# Uncertainty principle & simultaneous measurement

1. Jun 12, 2014

### normvcr

and would be interested in feedback to the comments that I make, below.

One of the points made in this paper is that the interpretation of the uncertainty relation
needs to be re-examined in its relation to the viability of having simultaneous measurements of A & B. For example, α(A)>0, or not, regardless of whether A & B are simultaneously measurable, and is the standard deviation of the possible outcomes of the operator A. So, this does not seem relevant to A & B being simultaneously measurable.

The paper also references and outlines von Neumann's proof that if A & B are simultaneously measurable, then [A,B]=0 . However, the proof makes the assumption that every observable corresponds to an operator, which validity the authors question.

Finally, the authors construct counter-examples to von Neumann's theorem, thus showing the inconsistency of the quantum mechanics postulates. I have worked through one of the counter examples (the one using Pauli spin matrices), but I am not convinced, as only a small set of states is considered, rather than an arbitrary state.

2. Jun 12, 2014

### bapowell

Right, but the uncertainty principle applies to observables that are not simultaneously measurable. If they are simultaneously measurable, then the minimum standard deviation of an observable is trivially zero, i.e. there is no fundamental limit imposed by quantum mechanics. Of course it can be larger, due to statistical fluctuations in the measurement, but these are not the types of uncertainties that the Heisenberg relation is concerned with.

It is generally accepted as an axiom that every observable corresponds to a Hermitian operator. Of course, one can consider alternative axioms.

3. Jun 12, 2014

### Staff: Mentor

Its a theorem. Either QM is wrong or the uncertainty relations are true - no ifs or buts. But you need to understand what it says. You can, in principle, measure any observable to any degree of accuracy. What you cant do is measure some to any degree of accuracy AT THE SAME TIME.

That's the assumption of the so called strong superposition principle. Most tend to accept it, its used in a lot of proofs, but we do not know if its valid. Nor do I know of anyone that has definitely disproven it either. BTW its an in principle assumption - that in practice you cant do it in some cases would not be enough to invalidate it, just like that you can never measure position exactly does not invalidate the theoretical use of the Dirac delta function for such states.

If anyone has shown that it would be BIG news, earning them an immediate Nobel prize. Rest assured it's in error, but most people would probably be like me, they don't like going through obviously silly claims to find the precise error they made. If anyone wants to do it - feel free - but I don't think such would appeal to too many.

Thanks
Bill

4. Jun 12, 2014

### hilbert2

One point that has to be remembered, is that even if two observables do not commute, it may be possible to simultaneously measure them in isolated cases. Consider a three level quantum system, and the two hermitian operators:

$A=\left[\begin{smallmatrix}a&0&0\\0&b&0\\0&0&c\end{smallmatrix}\right]$ $B=\left[\begin{smallmatrix}0&c&0\\c&0&0\\0&0&d\end{smallmatrix}\right]$,
where a,b,c,d are real numbers. The operators do not commute, but still the vector $\left[\begin{smallmatrix}0\\0\\1\end{smallmatrix}\right]$ is an eigenvector of both of them. The non-commutativity only means that it is impossible to form a complete basis that consists of common eigenvectors of $A$ and $B$.

5. Jun 12, 2014

### Staff: Mentor

I also just checked Ballentine.

I am pretty sure Von Neumann's proof, from memory, does use the strong superposition principle, but wasn't sure if its required. So checked Ballentine and his proof doesn't use it. In fact I think he makes a point of not assuming it all through his excellent book.

Thanks
Bill

6. Jun 12, 2014

### vanhees71

The uncertainty principle as usually stated, has nothing to do with the possibility to measure observables accurately. It is all about the possibility or impossibility to prepare quantum states where several observables are determined.

If the operators representing two observables do not commute, it is in general not possible to prepare the system in a state where both variables are determined sharply or with arbitrary precision. E.g., the famous position-momentum uncertainty relation, $\Delta x \Delta p_x \geq \hbar/2$ says that you have the choice to prepare one of them very accurately but that then the other is the more indetermined. It is impossible to localize a particle precisely at a given point in a strict geometrical sense, because this $\Delta x=0$ or $\Delta p_x=0$ would violate the uncertainty relation in any case.

Of course, you can always measure any observable on any system as accurately as is technically possible. In order to verify the uncertainty relation for position and momentum you have to prepare ensembles of particles independently in a given state and then measure either position or momentum and analyze the outcomes statistically. Then you can measure the position and momentum to any accuracy, including the variances of these quantities, you like. The accuracy of your measurement device must be better than the expected standard deviations of these quantities in order to really measure the quantum uncertainty and not the uncertainty due to your measurement devices.

7. Jun 12, 2014

### phinds

No, my understanding is that is NOT what it says. See post #6. My understanding is that THAT is what it says.

8. Jun 12, 2014

### Staff: Mentor

OK. Poor choice of words - fair cop. What it is, is a statement about measurements of similarly prepared systems. The correct statement is if you have a large number of similarly prepared systems and measure half with one observable and half with the other observable the distribution will have the variances predicted by the uncertainty relations.

Thanks
Bill

9. Jun 12, 2014

### phinds

I'll argue with your statement that it was just a "poor choice of words". I think it's much worse than that and here's why. For years, I was POSITIVE that the HUP said EXACTLY what you said ... that you can't make simultaneous measurements to an arbitrary degree of accuracy on a single quantum object. I think that most people believe that's what it says. It was only after having it explained to me here on this forum that I realized that I was wrong about that.

Pop-science information pretty much always presents it that way, so it's not just a poor choice of words, it's supporting a misleading view of what science actually says in this case.

I'm vehement about this not because I'm trying to give you a hard time but because I was embarrassed to realize that I had been suckered by pop-sci and I hate to see anything on this forum that contributes to that kind of misinformation.

10. Jun 12, 2014

### Staff: Mentor

Fair enough comment.

But I have to also say some of those pop-sci books are written by people like Brian Cox that do know what they are talking about. I suspect the issue is, like it was for me when I wrote it, was simply being slack. Of course QM is the LAST subject you should do that for, but all of us can slip into bad habits.

I have Brian Coxes book so decided to see what it said. Its generally quite good, but yes, he falls into the same trap.

Thanks
Bill

Last edited: Jun 12, 2014
11. Jun 12, 2014

### phinds

Also, a fair comment.

12. Jun 12, 2014

### normvcr

In response to Post #5, I agree that almost all authors make the strong assumption about observables i.e. that every observable corresponds to an operator. In "Math. Foundations of QM", von Neumann uses the strong assumption in his proof that, given two simultaneously measurable dynamic variables,R and S, the corresponding operators, A & B, must satisfy [A,B] = 0. The strong assumption is used to assign an operator, C, to the dynamic variable R+S. He then shows that C = A + B (pp 225-226). (I did not find where in Ballentine's book this theorem is addressed).

13. Jun 12, 2014

### WannabeNewton

The concept of "simultaneous measurement", non-commutativity of observables, Von Neumann's proof, and Park/Margenau's paper are all discussed in detail in "Lectures on Quantum Theory"-Isham; see sections 5.2 and 6.3.

14. Jun 12, 2014

### Staff: Mentor

I thought he did on page 223. But had a more careful look and that isn't what he does - he is proving the uncertainly relations and I thought it would be a simple consequence of that proof. It isn't - it uses inequalities and the argument isn't reversible - drats.

I also tried to find the page where he mentioned, and least I seem to recall he mentioned, he didn't want to get into the issue of if every observable corresponded to a realisable observation, but couldn't find it.

Personally I accept the strong superposition principle - it seems pretty obvious to me.

Thanks
Bill

15. Jun 13, 2014

### atyy

Except for some special cases, non-commutation does prevent accurate joint/sequential measurements of two observables, for certain measurement procedures:

http://arxiv.org/abs/1304.2071
How well can one jointly measure two incompatible observables on a given quantum state?
Cyril Branciard

http://arxiv.org/abs/1306.1565
Proof of Heisenberg's error-disturbance relation
Paul Busch, Pekka Lahti, Reinhard F. Werner

Other measurement procedures allow different inequalities:

http://arxiv.org/abs/1212.2815
Correlations between detectors allow violation of the Heisenberg noise-disturbance principle for position and momentum measurements
Antonio Di Lorenzo

One special case in which accurate sequential measurements of an observable A followed by its conjugate B is possible is when the state is an eigenstate of of A. This is because measuring A will not disturb the state, leaving it still available for B to be measured. A second special case in which it may be argued that accurate simultaneous measurement is possible is the EPR experiement, due to the probe and system being initially correlated. This is discussed in the above paper by di Lorenzo. The special cases in which joint accurate measurement is possible are also discussed in Ozawa's http://arxiv.org/abs/0911.1147 (Theorems 10 - 14).

http://arxiv.org/abs/1211.4169
Uncertainties in Successive Measurements
Jacques Distler, Sonia Paban

Last edited: Jun 13, 2014
16. Jun 13, 2014

### atyy

No, you are wrong, and bhobba was right. bhobba referred to QM. Post #6 is referring to the "uncertainty principle", which is only part of QM. Except for special cases, the commutation relations do prevent joint accurate measurement. See the references in post #15.

17. Jun 27, 2014

### normvcr

Further to post #13, I went through the very interesting book by Isham. He clarifies the situation by explaining that the necessity of commutativity of the operators of simultaneously measurable quantities,
One of the examples from the Park-Margenau paper is a setup where the position and momentum of a particle can be measured to arbitrary accuracy. The catch is that the accuracy is asymptotic with the wait time (the longer you wait, the more accurately you can estimate the constant value of the momentum). The conclusion seems to be that, although clever measurement setups are certainly to be encouraged, there is not sufficient reason to tweak the postulates of quantum theory.

18. Jun 27, 2014

### DrChinese

You can see the underlying issues when you measure non-commuting observables of an entangled particle pair. What have you learned when you have the values in your hand?

19. Jun 27, 2014

### stevendaryl

Staff Emeritus
Actually, I don't think that discovering a physical theory is inconsistent would be that big a deal, provided that there were rules of thumb for working with the theory that gave consistent results. For example, I think that the way that many physicists used delta-functions was inconsistent, but that wasn't really a problem, because they developed rules for when they could be used safely and when they couldn't. (I'm not saying that there isn't a consistent theory of delta functions, just that many people worked with delta functions without knowing the consistent way of doing it.)

20. Jun 27, 2014

### stevendaryl

Staff Emeritus
What is the strong superposition principle? Googling did not turn up much.

21. Jun 27, 2014

### normvcr

The individual measurements on an entangled particle pair are of the form $$A \otimes I$$ and and $$I \otimes B$$ and these operators commute.

22. Jun 27, 2014

### Staff: Mentor

Its that any observable, at least in principle, corresponds to a physically realizable observation.

Its a tacit assumption in a lot of proofs eg Gleason and its variants.

Thanks
Bill

23. Jun 27, 2014

### stevendaryl

Staff Emeritus
But what's the definition of "observable" here? Does that mean any Hermitian operator?

24. Jun 27, 2014

### Staff: Mentor

I think it would be devastating (and a huge opportunity as well), just like the inconsistencies discovered in classical physics are devastating. Its just that some of those, eg acausual runaway solutions in EM, were discovered after QM so we knew the cause - it was the point particle idea which QFT does away with:
http://arxiv.org/abs/gr-qc/9912045
'The problems of runaways and pre-acceleration cast a serious doubt on the validity of the Lorentz-Dirac equation. The root of the problem resides with the fact that we are trying to describe the motion of a point particle within a purely classical theory of electromagnetism. This cannot be done consistently. Indeed, a point particle cannot be taken too literally in a classical context; it must always be considered as an approximation to a nonsingular, and extended, charge distribution. Essentially, the difficulties of the Lorentz-Dirac equation come from a neglect to take this observation into account.'

You should get your hands on Von Neumann's Mathematical Foundations of QM where he really has a go at the use of the Dirac delta function. That spurred the mathematical theory to make it valid, but it was a big issue before. Its the reason I did a detour those many moons ago into Rigged Hilbert spaces to see how it was corrected.

And until Wilson sorted it out with the effective field theory approach that takes a cut-off seriously renormalisation was a great big issue with QFT.

Inconsistencies are always of concern. Many times they turn out to be resolvable without paardigm changing new physics, but every now and then they lead to leaps.

Thanks
Bill

Last edited: Jun 27, 2014
25. Jun 27, 2014

### atyy

If I understand Dr Chinese correctly, what he's saying is that if you measure observable A on one particle in an entangled pair, you will immediately know the result of a measurement of observable A for the second particle in that pair. Now you can measure non-commuting B on the second particle, so that you know A and B simultaneously for the second particle.

The other case in which one can measure non-commuting observables simultaneously is when one knows the system is in an eigenstate of one observable. Measuring that observable does not change the state, leaving the same state available for a non-commuting observable to be accurately measured.

I think his point is that the cases where it can be argued that it is possible to simultaneously measure non-commuting observables accurately are those in which one gains limited knowledge from the measurement. Suppose we define an accurate measurement as one that produces the same distribution of outcomes as a single projective measurement. If we know the wave function, then we can generate pairs of outcomes that meet the definition of simultaneous accurate measurements. But we don't get any information, since we already knew the state. I don't have access to the Park and Margenau paper, but my guess is that their example is such a case.

Can this be formalized? Some possibilities are:
Busch, http://arxiv.org/abs/0706.3526
Fuchs, http://arxiv.org/abs/quant-ph/9611010
Leifer and Spekkens, http://arxiv.org/abs/1107.5849 (Theorem V.2, p33)

Last edited: Jun 27, 2014