2
$\begingroup$

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.

enter image description here

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

$\endgroup$
3
$\begingroup$

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:

Quantum phase estimation for Z gate

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.

$\endgroup$
  • $\begingroup$ It does help with phase estimation. This trick is called the semi-classical Fourier transform. $\endgroup$ – DaftWullie Aug 5 at 5:39
  • $\begingroup$ @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. $\endgroup$ – Suslik Aug 5 at 10:45
  • 2
    $\begingroup$ @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 $\endgroup$ – DaftWullie Aug 5 at 10:52
  • $\begingroup$ Thank you for the paper. $\endgroup$ – Suslik Aug 5 at 10:54
  • $\begingroup$ @DaftWullie I didn't know that, thank you for correcting me! $\endgroup$ – Mariia Mykhailova Aug 5 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.