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:


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

  • $\begingroup$ What's your input state? $\endgroup$ Commented Aug 24, 2020 at 2:13
  • $\begingroup$ It’s just a superposition of all states with equal probability $\endgroup$ Commented Aug 24, 2020 at 2:25

2 Answers 2


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.

  • $\begingroup$ Thank you! I tried that and didnt get what I was expecting. I think am not sure what Im(expectation) may mean then $\endgroup$ Commented Aug 24, 2020 at 16:19
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Aug 24, 2020 at 17:09
  • 1
    $\begingroup$ Thank you! I was missing something when i read about the test but your explanation helped me understand it. $\endgroup$ Commented Aug 25, 2020 at 4:53

Here, I think you wanted this link

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];
  • $\begingroup$ 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. $\endgroup$ Commented Aug 24, 2020 at 7:26

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