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

  • $\begingroup$ 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 '19 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 '19 at 16:40
  • $\begingroup$ Yes you apply the CNOT gate and then measure in the computational basis. $\endgroup$ – Frank Schroer IV Jan 2 '19 at 22:46

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.

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