# 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 unitrary operation $M$ with real amplitude on one bit is the same as doing the inverse operation $M^\dagger$ on the other.

• doing an unitrary operation $M$ with real amplitude 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 $\theta + 90^\circ$, which loses all meanings.
2. So we have to measure first and give instruction to receiver on how to interpret the result, since the outcome of our measurement (hence the state it actually collapse to) is random.
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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