Lecture 014 - Maximally Entangled Pair

Maximally Entangled Pair of Qubits

Maximally Entangled Pair of Qubits

Make 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:

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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