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I am reading about Entanglement-assisted Quantum Error Correction Codes from Quantum Information Processing and Quantum Error Correction: An Engineering Approach (Chapter 9) . It is a scheme that allows the usage of any classical error correction code. Furthermore, it uses entangled qubits (ebits) which is the Bell state $\vert \Phi^+ \rangle$. It is assumed to be error-free and shared between the sender Alice and receiver Bob, prior to the start of the communication.

Alice will encode her state $\vert \psi \rangle$ with the help of local ancillary qubits $\vert 0 \rangle$ and her half of shared ebits. She will then sends the encoded qubits over a noisy quantum channel.

The book then states:

Notice that the channel does not affect the receiver’s half of shared ebits at all.

How is that even possible? I thought if one entangled subsystem changes, then the other changes automatically.

Furthermore, consider the following (to make it simple, assume one ebit is used which is shared between Alice and Bob before communication begins). Let us say Alice will send her half of the ebit over the noise channel and a bit-flip occurred, what happens to the other half (Bob's half)?

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If you have a noisy channel $$ \rho\mapsto \mathcal{E}(\rho) $$ for a single qubit, then if you have a Bell pair $$ \rho_B=\frac12\left(|00\rangle+|11\rangle\right)\left(\langle 00|+\langle 11|\right), $$ then you can describe the effect of sending the first qubit through the channel using $$ (\mathcal{E}\otimes I)\cdot\rho_B. $$ Thus, mathematically, you see the operator is only acting on the first qubit. I think this is primarily what the authors were meaning. Now, it depends on what the noise actually does as to whether the state of the second qubit (not sent through the channel) actually changes. If the action of the channel is just a unitary, it really doesn't change the second qubit. This is because $$ \text{Tr}_1(U_1\otimes I\cdot\rho_{12})=\text{Tr}_1(\rho_{12}) $$ for all possible states $\rho_{12}$ and and possible unitaries $U_1$. On the other hand, a measurement on the first qubit would change the second qubit. However, even in that case, unless the person holding the second qubit knows what measurement was performed, and what the result was, their description of that qubit (in terms of a density matrix, for example), does not change.

For example, if your channel applies $X$ on the first qubit of an entangled pair (I won't use the Bell pair, because using a maximally entangled state creates some ambiguity), the overall state changes to $$ \alpha|00\rangle+\beta|11\rangle\rightarrow\alpha|10\rangle+\beta|01\rangle. $$ It's not entirely a fair question, really, to be asking "has the second qubit changed?". The state is a state of both qubits and you cannot talk about just the state of one qubit. However, it is certainly true that (i) the action can always be described as an action just on the first qubit (as it happen, in this basis, you can describe it by an action on the second qubit instead, but that's not universally true) and (ii) the reduced density matrix of the second qubit does not change. It is $|\alpha|^2|0\rangle\langle 0|+|\beta|^2|1\rangle\langle 1|$ both before and after the error.

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  • $\begingroup$ Also, figure 1 in here explicitly states that the noisy channel affects only Alice’s halves. $\endgroup$ – M. Al Jumaily Feb 6 at 22:09
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The fact that the channel does not affect to half of the ebits is an assumption that is made when entanglement-assisted quantum error correction is considered. It is usually thought as if the sender and the receiver pre-share this EPR pairs beforehand, and so then only half of the ebits (the ones from the sender) do go through the channel and suffer its action.

Obviously, this will not be practical nor realizable in reality, and so different entanglement distribution protocols will be needed so that the "noiseless" ebits reach the receiver (i.e. entanglement distillation, protecting such half with unassisted QECCs ...). This paradigm is studied in Entanglement-Assisted Quantum Error-Correcting Codes with Imperfect Ebits, I give the reference so that you can further deepen your knowledge about QECCs operating with noisy shared entanglement.

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  • $\begingroup$ Thank you for the clarification! $\endgroup$ – M. Al Jumaily Feb 6 at 22:07

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