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

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 $$|1\rangle \otimes U |\Psi\rangle$$

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.

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.

• It does help with phase estimation. This trick is called the semi-classical Fourier transform. – DaftWullie Aug 5 '19 at 5:39
• @DaftWullie Thank you for your answer. In a standard implementation I would initialize, apply hadamard Gate, perform controlled evolution, perform inverse FT and then i would measure (like in Mariias figure). What will change in my implementation if i want to use the above trick to implement a single-qubit phase estimation? Do i need to apply a final Hadamard Gate after inverse FT and then measure the qubits. – Suslik Aug 5 '19 at 10:45
• @Suslik Apply Hadamard gate, measure that qubit. Apply phase gates (rotations same as angles of the controlled-gates) on the target qubits. See arxiv.org/abs/quant-ph/9511007 – DaftWullie Aug 5 '19 at 10:52
• Thank you for the paper. – Suslik Aug 5 '19 at 10:54
• @DaftWullie I didn't know that, thank you for correcting me! – Mariia Mykhailova Aug 5 '19 at 16:02