Heisenberg uncertainty - uncertainty about its meaning

[studious]

HUP states that "certain physical quantities, like the position and momentum, cannot both have precise values at the same time. The narrower the probability distribution for one, the wider it is for the other." (Wikipedia)

I am not a physicist, but I have been pondering this question:
In this particular case, where we are concerned about position and momentum, does HUP mean that humans can not possibly find out precise values for position and momentum, or that the values themselves don't even exist (or "exist as a continuum" whatever that means)?

I have a particle it has some momentum and some position at some particular time. I may not/can't know both those values, but they do exist?... the particle can not be without one of momentum or position, right?

A physics students has been trying to convince me that at a particular time, the two pieces of "information" (a particle's position or momentum) don't even exist (much less humans being able find out both, which HUP is concerned with?).

(In fact, the Wikipedia page itself says that "position and momentum, cannot both have precise values at the same time." Does that mean the values don't even exist or that humans can't KNOW them precisely... if the second... why the HELL IS THERE the word 'cannot' in that Wikipedia statement?)

 

[jtbell]

I have a particle it has some momentum and some position at some particular time. I may not/can't know both those values, but they do exist?... the particle can not be without one of momentum or position, right?

We don't know the answer to this question. The mathematics of QM makes predictions for what we will find the momentum and position of the particle to be, when we observe/measure it. It does not address the question, what the position and momentum "really are" before we measure them. This is the subject of interpretations of QM, of which there are several. There is no general agreement on which one is "correct," and there is (as yet) no way to distinguish among them by experiment, even in principle. So people argue about it a lot.

 

[alxm]

In this particular case, where we are concerned about position and momentum, does HUP mean that humans can not possibly find out precise values for position and momentum, or that the values themselves don't even exist (or "exist as a continuum" whatever that means)?

Well, they exist in the sense that any particle will have those properties. But they haven't assumed a value before it's measured. The definition of 'measure' is narrow here. Any interaction that can, if only in theory, provide that information, counts as a 'measurement'. It's not inaccessible only to humans. It's truly not possible to measure.

This doesn't mean that all values are equally likely though. Or that these values are unknowable. It means that for a single particle, you cannot predict the exact result of a single measurement. But you can know the statistical average of many measurements, and the likelihood of finding a deviation by any given amount from that value.

The big 'weirdness' here is that we're accustomed to such statistical things as coming from our own not-knowing-about-it. I.e. if you pick a random card from a deck, there's a fifty-fifty chance of getting red or black. This is just because you don't have knowledge of the card - but the card itself was always either red or black.

In quantum mechanics, it simply did not have a value until you measured it. The card itself was fifty-fifty! (at least in the most common interpretations)

 

[studious]

Hmm... the analogies seemed very interesting, but I'm still confused.

1. Does every particle (wave or whatever it is) have properties like momentum and position at every instance regardless of human "measurability"? (is this the question which is up to debate according to jtbell's reply?)

2. Does the actual act of measuring properties like position and momentum have any effect on their values at an instance in time? (take off from the card analogy of alxm)

3. Extension of previous question. Can there be >1 values for either position or momentum at any particular instance (e.g. two conflicting answers to the same question... two positions at the same time, etc.)? If yes and if we regard position or momentum as functions, then they can not be single-variable functions... what can they be seen as?

 

[DrChinese]

Hmm... the analogies seemed very interesting, but I'm still confused.

1. Does every particle (wave or whatever it is) have properties like momentum and position at every instance regardless of human "measurability"? (is this the question which is up to debate according to jtbell's reply?)

jtbell has provided an accurate answer: we actually do not know. This leaves you in a position to select from several different interpretations which provide slightly different answers to this question. However, before you come to a conclusion, you need to know about some key papers and experiments relating to this subject. Try reading up on the EPR Paradox (which was an attempt to circumvent the HUP); Bell's Theorem; and the experiments of Alain Aspect et al (which show more clearly that the HUP is fundamental and does not relate to experimental limits).

Specifically: 2 "entangled" particles are essentially clones of each other. So you would think that you could beat the HUP by checking the position of one and the momentum of the other - since they are clones and should have identical values. Turns out that even in this situation, measuring one wreaks your ability to get a precise read on the other in such a way as to beat the HUP.

The lesson is: don't think of the HUP as an experimental phenomenon. It is deeper than that. But we don't understand why that is the case.

 

[$tefan]

Heisenberg uncertainty is caused by photons hitting elemental particles, so just looking at the particle will change it's position. Thats why you cannot determine its place and speed. Or is this a wrong summary of Heisenberg uncertainty?

 

[malawi_glenn]

Heisenberg uncertainty is caused by photons hitting elemental particles, so just looking at the particle will change it's position. Thats why you cannot determine its place and speed. Or is this a wrong summary of Heisenberg uncertainty?

well, yes, since then there is no intrinsic indeterminism of the state, only experimental resolution. When one derives the HUP, only the non-vanishing commutator of P and X observables is used.

The 'disturbing the system' analogy is for classical, Newtonial mechanical, systems only.

 

[WaveJumper]

The 'meaning' of the HUP and why it's there at all, has more to do with philosophy than physics. The HUP is a limitation that prevents us from digging much deeper into the quantum world.

 

[dpk]

actually particle and wave both have momentum & position but we canot find mathematically

 

[DrChinese]

actually particle and wave both have momentum & position but we canot find mathematically

Welcome to PhysicsForums, dpk!

Your viewpoint is considered an acceptable interpretation of QM, but is not generally accepted as a fact. Other acceptable interpretations deny that particles/waves have well-defined attributes at all times.

 

[mn4j]

Usually, a clear understanding of uncertainty principles is muddied by wanderings into the realm of interpretations of QM or philosophy. However, it is a very simple principle, not limited to Quantum systems or to position and momentum for that matter.

The uncertainty between the two comes from the way they are defined. One is based directly on a variable, while the other is based on the derivative of the variable. For every two operators for which one operates on a variable and the other on the derivative of the variable, you will have an uncertainty relationship. Position and momentum are just one example.

In classical wave mechanics, you have uncertainty between
- Wavelength and Position
- Frequency and time ( given ONLY a very short pulse in time, the frequency will be very uncertain and vice versa)

Another example, consider a soccer ball moving through air. If you are presented ONLY with a snapshot of the scene at a precise point in time, you will only be able to give a precise position of the ball, not it's momentum. This is because, by definition, you need more than one position to calculate the momentum. In other words, the momentum is based on $$\Delta{x}$$

NO soccer ball has a well defined momentum at all positions. This says nothing about the soccer ball, but about the ACTUAL meaning of momentum and position. Momentum by definition must cover a range of positions. This is where the uncertainty comes in. It has nothing to do with Quantum mechanics at all.

The uncertainty principle is not about metaphysics or what can or can not be measured. It's simply derived from the mathematical definition of the terms or operators, in the same way as when we say there is uncertainty between an ocean and a water molecule. In one case the definition implies a multitude of the other, except in the case of position and momentum, one implies the differential of the other with respect to a common variable.

 

[DrChinese]

NO soccer ball has a well defined momentum at all positions. This says nothing about the soccer ball, but about the ACTUAL meaning of momentum and position. Momentum by definition must cover a range of positions. This is where the uncertainty comes in. It has nothing to do with Quantum mechanics at all.

This is demonstrably wrong. Experiments on soccer balls will NOT reveal the HUP as your description implies. This is a quantum phenomenon and you must be looking a quantum-size objects to see it. Many well-documented quantum spin experiments - in which particle characteristics can be observed without disturbing them in any way - demonstrate the non-classical nature of the HUP.

It would really make more sense if you would simply denote your personal opinions - especially the misleading ones - as such. That way the OP and other readers can judge the value of your post for themselves.

 

[Naty1]

For every two operators for which one operates on a variable and the other on the derivative of the variable, you will have an uncertainty relationship

That is also not correct... the correct formulation: Robertson–Schrödinger relation
at
http://en.wikipedia.org/wiki/Uncertainty_principle#Robertson.E2.80.93Schr.C3.B6dinger_relation

which leads to a specific list of uncertainty relations:(like position, momentum)

http://en.wikipedia.org/wiki/Uncertainty_principle#Other_uncertainty_principles

I hadn't seen that list before...

 

[mn4j]

I think you should check your facts before you speak:

If the velocity of the electron is at first known, and the position then exactly measured, the position of the electron for times previous to the position measurement may be calculated. For these past times, δpδq is smaller than the usual bound. (Heisenberg 1930, p. 15)​

These reciprocal uncertainty relations were given in a recent paper of Heisenberg as the expression of the statistical element which, due to the feature of discontinuity implied in the quantum postulate, characterizes any interpretation of observations by means of classical concepts. It must be remembered, however, that the uncertainty in question is not simply a consequence of a discontinuous change of energy and momentum say during an interaction between radiation and material particles employed in measuring the space-time coordinates of the individuals. According to the above considerations the question is rather that of the impossibility of defining rigourously such a change when the space-time coordination of the individuals is also considered. (Bohr, 1985 p. 93)​

 

[mn4j]

That is also not correct... the correct formulation: Robertson–Schrödinger relation
at
http://en.wikipedia.org/wiki/Uncertainty_principle#Robertson.E2.80.93Schr.C3.B6dinger_relation

which leads to a specific list of uncertainty relations:(like position, momentum)

http://en.wikipedia.org/wiki/Uncertainty_principle#Other_uncertainty_principles

I hadn't seen that list before...

Apparently, you did not read the derivation section of the link you posted where it says:

When linear operators A and B act on a function ψ(x), they don't always commute. A clear example is when operator B multiplies by x, while operator A takes the derivative with respect to x.​

Or do you claim this statement to be false as well?
Do you agree that this is not limited to quantum systems?

 

[DrChinese]

...

Do you agree that this is not limited to quantum systems?

The HUP is essentially limited to quantum systems, does not apply to classical objects as you imply. (Of course, when progressively more complex/larger quantum objects are involved, there is an element of uncertainty still present.)

So let's get specific about something which is not ambiguous: particle spin. There are spin components which are non-commuting, *and* one does not involve a derivitive of the other as you claim. These obey the HUP, and do not exist in the classical world. This is a clear sign that the HUP does not rest on a classical principle as you seem to claim.

You are taking time and diverting the attention of readers of this forum by twisting the meanings of words and harping on points that don't really apply to the subject at hand. This is not fair to those who are not aware of your agenda. I come here to participate in legitimate discussion of science.

Why don't you operate in a straight-forward manner instead of hiding - here and in other threads - from your real agenda: pushing a variety of classical and/or realist positions. Please come out from behind the bushes with your views and label them appropriately - instead of creating rabbit trails to divert us under the pretense of debate.

 

[mn4j]

Despite your diatribe and accusations, you have failed to specify what is wrong in my statements. It appears you are the one with a hidden agenda trying to project your intentions on others. I wonder why you would feel threatened if your views were based on a solid foundation.

So let's get specific about something which is not ambiguous: particle spin. There are spin components which are non-commuting, *and* one does not involve a derivitive of the other as you claim. These obey the HUP, and do not exist in the classical world. This is a clear sign that the HUP does not rest on a classical principle as you seem to claim.

Show me where I claimed that spin components are derivatives of each other?! Show me where I claimed that the relationship must be a derivative?! Are you utterly unable to understand the English language? Just because I mention the derivative as one example does not mean it must be a derivative relationship in every case. By your argument, it would be sufficient for someone to show energy conservation in a quantum system, and then then argue that "this proves energy conservation does not exist in the classical world".

Since you are bent on proving me wrong, explain how the momentum and position of a soccer ball are both well defined at an instant in time as in the example I gave above. Or explain how the frequency of a classical wave is well defined at an instant in time. If you are truly interested in real scientific debate, rather than religious assertions, you will not shy away from this challenge which is very relevant to the issue at hand.

I will repeat my main point in more generic terms. Feel free to explain why it is wrong.

Every pair of conjugate observables have an uncertainty relationship which derives solely from the fact that they are by definition a Fourier transform pair. Momentum and position are simply one example of such pair. There are many others. This is not limited to quantum systems as some naively think.

 

[Naty1]

Apparently, you did not read the derivation section of the link you posted where it says:

When linear operators A and B act on a function ψ(x), they don't always commute. A clear example is when operator B multiplies by x, while operator A takes the derivative with respect to x.​

Or do you claim this statement to be false as well?
Do you agree that this is not limited to quantum systems?

mn: ...the quote above doesn't seem related my point... if uncertainty derives merely from a function and its derivative, how do you explain he uncertainty of two orthogonal components of angular momentum,for example,...doesn't seem like either is the derivative of the other...but I am always willing to learn...

Ooops, I just saw:
"Every pair of conjugate observables have an uncertainty relationship which derives solely from the fact that they are by definition a Fourier transform pair." I thought you said uncertainty derived solely from a variable and its derivative..or is it both?? Never came across either claim...if only life/physics were so simple..

 

[mn4j]

I thought you said uncertainty derived solely from a variable and its derivative..

Where did I say that?

See http://en.wikipedia.org/wiki/Conjugate_variables
See also:
* R. Gilmore, Uncertainty relations of statistical mechanics, Phys. Rev. A 31(5), 3144-3146 (1985).

 

[DrChinese]

I will repeat my main point in more generic terms. Feel free to explain why it is wrong.

Every pair of conjugate observables have an uncertainty relationship which derives solely from the fact that they are by definition a Fourier transform pair. Momentum and position are simply one example of such pair. There are many others. This is not limited to quantum systems as some naively think.

Are photon conjugate (non-commuting) spin observables by definition a Fourier transform pair? They aren't continuous. Seems to me these are not related in the way position and momentum are. They follow the HUP too.

What are some non-quantum observables that obey the HUP? (That is, if you are saying the HUP is not a quantum phenomena.)

In other words: the similarity of the math is not what makes something respect the HUP. Do you think Planck's constant might be a fundamental component of the equation as well?

 

[mn4j]

Are photon conjugate (non-commuting) spin observables by definition a Fourier transform pair? They aren't continuous. Seems to me these are not related in the way position and momentum are. They follow the HUP too.

What are some non-quantum observables that obey the HUP? (That is, if you are saying the HUP is not a quantum phenomena.)

In other words: the similarity of the math is not what makes something respect the HUP. Do you think Planck's constant might be a fundamental component of the equation as well?

Apparently you did not read the link I provided about the meaning of conjugate variables. Fourier transforms not limited to continuous variables. If you want to prove that the spin observables you are talking about are not Fourier transform pairs, be my guest.

I already gave you examples of the uncertainty principle in non quantum systems. See post #17. Note that the time and frequency example is mentioned also in the wikipedia article on Conjugate variables:

- Time and frequency: the longer a musical note is sustained, the more precisely we know its frequency (but it spans more time). Conversely, a very short musical note becomes just a click, and so one can't know its frequency very accurately.

Do you disagree with this? Unless you believe the musical note is a quantum system.

 

[DrChinese]

- Time and frequency: the longer a musical note is sustained, the more precisely we know its frequency (but it spans more time). Conversely, a very short musical note becomes just a click, and so one can't know its frequency very accurately.

Do you disagree with this? Unless you believe the musical note is a quantum system.

There is something seriously wrong with you, and you are wasting my time and (more importantly) that of others. :grumpy:

There is nothing that prevents us from measuring these two simultaneously to whatever precision we want. That is typical of classical systems but not of quantum systems. In quantum systems, the HUP is obeyed. In classical systems, it is not. That is the difference, I have no idea why you deny this. In addition, measuring more precisely one conjugate observable of a classical object does NOT cause the other to take on a increasingly random value, as occurs in quantum objects. That is a critical component of the HUP.

NOTE TO EVERYONE ELSE: mn4j likes to debate meaningless (and often completely wrong) points like this until you are ready to leave. He already more or less did this to another serious poster (kobak - hope you come back). Continue at your own risk. I am tired of repeating the obvious.

 

[mn4j]

There is something seriously wrong with you, and you are wasting my time and (more importantly) that of others. :grumpy:

There is nothing that prevents us from measuring these two simultaneously to whatever precision we want. That is typical of classical systems but not of quantum systems. In quantum systems, the HUP is obeyed. In classical systems, it is not. That is the difference, I have no idea why you deny this. In addition, measuring more precisely one conjugate observable of a classical object does NOT cause the other to take on a increasingly random value, as occurs in quantum objects. That is a critical component of the HUP.

NOTE TO EVERYONE ELSE: mn4j likes to debate meaningless (and often completely wrong) points like this until you are ready to leave. He already more or less did this to another serious poster (kobak - hope you come back). Continue at your own risk. I am tired of repeating the obvious.

Like I thought, you have no response other than ad-hominem attacks. You are unable to show how the frequency of a musical note is well defined at an instant in time because it is not. Rather than just admit that fact, you resort to accusations. This is unbecoming of a so called "science advisor".

 

[malawi_glenn]

mn4j, I don't think you are the one who shall judge what a sicence advisor should be able to do and say.

mn4j, you are mixing apples and pears as DrChinese points out. The uncertainty in quantum physics is intrinsic. Uncertainty relations in QM results from the postulates of QM, first quantization of operators and the imaginary valued quantum canonical commutator of position and momentum operator. You also obtain uncertainty relations for other pairs of non commuting observables, even tough they are not related by a derivative.

 

[mn4j]

mn4j, I don't think you are the one who shall judge what a sicence advisor should be able to do and say.

You don't have to agree with me but my response was warranted by DrChinese's off-topic and vendetta-based statements about me.

mn4j, you are mixing apples and pears as DrChinese points out.

If you actually read this thread, you will understand that DrChinese has not pointed out anything of substance.

The uncertainty in quantum physics is intrinsic. Uncertainty relations in QM results from the postulates of QM, first quantization of operators and the imaginary valued quantum canonical commutator of position and momentum operator.

So what? If you try to actually understand what I am talking about you will realize that it is more fundamental that.

1 - Quantization of operators does not result in uncertainty.
2 - The canonical commutation relation of which you speak is an entirely mathematical result which originates from the fact that the conjugate variables or operators are defined (mathematically) in such a way that they are a Fourier pair. The uncertainty arises because the more concentrated a function, the more spread out it's Fourier transform MUST be. Thus it is not possible to arbitrarily concentrate both a function and its Fourier transform.
3 - Just because there are canonical commutator relationships in QM for which an uncertainty relationship exists does not change the fact that the source of uncertainty is from their mathematical definition. Therefore, by definition, there exists an uncertainty relationship between every such pair of conjugate variables whether they be in QM or not. I have given examples in this thread.

You also obtain uncertainty relations for other pairs of non commuting observables, even tough they are not related by a derivative.

It appears from this statement that you think I have claimed there must be a derivative involved. Please read the thread again and see if you can find where I made such a claim. I didn't.

Please do not confuse the uncertainty in QM which originates from the fact that QM is intrinsically probabilistic with the uncertainty principle. They are different.

 

[malawi_glenn]

I have read this thread carefully.

I do know what you are talking about. But I am wondering where you have learned quantum mechanics and how to derive the HUP.

The QM canonical commutation relation between X and P is POSULATED by Dirac, one takes the classical canonical commutator (the Poisson bracket, which has nothing to do with measurements of observables or Fourier transforms either) multiply it with i/hbar. Use inner product space of complex valued differentiable functions. Then you derive the HUP without caring if x and p are

L_x and L_z have an uncertainty relation witch is the same as for x and p, and L_x and L_y is not related by a Fourier transform? Instead, they satisfy a non vanishing commutator, which is the source for the uncertainty relation.

Yes, Fourier pairs also leads to an uncertainty relation and one can follow Heisenberg's approach witch is the one that you use Fourier pair to deduce the commutator relation, which leads one to the uncertainty relation in the same way but using the formalism of quantum mechanics. I.e:
$$< Q > = \int _{\text{All space}} \psi^*(x)Q(x)\psi(x) dx$$
etc.

Recall that in QM you quantize the operators and they must act on wave functions, you don't deduce uncertainty relations from Fourier transforms of operators, but instead, due to the formalism (inner product space of complex valued functions), you use the commutation relations.

The uncertainty relation in QM, for two operators A and B is:
$$\langle A^2 \rangle \langle B^2 \rangle \ge \dfrac{1}{4}\,\langle [A,B]\rangle|^2$$
.. due to the formalism of inner product space of complex valued functions.

Then you can always argue that the commutator between x and p comes from the fact that they are Fourier pairs, but you don't have to do that as Dirac pointed out, you can use the classical poisson bracket and take it times i.

So, I think that I now have shown you that the essential feature of Uncertainty relations in QM arises from commutation relations and the formalism of inner product space of complex valued functions (Cauchy Schwarz inequality), and not solely Fourier Pairs, which is more restrictive.

You can obtain an uncertainty relation for any pair of non commuting operators A and B, not just solely those quantities which are Fourier pairs.

 

[mn4j]

The QM canonical commutation relation between X and P is POSULATED by Dirac.

No. It is not a postulate. It derives naturally from the mathematical definitions of position and momentum.

one takes the classical canonical commutator (the Poisson bracket, which has nothing to do with measurements of observables or Fourier transforms either) multiply it with i/hbar. Use inner product space of complex valued differentiable functions. Then you derive the HUP without caring if x and p are

Apparently you do not realize how intimately coupled the Fourier transform is to 'inner product spaces', which again is not exclusive to QM. Check Parseval's theorem for more details about this.

L_x and L_z have an uncertainty relation witch is the same as for x and p, and L_x and L_y is not related by a Fourier transform? Instead, they satisfy a non vanishing commutator, which is the source for the uncertainty relation.

Are you sure? The angular momentum eigenbases of orthonormal spin directions ARE related by a discrete Fourier transform. This is not difficult to see if you consider that both bases are maximally different orthogonal bases. This is the source of the uncertainty. Therefore the commutator relationship originates from the fact that they are conjugate variables. In any case, there is nothing exclusively quantum about this. It is an entirely mathematical result as I have repeatedly mentioned.

Yes, Fourier pairs also leads to an uncertainty relation and one can follow Heisenberg's approach witch is the one that you use Fourier pair to deduce the commutator relation, which leads one to the uncertainty relation in the same way but using the formalism of quantum mechanics. I.e:
$$< Q > = \int _{\text{All space}} \psi^*(x)Q(x)\psi(x) dx$$
etc.

Recall that in QM you quantize the operators and they must act on wave functions, you don't deduce uncertainty relations from Fourier transforms of operators, but instead, due to the formalism (inner product space of complex valued functions), you use the commutation relations.

The uncertainty relation in QM, for two operators A and B is:
$$\langle A^2 \rangle \langle B^2 \rangle \ge \dfrac{1}{4}\,\langle [A,B]\rangle|^2$$
.. due to the formalism of inner product space of complex valued functions.

Then you can always argue that the commutator between x and p comes from the fact that they are Fourier pairs, but you don't have to do that as Dirac pointed out, you can use the classical poisson bracket and take it times i.

The derivation of the equation for the uncertainty relation has no bearing on the origin of the relation. It follows naturally from the mathematical definition of the terms. You can derive the equation any way you want. It doesn't change the fact that the source is from their mathematical definition.

So, I think that I now have shown you that the essential feature of Uncertainty relations in QM arises from commutation relations and the formalism of inner product space of complex valued functions (Cauchy Schwarz inequality), and not solely Fourier Pairs, which is more restrictive.

If you consider that inner product spaces and Fourier transform pairs are not unrelated, then in effect you have helped to prove my point that there is nothing inherently quantum about the uncertainty principle. It is a entirely mathematical consequence.

You can obtain an uncertainty relation for any pair of non commuting operators A and B, not just solely those quantities which are Fourier pairs.

Find any pair of such non commuting operators and you will realize that their commutation relation originates from the basis transformation inherent in their mathematical definition. What again is the Fourier transform?

 

[malawi_glenn]

so you mean that J_x and J_y are Fourier pairs? The uncertainty relation in J_x and J_y come from the commutator relation which comes from SO(3) Lie Algebra.

Since quantum mechanics IS a mathematical framework in order to describe nature, then we would not be surprised that the phenomenon boils down to the math.. So one can then in the angular momentum uncertainty relation argue that the uncertainty relation just comes from the lie-algebra criterion and not quantum mechanics?

And if you don't know Dirac postulated that quantum commutators should be as the classical poission bracket times i/hbar, read Sakurai - Modern Quantum mechanics for instance.

The thing is we are discussing physics here, and physics has math as language. So we can't just say that "it is all math". We want to compare classical physics and quantum physics.

 

[DrChinese]

so you mean that J_x and J_y are Fourier pairs? The uncertainty relation in J_x and J_y come from the commutator relation which comes from SO(3) Lie Algebra.

Since quantum mechanics IS a mathematical framework in order to describe nature, then we would not be surprised that the phenomenon boils down to the math.. So one can then in the angular momentum uncertainty relation argue that the uncertainty relation just comes from the lie-algebra criterion and not quantum mechanics?

And if you don't know Dirac postulated that quantum commutators should be as the classical poission bracket times i/hbar, read Sakurai - Modern Quantum mechanics for instance.

The thing is we are discussing physics here, and physics has math as language. So we can't just say that "it is all math". We want to compare classical physics and quantum physics.

So if I understand: there is x spin and x angular momentum; and y spin and y angular momentum. x spin and x angular momentum form a conjugate pair, correct? Am I saying this correctly?

 

[Count Iblis]

[DrChinese, mn4j] = i hbar

 

[malawi_glenn]

So if I understand: there is x spin and x angular momentum; and y spin and y angular momentum. x spin and x angular momentum form a conjugate pair, correct? Am I saying this correctly?

i give up:P

 

[DrChinese]

i give up:P

 

[malawi_glenn]

you know J is not always L + S, J is also general angular momentum in SO(3). See for instance Sakurai.

We can specify to orbital angular momentum, L, L_x and L_y does not commute and we have an uncertainty relation for those two.

I would say that this is a quantum mechanical property, observables are assigned o quantized hermitian operators on a Hilbert space. If operators for two observables don't commute, there is an intrinsic limit for how good we can measure these two observables.

mn4j would say that this is just math, it has nothing to do with quantum mechanics, the uncertainty relation for L_x and L_y boils down to so(3) Lie algebra relation, nothing strange about it he would say (this is how how explained that HUP was not strange for position and momenta).

But quantum mechanics IS a MATHEMATICAL framework, to describe nature, so we should not be surprised that the properties of QM lies in it's math. Tautology I would say. The interesting thing is how this mathematical framework differs from the framework we choose for describing classical (Newtonian, non-quantum) mechanics.

 

[DrChinese]

... I would say that this is a quantum mechanical property, observables are assigned o quantized hermitian operators on a Hilbert space. If operators for two observables don't commute, there is an intrinsic limit for how good we can measure these two observables.

mn4j would say that this is just math, it has nothing to do with quantum mechanics, the uncertainty relation for L_x and L_y boils down to so(3) Lie algebra relation, nothing strange about it he would say (this is how how explained that HUP was not strange for position and momenta).

But quantum mechanics IS a MATHEMATICAL framework, to describe nature, so we should not be surprised that the properties of QM lies in it's math. Tautology I would say. The interesting thing is how this mathematical framework differs from the framework we choose for describing classical (Newtonian, non-quantum) mechanics.

Thanks. I can follow the idea that the standard formalism describes the world according to QM. (And I see your point about the tautology.) I don't follow the idea that those same ideas describe the classical world (and evidently you don't either, since you remark that the framework is different).

Thanks for your time.

 

[ajw1]

I must confess I do not understand everything that has been said in this thread, but I'm very interested in whether or not there is a demonstrable difference between classical and Heisenberg's quantum uncertainty (and of course, what this difference really is).

If I understand correctly the original reasoning from Heisenberg was quite 'classical', being in laymen language that fotons would disturb the particle to much to get a proper reading of it's position and momentum.