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?

  • 1
    $\begingroup$ Is the answer expected in the form of a density operator? $\endgroup$
    – ahelwer
    May 13 '19 at 20:46
  • $\begingroup$ @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. $\endgroup$
    – 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 $$ |\beta\rangle\rightarrow(1-p)|\beta\rangle\langle\beta|+p(\mathbb{I}\otimes X)|\beta\rangle\langle\beta|(\mathbb{I}\otimes X). $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.