# In quantum teleportation, does Bob's qubit depend on Alice's $\alpha$ and $\beta$ coefficients?

Does applying an operation to one half of an entangled pair affect the other? As in the case of quantum teleportation when Alice applies CNOT to her half of the pair and the qubit she wants to send ($$\alpha|1\rangle +\beta|1\rangle$$). Does Bob's half also depend on $$\alpha$$ and $$\beta$$? How?

The answer is negative: Bob's half of the entangled pair is unaffected by operations performed by Alice on her half. However, there is a caveat. It is the hallmark of entanglement that the state of either half is an incomplete description of the pair. The complete description is given by the joint state. The joint state is very much affected by what Alice does to her half of the pair. It is the effect on the joint state that enables teleportation.

## Proof for unitary operations

Perhaps the easiest way to see this is in the case where the operation Alice performs on her half of the pair is unitary. Let $$|\psi\rangle_{AB}$$ be the joint initial state of the composite system $$AB$$ where $$A$$ denotes Alice's half of the pair and $$B$$ denotes Bob's half. Consider the Schmidt decomposition $$|\psi\rangle_{AB} = \sum_i \lambda_i |i_A\rangle|i_B\rangle$$. The state of Bob's subsystem is $$\rho_B = \mathrm{tr}_A(|\psi\rangle_{AB}\langle\psi|) = \sum_i\lambda_i^2|i_B\rangle\langle i_B|$$. Any unitary operation on Alice's half of the pair merely changes the states $$|i_A\rangle \to |i_A'\rangle$$ and so we see that Bob's state remains unaffected.

## Proof for CPTP maps

More generally, suppose that Alice applies a completely positive trace-preserving map $$\mathcal{E}_A$$ with Kraus operators $$E_{i,A}$$ to her half of the pair and Bob does nothing to his. The resulting joint state is $$\rho'_{AB} = (\mathcal{E}_A\otimes \mathcal{I}_B)(|\psi\rangle_{AB}\langle\psi|)$$. The resulting state of Bob's half is

\begin{align} \rho'_B &= \mathrm{tr}_A\left((\mathcal{E}_A\otimes \mathcal{I}_B)(|\psi\rangle_{AB}\langle\psi|)\right) \\ &= \sum_i\mathrm{tr}_A\left((E_{i,A}\otimes I_B)|\psi\rangle_{AB}\langle\psi|(E_{i,A}^\dagger \otimes I_B)\right) \\ &= \mathrm{tr}_A\left(|\psi\rangle_{AB}\langle\psi|\sum_i E_{i,A}^\dagger E_{i,A}\right) \\ &= \mathrm{tr}_A(|\psi\rangle_{AB}\langle\psi|) \\ &= \rho_B \end{align}

where we used the cyclic property of the partial trace for operators acting on the subsystem being traced over and the completeness relation for Kraus operators

$$\sum_i E_{i,A}^\dagger E_{i,A} = I_A.$$

## Example

Suppose Alice and Bob share an entangled pair in the state

$$|\psi\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}.$$

The initial state of Bob's half of the pair is

\begin{align} \rho_B &= \mathrm{tr}_A(|\psi\rangle\langle\psi|) \\ &= \frac{1}{2}\mathrm{tr}_A(|00\rangle\langle 00| + |00\rangle\langle 11| + |11\rangle\langle 00| + |11\rangle\langle 11|) \\ &= \frac{1}{2}(|0\rangle\langle 0| + |1\rangle \langle 1|). \end{align}

Now suppose that Alice applies Hadamard gate to her half of the pair

$$|\psi'\rangle = (H \otimes I) \frac{|00\rangle + |11\rangle}{\sqrt{2}} = \frac{|+\rangle|0\rangle + |-\rangle|1\rangle}{\sqrt{2}}.$$

The new state of Bob's half of the pair is

\begin{align} \rho'_B &= \mathrm{tr}_A(|\psi'\rangle\langle\psi'|) \\ &= \frac{1}{2}\mathrm{tr}_A(|+\rangle|0\rangle\langle +|\langle 0| + |+\rangle|0\rangle\langle -|\langle 1| + |+\rangle|0\rangle\langle -|\langle 1| + |-\rangle|1\rangle\langle -|\langle 1|) \\ &= \frac{1}{2}(|0\rangle\langle 0| + |1\rangle \langle 1|) \end{align}

i.e. the same as before Alice applied Hadamard.

## Relationship to teleportation

Teleportation entails the following steps

1. Alice interacts the input qubit $$|\phi\rangle$$ with her half of an entangled pair.
2. Alice performs a measurement on the input qubit and her half of the entangled pair.
3. Alice transmits classical measurement outcome to Bob.
4. Bob uses the classical information to recover $$|\phi\rangle$$.

Note that Bob is unable to recover $$|\phi\rangle$$ until he receives classical information about measurement outcomes from Alice. This reflects the fact that the state of his half of the pair is not affected by the operations that Alice performs in steps 1 and 2 and therefore is insufficient to recover $$|\phi\rangle$$. However, it is possible to recover $$|\phi\rangle$$ from the joint state of the three qubits. The measurement outcomes transmitted classically by Alice together with the state of Bob's half of the pair together supply him with the necessary knowledge of the joint state and enable him to recover $$|\phi\rangle$$.

## Consistency with special relativity

The fact that the state of Bob's half of the pair is unaffected by operations performed by Alice on her half denies the use of entanglement for faster-than-light communication. It is also reassuring because it means that quantum mechanics does not come into conflict with special relativity over the fact that if Alice's and Bob's actions are spacelike separated then different observers may disagree on the order in which the actions take place due to relativity of simultaneity.