1

I've the following circuit which gives an output of 1 with a phase angle of 3π/4. When we measure the circuit all we get is the state 1 (On actual QC, it's 90% of the time 1). Is the information about the phase angle lost by measuring the state? Could the phase angle also be part of the output?

Circuit

OPENQASM 2.0;
include "qelib1.inc";

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


u(pi/2,pi/2,0) q[0];
h q[0];
u(pi/2,pi/2,pi/2) q[0];
measure q[0] -> c[0];
Van Peer
  • 577
  • 5
  • 13
  • 2
    As you said, the state that you have is $|\psi \rangle = e^{i 3\pi/4} |1\rangle$. This state is indistinguishable from the state $|\psi \rangle = |1\rangle$. States that are identical up to a global/overall phase are indistinguishable from each other. – KAJ226 May 25 '21 at 14:28
  • 1
    This depends on if the phase is global or relative. Here's some more discussions about this : https://physics.stackexchange.com/questions/257824/global-and-relative-phases-of-kets-in-qm , https://physics.stackexchange.com/questions/401523/global-phases-and-indistinguishable-quantum-states-mathematical-understanding, https://quantumcomputing.stackexchange.com/questions/5125/what-is-the-difference-between-a-relative-phase-and-a-global-phase-in-particula, https://quantumcomputing.stackexchange.com/questions/12448/understanding-global-phase – Purva Thakre May 26 '21 at 13:00

1 Answers1

1

When we make a measurement we project the state of the qubit to the z-basis (i.e 0 or 1) so in general it is not possible to measure a global phase.

That said, Measuring such global phases is an important subroutine in many quantum algorithms such as Shor's algorithm.

This can be done using the Quantum phase estimation algorithm. For details, check https://qiskit.org/textbook/ch-algorithms/quantum-phase-estimation.html and the following chapter on Shor's algorithm. The idea is roughly that of performing an interference experiment.

R K Rupesh
  • 66
  • 3