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?


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.

  • $\begingroup$ So for a given state $|\Psi\rangle$, $\Delta A$ and $\Delta B$ have fixed values? @ACuriousMind $\endgroup$ – SRS Aug 24 '17 at 14:03
  • $\begingroup$ @SRS Yes, that is precisely what I said. Of course you have to understand that in the context of discussing an actual measurement process, "$\Delta A$" may refer to a different uncertainty (e.g. the actual standard deviation of data obtained so far, not the standard deviation as defined by my displayed equation), but these uncertainties are not the uncertainties that appear in the standard version of the HUP. $\endgroup$ – ACuriousMind Aug 24 '17 at 14:04
  • $\begingroup$ Are those uncertainties correlated? Are they due to the fact that measurements disturb the system? @ACuriousMind $\endgroup$ – SRS Aug 24 '17 at 14:14
  • $\begingroup$ @SRS I don't know what it means for uncertainties to be "correlated", and I don't understand the second question at all. $\endgroup$ – ACuriousMind Aug 24 '17 at 14:18
  • $\begingroup$ By the first question, I mean is it true in some sense that if one measure A more and more accurately the measurements of B becomes more and more scattered? $\endgroup$ – SRS Aug 24 '17 at 14:27

After measuring $A$ the system is no longer in the state $\Psi$ but instead in $\psi_A=P_A\Psi$ (up to normalization), where $\psi_A$ is one possible eigenstate of $A$. Hence measuring $B$ as you suggest would not produce the statistics for $\Psi$. The same applies when you first measure $B$ on $\Psi$, followed by $A$.

Side note: an interesting discussion of this can be found in Raymer, M. G. "Uncertainty principle for joint measurement of noncommuting variables." American journal of physics 62.11 (1994): 986-993.

  • $\begingroup$ Do you mean the probability distribution changes and there is no meaning in comparing variances calculated from different probability distributions?@ZeroTheHero $\endgroup$ – SRS Aug 24 '17 at 12:32
  • $\begingroup$ Basically yes since the operator $A$ acting on your state $\Psi$ results in a different wave function. Clearly average values for instance need not be the same for $\psi_A=A\Psi$ and $\psi_B=B\Psi$. $\endgroup$ – ZeroTheHero Aug 24 '17 at 13:07
  • $\begingroup$ I have added an addendum which really creates a confusion with this way of measuremnet. $\endgroup$ – SRS Aug 24 '17 at 13:44
  • $\begingroup$ @SRA I've added a link to a paper that deals with some of the issues of your post. $\endgroup$ – ZeroTheHero Aug 24 '17 at 14:20

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