Difference between different Quantum states with the same result - quantum-computing

Question

What is the practical and theoretical difference is between these 3 states, which ultimately produce the same output result.
Could you tell me some examples of different results obtained starting from these 3 states and doing the same operations below.
The concept is unclear to me.
Thank you
|0> -> RY(pi/2) -> RX(pi) -> cnot q[0] q[1]
|0> -> RX(pi/2) -> cnot q[0] q[1]
|0> -> H -> cnot q[0] q[1]

Answer 1

Not all of these states are the same, assuming that you're talking about the single-qubit states obtained before application of the CNOT gate (otherwise please specify which single-qubit gates are applied to which qubit in the 2-qubit state).
The last state is H|0⟩ = 1/sqrt(2) (|0⟩ + |1⟩).
The first state ends up being the same state, up to a global phase, which means there is no way to observe a difference between these two states.
But the second state is 1/sqrt(2) (|0⟩ - i|1⟩), which behaves differently.
To observe the difference between the second and the last states, apply a Hadamard gate to both and measure them multiple times: you'll always get 0 result for the last state, but you'll get both 0 and 1 for the second state.
To quickly run this experiment, you can use Q#: running the following snippet will give you ~50 0 measurements for the state prepared using Rx and 100 0 measurements for the state prepared using H.
open Microsoft.Quantum.Diagnostics;
open Microsoft.Quantum.Math;
operation RunTests (prep : (Qubit => Unit)) : Unit {
mutable n0 = 0;
use q = Qubit();
for _ in 1 .. 100 {
// Prepare the qubit in the given state.
prep(q);
// Apply Hadamard gate and measure.
H(q);
if M(q) == Zero {
set n0 += 1;
}
Reset(q);
}
Message($"{n0} zeros measured");
}
operation QubitsDemo () : Unit {
RunTests(Rx(PI() / 2.0, _));
RunTests(H);
}