2
$\begingroup$

Given a two-qubit state of equal superposition, what is the post-measurement state (should be the same number of qubits that have changed as a result of the measurement) on one of the two qubits and the probabilities that the state will be in a given state?

Input state $\to$ CNOT $\to$ Measurement

$\endgroup$
  • $\begingroup$ Hi, Frank. Welcome to Quantum Computing SE! The tags you were using are not appropriate for this question. Please review What are tags, and how should I use them? & the list of existing tags. I've edited the tags this time. $\endgroup$ – Sanchayan Dutta Jan 2 at 12:03
  • $\begingroup$ you have to specify what measurement is being performed. Does the last bit mean that you apply a CNOT gate and then measure in the computational basis? $\endgroup$ – glS Jan 2 at 16:40
  • $\begingroup$ Yes you apply the CNOT gate and then measure in the computational basis. $\endgroup$ – Frank Schroer IV Jan 2 at 22:46
1
$\begingroup$

If I interpret the question correctly, we start with a state $\frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$ (an equal superposition of all basis states on 2 qubits).

After we apply a CNOT, the state doesn't actually change (if the first qubit is the control, the $|10\rangle$ and $|11\rangle$ components swap amplitudes but they are equal, so nothing changes).

After we measure one of the qubits, we get 0 with probability 50% and 1 with probability 50%, and the state of the second qubit is guaranteed to be $\frac{1}{\sqrt2}(|0\rangle + |1\rangle)$. You can see this easily if you use the fact that the state is separable: $\frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle) = \frac{1}{\sqrt2}(|0\rangle + |1\rangle) \otimes \frac{1}{\sqrt2}(|0\rangle + |1\rangle)$, so the measurement of one qubit doesn't affect the state of the second one.

$\endgroup$

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.