quantum mechanics - Uncertainty principle and measurement

I would like to really understand how the uncertainty principle in QM works, from a practical point of view.

So this is my narrative of how an experiment goes, and I'm quickly in trouble: we prepare a set of many particles in the same state $\psi$ as best we can, then we start measuring 2 observables A and B that don't commute on... each of the particles (?). When we measure A, the wave function collapses to an eigenstate of A with some probability. By accumulating measurements with A, we obtain statistics, and in particular $\langle\psi|A|\psi\rangle$, the expected value of $A$ w.r.t to state $\psi$. But how do I get $\langle\psi|B|\psi\rangle$? Can I measure A and B "simultaneously" on one particle, even if $\psi$ has collapsed to an eigenstate of A, which is not an eigenstate of B, and A and B don't commute... What happens? How do I measure B? Do I need to pull in another particle, on which I'll measure B, but not A this time?

Answer

There are many steps:

Step 1, select a state $\Psi$.

Step 2, prepare many systems in same state $\Psi$

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

Now if you analyze the results, assuming strong (not weak) measurements then every time you measured A, you got an eigenvalue of A, and every time you measure B you got an eigenvalue of B. Each eigenvalue had a probability (which is equal to the ratio of the squared norm of the projection onto the eigenspace divided by the squared norm before you projected onto the eigenspace). So your eigenvalues of A come from a probability distribution that often has a mean $\langle A\rangle=\langle \Psi|A|\Psi\rangle $ and a standard deviation $\Delta A=\sqrt{\langle \Psi|\left(A^2-\langle \Psi|A|\Psi\rangle^2\right)|\Psi\rangle}$. And your eigenvalues of B come from a probability distribution that often has a mean $\langle B\rangle=\langle \Psi|B|\Psi\rangle $ and a standard deviation $\Delta B=\sqrt{\langle \Psi|\left(B^2-\langle \Psi|B|\Psi\rangle^2\right)|\Psi\rangle}$. You never get those from a measurement, or even from a whole bunch, but from steps 4a and 4b you do get a sample mean and a sample standard deviation, and for a large sample these are likely to be very close to the theoretical mean and the theoretical standard deviation.

The uncertainty principle says that way back in step 1 (when you selected $\Psi$) you could select a $\Psi$ that gives a small $\Delta A$, or a $\Psi$ that gives a small $\Delta B$ (in fact if $\Psi$ is an eigenstate of A then $\Delta A=0$, same for $B$). However, $$\Delta A \Delta B \geq \left|\frac{\langle AB-BA\rangle}{2i}\right|=\left|\frac{\langle\Psi| AB-BA |\Psi\rangle}{2i}\right|,$$

So in particular noncommuting operators often (i.e. if the expectation value of their commutator does not vanish) have a tradeoff, if the state in question has really low standard deviation for one operator, then the state in question must have a higher standard deviation for the other.

If the operators commute, not only is there no joint limit to how low the standard deviations can go, but measuring the other variable keeps you in the same eigenspace of the other operator. However that is a completely different fact since the uncertainty principle is about the standard deviations of two probability distributions for two observables applied to one and the same state, and thus approximately applies to the sample standard deviations generated from identically prepared states.

If you have a system prepared in state $\Psi$ and you measure A on it then you generally have to use a different system also prepared in $\Psi$ to measure B. That's because when you measure A on a system it projects the state onto an eigenspace of A, which generally changes the state. And since the probability distribution for B is based on the state, now that you have a different state you will have a different probability distribution for B. You can't find out $\Delta B=\sqrt{\langle \Psi|\left(B^2-\langle \Psi|B|\Psi\rangle^2\right)|\Psi\rangle}$ if you don't have $\Psi$ and only have $\Psi$ projected onto an eigenspace of A.