Question1. If there is a state $|\phi\rangle=\frac{1}{\sqrt{2}}(|0\rangle+e^{i\theta}|1\rangle)$, and I want to know the angle $\theta$. What kind of measurement should I do? Could somebody give me the quantum circuit?

Question2. How to perform a measurement with base $M{{({{\theta }_{k}})}_{\pm }}=\left\{ 1/\sqrt{2}\left( |0\rangle \pm {{e}^{-i{{\theta }_{k}}}}|1\rangle \right) \right\}$ on IBMQ?

  • 2
    $\begingroup$ Do you have many copies of the state $|\phi\rangle$, or just one? Can you access them all at once, or do you have them just one at a time? $\endgroup$
    – DaftWullie
    Mar 31, 2020 at 9:34
  • $\begingroup$ @WilliamYang in my answer for $\phi$ I assumed that we have many copies of the state $\phi$ (we can prepare them as many as we want). I will add this to my answer. $\endgroup$ Mar 31, 2020 at 9:55
  • $\begingroup$ @DavitKhachatryan but you also assumed you just get them one at a time. If you have them all at the same time, you can get a square root improvement I believe. $\endgroup$
    – DaftWullie
    Mar 31, 2020 at 11:11
  • 1
    $\begingroup$ @DavitKhachatryan I’ll write it up as a (partial) answer when I get the time. Basically, it’s the Fourier transform. $\endgroup$
    – DaftWullie
    Mar 31, 2020 at 11:34
  • 1
    $\begingroup$ @DavitKhachatryan hmmm maybe it doesn't quite work being given just the state $|\phi\rangle$ as compared to having an oracle that applies the unknown phase. I thought I'd seen something like it before, but cannot instantly reconstruct anything that out-performs your answer. $\endgroup$
    – DaftWullie
    Mar 31, 2020 at 14:35

2 Answers 2


Answer to the first question:

As mentioned in the comments of the question I assume that we can prepare $|\phi \rangle$ as many as we want. Let's calculate the relative phase for this one qubit pure state:

$$|\psi \rangle = \frac{1}{\sqrt{2}} \left( |0\rangle + e^{i\theta}|1\rangle\right)$$

We are going to execute $2$ different experiments in order to estimate $\theta$. In the first experiment we apply this circuit:

circuit_experiment_1.measure(q[0], c[0])

The state after Hadamard gate:

$$H \frac{1}{\sqrt{2}} \left( |0\rangle + e^{i\theta}|1\rangle\right) = \frac{1}{2}\left[(1 + e^{i\theta})| 0 \rangle + (1 - e^{i\theta})| 1 \rangle \right]$$

The probabilities of $|0\rangle$ and $|1\rangle$ states:

\begin{align*} P(0) = \frac{1}{4}\left| 1 + e^{i\theta} \right|^2 = \frac{1}{2}(1 + \cos(\theta)) \\ P(1) = \frac{1}{4}\left| 1 - e^{i\theta} \right|^2 = \frac{1}{2}(1 - \cos(\theta)) \end{align*}

From here we can see that:

$$\theta = \pm \arccos\big(P(0) - P(1)\big)$$

because the range of usual principal value arccosine function is equal to $[0, \pi]$. So we will need the second experiment in order to estimate the $sign(\theta)$. But, before that, how to find $P(0)$ and $P(1)$ with the described experiment? We will need to execute the circuit $N$ times (bigger $N$ gives better precision) and take into accout these relations between measurement outcomes and probabilities:

\begin{align*} P(0) = \lim_{N \rightarrow \infty} \frac{N_{0}}{N} \qquad P(1) = \lim_{N \rightarrow \infty} \frac{N_{1}}{N} \end{align*}

where $N_{0}$ is the number of $|0\rangle$ measurement outcomes and $N_{1}$ is the number of $|1\rangle$ measurement outcomes. Also, note that:

$$\langle X \rangle = \langle \psi | X | \psi \rangle = \langle \psi |H Z H| \psi \rangle = P(0) - P(1)$$

So, the formula can be written in this way:

$$\theta = \pm \arccos \big( \langle X \rangle \big)$$

The sign of the $\theta$

Now we should determine the $sign(\theta)$ with this circuit:

circuit_experiment_2.measure(q[0], c[0])

The state after applying $S^{\dagger}$ and $H$ gates:

$$H S^{\dagger} \frac{1}{\sqrt{2}} \left( |0\rangle + e^{i\theta}|1\rangle\right) = \frac{1}{2}\left[(1 - i e^{i\theta})| 0 \rangle + (1 + i e^{i\theta})| 1 \rangle \right]$$

with the same logic:

\begin{align*} P'(0) = \frac{1}{4}\left| 1 - ie^{i\theta} \right|^2 = \frac{1}{2}(1 + \sin(\theta)) \\ P'(1) = \frac{1}{4}\left| 1 + ie^{i\theta} \right|^2 = \frac{1}{2}(1 - \sin(\theta)) \end{align*}

So after determining the $P'(0)$ and $P'(1)$ from the second experiment we will find the sign of the $\theta$:

$$sign(\theta) = sign(\arcsin\left(P'(0) - P'(1)\right)) = sign(P'(0) - P'(1))$$

because the range of usual principal value of arcsine function is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Also, note that for the expectation value of the $Y$ operator (as can be seen from this answer) we have this relation:

$$\langle Y \rangle = \langle \psi| Y | \psi\rangle = \langle \psi| S H Z H S^{\dagger} | \psi\rangle = P'(0) - P'(1)$$

By taking this into account and combining two results:

\begin{align*} \theta = sign \big(\langle Y \rangle \big) \arccos \big(\langle X \rangle \big) \end{align*}

An approach for finding the relative phase of an arbitrary pure state is described in this answer.

Answer to the second question:

Here is the circuit for measuring in $M{{({{\theta }_{k}})}_{\pm }}=\left\{ 1/\sqrt{2}\left( |0\rangle \pm {{e}^{-i{{\theta }_{k}}}}|1\rangle \right) \right\}$ basis. I assume here that $\theta_k$ is given:

circuit.u1(theta_k, q[0])    # q[0] is one of the qubits
circuit.measure(q[0], c[0])   #c[0] is a classical bit

If the state was $M(\theta _k)_+= 1/\sqrt{2}\left( |0\rangle + e^{-i\theta _k}|1\rangle \right)$, then the outcome of the circuit will be $|0\rangle$, and if it was $M(\theta _k)_-= 1/\sqrt{2}\left( |0\rangle - e^{-i\theta _k}|1\rangle \right)$, then the outcome of the circuit will be $|1\rangle$. So this way we will be able to measure in $M{{({{\theta }_{k}})}_{\pm }}$ basis.

  • 1
    $\begingroup$ Thank you so much! You do help me a lot! $\endgroup$ Mar 31, 2020 at 9:41
  • 2
    $\begingroup$ Nice answer. Just one question, why not to calculate an angle as $\theta = \arccos(2P(|0\rangle)-1)$ or $\theta = \arccos(1-2P(|1\rangle))$? I supposed you subtracted probabilities $P(|0\rangle)$ and $P(|1\rangle)$ from each other, right? $\endgroup$ Mar 31, 2020 at 9:52
  • 2
    $\begingroup$ Thanks Martin. You are right. I will add also that expression to the equation in the answer. I think $P(0) - P(1)$ is a more easily understandable expression, so I will keep it :) $\endgroup$ Mar 31, 2020 at 10:10
  • $\begingroup$ @WilliamYang, I did a mistake, it is corrected now. In the previous version of the answer, I didn't calculate $\theta$, there was a sign ambiguity that comes from the arccosine function that I haven't taken into account. $\endgroup$ Jul 26, 2020 at 18:59
  • $\begingroup$ Sorry,but what if I have only one state $|\varphi\rangle$ at a time?What kind of measurement should I do to get $\theta$? $\endgroup$ Nov 26, 2020 at 8:26

I would just like to share a code for testing a phase measurement on IBM Q:

include "qelib1.inc";

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

//measuring theta in
//(|0> + |1>*exp(i*theta))

h q[0]; //(|0> + |1>)
t q[0]; //(|0> + |1>*exp(i*pi/4))
//s q[0]; //(|0> + |1>*exp(i*pi/2))
//u1 (pi/8) q[0]; //(|0> + |1>*exp(i*pi/8))

h q[0]; //measurment in Hadamard basis

measure q[0] -> c[0];

Tested on IBM Q Armonk (1 qubit processor).

EDIT (based on Davit comment): To infer a sign of the phase, a measurement in circular basis (i.e. adding $S^\dagger$ gate before Hadamard gate) has to be done as well. Combining results from measurement in Hadamard basis and circular basis gives complete knowledge about the phase.

  • 1
    $\begingroup$ Thank you Martin :) $\endgroup$ Mar 31, 2020 at 10:30
  • 1
    $\begingroup$ @DavitKhachatryan: I think that the circuit work correcly but only for phases between $0$ and $\pi$ as these values are returned by arccos. $\endgroup$ Jul 27, 2020 at 7:19
  • 1
    $\begingroup$ yes they work correctly for that values, but not for the relative phases between $\pi$ and $2\pi$. $\endgroup$ Jul 27, 2020 at 7:21
  • 1
    $\begingroup$ @DavitKhachatryan: Hi, I edited my answer. Just one question. I think we can replace $S$ with $S^\dagger$, right? Only one difference will be a sign before $i$ and switching formulas for $P'(0)$ and $P'(1)$. $\endgroup$ Jul 29, 2020 at 8:17
  • 1
    $\begingroup$ Martin, yes we did QST. BTW I was trying to avoid that :). Nevertheless, if for each expectation value we need to run let say $1000$ experiments for the precision, then the overall number of experiments will be $N = 3000$, but I think with this technique for the same precision $2000 < N \le 3000$ experiments will be needed. When the phase is near to $0$ we can execute QST, otherwise, we will do fewer experiments in order to estimate only the sign of the phase. $\endgroup$ Jul 31, 2020 at 7:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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