The answers here about experimentally verifying Heisenberg's uncertainty principle says the following.
Step 3, select two operators A and B
Step 4a, for some of the systems prepared in state $\Psi$, measure A
Step 4b, for some of the systems prepared in state $\Psi$, measure B
Why not measure $A$ on all members of the ensemble followed by a measurement of B on each? What is the problem if the first measurement (say, A) collapses the state $\Psi$ to an eigenstate of A?
Addendum If this is the case i.e., A and B are measured on different members of the ensemble, there is apparently no correlation between the measurments. Then what does it mean to say as $\Delta A$ decreases by measuring A more and more accurately, $\Delta B$ increases when we measure B?
Does any textbook explain these steps of measurement?
Answer
The Heisenberg uncertainty principle is not about the precision of the measurement. Case in point: If I hand you the spin-1/2 state $\lvert \psi\rangle = \lvert \uparrow_z\rangle + \lvert \downarrow_z\rangle$ and tell you to measure $z$, you will always measure spin-up 50% of the time and spin-down the other 50% of the time, even with a completely "accurate" measurement device. The standard deviation is $$ \Delta S_z(\lvert \psi\rangle) = \sqrt{\langle\psi\vert S_z^2\vert \psi\rangle - \langle \psi\vert S_z\vert \psi\rangle^2} = \sqrt{\frac{1}{4} - 0} = \frac{1}{2}.$$ This is not a property of the measurement device or the number of measurements, but an inherent feature of the quantum state $\lvert \psi\rangle$. This is what you find as the standard deviation of your measurement results after you have measured infinitely often with infinite precision. Quantum "uncertainties" are not about the accuracy of the measurement, or how often you have repeated it. They are not about correlations between measurements. You cannot "decrease $\Delta A$" by measuring $A$ since $\Delta A$ is a property of the state you measure, not of the measuring process.
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