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

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

• 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. – Sanchayan Dutta Jan 2 '19 at 12:03
• 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 Jan 2 '19 at 16:40
• Yes you apply the CNOT gate and then measure in the computational basis. – 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.