# On what qubit does Bob perform Pauli Correction in Quantum teleportation protocol?

In the lecture notes it is said that Bob applies the Pauli correction operation $X^x \cdot Z^z$ on his qubit. My questions:

1) Am I correct that Pauli correction means to multiply the qubit Bob has by a matrix $X^x \cdot Z^z$?

2) Am I correct that the entangled qubit $\beta_{00}$ can be created far in advance before any quantum teleportation happens?

3) If Alice and Bob share only one entangled qubit $\beta_{00}$, which is not related to $|\psi\rangle$ (or is it?), why a measurement on Alice's side allows Bob to infer something about $|\psi\rangle$? Shouldn't a measurement on Alice's side allow Bob to infer only something about $\beta_{00}$?

I think your lecture notes are not so clear - I would instead just look at Wikipedia.

1) You are correct that the 'Pauli correction' is to apply a correction operator (either $X$, $Z$, or some combination) to Bob's qubit.

2) The initial setup of this problem is very important. Alice starts with one qubit with state $|\psi\rangle$, and in addition Alice and Bob each have one part of an entangled Bell pair. Alice performs a projective measurement in the Bell basis on her pair of qubits - the pair consisting of the $|\psi\rangle$ qubit and one half of the entangled pair. After her measurement, Bob's qubit is also affected - and depending on the result of Alice's measurement, Bob knows to apply this correction matrix to rotate his qubit to $|\psi\rangle$.

Note the key here is that Alice and Bob share an entangled pair, and that Alice's measurement affects the state of Bob's qubit. The only information transmitted from Alice to Bob is the classical information corresponding to the result of Alice's projective measurement - ie., which of the four Bell pairs did her measurement yield.

2) Yes, the entangled pair of qubits shared by Alice and Bob can be created far in advance of any teleportation.

3) The state of the entangled pair shared by Alice and Bob is completely unrelated to the state $|\psi\rangle$ - indeed, the entangled pair is in a well defined Bell state in any teleportation measurement, such as $|\phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$.

I will work through the key step here: suppose Alice and Bob have the entangled qubits $|\phi^+\rangle_{AB}$, where AB denotes the qubits A and B. Suppose further that Alice has the information qubit $|\psi\rangle_C = \alpha|0\rangle_C + \beta|1\rangle_C$ corresponding to a third qubit C. The initial state of all three qubits could be fully written as: $$|\phi^+\rangle_{AB} |\psi\rangle_C = |\phi^+\rangle_{AB} (\alpha|0\rangle_C + \beta|1\rangle_C)$$ The key step is now to change the basis in which we write this initial state. In the above expression, we write the state of A and B in the Bell basis over qubits A and B, and we write the state of C in the $|0\rangle, |1\rangle$ basis. We will now change bases to express A and C in the basis of Bell states of qubits A,C, and we express B instead in the basis of $|0\rangle, |1\rangle$. $$\begin{eqnarray*} \frac{1}{2}\bigg[&|\phi^+\rangle_{AC} \big(\alpha|0\rangle_B +\beta|0\rangle_B\big) + \\ &|\phi^-\rangle_{AC} \big(\alpha|0\rangle_B - \beta|0\rangle_B\big) + \\ &|\psi^+\rangle_{AC} \big(\beta|0\rangle_B + \alpha|0\rangle_B\big) + \\ &|\psi^-\rangle_{AC} \big(\beta|0\rangle_B - \alpha|0\rangle_B\big) \bigg] \end{eqnarray*}$$

Note: it is not obvious that this is just a change of basis to the previous state, but it is true. To convince yourself that this really is just a change of basis, it may be necessary to work through the substitution of each term as shown on Wikipedia.

In this new basis, we see what happens when Alice measures A and C in their Bell basis - one of the four Bell states result, and for each of them B is left in a state where $\alpha$ and $\beta$, the initial amplitudes of the state $|\psi\rangle$, play a role. If we want to get B back to the exact state $|\psi\rangle$, we may need to apply a rotation to the qubit B to account for an extra minus sign (apply $Z$ operator) or a swap between $\alpha$ and $\beta$ ($X$ operator), or something along those lines. This is the Pauli correction.

• I have rewritten a 2nd question to reflect better what I wanted to ask – mercury0114 Dec 18 '16 at 18:41
• I think your lecture notes are not so clear -- You are aware that the OP only posted one page from a 10-page lecture note? – Norbert Schuch Dec 18 '16 at 21:56
• mercury, I am updating my answer to address your updated questions. – Harry Levine Dec 18 '16 at 21:58
• Norbert, I was not aware that the linked lecture notes were part of a longer set. – Harry Levine Dec 18 '16 at 22:00
• @HarryLevine It takes 30 seconds to find out (and it would be surprising if they wouldn't have been). On an unrelated note: Please use @[username] to ping me when you answer. I otherwise don't get notified. – Norbert Schuch Dec 18 '16 at 22:55

Since previous answer explains (1) and (2) well, I would like to elaborate on question (3):

WHY a measurement on Alice's side allows Bob to infer something about |ψ⟩?

This is just a magical property of entanglement. Even if initially $|\psi\rangle$ is not related to the entangled qubit pair, it does become after the measurement.

Note, that initially entangled qubits $\beta_{00}$ are of the form $(|00\rangle + |11\rangle) \cdot scalar$. If Alice were to perform a measurement on $\beta_{00}$ (using standard 2-dimensional basis), then whichever output state she gets would transform Bob's qubit to the same output state.

Hence, the same has to happen even when extra 3rd qubit $|\psi\rangle$ is introduced. The same state that Alice observes by making a measurement on 3 qubits (wrt. 4-dimensional basis) has to appear on Bob's side. And since for measurement we're deliberately choosing a basis such that the post-measurement state becomes related to $|\psi\rangle$, the information about $|\psi\rangle$ gets teleported to Bob.