# State of a system after the second qubit of a Bell state sent through a bit flip error channel

The second qubit of a two-qubit system in the Bell state
$$|\beta_{01}\rangle= \frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)$$ is sent through an error channel which introduces a bit flip error with probability $$p$$. I want to calculate the state of the system after the second qubit exits the error channel.

I know how to do this for a single qubit, but there isn't anything in my notes about doing it on a single qubit of a two-qubit system (or on two qubits for that matter).

This question is from a past exam paper (my exam is next week) could anyone show me the method of how to perform this operation?

• Is the answer expected in the form of a density operator? – ahelwer May 13 '19 at 20:46
• @ahelwer I don't think it need to be in the form of a density matrix, but the next question says using the density matrix formalism consider a projective measurement of the first qubit and determine the probability of obtaining the measurement result 0. So I would say judging from the question the part in my main question could be done in density matrix formalism or we could find the state $|\psi\rangle$ and find the density matrix from this. – bhapi May 13 '19 at 20:52

If you're told about an operation on a single qubit, then to convert it into an operation on both qubits, you just include the identity matrix on the other qubit. So, contrast the single qubit state $$|\psi\rangle$$ going through the bit-flip channel $$|\psi\rangle\rightarrow(1-p)|\psi\rangle\langle\psi|+pX|\psi\rangle\langle\psi|X$$ with what happens on two qubits $$|\beta\rangle\rightarrow(1-p)|\beta\rangle\langle\beta|+p(\mathbb{I}\otimes X)|\beta\rangle\langle\beta|(\mathbb{I}\otimes X).$$