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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?

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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.

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    $\begingroup$ This is an exceptionally good answer, not just to this question but to a lot of other similar questions that are frequently asked on this site. I hope a lot of those questioners get pointed to this. $\endgroup$ – WillO Mar 11 '15 at 20:14
  • $\begingroup$ Timaeus - I totally get it - but I have one more question. If I measure B on a $\Psi$ collapsed to an eigenstate of A, i.e. after having measured A, what do I get? I'm not expecting it to be meaningful - I'm just curious about what values are obtained: are they uniformly random, without any trace of the distribution for B? $\endgroup$ – Frank Mar 11 '15 at 23:59
  • $\begingroup$ @Frank If $\Psi$ projected to $\Phi$, then you get a distribution for $B$ eigenvalues as if you had originally prepared the state to be in $\Phi$ instead of $\Psi$. In fact, that is one way to prepare states to be in $\Phi$. If the operators commute, the eigenvector will be a common eigenvector to both so you can measure both over and over again always getting the same answers. If the operators do not commute, that might not happen. $\endgroup$ – Timaeus Mar 12 '15 at 1:17
  • $\begingroup$ OK - I get it and it makes sense. But in the case they don't commute, let's take $\hat{x}$ and $\hat{p}$, if $\Psi$ is in an eigenstate of $\hat{x}$, doesn't the uncertainty principle say that, because the position is infinitely precise, the measurements on $\hat{p}$ will have infinite variance? Which should mean that the $\hat{p}$ measurements now have a uniform distribution, no? $\endgroup$ – Frank Mar 12 '15 at 5:11
  • $\begingroup$ @Frank I'm not sure you could do a perfect strong measurement of $\hat{x}$ or $\hat{p}_x$ in practice, and even if you did I'd rather say there isn't a standard deviation (not every probability distribution has a standard deviation, or even a mean), but definitely there is no finite standard deviation. In your case, the distribution of $\hat{p}_x$ for an $\hat{x}$ eigenstate is indeed uniform, but we can't conclude that it is uniform from the lack of a finite standard deviation, there are nonuniform probability distributions that fail to have a finite standard deviation. $\endgroup$ – Timaeus Mar 12 '15 at 15:23
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One of the main problem about the uncertainty principle as it is usually told in quantum mechanics is that it is always told in the historical context, giving what Heisenberg thought about it or Feynman. For once (at least), this is not very clever.

In today's literature, we distinguish different types of "uncertainty relations", based on what they actually refer to. The first set of uncertainty relations is about state preparation. Recall that you can see a state as an abstract version of an experimental preparation procedure. It is not a particular photon, but it actually describes how to produce photons with particular properties. In objective programming languages, a "state" would be the same as a class, not an instance of a class. If you don't like this particular view of a state [this is part of what one could call the Ludwig school], you can also say that a state is the well-defined state for an ensemble. All of this should be equivalent.

In any case, the state describes the properties of this ensemble/preparation procedure and an uncertainty relation for state preparation tells us something about this. The usual Robertson-Schrödinger uncertainty relation of which the Heisenberg relation is but a special case are such state preparation uncertainty relations. The relations are expressed as expectation values and variances, but they do not really encompass any measurement. Therefore, they are about state preparation.

So then, what does the Heisenberg uncertainty tell us? Given a state, if I create instances of this state and measure their momentum, this will give me a distribution. Instead of measuring their momentum, I can also measure their position and this will give me another distribution. I could, for example, produce a stream of instances of this state (or an ensemble) and measure half of them for momentum and half of them for position, then the spread of this distribution is lower bounded by $\hbar/2$. The beauty of the Heisenberg uncertainty relation is that this statement holds true regardless of how I prepare my state. There is no experimental preparation procedure such that the spread of momentum times spread of position is not lower bounded by $\hbar/2$.

The other type of uncertainty relation is about measurements. They are also called "error-disturbance relations", because they try to quantify the oft quoted "if you measure a state, you necessarily disturb it".

Here, you have to consider measurements and you have to define what it means to measure the same instance of a state first with one observable, second, with another. Here, "simultaneous measurement" would have to be defined, but none of this is necessary for the usual Heisenberg uncertainty principle. There is a big literature on error-disturbance relations and a lot of active discussion, to quote just two, you could have a look at Ozawa or Busch, Lahti, Werner for two opposing views.

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  • $\begingroup$ Very interesting! A real hornet's nest... $\endgroup$ – Frank Mar 11 '15 at 16:55
  • $\begingroup$ Foundational issues in quantum mechanics are usually way more difficult than they sound. But the way this is fought about is also quite interesting (see the second paper I linked to)... $\endgroup$ – Martin Mar 11 '15 at 17:01
  • $\begingroup$ Mmm … how do you prepare without measuring? $\endgroup$ – Hector Mar 11 '15 at 19:45
  • $\begingroup$ @Hector: there are certainly procedures that do not use measurements (not for all states, though), many would usually need it. But I don't see any relevance for the question at hand, which is why I said "do not really encompass any measurements" instead of "do not encompass any measurements". $\endgroup$ – Martin Mar 11 '15 at 20:12
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You cannot get both $\langle \psi \rvert A \lvert \psi \rangle$ and $\langle \psi \rvert B \lvert \psi \rangle$ if you only have one state $\lvert \psi \rangle$, or if you always measure $A$ on your states.

What you need to do to check the uncertainty principle experimentally is preparing an ensemble of identical states and then measure $A$ on one half and $B$ on the other half.

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  • $\begingroup$ That's what I thought. Feynman has a statement in his Lectures under "Uncertainty principle" which requires care in understanding, because he says "If you make the measurement on any object, and you can determine the x-component of its momentum with an uncertainty Δp, you cannot, at the same time, know its x-position more accurately than Δx≥ℏ/2Δp, where ℏ is a definite fixed number given by nature." but "at the same time" really means on the same population of particles in the same state, right? $\endgroup$ – Frank Mar 11 '15 at 16:26
  • $\begingroup$ ACuriousMind - but wait - can I still ask what is the value of B when I measure A? The value of B will not be an eigenstate of B, of course, but it might be a superposition - does "measurement" exclusively mean "collapse to eigenvector of Hermitian operator"? If I have a A detector and a B detector, I "cannot" put the B detector on the particle if I put the A detector on it? I thought I might still get values for B, just smeared out a lot. $\endgroup$ – Frank Mar 11 '15 at 16:31
  • $\begingroup$ @Frank: To the Feynman quote: Yes. Although the waters get muddy here, because it is not clear whether "measuring with uncertainty $\Delta p$" is meant to actually collapse the state into a momentum eigenstate, or whether it is meant that the state is a momentum wavepacket with width $\Delta p$. In the latter case, the statement means that the equivalent position wavepacket has width $\frac{\hbar}{2\Delta p}$. $\endgroup$ – ACuriousMind Mar 11 '15 at 16:32
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    $\begingroup$ @Frank: Measurement is not easily understood in QM. We don't really have a clear, univserally accepted picture what happens there. It's not even clear that anything "collapses", the decoherence/einselection approach knows no such thing as collapse. That said, of course you can measure $B$ after $A$, but then you are not measuring $B$ on $\psi$, but on one of the eigenstates of $A$ of which $\psi$ was a superposition (since measurement with $A$ "collapses" the state). Furthermore, "the value of $B$" is an ill-defined thing for a quantum state that is not an eigenstate of $B$. $\endgroup$ – ACuriousMind Mar 11 '15 at 16:37
  • $\begingroup$ Ah - so I stumbled on a "real" issue in fact :-) Do you have references for further reading to educate myself in this area? $\endgroup$ – Frank Mar 11 '15 at 16:38

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