# Lecture 014 - Maximally Entangled Pair

## Maximally Entangled Pair of Qubits

### Maximally Entangled Pair of Qubits

Make A, B
H A
CNOT A, B


The above code will produce

\sqrt{\frac{1}{2}}|00\rangle + \sqrt{\frac{1}{2}}|11\rangle = \sqrt{\frac{1}{2}} \begin{bmatrix} 1\\ 0\\ 0\\ 1\\ \end{bmatrix}

The above state is a "Bell Pair" or "The Einstein-Podolsky-Rosen (EPR) Pair", or "Maximally Entangled Pair"

There are many maximally entangled pair of qubits, but the above is the most famous one. Although isn't a way to quantify how "entangled" a pair is, but the $|MEP\rangle$ is definitely as most entangled as a pair can be.

Property of MEP:

• doing an operation $M$ on one bit is the same as doing the inverse operation $M^\dagger$ on the other.

• doing an operation $M$ twice on different qubits will cancel out the effect.

• any rotation of individual bits of $|MEP\rangle$ is still a $|MEP\rangle$ (only joint operations can get the state out of $|MEP\rangle$)

• measuring any bits in $|MEP\rangle$ will give $\frac{1}{2}$ chance of being $1$ and $\frac{1}{2}$ chance of being $0$.

### Super-dense Coding

We saw, in one homework, that you can hide one extra bit of information in the amplitude. Sending one of the qubit in $|MEP\rangle$ can convey 2 classical bits of information.

### Remote State Preparation

We use the fact that Alice doing operation $M$ on a joined state is the same as Bob doing operation $M^\dagger$ on the joined state. So we can transmit information by "helping" Bob to do certain operation.

Sending 1 classical bit after preparing $|MEP\rangle$ can convey 1 qubit of information. Here is how we do it:

1. We encode message in $\theta = 67.79661016...$.
2. Our goal is to send a state $Rot_{\theta}|0\rangle$ where $\theta$ is our sample space.
3. Prepare $|MEP\rangle$ and distribute to message sender and receiver.
\begin{align*} &\sqrt{\frac{1}{2}}(|00\rangle + |11\rangle) \tag{original MEP}\\ \to&\sqrt{\frac{1}{2}}(|0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle) \tag{separate}\\ \to&\sqrt{\frac{1}{2}}(|0\rangle \otimes Rot_{\theta}|0\rangle + |1\rangle \otimes Rot_{\theta}|1\rangle) \tag{sender rotate his qubit}\\ \end{align*}
1. If we now let the receiver measure the qubit, the receiver would not be able to know which angle we have sent $\theta$ or $\theta + 90\degree$, which loses all meanings. // QUESTION: is it meaning less? how do we extract information in theta?
2. So we have to measure first and give instruction to receiver on how to interpret the result.
3. With $Pr\{A = 0\} = \frac{1}{2}$ state collapse to $|0\rangle \otimes Rot_\theta|0\rangle$. With $Pr\{A = 1\} = \frac{1}{2}$ state collapse to $|1\rangle \otimes Rot_\theta|1\rangle$
4. So we send 1 bit information to receiver let him know which one is intended.

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