# Quantum Teleportation

Quantum teleportation is the idea that entangling states can cause very fast information transfer. The name however is misleading as this information transfer is not instantaneous but simply travels at the speed of light. The basis of this idea is based on a very interesting trick that arises out of some of the math of quantum computing. Let’s examine a simple case in which quantum teleportation can be used.

The figure below describes a common method for quantum teleportation where the upper dashed box represents the system of person A, the one sending information, and the lower box represents the system of person B, the one receiving information.

Person A has a qubit in the state $\inline&space;|\psi\rangle&space;=&space;\alpha|0\rangle+\beta|1\rangle$ and wants to send it to person B. A can’t simply send numbers to B describing $\inline&space;\alpha$ and $\inline&space;\beta$ because it would require an infinite amount of information to describe every decimal. They must use quantum teleportation and somehow send the qubit to B.

In order to do this, before they separate, they both take a qubit with state $\inline&space;|0\rangle$ and B applies a Hadamard gate (description of all gates used throughout the post are at the bottom of the post). to his.  Now, his qubit is in the state $\inline&space;\frac{|0\rangle+|1\rangle}{\sqrt{2}}$resulting in a system in the state

$|\Psi\rangle=(\alpha|0\rangle+\beta|1\rangle)(|0\rangle)\frac{|0\rangle+|1\rangle}{\sqrt{2}}=(\alpha|0\rangle+\beta|1\rangle)\frac{|00\rangle+|01\rangle}{\sqrt{2}}$

Applying a CNOT gate with B’s qubit as the control and A’s as the target results in the following state (Note: these last two actions simply turned the two ground state bits into an EPR pair).

$|\Psi\rangle=(\alpha|0\rangle+\beta|1\rangle)\frac{|00\rangle+|11\rangle}{\sqrt{2}}$

In order to do the next step described in the figure (a CNOT), one must distribute the states of $\inline&space;|\psi\rangle$ as shown.

$|\Psi\rangle=\alpha\frac{|000\rangle+|011\rangle}{\sqrt{2}}+\beta\frac{|100\rangle+|111\rangle}{\sqrt{2}}$

From here, a CNOT gate can again be applied keeping the target qubit the same but changing the control qubit to $\inline&space;|\psi\rangle$ (Note: all that is done in this step is applying the transformation $\inline&space;|a,b,c\rangle&space;\rightarrow&space;|a,&space;b\oplus_2a,&space;c\rangle$ where the 2 indicates modulo 2).

$|\Psi\rangle=\alpha\frac{|000\rangle+|011\rangle}{\sqrt{2}}+\beta\frac{|110\rangle+|101\rangle}{\sqrt{2}}$

Applying a Hadamard gate to $\inline&space;|\psi\rangle$ and factoring out results in the following.

$|\Psi\rangle=\alpha\frac{|0\rangle+|1\rangle}{\sqrt{2}}\frac{|00\rangle+|11\rangle}{\sqrt{2}}+\beta\frac{|0\rangle-|1\rangle}{\sqrt{2}}\frac{|10\rangle+|01\rangle}{\sqrt{2}}$

This is distributed and rearranged below.

$|\Psi\rangle=\alpha\frac{|0\rangle+|1\rangle}{\sqrt{2}}\frac{|00\rangle+|11\rangle}{\sqrt{2}}+\beta\frac{|0\rangle-|1\rangle}{\sqrt{2}}\frac{|10\rangle+|01\rangle}{\sqrt{2}}$

$=\frac{\alpha(|000\rangle&space;+&space;|100\rangle+|011\rangle+|111\rangle)+\beta(|010\rangle+|001\rangle-|110\rangle-|101\rangle)}{2}$

$=\frac{|00\rangle(\alpha|0\rangle+\beta|1\rangle)+|10\rangle(\alpha|0\rangle-\beta|1\rangle)+|01\rangle(\alpha|1\rangle+\beta|0\rangle)+|11\rangle(\alpha|1\rangle-\beta|0\rangle)}{2}$

Notice the four distinct possible states that form. If the first two qubits are both in the ground state, the last qubit will be in state $\inline&space;|\psi\rangle$. If the first is excited and the second is ground, then if one applies a phase flip gate to the last state qubit, they get $\inline&space;|\psi\rangle$. Excited and ground would require a not gate. Two excited states would require both. This means that if A measures the first two qubits after all these gates and sends the information to B and B applies the appropriate transformations, he successfully transformed his regular ground state qubit into the one that A originally had. These “appropriate transformations” are simply shown in the diagram as the G gate.

So as shown by this method, it is possible to send information through the transformation of a qubit. Although B’s qubit instantly changes upon the measurement of A’s, B needs the information sent by A in order to perform the G gate and this information can only be sent at the speed of a light. As a result, it is not instantaneous but is still very fast and can send information complex messages like this qubit very accurately.

If you want to know more or see where I learned this from, read the book “Quantum Computation and Information” by Isaac Chuang and Michael Nielsen.It should be in the “Books” sections of this site.

Gates

$\inline&space;X$ : $\begin{matrix}&space;|0\rangle\rightarrow&space;|1\rangle\\&space;|1\rangle\rightarrow&space;|0\rangle&space;\end{matrix}$ or $\begin{bmatrix}&space;0&space;&&space;1\\&space;1&space;&&space;0&space;\end{bmatrix}$ (NOT gate)

$Z:$ $\begin{matrix}&space;|0\rangle\rightarrow|0\rangle\\&space;|1\rangle\rightarrow-|1\rangle&space;\end{matrix}$ or $\begin{bmatrix}&space;1&space;&&space;0\\&space;0&space;&&space;-1&space;\end{bmatrix}$ (Phase-flip gate)

$G(I_1,&space;I_2)=X^{I_1}Z^{I_2}$

$H:$ $\inline&space;\begin{matrix}&space;|0\rangle\rightarrow&space;\frac{|0\rangle+|1\rangle}{\sqrt{2}}\\&space;|1\rangle\rightarrow&space;\frac{|0\rangle-|1\rangle}{\sqrt{2}}&space;\end{matrix}$ or $\frac{1}{\sqrt{2}}\begin{bmatrix}&space;1&space;&&space;1\\&space;1&space;&&space;-1&space;\end{bmatrix}$ (Hadamard gate)

$CNOT:$ $|n_1,n_2\rangle\rightarrow|n_1,n_2\oplus_2&space;n_1\rangle$ or $\begin{bmatrix}&space;1&0&0&0\\&space;0&1&0&0\\&space;0&0&0&1\\&space;0&0&1&0&space;\end{bmatrix}$ (Controlled-NOT gate)