Uncertainty principle & simultaneous measurement

[normvcr]

I have been reading

SIMULTANEOUS MEASURABILITY IN QUANTUM THEORY by JAMES L. PARK AND HENRY MARGENAU, International.]ournal oj'TheoreticalPhysics, Vol. 1, No. 3 (1968), pp. 211-283.

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

σ(A)σ(B) ≥ 1/2 |μ([A,B])|

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.

 

[bapowell]

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.

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.

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.

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

 

[bhobba]

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.

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 can't do is measure some to any degree of accuracy AT THE SAME TIME.

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.

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 can't 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.

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.

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

 

[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 $B$.

 

[bhobba]

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

 

[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.

 

[phinds]

What you can't do is measure some to any degree of accuracy AT THE SAME TIME.

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

 

[bhobba]

The uncertainty principle as usually stated, has nothing to do with the possibility to measure observables accurately.

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

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

 

[phinds]

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

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.

 

[bhobba]

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.

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.

Added Later:
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

 

[phinds]

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.

Thanks
Bill

Also, a fair comment.

 

[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).

 

[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.

 

[bhobba]

(I did not find where in Ballentine's book this theorem is addressed).

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

 

[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).

See also the interesting paper:

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

 

[atyy]

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

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.

 

[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,

...depends on precisely what constitutes a 'measurement.'

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.

 

[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?

 

[stevendaryl]

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.

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.)

 

[stevendaryl]

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

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

 

[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.

 

[bhobba]

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

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

 

[stevendaryl]

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

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

 

[bhobba]

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.)

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.'

But others such a black body radiation lead to paradigm changing research.

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

 

[atyy]

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?

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

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)

 

[DrChinese]

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.

Yes, and that eliminates all the questions about performing simultaneous non-commuting measurements on a single particle (which is something ZapperZ insists is theoretically feasible).

But the resulting values are still subject to the uncertainty principle. If you attempted to check those values again, you would get a mixture of results over a series of trials. So now you are left wondering, "what is the meaning of the two values in my hand"? They are numbers, but what do they actually tell you about the underlying particle?

 

[normvcr]

Referring to Posts #25 & 26:

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 two values that you measure on the second particle are not on the same state, so we may not be able to call the two measurements simultaneous. I am assuming that the second value is not pre-existing, and is meaningful only as a consequence of the 2nd measurement.

Of course, we have not strictly defined what we mean by, "simultaneous measurement".

Here is the Park-Margenau paper:

http://www.statintquant.net/siq/pdf/park_simultaneous_measurability.pdf

 

[DrChinese]

Of course, we have not strictly defined what we mean by, "simultaneous measurement".

Here is the Park-Margenau paper:

The cited paper is from 1968. That's a bit outdated, don't you think? It's essentially pre-Bell (since Bell was new at that time and is not cited) and only touches on product states lightly.

And product states are what you need to have if you are trying to extract 2 simultaneous non-commuting measurement values out of a system with either 1 or 2 particles. And if you hope those to have any meaning.

 

[atyy]

Here is the Park-Margenau paper:

Thanks for the link to the Park-Margenau paper! My guess about what is going on is that they are able to demonstrate simultaneous accurate measurement of conjgate observables only on a restricted set of states, namely those attainable at large times. The inequalities for simultaneous measurement given in eg. the Branciard paper in post #15 are state dependent, so simultaneous accurate measurement of conjugate observables is possible in some states. For example, if the state is an eigenstate of one observable, that observable can be measured without disturbing the state, leaving the same state available for subsequent measurement. As far as I know, simultaneous accurate measurement of non-commuting observables on an arbitrary unknown state is not possible.

Edit: I see you made the same point about the small set of states in your OP. So I guess my only additional point is the references in #15, which using standard QM allow joint accurate measurement of non-commuting observables in some states.

 

[bhobba]

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

Basically yes.

It's an operator O = ∑ yi |bi><bi|, yi real. It means the resolution of the identity |bi><bi| describes some observation such that the probability of outcome i depends only on the |bi><bi|. Given any resolution of the identity we can in principle find an observation that describes it. The yi are the arbitrary real numbers associated with each outcome and the operator from that association is O = ∑ yi |bi><bi|. Via the spectral theorem the two are in 1-1 correspondence meaning each implies the other.

In fact, via Gleason the above implies the Born rule.

For foundational issues its easier to consider finite spaces and extend it via the RHS formalism. That way subtle issues of what exactly is a Hermitian operator is avoided.

Thanks
Bill

 

[stevendaryl]

Basically yes.

It's an operator O = ∑ yi |bi><bi|, yi real. It means the resolution of the identity |bi><bi| describes some observation such that the probability of outcome i depends only on the |bi><bi|. Given any resolution of the identity we can in principle find an observation that describes it. The yi are the arbitrary real numbers associated with each outcome and the operator from that association is O = ∑ yi |bi><bi|. Via the spectral theorem the two are in 1-1 correspondence meaning each implies the other.

In fact, via Gleason the above implies the Born rule.

For foundational issues its easier to consider finite spaces and extend it via the RHS formalism. That way subtle issues of what exactly is a Hermitian operator is avoided.

Thanks
Bill

Don't superselection rules imply that not all Hermitian operators are observables? For example, if $O$ is an operator whose eigenstates are superpositions of states with different total charges, then those eigenstates are not physically realizable, and $O$ is not observable.