# Bell state: simultaneous measurement of two entangled qubits

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.

• 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. Nov 20, 2019 at 9:15
• 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)? Nov 20, 2019 at 13:13
• @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. Nov 20, 2019 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.

• That means it can become a test for measuring Goodness of Quantum Hardware... Nov 20, 2019 at 14:46
• Yes, you are right. Nov 20, 2019 at 15:48

This is a great question! Naively, one might get quite puzzled because it would be natural (although incorrect) to think in the following way:

If we first measure the Alice qubit and it turns out $$1$$ (or $$0$$) then we know that the Bob qubit instantaneously collapses to $$1$$ (or $$0$$) and we can confirm this when we measure it. Similarly, if we first measure the Bob qubit and it turns out $$1$$ (or $$0$$) then we know that the Alice qubit instantaneously collapses to $$1$$ (or $$0$$) and we can confirm this when we measure it. However, if we are measuring simultaneously then if the Alice one turns up $$1$$ then Bob one would also have to turn up $$1$$ but unless the other qubit has been measured, there is a $$50\%$$ probability of getting either outcome on each side. So, is there a chance that the Alice one turns out $$1$$ but the Bob one turns out $$0$$?

Of course, what is wrong with this reasoning is thinking that there is an independent $$50\%$$ chance of getting either outcome at each end. What quantum mechanics tells us is that there is a conjoint probability distribution of either getting $$1$$ on both ends with a 50% probability or of getting $$0$$ on both ends with a $$50\%$$ probability. So, there is a 0% chance of getting an anti-correlated outcome in such a measurement. Mathematically, you can see this by noticing that the inner-product of an anti-correlated outcome (either $$\vert01\rangle$$ or $$\vert 10\rangle$$) with the given state vanishes, i.e., via the Born rule, the probability of getting an anti-correlated outcome is zero.

This also illustrates another great point: it is not that when we first measure the Alice qubit, there is some active link that conveys this information about the measurement result to the Bob qubit so that the Bob qubit knows to turn out a certain way. No, because when we measure the two ends simultaneously, they also give perfectly correlated results. So, the correlation is already built into the prepared state.