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Heisenberg's uncertainty principle for dummies?


To_Mars_and_Beyond
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At school, we were teached that you can understand Heisenberg's uncertainty principle by considering the measurement event: To measure the particle, you have to interact with it, and this is typically done by throwing a photon to it. Even a single photon will give some momentum to the particle, and thus, the state of the particle is not same anymore, and thus, you can never measure it exactly.

I have become to question this explanation.

Let us think in purely classical terms. We have a target particle with disposition x (a 3d vector) and velocity v (also a 3d vector). So, we have six unknowns. We poke it with a photon whose disposition changes by dx (a 3d vector) and the photon's wavelength changes by df (a 3d vector). So, we have six equations. Problem solved???

Or, maybe just use two photons? My question: Aren't there sufficient degrees of freedom in particle collisions to solve the state of the target particle exactly? Was my physics teacher kidding me? 

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The particle simply does not have a specific coordinate and a specific momentum at a given time. This is what the Heisenberg uncertainty relation says. Until a particle is detected, it passes through all the slits of the diffraction grating at once.

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15 minutes ago, SergUpstart said:

The particle simply does not have a specific coordinate and a specific momentum at a given time. This is what the Heisenberg uncertainty relation says. Until a particle is detected, it passes through all the slits of the diffraction grating at once.

I'm not certain, you're right...

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The most straightforward “for-dummies” elucidation  of the HUP is:  

The location and velocity of a microscopic particle can never be determined simultaneously together. Of course , an experiment might  be structured such that the location of a particle can be estimated. You may also estimate its speed from a separate experiment. But you can never plan things  so that both their location and their pace can be precisely determined. 

 

If you possibly take the decision to go a bit beyond the "for-dummies" understanding , then I suggest the following link to be followed. I have not edited it. Nor have my PhD students. However , we all found it helpful in our weekly casual debates : 

 

https://www.scientificamerican.com/article/common-interpretation-of-heisenbergs-uncertainty-principle-is-proven-false/

Edited by Prof Reza Sanaye
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20 minutes ago, Prof Reza Sanaye said:

“common” in this case means “incorrect” It’s a shame the article only hints at this.

It is common to confuse the HUP with the measurement effect, but they are distinct phenomena.

1 hour ago, To_Mars_and_Beyond said:

At school, we were teached that you can understand Heisenberg's uncertainty principle by considering the measurement event: To measure the particle, you have to interact with it, and this is typically done by throwing a photon to it. Even a single photon will give some momentum to the particle, and thus, the state of the particle is not same anymore, and thus, you can never measure it exactly.

I have become to question this explanation.

The explanation is wrong. What is being described here is the measurement effect. Heisenberg himself made this mistake, in trying to discern why there would be this uncertainty 

The uncertainty arises because the wave functions of the associated pair of variables are Fourier transforms of each other. i.e. the wave function in momentum-space is a Fourier transform of the WF in position-space. The uncertainty occurs because of this relation.

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8 minutes ago, swansont said:

“common” in this case means “incorrect” It’s a shame the article only hints at this.

It is common to confuse the HUP with the measurement effect, but they are distinct phenomena.

The explanation is wrong. What is being described here is the measurement effect. Heisenberg himself made this mistake, in trying to discern why there would be this uncertainty 

The uncertainty arises because the wave functions of the associated pair of variables are Fourier transforms of each other. i.e. the wave function in momentum-space is a Fourier transform of the WF in position-space. The uncertainty occurs because of this relation.

Excuse me Sir ,        Do you mean HP is or is NOT in the eyes of the beholder  ??

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3 minutes ago, Prof Reza Sanaye said:

Momentum vectors  vs  position vectors have their own space functions . Do you agree ?

I think this is irrelevant, since I’m talking about wave functions, not vectors. I don’t know what you mean by space function.

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Even if you could devise an experiment which didn't involve an interaction, you could never measure momentum and position of a quantum particle to arbitrary precision.
The uncertainty is inherent to the model, because the model, QM, necessarily treats them as waves.
( the clue is in the name, wave function )

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17 hours ago, To_Mars_and_Beyond said:

At school, we were teached that you can understand Heisenberg's uncertainty principle by considering the measurement event: To measure the particle, you have to interact with it, and this is typically done by throwing a photon to it. Even a single photon will give some momentum to the particle, and thus, the state of the particle is not same anymore, and thus, you can never measure it exactly.

Like has been pointed out by other posters here, this is called the measurement effect, which is not the same as the HUP. 
The fact that certain pairs of observables cannot be determined simultaneously with arbitrary precision is something that is intrinsic to the quantum nature of the system - it is not something that arises as an artefact of the measurement process.

As swansont has stated, this is because these observables aren’t independent quantities, they are Fourier transforms of one another. In more technical terms, these pairs of observables do not commute, and any pair of non-commuting quantities is always subject to some uncertainty relation.

17 hours ago, To_Mars_and_Beyond said:

So, we have six equations. Problem solved???

No, because what you are describing is a classical system, and one of the defining characteristics of classicality is precisely the fact that all observables always commute. This is not true in the case of quantum systems, though.

17 hours ago, To_Mars_and_Beyond said:

Aren't there sufficient degrees of freedom in particle collisions to solve the state of the target particle exactly?

Yes, but to do so you need to decide on a choice of basis representation. So you can either determine the state function in position representation, or in momentum representation - and these are again Fourier transforms of one another, so the HUP still applies.

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4 minutes ago, To_Mars_and_Beyond said:

If what my teacher said was true, you couldn't actually measure a classical system precisely. And that seems unlikely.

Well, even for a classical system there will be limitations due to the limited sensitivity of the measurement apparatus - e.g. you couldn’t weigh a grain of sand using a kitchen scale, since it’s not nearly sensitive enough. But that’s due to the apparatus, not due to anything inherent in the grain of sand. So that’s a different phenomenon than HUP.

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5 minutes ago, To_Mars_and_Beyond said:

Thank you guys! This is awesome! So measurement effect and HUP are two distinct phenomena.

If what my teacher said was true, you couldn't actually measure a classical system precisely. And that seems unlikely.

Yes that is correct the HUP and Measurement effects are different phenomena.

There is also a a non quantum uncertainty associated with certain classical phenomena, that is not due to measurement but arises from the same physical process as HUP.

 

The simplest and easiest physical explanation of the HUP I know arises in spectroscopy from the energy time version of the HUP.

Essentially it takes a finite time for the electron energy transition due to a photon interaction.
This is observed as a small, but measurable and also calculable using the HUP,  'blurring' of the spectral lines.

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Yeah nice. The blurring of spectral lines is a good demonstration that HUP is real. (Had I thought about this before my spectroscopy lab, it would have been one of the great revelations of my life, like seeing Andromeda or human karyotype for the first time!)

Non-quantum uncertainty associated with certain classical phenomena... Do you mean turbulence? Chaotic systems?

 

 

 

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18 hours ago, Prof Reza Sanaye said:

Do you mean HP is or is NOT in the eyes of the beholder  ??

The eyes of the beholder can be a chunk of iron at room temperature. Any system that interacts with the quantum system under observation and makes it decohere in the particular variables under consideration. Beholding is not that special.

I basically agree with everything else that's been said about the difference between HUP and the observer effect.

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27 minutes ago, To_Mars_and_Beyond said:

Yeah nice. The blurring of spectral lines is a good demonstration that HUP is real. (Had I thought about this before my spectroscopy lab, it would have been one of the great revelations of my life, like seeing Andromeda or human karyotype for the first time!)

Non-quantum uncertainty associated with certain classical phenomena... Do you mean turbulence? Chaotic systems?

 

 

 

Neither of these. Just simple classical mechanics.

Classically when we talk of the postion of an object we mean the position of its centre of mass.

Now take an old fashioned balance scale with a rider or slider.

At what point on the balance arm is the rider's weight applied ?

The rider has physical dimensions with a leading edge, a trailing edge and a COM somewhere in between.
So the question arises "Where do we pinpoint the application of the rider's weight?"
The leading edge, the trailing edge, the COM ?
For most of the length of the balance arm the COM is exact.
But as the leading edge begins to pass over the knife edge, this chages introducing uncertainty.

 

This question of where does it start and where does it end and at what point between do you apply the second variable is, as already noted, the basis of the HUP.
There are also classical wave packet systems that act in this manner. The Fourier transform, swansont mentioned doesn't only apply to quantum theory.

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2 hours ago, joigus said:

The eyes of the beholder can be a chunk of iron at room temperature. Any system that interacts with the quantum system under observation and makes it decohere in the particular variables under consideration. Beholding is not that special.

I basically agree with everything else that's been said about the difference between HUP and the observer effect.

Thank you Sir , for your explanation.  

And for your deciding to get involved in the clarification discussion. 

 

From your statement ,and  from Moderator Swansonts' , applying my own understanding of modern physics , I gather that  knowledge as such plays no active role in natural inanimate systems, whether classical or quantum – this term enters physics only in conjunction with the observer, experimenter or thinker. Which , in turn , spells that, in theory, calculated/measured values have no relation to the subjective significance level  of the result to the observer; that is impossible to evaluate. But we should note that we have potential knowledge, i.e. an expression of the results that an analyst does not have…. but hopes, on average, to obtain  from the result(s) PLUS the assumption that full new information will be received after the action. And, it is equal to maximum uncertainties (ie, maximum entropy) that can be predicted to arrive at  new knowledge following calculation. It may , furthermore , be noted the time order here becomes of vital importance: The degree of complexity (entropy) of the system can be measured by potential knowledge before its final condition(s) is determined in a calculation. 

 

I expect to be corrected by any one of the participants in this seemingly plain , but ( in my opinion ) deep discussion on this specific niche of modern physics  IF, OF COURSE , I AM ANYWHERE WRONG.

 

 

 

Quote from Studiot

There are also classical wave packet systems that act in this manner. The Fourier transform, swansont mentioned doesn't only apply to quantum theory. " 

 

Are you here propounding an equivalency of physics rules ( at least , one physics rule ) in Micro-  and  Macro-physics ? 

I am very keen on knowing about it a little bit more precisely . . .. .as this is one of my most favorite areas in teaching philosophy of science . . . .. . .

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14 minutes ago, Prof Reza Sanaye said:

Quote from Studiot

There are also classical wave packet systems that act in this manner. The Fourier transform, swansont mentioned doesn't only apply to quantum theory. " 

 

Are you here propounding an equivalency of physics rules ( at least , one physics rule ) in Micro-  and  Macro-physics ? 

I am very keen on knowing about it a little bit more precisely . . .. .as this is one of my most favorite areas in teaching philosophy of science . . . .. . .

It is a question of mathematics, not physics.

Or more exactly it is a question to the application(s) of Mathematics in Physics.

 

It occurs because of the dfference between addition and multiplication in mathematics.

In Physics, if you take two quantities say length and width and add them together you still have the same physical quantity, viz a longer length.
But if you multiply them together you generate a new quantity viz area.

It may be that the two quantities have different physical significance, for example 10 miles per hour and 3 hours.

Mathematically you can add 10 and 3 without a problem.

But you cannot add 10mph to 3hours in Physics and obtain anything sensible.

But you can multiply them, and obtain a new quantity 30 miles, which is different from either of the original quantities.

An uncertainty principle applies when the two quantities are distributed or spread out to some extent along their scale.

To go back to my area example it does not matter which order you multiply the pair of quantities you will arrive at the same answer.

So length x width = width x length = the area and the uncertainty is zero.

Or there is no difference between A x B and  B x A  or   (BA - AB) = 0

(BA - AB) is called the commutator of this product AB

If one of the quantities being multiplied has an extent in terms of the other quantity this commutator will not in general be zero.

It's easier to understand this last sentence in Heisenberg's original pair viz position and momentum.

In order to answer the question "what is the momentum of a particle when it passes point x?" you have to answer first the question "Which part of the particle passes x ?", since it does not all appear at x at once.
 

Does this help ?

Quote

 

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The best example I can think of involves classical wave motion, and, apparently, was an example used by R Feynman.

To localize a wave packet, and so determine its position to arbitrary precision, it has to be built up of many ( infinite ? ) harmonics, such that wavelength/frequency, and therefore momentum, hv, are increasingly indeterminate.
The alternate, where the wave has an exact wavelength/frequency ( and momentum ), sees a wave of infinite length.

Edited by MigL
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52 minutes ago, studiot said:

It is a question of mathematics, not physics.

Or more exactly it is a question to the application(s) of Mathematics in Physics.

 

It occurs because of the dfference between addition and multiplication in mathematics.

In Physics, if you take two quantities say length and width and add them together you still have the same physical quantity, viz a longer length.
But if you multiply them together you generate a new quantity viz area.

It may be that the two quantities have different physical significance, for example 10 miles per hour and 3 hours.

Mathematically you can add 10 and 3 without a problem.

But you cannot add 10mph to 3hours in Physics and obtain anything sensible.

But you can multiply them, and obtain a new quantity 30 miles, which is different from either of the original quantities.

An uncertainty principle applies when the two quantities are distributed or spread out to some extent along their scale.

To go back to my area example it does not matter which order you multiply the pair of quantities you will arrive at the same answer.

So length x width = width x length = the area and the uncertainty is zero.

Or there is no difference between A x B and  B x A  or   (BA - AB) = 0

(BA - AB) is called the commutator of this product AB

If one of the quantities being multiplied has an extent in terms of the other quantity this commutator will not in general be zero.

It's easier to understand this last sentence in Heisenberg's original pair viz position and momentum.

In order to answer the question "what is the momentum of a particle when it passes point x?" you have to answer first the question "Which part of the particle passes x ?", since it does not all appear at x at once.
 

Does this help ?

My Lady* 

You were very kind indeed , going to the extent of typing so in-detail to make a point clear to me ( and , I hope , to others). As for commutativity of extensionalities , I gather that you have remained faithful to the traditional sense of the word in common physics literature. the major moot point , I suppose , are the  last two lines

In order to answer the question "what is the momentum of a particle when it passes point x?" you have to answer first the question "Which part of the particle passes x ?", since it does not all appear at x at once. "  

Here , you have actually brought in the notion of a (wave) packet acting as an (Macro) object. Very interesting .. .. Of course, , , , ,  the packets themselves move according to some sort of "mechanics". In case they acted within a Field , then they could not have behaved Quantum-ly , to coin a word by your permission. Thence , A question arises because the assumption is that the Schrödinger equation must needs be solved in Minkowskian  spacetime. 

This Problematic is removed  in Francis (2014) since the wave function is viewed as a mathematical model rather than an ontological construct, and it evolves in a non-physical affine Minkowski spacetime rather than curved spacetime. When states are measured, they are projected back to real spacetime as and when wavefunction collapses. 

 

Yet another point is the fact that QT works without a priori data. It simply has NOT gotten any datum of that kind. How , then , do packets take time to pass , say , point x OR point y or z or so on and so forth ? 

 

I should hardly think I am illogical both in following you and in bringing things to a more coherent cogent  ship"shape as you always insisted on . . .Am I  ??

 

 

6 minutes ago, MigL said:

The best example I can think of involves classical wave motion, and, apparently, was an example used by R Feynman.

To localize a wave packet, and so determine its position to arbitrary precision, it has to be built up of many ( infinite ? ) harmonics, such that wavelength/frequency, and therefore momentum, hv, are increasingly indeterminate.
The alternate, where the wave has an exact wavelength/frequency ( and momentum ), sees a wave of infinite length.

Of infinite length ,  Sir  . . . . . So how and where can it collapse according to physics'  common sense ?

Edited by Prof Reza Sanaye
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Frankly, I haven't a clue what you're referring to, half the time.
I gave an example of a classical wave  which demonstrates uncertainty.
And you started talking about 'collapse'.

What 'collapse' are you talking about here ?

1 hour ago, Prof Reza Sanaye said:

Of infinite length ,  Sir  . . . . . So how and where can it collapse according to physics'  common sense ?

Or is it simply nonsense, as usual ?

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