If we have this given circuit:
So the output for $|0\rangle$ will be: $\frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right)$
And we have this given circuit:
What will be the output for $|00\rangle$ and why?
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1 Answer
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The Hadamard gate is its own inverse. Hence, the two Hadamards on the first qubit cancel, leaving the control qubit in state $|0 \rangle$. The bottom Hadamard gate will put the target qubit in the state $\frac{1}{\sqrt{2}} (|0\rangle + |1\rangle)$. Since the control qubit is still in state $|0\rangle$, the CNOT will do nothing to the target. Your output state is therefore $\frac{1}{\sqrt{2}} (|00\rangle + |01\rangle)$.
