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I am trying to understand the Hadamard Test by finding the average value of $U_1$, which is a diagonal matrix with $1$ everywhere except on the first element.

I performed the regular Hadamard Test as presented in the wiki page:

Text

and so far so good, everything works as it should.

What is the circuit variant to perform the Hadamard Test to calculate the imaginary part? The wiki page says you only have to start with $\frac{1}{\sqrt{2}}(|0\rangle-i |1\rangle)$ instead of $H|0\rangle$, I attempted doing the test adding a phase shift of -i on $|0\rangle$ by applying $U_s$ = $\begin{bmatrix}\ 1 & 0 \\ 0 & -i \end{bmatrix}$ on the control qubit after $H$, but all im getting is random measurements

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  • $\begingroup$ What's your input state? $\endgroup$ Aug 24 '20 at 2:13
  • $\begingroup$ It’s just a superposition of all states with equal probability $\endgroup$ Aug 24 '20 at 2:25
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Here is a circuit for calcualating $Im(\langle\psi|U |\psi \rangle)$ (circuit composer from IBM):

enter image description here

Initial state: $$|\Psi_0 \rangle=|0\rangle |\psi\rangle$$

After $S^{\dagger} H$ on the first qubit:

$$|\Psi_1 \rangle=\frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle) |\psi\rangle$$

Controlled $U$

$$|\Psi_2 \rangle=\frac{1}{\sqrt{2}}(|0\rangle |\psi\rangle - i|1\rangle U |\psi\rangle)$$

After final Hadamard on the control qubit:

\begin{align*} |\Psi_3 \rangle &=\frac{1}{2} \big[(|0\rangle + |1\rangle) |\psi\rangle - i(|0\rangle - |1\rangle) U |\psi\rangle \big] = \\ &=\frac{1}{2} \big[|0\rangle (|\psi\rangle - i U |\psi\rangle) + |1\rangle(|\psi\rangle + i U |\psi\rangle) \big] \end{align*}

The probability of measuring $|0\rangle$ and the probability of measuring $|1\rangle$:

$$p_0 = \frac{1}{4}\big[(\langle \psi | + i \langle \psi | U^{\dagger})(|\psi\rangle - i U |\psi\rangle) \big]= \frac{1}{4}\big[2 - i \langle\psi|U|\psi\rangle + i \langle\psi|U^{\dagger}|\psi\rangle \big] \\ p_1 = \frac{1}{4}\big[(\langle \psi | - i \langle \psi | U^{\dagger})(|\psi\rangle + i U |\psi\rangle) \big]= \frac{1}{4}\big[2 + i \langle\psi|U|\psi\rangle - i \langle\psi|U^{\dagger}|\psi\rangle \big]$$

because $U^\dagger U = I$ and $\langle \psi|\psi \rangle = 1$. Calculating the expectation value of $\sigma_z$:

$$\langle \sigma_z \rangle = p_0 - p_1 = -i \frac{\langle\psi|U |\psi \rangle - \langle\psi| U^{\dagger} |\psi \rangle}{2} = Im(\langle\psi|U |\psi \rangle)$$

So the circuit works as was described in the Wikipedia page about the Hadamard test.

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  • $\begingroup$ Thank you! I tried that and didnt get what I was expecting. I think am not sure what Im(expectation) may mean then $\endgroup$ Aug 24 '20 at 16:19
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    $\begingroup$ @CésarLeonardoClementeLópez, what you were expecting? How did you try it? With QIskit or something else? $\langle \psi| U |\psi\rangle$ is a complex number that has imaginary and real parts, it is the overlap between $|\psi\rangle$ and $U|\psi \rangle$ states. $|\langle \psi| U |\psi\rangle|^2$ is the probability of measuring $|\psi \rangle$ after applying $U$ on $|\psi\rangle$. Note that the circuit is not doing the whole job, one also should do classical post-processing of the measured outcomes in order to calculate what is needed. $\endgroup$ Aug 24 '20 at 17:09
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    $\begingroup$ Thank you! I was missing something when i read about the test but your explanation helped me understand it. $\endgroup$ Aug 25 '20 at 4:53
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Here, I think you wanted this link

OPENQASM 2.0;
include "qelib1.inc";

qreg q[2];
creg c[1];

x q[0];
x q[1];
h q[0];
s q[0];
cu1(pi) q[0],q[1];
h q[0];
measure q[0] -> c[0];
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  • $\begingroup$ I pre-assumed that you know how to get expectation value from counts, in case you don't , here we are measuring in z-basis and only one qubit, just subtract counts[0] from count[1] and divide it by total no. of counts. $\endgroup$ Aug 24 '20 at 7:26

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