What will happen if the two qubits in Bell state for a 2-qubit system, $\frac{1}{\sqrt2}(\left|00\right\rangle+\left|11\right\rangle)$ are simultaneously measured at the same instant? Two scenarios: Qubits are measured in the same states or different states. Has anyone tried to measure the qubits experimentally in IBM-Q? What is the answer from Quantum Mechanics?


In theory, there is no difference between this and measuring one qubit and then the other. One way of thinking about this is from the perspective of inertial frames of reference in special relativity - for two separated parties, there is no global definition of simultaneous. Different observers can observe different orderings of space-like separated events. While this is not strictly built in to quantum mechanics, it does respect that structure.

  • $\begingroup$ So if quantum mechanics is correct, then both measurements should be correlated or Computer claiming to be quantum, should exhibit this property and this test would be physically more rigorous as compared to the current test of entanglement in which measurement is performed one after the other. $\endgroup$ – Ashish Nov 20 '19 at 9:15
  • $\begingroup$ This is an interesting observation. Though, why exactly would quantum measurement respect the structure imposed by special relativity? Any specific reason (otherwise, it might as well be a coincidence)? $\endgroup$ – Sanchayan Dutta Nov 20 '19 at 13:13
  • $\begingroup$ @Ashish what i’m saying is that that test should be no different to any other pair of measurements separated by a space-like interval and, indeed, this is exactly what people do when closing the “locality loophole” in, for example, a Bell test. $\endgroup$ – DaftWullie Nov 20 '19 at 17:01

I tried to prepare and measure this Bell state on IBM Q. There is a low probablity, around few percent, that you measure states $|01\rangle$ or $|10\rangle$. But this is caused by a thermal noise and spontaneous break of the entangled state.

  • 1
    $\begingroup$ That means it can become a test for measuring Goodness of Quantum Hardware... $\endgroup$ – Ashish Nov 20 '19 at 14:46
  • $\begingroup$ Yes, you are right. $\endgroup$ – Martin Vesely Nov 20 '19 at 15:48

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