# Can entanglement allow to predict deterministically results of incompatible measurements?

In my notes regarding entanglement and CHSH inequality it says:

Further, this is the case even for incompatible observables: measurements on one system allow us to predict with certainty the outcomes of measurement of incompatible observables on the other system. However, the uncertainty principle tells us that we cannot predict with arbitrary accuracy the results of measurement of incompatible observables.

How is the first sentence allowed. My understanding was that if you measure one set of observable, then the other incompatible observables are still in superposition.

• @Viloria My modest understand will say that the uncertainty principle is related to conjugated variables even for one body and the entanglement can act on the same observable of a many body system. May 12 at 23:29
• What is "the case even for incompatible observables"? The question/post as such is incomplete - and your last sentence also does not really make sense. Other than that, the quote is very confusing. Local observables in QI are of the form $A\otimes \mathbb I$ or $\mathbb I\otimes B$ and both commute, irrespective of the commutator of $A$ and $B$. May 13 at 13:55
• "My understanding was that if you measure one set of observable, then the other incompatible observables are still in superposition." I believe this is backwards. It's only true for compatible observables. May 13 at 14:55
• It's probably supposed to mean that an X measurement on qubit 1 can uniquely determine the outcome of a Z measurement on qubit 2 (that's true, for a suitable entangled state) -- but it is poor wording, those measurement were only incompatible if they were performed on the same qubit. May 13 at 23:01

More concretely, consider a simple example such as a two-qubit maximally entangled state of the form $$|00\rangle+|11\rangle$$. In this case, measuring the first qubit in the $$Z$$ basis means to perform the projective measurement with possible outcomes $$|0\rangle,|1\rangle$$. If you find $$|0\rangle$$, then the second qubit collapses to $$|0\rangle$$, and same for $$|1\rangle$$. This means that if you also measure the second qubit in the $$Z$$, you get a deterministic outcome (always $$0$$ if the first measurement gave $$0$$, and same for $$1$$). However, if you measure instead the second qubit in the $$X$$ basis, i.e. perform the projective measurement with outcomes $$|\pm\rangle$$, then regardless of what you found in the first measurement, you have total uncertainty about the second measurement outcome.
But the above has nothing to do with $$Z$$,$$X$$ being incompatible observables. In fact, if you repeat the same exact reasoning but with initial state $$|0,+\rangle+|1,-\rangle$$, you'll find the opposite result: measuring $$Z$$ on one side determines the result of $$X$$ on the other size.
There is no contradiction with uncertainty principles or the likes, because the measurements that are being performed in this scenario are really $$Z\otimes I$$ and $$I\otimes X$$, which commute.