This forum gives conflicting info on the HUP

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In summary: Bottom two rows are in a superposition.In summary, the article discusses the uncertainty principle and how it applies to the single slit experiment. The article states that the uncertainty principle applies to simultaneous or successive measurements of the same characteristic on different ensembles.
  • #1
phinds
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I am not going to look up specific threads to support that assertion because I assume they have been seen by all those who regularly read threads on the HUP. If necessary, I'll dig up a few.

The two points of view expressed in various threads are as follows

1) Simultaneous measurements for two characteristics that are governed by the HUP can give arbitrarily precise (including exact) values but successive measurements from the same setup will not give the same results. That is the fundamental meaning of the HUP. Single measurements are not constrained by the HUP but repeated measurements based on the same setup do not give consistent results, they give a probability spread like the single slit experiment.

2) Simultaneous measurements of the exact values of two such characteristics of one quantum object at one time is not possible. Period. That is the fundamental meaning of the HUP

As I said, I have seen threads that say one or the other of these is true and the implication is that the other is not.

How about the mods do a FAQ answer on this and pin it in the Quantum Physics Forum?

I do not really want to start that argument here in this thread because the points of view have already been expressed in various threads. Again, what I would like, if possible and reasonable, is for the moderators and Science Advisers to agree on which is true and pin the answer to this forum for all to see.

Thanks,

Paul
 
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  • #2
The fundamental underlying principle is the non-commutation of conjugate observables.

There are several "uncertainty principles" that can be derived from the underlying non-commutation. The most common textbook form of "HUP" is the Kennard-Robertson form which has nothing to do with simultaneous or successive measurements on the same ensemble. The Kennard-Robertson uncertainty principle states that if one makes precise measurement of one observable on one ensemble, and non-simultaneous and non-successive precise measurement of the conjugate observable on a different, identically prepared ensemble, then the resulting variances from the two sets of precise measurements obey the textbook inequality.

There are other "uncertainty principles" which have to do with simultaneous or joint measurements, or non-simultaneous successive measurements.
http://arxiv.org/abs/1304.2071
http://arxiv.org/abs/1312.1857
http://arxiv.org/abs/1306.1565

You may find this hilarious.
http://physicsworld.com/cws/article...y-reigns-over-heisenbergs-measurement-analogy
 
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  • #3
atyy, I'm sure it's my own ignorance but I don't see how you have in any way answered my question. I THINK what you are saying is that there IS no answer to my question because both points of view are wrong. If that's the case, then THAT should be the subject of a pinned article by the mods and science advisers. Perhaps the subject line should be "there is no the HUP"

As for the "hilarious" article, I didn't find it even humorous, it just tells me that I'm not the only one who is confused. It DOES reaffirm the point of view (true in both of 1 and 2 above) that the HUP is not a measurement problem but a statement about the nature of quantum objects.
 
  • #4
phinds said:
1) Simultaneous measurements for two characteristics that are governed by the HUP can give arbitrarily precise (including exact) values but successive measurements from the same setup will not give the same results. That is the fundamental meaning of the HUP. Single measurements are not constrained by the HUP but repeated measurements based on the same setup do not give consistent results, they give a probability spread like the single slit experiment.

2) Simultaneous measurements of the exact values of two such characteristics of one quantum object at one time is not possible. Period. That is the fundamental meaning of the HUP

As I said, I have seen threads that say one or the other of these is true and the implication is that the other is not.

I'm not an expert - so please correct me if I'm wrong – but I do think that 1) is not correctly expressed, and I somewhat doubt that anyone has claimed the possibility of exact* single measurements of complementary variables...

If we take the single-slit experiment, it would be hard to have "a probability spread", unless all particles in the ensemble obey the same (probability) principles, right?

Afaik, there is no separate set of QM rules for "the first particle", right? (and who defines "the first particle"... the whole universe is one big "QM experiment" afaik...)

Interference, diffraction and superposition are basically different names of the same phenomena, and diffraction occurs when the wavelength is comparable to the dimensions of the (diffracting) slit. Hence, when the slit gets narrower disturbance, in form of diffraction, gets bigger.

220px-Wavelength%3Dslitwidthblue.gif
.
220px-6wavelength%3Dslitwidthblue.gif


If we talk about the wavefunction and different states; I think it's obvious that an 'undisturbed' stationary state, that passes through a wide enough slit, generates a 'fixed' probability density. While a 'disturbed' superposition state ('squeezed' through the slit) generates a 'wobbling' probability density, hence greater uncertainty.

StationaryStatesAnimation.gif

Top two rows are stationary and bottom is superposition

I think it all boils down to the inherent properties of all wave-like systems.

https://www.youtube.com/watch?v=a8FTr2qMutA
http://www.youtube.com/embed/a8FTr2qMutA

https://www.youtube.com/watch?v=7vc-Uvp3vwg
http://www.youtube.com/embed/7vc-Uvp3vwg


*If exact values is possible, then for example the 1935 EPR paper would have been a total waste of time, since Einstein & Co tried to prove that you can know the exact values for x and p through entanglement, which then led to a 20 year long debate with Bohr.
 
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  • #5
phinds said:
atyy, I'm sure it's my own ignorance but I don't see how you have in any way answered my question. I THINK what you are saying is that there IS no answer to my question because both points of view are wrong. If that's the case, then THAT should be the subject of a pinned article by the mods and science advisers. Perhaps the subject line should be "there is no the HUP"

As for the "hilarious" article, I didn't find it even humorous, it just tells me that I'm not the only one who is confused. It DOES reaffirm the point of view (true in both of 1 and 2 above) that the HUP is not a measurement problem but a statement about the nature of quantum objects.

I'm pretty sure there is no controversy that the textbook HUP is the Kennard-Robertson relation, and that has nothing to do with simultaneous or successive measurements. So let's call that the HUP.

So the basic answer to your question is there is no controversy, and a question about simultaneous measurement of anything is irrelevant to the HUP.

Where there may be a controversy is the answers to the irrelevant questions (eg. can conjugate observables be simultaneously and precisely measured?). One problem is that Ballentine gave a misleading answer to one of these questions in his famous 1970 review. I also don't really know whether ZapperZ's blog post on some aspects of these issues are correct. I think the basic correct answers are given by Dr Chinese in #2 with the correction by kith in #7 of https://www.physicsforums.com/showthread.php?t=746607. For more answers to the irrelevant questions, you can look at the links above. The quantitative answer depends on how you choose to ask the question in quantitative terms.
 
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  • #6
Asher Peres has made a statement of HUP which has to do basically with the building of a set of systems which present very close values of one observable (belonging to the conjugate pair). He says that once this set is mounted, one observes unavoidable statistical dispersion on measurements of the other observable made on all the systems of this set.

Best wishes,

DaTario
 
  • #7
It sounds to me like there is a "safe" version of the HUP, the "textbook version" that atyy refers to, whereby we use it to make testable predictions about observations on carefully prepared ensembles. Typically speaking, when QM is used in that way, it is uncontroversial, which is why many prefer the "ensemble interpretation" (it skirts the murkier waters of "what does QM tell us is really going on", if you will). But the OP talked about the "fundamental meaning" of the HUP, which can be interpreted as a question about that murky issue: what is the lesson that the HUP is trying to teach us about what is possible to know about some reality? So to me, the resolution of any controversy here is that atyy is right about what the HUP refers to when it is treated uncontroversially, and anyone who wants to take it further and use it to learn something about the "true nature" of some individual system, and what is possible to simultaneously "know" about such a system, they are inevitably entering the area of interpretations of quantum mechanics. We do like to enter those murky waters, because we feel that QM is indeed trying to tell us something, but we are not surprised, as we wade into those depths, that consensus will be lost along the way!
 
  • #9
phinds said:
1) Simultaneous measurements for two characteristics that are governed by the HUP can give arbitrarily precise (including exact) values but successive measurements from the same setup will not give the same results. That is the fundamental meaning of the HUP. Single measurements are not constrained by the HUP but repeated measurements based on the same setup do not give consistent results, they give a probability spread like the single slit experiment.

2) Simultaneous measurements of the exact values of two such characteristics of one quantum object at one time is not possible. Period. That is the fundamental meaning of the HUP

As everyone agrees, these questions about simultaneous measurements have nothing to do with the textbook Heisenberg uncertainty relation, or more precisely, the Kennard-Robertson uncertainty relation. The textbook Heisenberg uncertainty relation is not about simultaneous measurements or successive measurements on a given ensemble.

However, it is interesting to discuss these irrelevant questions. For a given state, can successive accurate measurements of conjugate observables be performed? If the state is an eigenstate of observable A, then an accurate measurement of A will not disturb the state, leaving the same state available for accurate measurement of the conjugate observable B. So an accurate measurement of conjugate observables A and B can be performed in this special case. However, this measuring device will be very inaccurate on an arbitrary state, since an accurate measurement of A will disturb the state, and make the following measurement of the conjugate observable B inaccurate. (Here a measurement of B is defined to be accurate in a successive measurement if the distribution of outcomes produced is the same as that produced by an accurate non-simultaneous, non-successive measurement of B. It doesn't mean that the system had simultaneously sharp values of A and B.)

Even the statement that a particle cannot have simultaneously sharp position and momentum has an interesting special case. One way of thinking about that statement is that there is no joint probability distribution for the particle's position and momentum. The closest thing to a joint distribution is the Wigner function, which is not a probability distribution because it has negative parts in general. However, for a free particle with a Gaussian wave function, the Wigner distribution is positive. Also the time evolution keeps the Gaussian shape of the wave function, and time evolution of the Wigner function is the same as the classical Liouville equation. So in this case the particle can be thought of as being an ensemble of particles with definite trajectories which obey the classical equations of motion. (In non-relativistic quantum mechanics, one can always think of particles having definite trajectories if a Bohmian interpretation is used, but these trajectories do not obey classical equations of motion.)
 
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  • #10
DevilsAvocado said:
*If exact values is possible, then for example the 1935 EPR paper would have been a total waste of time, since Einstein & Co tried to prove that you can know the exact values for x and p through entanglement, which then led to a 20 year long debate with Bohr.

This paper suggests that the EPR experiment allows a "Heisenberg noise disturbance relation" to be violated: http://arxiv.org/abs/1212.2815 (Phys. Rev. Lett. 110, 120403). As above, this has nothing to do with the textbook HUP, which is not about simultaneous or sequential measurements.
 
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  • #11
Interesting thanks atyy, I'll check it out.
 
  • #12
phinds said:
As I said, I have seen threads that say one or the other of these is true and the implication is that the other is not.

Okay, by hints from other comments I take it you're talking about Zz's blog, right?

[my bolding]
Misconception of the Heisenberg Uncertainty Principle said:
Take note that the measurement uncertainty in a single is still the same as in the classical case. If I shoot the particle one at a time, I still see a distinct, accurate "dot" on the screen to tell me that this is where the particle hits the detector. However, unlike the classical case, my ability to predict where the NEXT one is going to hit becomes worse as I make the slit smaller. As the slit and Delta(y) becomes smaller and smaller, I know less and less where the particle is going to hit the screen. Thus, my knowledge of its y-component of the momentum correspondingly becomes more uncertain.

I agree, this is somewhat confusing and the next paragraph (in my opinion) actually makes it 'worse'...

[my bolding]
Misconception of the Heisenberg Uncertainty Principle said:
What I am trying to get across is that the HUP isn't about the knowledge of the conjugate observables of a single particle in a single measurement. I have shown that there's nothing to prevent anyone from knowing both the position and momentum of a particle in a single mesurement with arbitrary accuracy that is limited only by our technology. However, physics involves the ability to make a dynamical model that allows us to predict when and where things are going to occur in the future. While classical mechanics does not prohibit us from making as accurate of a prediction as we want, QM does! It is this predictive ability that is contained in the HUP. It is an intrinsic part of the QM formulation and not just simply a "measurement" uncertainty, as often misunderstood by many.

I'm just a bum and Zz is a pro, but this looks like a contradiction to me... mixing of experimental determination in retrospect for the first particle, and QM/HUP predictive uncertainty for the following particles...

Must not (for same slit width) the same amount of predictive uncertainty and experimental determination in retrospect, comply with all the particles in the ensemble, or...?

What am I missing?? :bugeye:
 
  • #13
I think part of the issue must involve what is meant by a measurement, and how measurements convey knowledge about systems. In general physics, we do not need to be very precise about either of those things. We consider knowledge of a system to be pretty much anything that allows us to predict future measurements, and anything that tests our predictions based on previous measurements. And we consider a measurement to be anything that conveys said knowledge-- it can be the reading on a pointer, but it can also be something we infer, using our theories, that is based on the reading of a pointer. But when these "informal" meanings get used in quantum mechanics, a problem emerges because quantum mechanics has its own formal meanings for these things, and they might not be commensurate.

One example of this problem is time. We all know how to use clocks to make time "measurements", but our clocks are generally not carried with the systems we are "measuring time" on. Hence we are not actually doing time measurements on our systems, and time is not an operator in quantum mechanics. In formal QM, time is just a parameter, a label we use to organize the things we actually do measure, and we test that label by doing measurements on other systems, on clocks-- and we just hold that the clock measurement gives us the time label that we can then use to organize the measurements we are doing on our systems. So does a time measurement on a clock convey knowledge about a system that could apply to an uncertainty principle? We see that the situation is even worse in the E-t version of the HUP.

As for "simultaneous measurements" of the complementary observables x and p, we again have a formal version (which is meaningless), and an informal one. There's no formal meaning of that phrase because you cannot do two simultaneous formal measurements on complementary observables, a formal measurement in QM decoheres a system toward one eigenstate of the corresponding operator, and two simultaneous "measurements" would interfere with each other. Working at cross purposes, neither measurement could succeed in decohering to an eigenstate, so no measurement would occur at all. However, it can be held that two separate measurements that happened at different times could be combined to yield simultaneous "knowledge" in the informal sense, but that is not the same as actually performing two simultaneous measurements. As such, this kind of "knowledge" is debatable, and becomes subject to one's interpretation of the meaning of the quantum postulates (as well as how empiricist is one's bent toward the entire pursuit of doing physics). So if one veers from the precise meaning of the HUP that atyy described, one needs to bring some significant philosophical hardware along, and abandon hope of consensus in so doing.
 
  • #14
atyy said:
For a given state, can successive accurate measurements of conjugate observables be performed? If the state is an eigenstate of observable A, then an accurate measurement of A will not disturb the state, leaving the same state available for accurate measurement of the conjugate observable B. So an accurate measurement of conjugate observables A and B can be performed in this special case. However, this measuring device will be very inaccurate on an arbitrary state, since an accurate measurement of A will disturb the state, and make the following measurement of the conjugate observable B inaccurate. (Here a measurement of B is defined to be accurate in a successive measurement if the distribution of outcomes produced is the same as that produced by an accurate non-simultaneous, non-successive measurement of B. It doesn't mean that the system had simultaneously sharp values of A and B.)

I take this back, because it seems to contradict the inequality in http://arxiv.org/abs/quant-ph/0207121. I'll start a new thread about this.
 
  • #15
phinds said:
The two points of view expressed in various threads are as follows

1) Simultaneous measurements for two characteristics that are governed by the HUP can give arbitrarily precise (including exact) values but successive measurements from the same setup will not give the same results. That is the fundamental meaning of the HUP. Single measurements are not constrained by the HUP but repeated measurements based on the same setup do not give consistent results, they give a probability spread like the single slit experiment.

2) Simultaneous measurements of the exact values of two such characteristics of one quantum object at one time is not possible. Period. That is the fundamental meaning of the HUP

As I said, I have seen threads that say one or the other of these is true and the implication is that the other is not.
The correct statement is 1), and let me explain why.

Suppose you have two macroscopic measuring apparata, say one for measurement of position and the other for measurement of momentum. Nobody can stop you to turn on them both at the same time. So what will happen when you do do that? No, the universe will not disappear and the apparata will not explode. In fact, the apparata can do nothing else but to show some sharp numbers. That's how they are designed, to show some sharp numbers. So whatever these numbers will be, they are to be interpreted as measured values of position and momentum.

But the important question is - what will happen with the microscopic measured state during such a measurement? Will it collapse to a position eigenstate? Or to a momentum eigenstate? Or to both?

It, of course, cannot collapse to both because there is no simultaneous eigenstate of both position and momentum. In fact, it can be shown that it will collapse to a state which is neither a position eigenstate nor a momentum eigenstate. Instead, It will collapse to something in between, namely to a coherent state with a well defined AVERAGE position and momentum, in which neither position nor momentum are certain. That's why, in the subsequent measurement, you cannot predict neither the result of the measurement of position, nor that of momentum.
 
  • #16
Demystifier said:
The correct statement is 1), and let me explain why.

Suppose you have two macroscopic measuring apparata, say one for measurement of position and the other for measurement of momentum. Nobody can stop you to turn on them both at the same time. So what will happen when you do do that? No, the universe will not disappear and the apparata will not explode. In fact, the apparata can do nothing else but to show some sharp numbers. That's how they are designed, to show some sharp numbers. So whatever these numbers will be, they are to be interpreted as measured values of position and momentum.
I disagree with this. Consider "simultaneous measurements" on spin components of spin-1/2 particles instead. What do you get if you try to put two Stern-Gerlach devices in the same place? Now you have two magnets, but you still have only one magnetic field, and only one detector screen (possibly bigger than you're used to). What you would end up measuring isn't ##S_x## and ##S_y##. I haven't really thought about what the magnetic fields add up to, but my first guess would be that the result is something similar to the field from a device oriented in the (1,1,0) direction, so that we would actually end up measuring ##(S_x+S_y)/\sqrt{2}##. The universe didn't disappear, and nothing exploded, but we ended up measuring just one observable that isn't equal to either of the two that we were trying to measure simultaneously.

I think the lesson learned from this is that simultaneous measurements are possible if and only if the measuring devices can exist in the same place without interfering with each other. This is why very few pairs of observables can be measured simultaneously.

A momentum measurement is a sequence of approximate position measurements, from which we calculate the average momentum that the particle had between the first and the last of these approximate position measurements. If we insert another device meant to determine the position very accurately, at some time between the first and the last approximate position measurement, this just ends the momentum measurement early.

I don't think we can consider a measurement of average momentum from t1 to t2 and a measurement of position at t2 "simultaneous measurements". The problem is that these measurements are performed on particles in different states. The momentum measurement is performed on a particle that was prepared in some way before the first approximate position measurement was made, and the very accurate position measurement that ends the momentum measurement is performed on a particle in the state prepared by the last of the approximate position measurements.
 
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  • #17
Fredrik;4711901 Consider "simultaneous measurements" on spin components of spin-1/2 particles instead. [/QUOTE said:
How about the Einstein Podolsky Rosen (EPR) setup, then?
Suppose you have 2 spins coupled into a singlet and you measure s_x on particle 1 and s_y on particle 2. This is possible, but the result will be erratic.
 
  • #18
Fredrik said:
I disagree with this. Consider "simultaneous measurements" on spin components of spin-1/2 particles instead. What do you get if you try to put two Stern-Gerlach devices in the same place? Now you have two magnets, but you still have only one magnetic field, and only one detector screen (possibly bigger than you're used to). What you would end up measuring isn't ##S_x## and ##S_y##. I haven't really thought about what the magnetic fields add up to, but my first guess would be that the result is something similar to the field from a device oriented in the (1,1,0) direction, so that we would actually end up measuring ##(S_x+S_y)/\sqrt{2}##. The universe didn't disappear, and nothing exploded, but we ended up measuring just one observable that isn't equal to either of the two that we were trying to measure simultaneously.
What you presented is one particular example of a measuring apparatus which does not allow simultaneous measurement of noncommuting observables. But nothing in your argument shows that this is a general conclusion.

Let me try to describe a variation of SG device which, in principle, would allow simultaneous measurement.

Suppose you want to measure the spin of particle A. For that purpose you first make the particle A interact with another particle B, such that the two particles become entangled. The entanglement may be such that, by measuring spin s_x of particle B, you also know spin s_x of particle A. Then you separate the entangled particles and route particle B in a standard SG device. In this way you measure s_x of particle A without ever routing particle A in a standard SG device.

Similarly, you can measure spin s_y of particle A by entangling it with a third particle C and routing particle C in another standard SG device.

Finally, you can combine these two modified SG devices, such that you can simultaneously measure s_x via particle B and s_y via particle C.
 
  • #19
DrDu said:
How about the Einstein Podolsky Rosen (EPR) setup, then?
Suppose you have 2 spins coupled into a singlet and you measure s_x on particle 1 and s_y on particle 2. This is possible, but the result will be erratic.
This is a simultaneous measurement of ##S_x\otimes I## and ##I\otimes S_y## on a two-particle system, and those operators commute. I don't think it can be viewed as a simultaneous measurement of two non-commuting observables on a single particle. The single-particle theory isn't accurate anyway, since the particle isn't adequately isolated from its environment before the measurement(s) begin.

Demystifier said:
What you presented is one particular example of a measuring apparatus which does not allow simultaneous measurement of noncommuting observables. But nothing in your argument shows that this is a general conclusion.
Agreed. I'm not sure it can be proved in general, since we may have to be able to describe all measuring devices to do that. I will think about it.

Demystifier said:
Let me try to describe a variation of SG device which, in principle, would allow simultaneous measurement.

Suppose you want to measure the spin of particle A. For that purpose you first make the particle A interact with another particle B, such that the two particles become entangled. The entanglement may be such that, by measuring spin s_x of particle B, you also know spin s_x of particle A. Then you separate the entangled particles and route particle B in a standard SG device. In this way you measure s_x of particle A without ever routing particle A in a standard SG device.

Similarly, you can measure spin s_y of particle A by entangling it with a third particle C and routing particle C in another standard SG device.

Finally, you can combine these two modified SG devices, such that you can simultaneously measure s_x via particle B and s_y via particle C.
If you entangle A with B, and A with C, then we're dealing with a three-particle system. I don't see how to define a state vector that represents the preparation procedure you described. If there is one, I would expect that my answer is going to be similar to what I told DrDu: The single-particle theory doesn't apply, and we're just doing simultaneous measurements of commuting observables in the three-particle theory.
 
  • #20
Demystifier said:
The correct statement is 1), and let me explain why.

Suppose you have two macroscopic measuring apparata, say one for measurement of position and the other for measurement of momentum. Nobody can stop you to turn on them both at the same time. So what will happen when you do do that? No, the universe will not disappear and the apparata will not explode. In fact, the apparata can do nothing else but to show some sharp numbers. That's how they are designed, to show some sharp numbers. So whatever these numbers will be, they are to be interpreted as measured values of position and momentum.

Talking macroscopic: Phase & frequency for a wave in the time domain is analogous to position & momentum for a wave* in the spatial domain, right?

Question: How do you measure the exact pitch for a sound wave in an exact moment in time?

220px-Simple_harmonic_motion_animation.gif


*The wavefunction is definitely a wave, and I guess this is also true for the pilot wave... ;)

Demystifier said:
Instead, It will collapse to something in between, namely to a coherent state with a well defined AVERAGE position and momentum, in which neither position nor momentum are certain.

But... we are talking exact values, right... not average, no matter how well defined... and if neither position nor momentum are certain, it cannot reasonably be exact...

I'm completely lost... :uhh:
 
  • #21
DrDu said:
How about the Einstein Podolsky Rosen (EPR) setup, then?
Suppose you have 2 spins coupled into a singlet and you measure s_x on particle 1 and s_y on particle 2. This is possible, but the result will be erratic.

It won't work. If Alice measures +z we know that Bob will get -z, but if Bob measures x instead (trying to violate HUP) his result will be completely random 50/50 +x and -x.

I.e. HUP wins and entanglement is 'decoupled' for non-commuting observables.
 
  • #22
DevilsAvocado said:
Talking macroscopic:
...
Question: How do you measure the exact pitch for a sound wave in an exact moment in time?
You measure many "exact moments" according a time-resolution of your choice and after you have determined the pitch using information from from all those measurements , you will know precisely what the value was at one of those "exact moments".
 
  • #23
Fredrik said:
This is a simultaneous measurement of ##S_x\otimes I## and ##I\otimes S_y## on a two-particle system, and those operators commute.

Of course. I should think sometimes before writing.
 
  • #24
DrDu said:
Of course. I should think sometimes before writing.

There may be a variant of your argument that works. http://arxiv.org/abs/0911.1147 argues on p14 "In the EPR state of two particles, I and II, the momentum of particle I can be measured by directly and locally measuring the momentum of particle II taking into account the EPR correlation; this follows from the EPR original argument stating that the locality of measurement ensures that the predicted correlation determines the value of momentum of particle I. The locality of the momentum measurement of particle II also concludes that it does not disturb the particle I, and hence we can simultaneously measure the position of particle I by a direct measurement on particle I. Thus, the momentum and position of particle I are simultaneously measurable, so that both the measured values corresponds to elements of reality."

And a claimed proof of "Theorem 14. In any Hilbert space with dimension more than 3, there are nowhere commuting observables that are simultaneously measurable in a state that is not an eigenstate of either observable." is provided on p10.
 
  • #25
I think his argument may be running afoul of an unclear meaning of concepts like "simultaneous knowlege" or "simultaneous measurements." Consider a wave packet for a single electron fired from a source that reaches a detector a million light years away (by some very patient physicists). In advance of the electron comes photons carrying information about when the photon was fired, and the distance is also known. Cannot a detector now tell us the electron position with extreme accuracy, and we can calculate its speed by x/t, again to extreme accuracy (because x and t are known to high precision, given how large they are)? Why does that not violate the version of the HUP that talks about sigma(x)*sigma(p)? I would say it is not a violation because there is never a moment when you could predict the outcome of both an x and p measurement on the particle. If there is never a moment when you could predict such a measurement, how can you claim to know it? In other words, a measurement that destroys the information it is measuring cannot convey simultaneous knowledge of two observables. We are knowing something in a present that only applies to a past that no longer exists.

ETA: Put differently, let's say you build a laser that sends out photons with very definite momentum. You then walk across the room and start measuring their positions as they hit a wall, again to unlimited precision. Why would we not say this is like using an entangled pair to get simultaneous knowledge of the x and p of each photon, we are correlating knowledge of p from some other means with measured knowledge of x. But again, I would say it is because the p and the x are not simultaneously measured because they are not simultaneously known, there is never a moment when I could predict both the x and the p at the same moment, and so how can I claim I have simultaneous knowledge of x and p?
 
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  • #26
I agree with Ken G that these notions of measurement seem different from the traditional measurements in textbooks, which involve an apparatus that gives a "correct" result for any state. These "state-dependent" measurements only work for certain special states, and are inaccurate for all other states. It seems that in this spirit, one could say that if one knew the state, then a state dependent measurement of any observable could involve tossing the state in the garbage, and simply writing down an outcome such that the distribution of outcomes is the same as if an accurate measurement were made on the state. The Ozawa paper linked to in #24 seems to acknowledge this in the discussion (starting at bottom of p12).
 
  • #27
Yes, it should probably be said that any "measurement" that requires quantum mechanics be right to yield a valid result should not be regarded as a measurement at all, it's more of a calculation that we should be using measurements to test, and of course the measurements doing the testing have to be valid even if quantum mechanics is wrong.
 
  • #28
DevilsAvocado said:
But... we are talking exact values, right... not average, no matter how well defined... and if neither position nor momentum are certain, it cannot reasonably be exact...

I'm completely lost... :uhh:
I didn't use the word "exact". I used the word "sharp". It referred to the value on the macroscopic apparatus. For example, the apparatus may have a digital display which may show a sharp number 7 or a sharp number 8, but it cannot show some unsharp sign which resembles both 7 and 8.

Think like an experimentalist. Pretend that you know nothing about quantum theory, and just use two sophisticated gadgets for which you were told that they measure position and momentum. You don't even need to know how the gadgets work. All you need to know is how to use them, by pressing appropriate buttons. When you do that, the displays on the gadgets show some digital numbers which, you are told, are the measured position and momentum.

So when you turn on both gadgets at the same time, what do you expect to see? Do you expect that only one of the gadgets will display a number? If so, then which one?

No, you should not expect such a thing. There is no doubt that both gadgets will show some numbers. As long as you think like an experimentalist without any theoretical prejudices, there is nothing more natural than to interpret these two numbers as simultaneous measurement of position and momentum. That is all.
 
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  • #29
Fredrik said:
The single-particle theory doesn't apply, and we're just doing simultaneous measurements of commuting observables in the three-particle theory.
You are right that it requires at least a three-particle theory. But in fact, you should have in mind that there is no such thing as one-particle measurement. To perform a measurement, you always need a macroscopic apparatus which is made of many particles. Moreover, from the von Neumann theory of meaurement, or from the modern development of this theory based on decoherence, you should know that some macroscopic degrees (made of many particles) are indeed ENTANGLED with the single-particle object you want to measure.

In my case of simultaneous measurement of s_x and s_y, you are also right that what you REALLY observe are simultaneous values of two commuting observables. These are in fact two macroscopic observables, corresponding to the pointer positions of two different macroscopic apparata.

But from the point of view of an experimentalist, who may be partially agnostic about the underlying quantum theory, it is natural to interpret these two measurements as simultaneous measurements of s_x and s_y of a single particle. It is from HIS point of view correct to say that a simultaneous measurement of non-commuting observables is possible.

But ultimately, the answer to the question whether 1) or 2) is true depends on the definition of "measurement". Essentially, I used an operational definition of measurement according to which 1) is true, but with a different definition of measurement one may also arive at the conclusion that 2) is true.
 
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  • #30
Fredrik said:
If you entangle A with B, and A with C, then we're dealing with a three-particle system. I don't see how to define a state vector that represents the preparation procedure you described.
That is actually quite easy. If you know
1) how A entangles with B when C is not present, and
2) how A entangles with C when B is not present,
then linearity alone is sufficient to determine how A entangles with both B and C when both are present.
 
  • #31
DevilsAvocado said:
Question: How do you measure the exact pitch for a sound wave in an exact moment in time?
By observing a classical wave, you cannot measure its frequency in an exact moment of time.

Yet, in principle, in QM you can measure any observable in an arbitrarily short time. Since this is valid for any observable, this is valid also for the Hamiltonian. But by measuring Hamiltonian you measure energy, which means that you measure frequency.

So how is it possible that in QM you can measure frequency in an exact moment of time? Essentially, that is because in QM you never really measure frequency. Instead, you really measure something else (the position of some macroscopic pointer) which turns out to be ENTANGLED with frequency.

This is somewhat analogous to the following common sense example. How to make a picture of a very short event (say 1 milisecond) with a very slow camera (with exposition, say, 1 second)? Easy! First make a picture with another sufficiently fast camera, and then use your slow camera to take a picture of the picture made by the fast camera. In this case, by your slow camera you really take a picture of something else (of another picture), but this other picture is "entangled" with the short event, in the sense that it is strongly correlated with it.
 
  • #32
All my posts above can be summarized as follows:
Whenever you measure something in QM, you really measure something else.
 
  • #33
billschnieder said:
You measure many "exact moments" according a time-resolution of your choice and after you have determined the pitch using information from from all those measurements , you will know precisely what the value was at one of those "exact moments".

Okay Bill, but value and pitch is not the same thing, right? If you pick one off the "exact moments", all you will hear is a very short 'click', right? Or, if you send one "exact moment" to Schrödinger and ask him to calculate/predict which pitch it represents, he will be very mad at you, right? :wink:

This is digital signal processing (as in sampling), and in this form of quantization there will always be error/distortion, since the set of possible input values may be infinitely large (thus uncountable).

800px-Quantization_error.png


And even if you could build an extreme super cool quantum clock with a resolution of some yottahertz, you will ultimately run into the wall of Planck time ≈ 5.391 × 10−44 second (thus HUP) ...

[itex]t_P \equiv \sqrt{\frac{\hbar G}{c^5}}[/itex]
 
  • #34
Demystifier said:
I didn't use the word "exact". I used the word "sharp". It referred to the value on the macroscopic apparatus. For example, the apparatus may have a digital display which may show a sharp number 7 or a sharp number 8, but it cannot show some unsharp sign which resembles both 7 and 8.
[...]
There is no doubt that both gadgets will show some numbers. As long as you think like an experimentalist without any theoretical prejudices, there is nothing more natural than to interpret these two numbers as simultaneous measurement of position and momentum. That is all.

Okay thanks DM, think I understand now. Simultaneous "sharp" measurement of position and momentum is possible, but arbitrarily precise (including exact) values are not possible, right?
 
  • #35
Demystifier said:
By observing a classical wave, you cannot measure its frequency in an exact moment of time.

Thanks DM! I knew it! :smile:

Demystifier said:
Yet, in principle, in QM you can measure any observable in an arbitrarily short time. Since this is valid for any observable, this is valid also for the Hamiltonian. But by measuring Hamiltonian you measure energy, which means that you measure frequency.

So how is it possible that in QM you can measure frequency in an exact moment of time? Essentially, that is because in QM you never really measure frequency. Instead, you really measure something else (the position of some macroscopic pointer) which turns out to be ENTANGLED with frequency.

But... what about 'the wall' of Planck time...? :uhh:

[se post #33]

Demystifier said:
First make a picture with another sufficiently fast camera, and then use your slow camera to take a picture of the picture made by the fast camera.

But... if "the fast camera" is severely restricted by Mr. Planck, whom is a very close friend to Mr. Heisenberg... doesn't HUP always win in the end...?
 
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