What is the post measurement state given an input and the outcome measurement?

2

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

Frank Schroer IV

Posted 2019-01-02T04:59:25.463

Reputation: 71

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 [edit]ed the tags this time.

– Sanchayan Dutta – 2019-01-02T12:03:46.577

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? – glS – 2019-01-02T16:40:41.950

Yes you apply the CNOT gate and then measure in the computational basis. – Frank Schroer IV – 2019-01-02T22:46:32.343

Answers

1

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.

Mariia Mykhailova

Posted 2019-01-02T04:59:25.463

Reputation: 6 616