Sketch the quantum logic gates correctly and give a proof for the identity

Question

If I denote by $U^c$ the controlled version of the quantum operation $U$ $$U^c=|0\rangle \langle 0|\otimes \mathbb{1}+|1\rangle \langle 1|\otimes U$$

1. I can first apply $U^c$ and afterward measure the control qubit.
2. Or I can first measure the control qubit and then apply $U$ only if the measurement outcome was 1.

This is a trick to save qubits when performing quantum phase estimation. I tried to sketch this with quantum logic gates - I am not sure if I did it right.

Now I want to show that these two methods give the same result.

1. \begin{align*} &\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)U^c|\Psi\rangle\\ &= \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)(|0\rangle \langle 0|\otimes \mathbb{1}+|1\rangle \langle 1|\otimes U)|\Psi\rangle\\ &=\frac{1}{\sqrt{2}}(|0\rangle \langle 0|0\rangle\otimes 1 + |1\rangle \langle 1|0\rangle \otimes U +|0\rangle \langle 0|1\rangle\otimes 1 +|1\rangle \langle 1|1\rangle \otimes U)|\Psi\rangle\\ &=\frac{1}{\sqrt{2}}(|0\rangle \otimes 1 + |1\rangle \otimes U)|\Psi\rangle \end{align*}

   I hope my calculations are correct.

2. Now I can say, this is the same as measuring the control qubit and if it is $|1\rangle $ I calculate $|1\rangle \otimes U |\Psi\rangle$

Answer 1

These two circuits produce the same results - in both cases you'll get $|0\rangle \otimes |\psi\rangle$ with 50% probability or $|1\rangle \otimes U|\psi\rangle$ with 50% probability.

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But I don't think this is going to help you with the phase estimation algorithm. In quantum phase estimation application of $U^c$ is followed by inverse Fourier transform before the measurement is done.

Let's take a look at how QPE works for the case of 1 control qubit if we know that the eigenvalue of $U$ is either $+1$ or $-1$ (i.e., the phase estimated is either 0 or 0.5). We can use the $Z$ gate as $U$ to make our calculations more specific.

The inverse QFT for 1 qubit is just the Hadamard gate, so the circuit for QPE looks as follows:

You can check that the measurement result will be 0 if the second qubit starts in the $|0\rangle$ state and 1 if it starts in the $|1\rangle$ state - which allows you to estimate the phase you're looking for.

The circuits you started with give you a 50-50 chance of measuring 0 or 1 on the first qubit regardless of the starting state of the second qubit, and thus do not help you estimate the phase.