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How can I measure a qubit on a generic axis on Qiskit?

Quantum Computing Asked by Marco Gobbo on April 12, 2021

I’m trying to prove Bell’s inequality by following the one made by Wigner, that is:

$P(+_a,+_b) le P(+_a,+_c) + P(+_c,+_b)$

And that this proof is NOT satisfied for angles between the various axes $hat a$, $hat b$ and $hat c$.
In particular I have chosen the coplanar axes (plane x-y) and that the $hat c$ axis is the bisector of $hat a$ and $hat b$.

So I’m in the situation in the picture Graph

My idea is to take the singlet state (represented in this case by a specific state of the Bell state) and measure one qubit on one axis ($hat a$, $hat b$, $hat c$) and the other qubit on one of the remaining axes in in such a way as to make all possible combinations.

In the specific situation represented in the graph, the angle that I can choose between the various axes must be a value between $0$ and $frac pi2$.

Now I come to my question: my doubt is whether it is correct how to interpret the measurement phase after I have created the singlet state (entagled state).

The entangled state I create it like this:

circ = QuantumCircuit(2,2)
circ.x(0)
circ.x(1)

circ.h(0)
circ.cx(0,1)

Referring to the graph I put earlier, if I wanted to now measure a long qubit $hat a$ and a long $hat b$ I would have to measure along x (no rotation for the first case) and rotate the second theta qubit around z and then measure along x, correct?

I thought about doing it this way

# Long a-axis for qubit 1
circ.h(0) 

# Long b-axis for qubit 2
circ.rz(pi/3, 1) 
circ.h(1)


circ.measure(range(2),range(2))

I wonder if these steps are correct or I forgot something or I can do it in another way.
Because if I measured directly without applying the Hadamard operator before the measurement I would make a measurement along the z-axis I suppose, right?

I hope I was clear enough and that we can resolve my doubts!
Thank you for your time!

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