# How to get the relative phase of a qubit?

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?

• 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? – DaftWullie Mar 31 at 9:34
• @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. – Davit Khachatryan Mar 31 at 9:55
• @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. – DaftWullie Mar 31 at 11:11
• @DavitKhachatryan I’ll write it up as a (partial) answer when I get the time. Basically, it’s the Fourier transform. – DaftWullie Mar 31 at 11:34
• @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. – DaftWullie Mar 31 at 14:35

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)    # q is one of the qubits
circuit.h(q)
circuit.measure(q, c)   #c 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.

If we don't know $$\theta$$, then firstly let's apply to the qubit a Hadamard gate and see what we can do:

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

Now let's look at 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:

$$\theta = \arccos\left[P(0) - P(1)\right] = \arccos\left[1 - 2P(1)\right]$$

Now, how to find $$P(0)$$ and $$P(1)$$. We need to execute this circuit $$N$$ times. As mentioned in the comments of the question I assume that we have many copies of $$|\phi \rangle$$ state (we can prepare $$|\phi \rangle$$ as many as we want):

circuit.h(q)
circuit.measure(q, c)


The probabilities from the measurement outcome:

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

where $$N_{0}$$ is the number of $$|0\rangle$$ measurements and $$N_{1}$$ is the number of $$|1\rangle$$ measurements.

• Thank you so much! You do help me a lot! – WilliamYang Mar 31 at 9:41
• 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? – Martin Vesely Mar 31 at 9:52
• 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 :) – Davit Khachatryan Mar 31 at 10:10

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

OPENQASM 2.0;
include "qelib1.inc";

qreg q;
creg c;

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

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

h q; //measurment in Hadamard basis

measure q -> c;


Tested on IBM Q Armonk (1 qubit processor).

• Thank you Martin ：） – WilliamYang Mar 31 at 10:30