qubit is a NOT gate (or $X$ gate), which is arithmatically the same as an addition module $2$. $$|\psi\rangle \leftarrow \frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |11\rangle$$ That final state is a Bell state that shows some characteristics helpful in teleportation and other algorithms in general. You should explore by yourself what the eventual results would be if we change the initialized states to $|01\rangle$, $|10\rangle$ and $|11\rangle$ respectively.


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