Ciro Santilli Our example uses a Bell state circuit to illustrate all the fundamental Qiskit basics.
Sample program output,
counts are randomized each time.First we take the quantum state vector immediately after the input.
We understand that the first element of
input:
state:
Statevector([1.+0.j, 0.+0.j, 0.+0.j, 0.+0.j],
dims=(2, 2))
probs:
[1. 0. 0. 0.]Statevector is , and has probability of 1.0.Next we take the state after a Hadamard gate on the first qubit:
We now understand that the second element of the
h:
state:
Statevector([0.70710678+0.j, 0.70710678+0.j, 0. +0.j,
0. +0.j],
dims=(2, 2))
probs:
[0.5 0.5 0. 0. ]Statevector is , and now we have a 50/50 propabability split for the first bit.Then we apply the CNOT gate:
which leaves us with the final .
cx:
state:
Statevector([0.70710678+0.j, 0. +0.j, 0. +0.j,
0.70710678+0.j],
dims=(2, 2))
probs:
[0.5 0. 0. 0.5]Then we print the circuit a bit:
qc without measure:
┌───┐
q_0: ┤ H ├──■──
└───┘┌─┴─┐
q_1: ─────┤ X ├
└───┘
c: 2/══════════
qc with measure:
┌───┐ ┌─┐
q_0: ┤ H ├──■──┤M├───
└───┘┌─┴─┐└╥┘┌─┐
q_1: ─────┤ X ├─╫─┤M├
└───┘ ║ └╥┘
c: 2/═══════════╩══╩═
0 1
qasm:
OPENQASM 2.0;
include "qelib1.inc";
qreg q[2];
creg c[2];
h q[0];
cx q[0],q[1];
measure q[0] -> c[0];
measure q[1] -> c[1];And finally we compile the circuit and do some sample measurements:
qct:
┌───┐ ┌─┐
q_0: ┤ H ├──■──┤M├───
└───┘┌─┴─┐└╥┘┌─┐
q_1: ─────┤ X ├─╫─┤M├
└───┘ ║ └╥┘
c: 2/═══════════╩══╩═
0 1
counts={'11': 484, '00': 516}
counts={'11': 493, '00': 507}