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You probably think of the position of a point particle as something that can be represented by 3 numbers. That's only true in classical physics. In quantum mechanics the position must be represented by a function (the wave function), which also contains all the information about the particle's momentum.

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I don't think that's correct. I would say that the main idea is that the state of a physical system is represented by a wave function instead of a set of numbers that specify the position and momentum of every component, and this implies that neither position nor momentum is well-defined at any time.The main idea is that there is a smallest possible nudge.

Forget about the uncertainty principle for the moment, and focus on wave functions. The fact that classical theories don't work, and quantum theories do, is what implies that the particle is "smeared" over some region (the region where its wave function is non-zero).

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But then as a student I used to ask: "what if we don't measure it? what if we leave it undisturbed? why doesn't it have an exact value then?".

The answer I usually got was "if you don't measure it, it's like it doesn't exist for you so why bother?".

But it's hard for a student to think that something doesn't exist with a value if you just choose to not bother...

I don't know if it's wrong, but nowadays I am trying to see if I can think of it in slightly different terms, which (for me) may give back more sense to such answers. Instead of "measuring", can we use the word "interacting"? That helps me to forget about the image of a mad scientist trying to run after a particle with a microscope to measure it. Then perhaps the particle's property we are considering (or a couple of related properties) is defined only as through the possible interactions it can give with the rest of the universe or at least some part of it, such as another particle with the same properties.

So perhaps position+momentum make sense only if they affect the particle's interactions with others? I might be completely off... But do they manifest themselves in a particle that doesn't have a universe around it, no other particles to meet?

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I kind of liked the image of a smeared particle tho... A "small cloud" rather than a "hard pebble".

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After doing some quantum courses at university, I'm still not happy with how the HUP is used. The HUP expresses a limit of the standard deviation from the mean of a measurement on position or momentum.

One more thing that is buzzing in my mind (but still I haven't managed to grasp it) is that the HUP just doesn't say there is a limit, but it actually

I think it has something to do with the fact that e.g. electromagnetic interaction is carried by photons, and the

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I disagree with this. You're describing a hidden variable theory, and those typically imply Bell inequalities which are inconsistent with experiments. Is a superposition of many different positions really

If you are unfamiliar with Bell inequalities for spin, this is a short summary of what they're about:

When a two-particle system is in the state |+>|-> + |->|+>, and we measure the spin-z component of one of the particles, and get the result +1/2, then we know with certainty that a measurement of the other will yield the result -1/2. This doesn't

It might seem plausible that that the first particle was in a state that corresponds to the result +1/2, and the other in a state that corresponds to -1/2, but if we assume that they were, we can derive a mathematical relationship called a Bell inequality from this assumption, and experiments have shown that this Bell inequality doesn't hold in the real world.

I expect the same thing to apply to position. If a particle really

I think I have also seen an even better argument against the uncertainty principle for position and momentum to be about "nudges", but unfortunately i can't remember it. Maybe someone else can contribute with a good argument.

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I'd be worried if itI learned about this in class and some parts have been bugging me for a while.

Not exactly. That gives the impression that if, someday, we come up with better instruments that we'd be able to get more accurate measurements. That is not the case. The idea is that something likeI am to understand that because were dealing with such small and minute measurements, that if one were to attempt to measure values that the experiment would mess up the values read.

No. That seems to me to be a common misconception. It is quite possible to measure the position and momentum of a particle to arbitrary precision. The problem is that if you keep repeating the measurements under identical circumstances you'll get different results. Its only theThat it is "physically" impossible to measure both values at once.

Its also important to take into consideration that a measurement consists of two systems interacting. One system is the system whose properties you wish to measure while the other system is the actual measuring device.

Best regards

Pete

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To arbitrary precision, yes, but that doesn't mean that you can do it at the sameNo. That seems to me to be a common misconception. It is quite possible to measure the position and momentum of a particle to arbitrary precision.

The reason why you can't is of course that a position measurement changes the velocity. So

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Because this has to do with the physical definition of "having an exact value". I make an example. Take a metre and measure a table's lenght; let's say it's 1.025 metres. Repeat the measure after a minute: 1.025, same value; the same for many other subsequent measures. That's the exact value and we all agree about it. Now repeat the experiment with a strange kind of table: the very act of measuring its length disturb it so much that the subsequent measure can be completely different: the first is 1.025 m, the second is 0.001 m, the third is 100 m, ecc. Now you are tempted to think that it's only a technical problem and that making more refined measurements you are able to disturb less the table and to find less length variations; but then you learn from the theory that this is impossible, it's not just a technical problem, but something intrinsic to the system.For me one problem has always come from the use of the term "measurement". When teaching the basics of quantum physics, everything is described in terms of measuring a property (position, momentum, energy...) and the HUP sounds reasonable to a student's ears when explained in terms of "disturbance due to the measuring process".

But then as a student I used to ask: "what if we don't measure it? what if we leave it undisturbed? why doesn't it have an exact value then?".

Which would you say it is the "exact value" of the of the table's length ? Of course you could say it has an "exact value" without measuring it but, how would you *prove* it? We are talking about physics and the only way to prove something, in physics, is...(you have to complete the sentence)

You have a good intuition. Go on this way.The answer I usually got was "if you don't measure it, it's like it doesn't exist for you so why bother?".

But it's hard for a student to think that something doesn't exist with a value if you just choose to not bother...

I don't know if it's wrong, but nowadays I am trying to see if I can think of it in slightly different terms, which (for me) may give back more sense to such answers. Instead of "measuring", can we use the word "interacting"? That helps me to forget about the image of a mad scientist trying to run after a particle with a microscope to measure it. Then perhaps the particle's property we are considering (or a couple of related properties) is defined only as through the possible interactions it can give with the rest of the universe or at least some part of it, such as another particle with the same properties.

I don't want to tell you my opinion about it, at least for the moment; just think about this: can you define the absolute position of a point without a coordinate system? Can you define the absolute value of an object's velocity without referring to a reference frame?So perhaps position+momentum make sense only if they affect the particle's interactions with others? I might be completely off... But do they manifest themselves in a particle that doesn't have a universe around it, no other particles to meet?

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I disagree. What Heisenberg's Uncertainty relation speaks of is the relationship between quantities which are statistically defined. It does not pertain to individual measurements.To arbitrary precision, yes, but that doesn't mean that you can do it at the sametime.

An example will illustrate what I mean. Since I am unaware of your background may I'd like to ask you a question if that's okay? Your response will better help me phrase my responses. If a person hasn't taken a course in quantum mechanics at the college level then that person might not be aware of the precise definition of some of the terminology. If you are already familiar with the terminology then I appologize. Its difficult knowing how to phrase a response when you don't know if the person reading it knows the precise definitions of the terms one is using.

My question is this - Do you know the mathematical expression of the

Consider the case where you have a quantum mechanical system and are measuring two incompatible observables of that system for which the specturm of possible values is discreet. This means that an observable can only take on certain values, i.e. the observables are

Take a dice as an example. Even though when you roll the die and throw them it then there are only seven possibilities which can result, i.e. a "1" or a "2" or ... or a "7". But even though you know the exact value of the measurement you would not have a zero uncertainty in that value because uncertainty

This is an important point to understand otherwise one might get the impression that the uncertainty principle refers to the finite capabilities in the gadgets used to do the measurements rather then the

I hope that clarifies what I'm trying to say a bit better.

Pete

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I wrote most of this reply before you posted this, so some of what I'm saying overlaps with what you just said. I decided to try to explain why it doesn't make sense to say that it's possible to measure two incompatible observables at the same time:Regarding "at the same time" - Operationally, when this phrase is used it means that two measurements are carried out in rapid succession. If the observables are incompatible then measurement of one of the observables changes the state of the system so that it is no longer in a state corresponding to the state it was left in by the preceding measurement. I would imagine that what I just said is unclear. I'm having a hard time finding the words to describe what I mean. Let me get back to you on this.

Suppose that you measure an observable A. This will put the system in an eigenstate of A, and the system can now be said to have a well-defined value of A (the eigenvalue corresponding to that eigenstate). Suppose that you immediately after that measure a second observable B, that commutes with A. This will put the system in a state that's an eigenstate of

In this case, it makes

If A and B

OK, that was the part I wrote before you posted #16. A few more comments:

Yes, I have studied this stuff at the university, so you don't have to worry about my background. I don't have any problems following this for example.

According to what I wrote above, it makes sense to say that we can measure two observables at the same time, if and only if the observables commute, and since they do that if and only if the right hand side of the (uncertainty principle) inequality is zero, I think it

But I agree of course that the uncertainty principle is much more than that. I have tried to emphasize that in all my posts in this thread.

OK, I think this explains what you had in mind when you said that you can measure two incompatible observables at the same time. It sounded like a really strange claim to me. You're doing your simultaneous measurements on two different members of an ensemble of systems that are all in the exact same quantum state. Yes, I agree you can doOnly statistically is there an uncertainy, i.e. when succesive measurements are excecuted on systems which are identically prepared.

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I guess my main point was that [itex]\Delta x[/itex] doesn't address the precision of the instruments used to make a measurement. I think a lot of people misunderstand that.OK, I think this explains what you had in mind when you said that you can measure two incompatible observables at the same time. It sounded like a really strange claim to me.

Cool! Okay. We're on the same wavelength then.You're doing your simultaneous measurements on two different members of an ensemble of systems that are all in the exact same quantum state. Yes, I agree you can dothat.

I'd like to take this time to point out something that many people don't understand. [itex]\Delta x[/itex] is not something that describes the equipment that is being used to do the measuring. It refers to a property of the state that the system is in, i.e. give me a wavefunction and I can tell you what [itex]\Delta x[/itex] is.

Best regards

Pete

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Let me ask you a question before I respond to this. Do you know the precise mathematical definition offundamentally indeterminate; Heisenberg actually advocated the term "indeterminacy principle" to emphasise that this was not a question of our knowledge of the system, but of the system itself. Once an initial measurement has been performed, you can go on to measure both the momentum and precision of a particle to arbitrary accuracy; this has actually been carried out in the labs. But you cannot measure both at the same time. The point of the thought experiments involving electron microscopes and suchlike are NOT the physical basis of the uncertainty principle, but were Heisenberg's consistency check that no experiment could refute his theoretical deduction by simultaneously measuring both quantities of a previously undisturbed system to arbitrary accuracy. The fundamental indeterminacy of the system prior to the act of measurement manifests itelf in the statistical distribution of results from identically prepared systems.

Pete

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That is correct.My understanding of the issue is that prior to an act of measurement, the momentum and position of a single particle arefundamentally indeterminate;..

I assume that you meant "position" and not "precision," right? If so then yes. That is correct. Although no lab has access to arbitrary accuracy.Once an initial measurement has been performed, you can go on to measure both the momentum and precision of a particle to arbitrary accuracy; this has actually been carried out in the labs.

That is incorrect. One has to be careful here. Thinking of this as taking measurements "at the same time" can be confusing. After all its unpractical to take two measurements atBut you cannot measure both at the same time.

As to why its incorrect - Suppose we have a particle which is initially in the state |[itex]\Psi[/itex]>. The value of a particle's position is an eigenvalue of the position operator. The system is then left in an eigenstate of position. Since position and momentum are incompatible (i.e. the position operator does not commute with the momentum operator) they don't have the same set of eigenstates. As such measurement immediately after of the momentum of the particle will take the system out of an eigenstate of position and then place it into a eigenstate of momentum. If you were to then go back and measure the position again then the possible results are any of the position eigenvalues, not neccesarily the originaly measured eigenvalue of position. However, with all that said, it does not imply that we can't measure either the position nor the momentum to an arbitrary degree of precision.

The term

The term

As such there is no reason that one can't measure the position of a particle to an arbitrary degree of precison and then immediately measure the particle's momentum to an arbitrary degree of precison. Its too bad the texts books don't explain this that well.

Pete

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But not on the same particle, right?That is correct.

I assume that you meant "position" and not "precision," right? If so then yes. That is correct. Although no lab has access to arbitrary accuracy.

That is incorrect. One has to be careful here. Thinking of this as taking measurements "at the same time" can be confusing. After all its unpractical to take two measurements atexactlythe same time precise to a gazzilion decimal points. One is better off expressing this in terms of taking one measurement which isimmediatelyfollowed by another measurement.

As to why its incorrect - Suppose we have a particle which is initially in the state |[itex]\Psi[/itex]>. The value of a particle's position is an eigenvalue of the position operator. The system is then left in an eigenstate of position. Since position and momentum are incompatible (i.e. the position operator does not commute with the momentum operator) they don't have the same set of eigenstates. As such measurement immediately after of the momentum of the particle will take the system out of an eigenstate of position and then place it into a eigenstate of momentum. If you were to then go back and measure the position again then the possible results are any of the position eigenvalues, not neccesarily the originaly measured eigenvalue of position. However, with all that said, it does not imply that we can't measure either the position nor the momentum to an arbitrary degree of precision.

The termuncertaintydoesnotrefer to an imprecision in a measurement. E.g. suppose you were to measure the distance between two points and the result of that measruement was 1.02 m +- 1nm. The +- 1nm is not an uncertainty in case that is what you thought it was. In fact the uncertainty of a particular observable is completely determined by the state of the system and has absolutely nothing to do with the precision of your measuring devices. For that reason if you were to give me only the state that a particle is in then I can tell you what the uncertainty of any observable that you're intersted in is.

The termuncertaintyalso goes by another name -standard deviation(sd). If you ever took a course in statistics then you'd know that sd has no meaning for one datum. The sd is determined from asetof data.

As such there is no reason that one can't measure the position of a particle to an arbitrary degree of precison and then immediately measure the particle's momentum to an arbitrary degree of precison. Its too bad the texts books don't explain this that well.

Pete

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Yes. The same particle.But not on the same particle, right?

Its a little unclear to me where people get the idea that you can't measure the particle's position and momentum at the same time to an arbitrary precision. The uncertainty principle only states that both a particle's position and momentum do not have well defined values simultaneously. This is different than saying that they can't be measured. On the other hand I guess I can see how this can be confusing. My post above explains some of this.

Suppose you have a particle whose wave function has a Gaussian shape. The probability density is also Gaussian (The Gaussian function is bell shaped). What this means is that more measurements of the particles position will find it near the probability density is highest, i.e. near the top of the curve (or top of "bell"). The more you execute such an experiment, always starting with the exact same wave function, the more your measurements will find the position of the particle where the curve is large. There will also be measurements where the curve is lower but these will occur less often. Therefore the particle's position will tend to be around the middle. The width of the bell is known as the

I hope that clears some of my comments up?

Pete

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But the particle will be in a different state after the first measure; the HUP refers to uncertainties aboutYes. The same particle.But not on the same particle, right?

Pete

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Correct.But the particle will be in a different state after the first measure;

Yes. But you asked me whether I was speaking about the same particle and not the same state... the HUP refers to uncertainties aboutthe same sateof a quantum system, am I right?

What eactly do you think the relationship is between uncertainty and imprecision in a measurement??

Pete

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Maybe because they think of a particle asIts a little unclear to me where people get the idea that you can't measure the particle's position and momentum at the same time to an arbitrary precision.

I think it's you who have an unorthodox view of what it means to be able to do two things at the same time. Most people wouldn't take that to mean "having the option to do one

Do you also tell people that you are able to lift both of your legs off the ground at the same time?

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Certainly, I just wanted to clarify the concept to me, and now it's more clear.Correct.

Yes. But you asked me whether I was speaking about the same particle and not the same state.

None, if I have understood how you define them. If I measure a steel ball's diametre with a caliber, I have, let's say, a value of 10 mm with an imprecision of 0.01 mm (because of the instrument); if however I measure many balls made from a factory, I can find that, due to intrinsic variations in the productive process, those balls can have different diametres: 9.80 mm, 11.02, ecc. A large number of them will be (approximateley normal) distributed with an average and a standard deviation. Let's say the average is 10.00 mm and the standard deviation 2.00 mm. Then 2.00 mm is the uncertainty and 0.01 mm the imprecision. Correct?What eactly do you think the relationship is between uncertainty and imprecision in a measurement??

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That is correct. This would also be true if the measurements were taken with infinite precision. Even then there would be an uncertainty in the diameter.Certainly, I just wanted to clarify the concept to me, and now it's more clear.None, if I have understood how you define them. If I measure a steel ball's diametre with a caliber, I have, let's say, a value of 10 mm with an imprecision of 0.01 mm (because of the instrument); if however I measure many balls made from a factory, I can find that, due to intrinsic variations in the productive process, those balls can have different diametres: 9.80 mm, 11.02, ecc. A large number of them will be (approximateley normal) distributed with an average and a standard deviation. Let's say the average is 10.00 mm and the standard deviation 2.00 mm. Then 2.00 mm is the uncertainty and 0.01 mm the imprecision. Correct?

Pete

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Gentlemen;

Listening to you talk is better than any course I've had at University,(peer to my best course in Quantum ). Thanks

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You're welcome. We're glad to be of help.Gentlemen;

Listening to you talk is better than any course I've had at University,(peer to my best course in Quantum ). Thanks

Pete

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I learned about this in class and some parts have been bugging me for a while. I am to understand that because were dealing with such small and minute measurements, that if one were to attempt to measure values that the experiment would mess up the values read. That it is "physically" impossible to measure both values at once.

This is a common misconception of uncertainty principle that if one were to attempt to measure values that the experiment would mess up the values read.

It happens to classical systems too, Not just quantum systems. For example, if you measure the temperature of water, its temperature would change due to the measurement process.

This measurement has nothing to do with uncertainty principle.

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It appears that another common misconception is the definition of the uncertainty of an observable. ByThis is a common misconception of uncertainty principle that if one were to attempt to measure values that the experiment would mess up the values read.

It happens to classical systems too, Not just quantum systems. For example, if you measure the temperature of water, its temperature would change due to the measurement process.

This measurement has nothing to do with uncertainty principle.

[itex]\Delta A =\sqrt{\langle(A-\langle A\rangle)^2\rangle}[/itex]

Therefore the value of the uncertainty of a physical observable is therefore completely determined by the state that the system is in. In the case of the physical observable x we have

[itex]\Delta X =\sqrt{\langle(X-\langle X\rangle)^2\rangle}[/itex]

where X is the operator corresponding to the position eigenvalue x. Notice that this has absolutely nothing to do with the error [itex]\delta x[/itex] in the measured position of a particle. There is no reason why [itex]\delta x[/itex] can't be zero. Currently there is no known lower bound for [itex]\delta x[/itex]. Note that [itex]\delta x[/itex] and [itex]\Delta x[/itex] represent very different quantities and that they are unrelated to each other in all but the most cursory way (i.e. both address position, otherwise they are not related).

Pete

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I forgot to mention thatIt appears that another common misconception is the definition of the uncertainty of an observable. Bydefinitionthe uncertainty in the physical observable A is given by

[itex]\Delta A =\sqrt{\langle(A-\langle A\rangle)^2\rangle}[/itex]

Therefore the value of the uncertainty of a physical observable is therefore completely determined by the state that the system is in. In the case of the physical observable x we have

[itex]\Delta X =\sqrt{\langle(X-\langle X\rangle)^2\rangle}[/itex]

where X is the operator corresponding to the position eigenvalue x. Notice that this has absolutely nothing to do with the error [itex]\delta x[/itex] in the measured position of a particle. There is no reason why [itex]\delta x[/itex] can't be zero. Currently there is no known lower bound for [itex]\delta x[/itex]. Note that [itex]\delta x[/itex] and [itex]\Delta x[/itex] represent very different quantities and that they are unrelated to each other in all but the most cursory way (i.e. both address position, otherwise they are not related).

Pete

Pete

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I must be getting old because it took me until now to realize that there is a very simple example which will illustrate the point I've been trying to make.It appears that another common misconception is the definition of the uncertainty of an observable. Bydefinitionthe uncertainty in the physical observable A is given by

[itex]\Delta A =\sqrt{\langle(A-\langle A\rangle)^2\rangle}[/itex]

Therefore the value of the uncertainty of a physical observable is therefore completely determined by the state that the system is in. In the case of the physical observable x we have

[itex]\Delta X =\sqrt{\langle(X-\langle X\rangle)^2\rangle}[/itex]

where X is the operator corresponding to the position eigenvalue x. Notice that this has absolutely nothing to do with the error [itex]\delta x[/itex] in the measured position of a particle. There is no reason why [itex]\delta x[/itex] can't be zero. Currently there is no known lower bound for [itex]\delta x[/itex]. Note that [itex]\delta x[/itex] and [itex]\Delta x[/itex] represent very different quantities and that they are unrelated to each other in all but the most cursory way (i.e. both address position, otherwise they are not related).

Pete

Consider an electron which goes through a spin analyzer and which is initially in the state (let a = 1/sqrt(2))

|[itex]\Psi[/itex]> = a|+> + a|->

Where |+> is an eigenket corresponding to the operator S

Pete

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