7. Multi-Qubit: Entanglement

Entanglement

This is the point where your intuition will try to fail you. Do not let it.

You just learned about Tensor Products, which allow us to combine independent qubits into a larger system (e.g., $|\Psi\rangle = |a\rangle \otimes |b\rangle$).

Entanglement is simply the realization that not all states in the larger system can be created this way.

1. The Definition: Product vs. Entangled

In the 4-dimensional space of two qubits ($\mathbb{C}^4$), most vectors cannot be factored back into two smaller vectors.

• Product State: A state that can be written as $|\psi\rangle_1 \otimes |\phi\rangle_2$. The qubits are independent. Measuring one tells you nothing about the other.
• Entangled State: A state where no such factorization exists. The qubits have lost their individual identity. They are no longer "Qubit A" and "Qubit B"; they are a single system sharing a probability distribution.

2. The Bell State ($|\Phi^+\rangle$)

The canonical example of entanglement is the Bell State:

$$|\Phi^+\rangle = \frac{1}{\sqrt{2}} (|00\rangle + |11\rangle) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}$$

Look closely at this vector. It is a superposition of "Both Zero" and "Both One".

It contains zero probability for the states $|01\rangle$ and $|10\rangle$.

3. The Consequence: Correlation

If you measure the first qubit of the $|\Phi^+\rangle$ state:

1. Randomness: You have a 50% chance of measuring $0$ and a 50% chance of measuring $1$.
2. Collapse:
   • If you measure 0, the state collapses to $|00\rangle$. The second qubit instantly becomes $|0\rangle$.
   • If you measure 1, the state collapses to $|11\rangle$. The second qubit instantly becomes $|1\rangle$.

There is no time delay. There is no signal sent between them. The correlation is absolute. If you know one, you know the other.

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Your Task: The Proof of Impossibility

You need to prove to yourself that $|\Phi^+\rangle$ cannot be broken down.

Assume that $|\Phi^+\rangle$ could be written as a product of two independent qubits:

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

Expand the tensor product on the left:

$$\alpha\gamma|00\rangle + \alpha\delta|01\rangle + \beta\gamma|10\rangle + \beta\delta|11\rangle$$

Now, match the coefficients with the Bell State on the right:

1. $\alpha\gamma = \frac{1}{\sqrt{2}}$ (Must be non-zero)
2. $\beta\delta = \frac{1}{\sqrt{2}}$ (Must be non-zero)
3. $\alpha\delta = 0$ (Middle terms must vanish)
4. $\beta\gamma = 0$ (Middle terms must vanish)

The Logic Puzzle:

Look at equations 3 and 4. For $\alpha\delta$ to be 0, either $\alpha$ or $\delta$ must be 0.

• If $\alpha = 0$, then equation 1 ($\alpha\gamma$) becomes 0. Contradiction.
• If $\delta = 0$, then equation 2 ($\beta\delta$) becomes 0. Contradiction.

Conclusion: The system of equations has no solution.

Question for you:
Since you cannot describe the Bell State using individual qubit coefficients ($\alpha, \beta, \gamma, \delta$), what does this imply about the "state" of Qubit 1 before it is measured? Does Qubit 1 even have a state?