Quantum Teleportation: Measurement and Communication

Recall, so far, all operations in preparation for the teleportation of the state to Bob, have been done locally by Alice on her system.

Following Alice's local operations ($\text{CNOT}_{QA}$ gate then Hadamard gate on $Q$), the state of the 3-qubit system, $|\Psi_2\rangle$, which we derived in the last lesson is:

$|\Psi_2\rangle = \frac{1}{2} [ \alpha(|000\rangle + |100\rangle + |011\rangle + |111\rangle) + \beta(|010\rangle - |110\rangle + |001\rangle - |101\rangle) ]_{QAB}$

The next part of the teleportation protocol involves Alice performing measurements on her two qubits.

1. Rewriting the State for Measurement Insight: To understand what happens when Alice measures her qubits $Q$ and $A$, it's crucial to rewrite $|\Psi_2\rangle$ by grouping terms based on the four possible computational basis states of Alice's two qubits ($|00\rangle_{QA}, |01\rangle_{QA}, |10\rangle_{QA}, |11\rangle_{QA}$). This explicitly shows the corresponding state of Bob's qubit $B$ for each of Alice's potential measurement outcomes.

Let's collect terms for each state of Alice's qubits ($QA$):

• For $|00\rangle_{QA}$: The terms from $|\Psi_2\rangle$ are $\frac{1}{2}(\alpha|000\rangle + \beta|001\rangle)$. This can be factored as:

  $\frac{1}{2}|00\rangle_{QA} \otimes (\alpha|0\rangle_B + \beta|1\rangle_B)$.

• For $|01\rangle_{QA}$: The terms are $\frac{1}{2}(\beta|010\rangle + \alpha|011\rangle)$. This can be factored as:

  $\frac{1}{2}|01\rangle_{QA} \otimes (\beta|0\rangle_B + \alpha|1\rangle_B)$.

• For $|10\rangle_{QA}$: The terms are $\frac{1}{2}(\alpha|100\rangle - \beta|101\rangle)$. This can be factored as:

  $\frac{1}{2}|10\rangle_{QA} \otimes (\alpha|0\rangle_B - \beta|1\rangle_B)$.

• For $|11\rangle_{QA}$: The terms are $\frac{1}{2}(\alpha|111\rangle - \beta|110\rangle)$. This can be factored as:

  $\frac{1}{2}|11\rangle_{QA} \otimes (\alpha|1\rangle_B - \beta|0\rangle_B)$.

So, the full state $|\Psi_2\rangle$ can be rewritten as:

$$\begin{align*} |\Psi_2\rangle &= \frac{1}{2}\Big( |00\rangle_{QA} \otimes (\alpha|0\rangle_B + \beta|1\rangle_B) \\ &+ |01\rangle_{QA} \otimes (\beta|0\rangle_B + \alpha|1\rangle_B) \\ & + |10\rangle_{QA} \otimes (\alpha|0\rangle_B - \beta|1\rangle_B) \\ &+ |11\rangle_{QA} \otimes (\alpha|1\rangle_B - \beta|0\rangle_B)\Big) \end{align*}$$

Note that the terms like $(\beta|0\rangle_B + \alpha|1\rangle_B)$ can be rewritten as $(\alpha|1\rangle_B + \beta|0\rangle_B)$ for easier comparison later.

2. Alice's Measurement: Alice now measures her two qubits, $Q$ and $A$, in the computational basis. This measurement will yield one of four possible results: $00$, $01$, $10$, or $11$. Let $m_Q$ be the outcome for qubit $Q$ and $m_A$ be the outcome for qubit $A$.

3. Probabilities and State Update: From the regrouped form of $|\Psi_2\rangle$, we can see that each of the four states for Alice's qubits ($|00\rangle_{QA}, |01\rangle_{QA}, |10\rangle_{QA}, |11\rangle_{QA}$) is associated with a specific state of Bob's qubit $B$. Since the initial state $|\psi\rangle_Q$ was normalized (meaning $|\alpha|^2 + |\beta|^2 = 1$), each of the conditional states for Bob's qubit (e.g.

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