## How to measure a Bell inequality violation in IBM Q?

2

Note: Cross-posted on Physics SE.

I made some circuit to prepare a 2 qubit state, but I am having trouble understanding how to measure Bell's inequality. I know the inequality is of the form

$$|E(a,b)-E(a,b')+E(a',b)+E(a',b')| \leq 2$$

where for each $$E$$

$$E = \frac{N_{++} + N_{--} - N_{-+} - N_{+-}}{N_{++} + N_{--} + N_{-+} + N_{+-}}$$

My problem is, what would the different $$a,a',b,b'$$ be? With this question, I don't mean what their values would be (since IBM Q just outputs $$0$$ or $$1$$ in the $$01$$ basis), but how do I implement this?

Can I just do $$a,b$$ in $$01$$ basis and $$a',b'$$ in $$\pm$$ basis? And if so, how do I proceed with this? Do I just apply a Hadamard gate before the measurement and take whatever $$0$$ or $$1$$ value it outputs?

Not sure if this is exactly what you are looking for, but here is an article explaining Bell inequality and incorporating it inside of IBMQ Composer.

https://github.com/Qiskit/ibmqx-user-guides/blob/master/rst/full-user-guide/003-Multiple_Qubits_Gates_and_Entangled_States/002-Entanglement_and_Bell_Tests.rst

– Matthew Stypulkoski – 2019-05-01T15:28:24.637