| 1 "Akira" |
A Question about Uncertianty Principle | dinsdag 26 maart 2002 9:47 |
| 2 "franz heymann" |
Re: A Question about Uncertianty Principle | dinsdag 26 maart 2002 10:40 |
| 3 "Pmb" |
Re: A Question about Uncertianty Principle | dinsdag 26 maart 2002 19:40 |
| 4 "Robert Hamilton" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 5:25 |
| 5 "Mike Hanson" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 11:49 |
| 6 "Nicolaas Vroom" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 13:05 |
| 7 "Robert Hamilton" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 15:09 |
| 8 "franz heymann" |
Re: A Question about Uncertianty Principle | woensdag 27 maart 2002 23:47 |
| 9 "Nicolaas Vroom" |
Re: A Question about Uncertianty Principle | donderdag 28 maart 2002 19:02 |
| 10 "Mike Hanson" |
Re: A Question about Uncertianty Principle | vrijdag 29 maart 2002 18:13 |
| 11 "franz heymann" |
Re: A Question about Uncertianty Principle | vrijdag 29 maart 2002 20:47 |
| 12 "Nicolaas Vroom" |
Re: A Question about Uncertianty Principle | maandag 1 april 2002 10:49 |
It is a question confuses me long time.
That is the delta(x) * delta(p) >= h / (4*pi)
If the problem is following...
There is a rest body with its mass M which its position is defined, and its position is at X = a. According to the hypothesis,we have delta(x) is zero. and delta(p) is infinite!!
How do we explain this phenomenon? just said we can determine the delta(p) ? But back to the problem, we have a rest body then its momentum is ZERO. How can I say its delta(p) is infinite?
Akira
It is a question confuses me long time.
That is the delta(x) * delta(p) >= h / (4*pi)
If the problem is following...
There is a rest body with its mass M which its position is defined, and its
position is at X = a.
According to the hypothesis,we have delta(x) is zero. and delta(p) is
infinite!!
That is correct. If you know the position of a particle precisely,
you can say nothing at all about its momentum
How do we explain this phenomenon?
What do you mean by "explain"?
The HUP is derivable from the postulstes of quantum mechanics. These
postulates have never made a prediction which has been contradicted
by experiment.
It sounds weird, but no actual observation has ever given results
which are in disagreement with the predictions of the HUP.
It simply means that you know nothing at all about the momentum under
circumstances when you have determined the position to indefinitely
small presision.
Roughly speaking, if you want to see with precision where a particle
is, you have to illuminate it with light which has a wavelength
smaller than the precision you wish to achieve. Such light consists
of photons which will impart a large momentum to the particle when a
photon hits it. So you have disturbed its original momentum by a
correspondingly large amount.
Franz Heymann
"Akira"
That is the delta(x) * delta(p) >= h / (4*pi)
If the problem is following...
There is a rest body with its mass M which its position is defined, and its
position is at X = a.
According to the hypothesis,we have delta(x) is zero. and delta(p) is
infinite!!
How do we explain this phenomenon? just said we can determine the delta(p) ?
But back to the problem, we have a rest body then its momentum is ZERO.
How can I say its delta(p) is infinite?
Well first off let's clear up what the "delta" means. It refers to a
statistical quantity. It has to do, not with one single measurement,
but with many measurements of a system in a given (i.e. known) state.
so when you measure "x=a" it means that you go and measure the
position of the particle and the result of the measurement was x=a.
Now you measure the momentum of the particle at the same time and you
get P=b. you repeat this many many many times. You calculate the
quanties delta(x) and delta(p) and you'll find that
delta(x)*delta(p) >= hbar/2 (I think that's the lower limit .. I
always have to look that up).
Given the definition of dx and dp and the popstulates of quantum
mechanics you can actually derive the expression dx*dp >= hbar/2
Pmb
On 26 Mar 2002 09:40:57 -0800, Pmb
It is a question confuses me long time.
That is the delta(x) * delta(p) >= h / (4*pi)
If the problem is following...
There is a rest body with its mass M which its position is defined, and its
position is at X = a.
According to the hypothesis,we have delta(x) is zero. and delta(p) is
infinite!!
How do we explain this phenomenon? just said we can determine the delta(p) ?
But back to the problem, we have a rest body then its momentum is ZERO.
How can I say its delta(p) is infinite?
Well first off let's clear up what the "delta" means. It refers to a
statistical quantity. It has to do, not with one single measurement,
but with many measurements of a system in a given (i.e. known) state.
This is not true
delta(x)*delta(p) >= hbar/2 (I think that's the lower limit .. I
always have to look that up).
Given the definition of dx and dp and the popstulates of quantum
mechanics you can actually derive the expression dx*dp >= hbar/2
This is true.
Pmb
"franz heymann"
Absolutely. However:
Absolutely not! Or rather, although this would certainly happen if
that's what you did, it has nothing whatsoever to do with the
uncertainty principle.
I would imagine that you yourself know better than to see HUP as
'unavoidable disturbance during measurement'. And I know that such an
explanation is commonly given for the sake of simplicity (even Asimov
explains it in these terms). But that does not justify giving out
misleading information which subsequently needs to be corrected once a
higher level of understanding is achieved.
If this is the way that HUP is introduced to people, then it takes an
awful lot of effort for them to unlearn this misconception and fully
appreciate the implications of uncertainty.
Mike.
"Robert Hamilton"
It is a question confuses me long time.
That is the delta(x) * delta(p) >= h / (4*pi)
If the problem is following...
There is a rest body with its mass M which its position is defined, and its
position is at X = a.
According to the hypothesis,we have delta(x) is zero. and delta(p) is
infinite!!
How do we explain this phenomenon? just said we can determine the delta(p) ?
But back to the problem, we have a rest body then its momentum is ZERO.
How can I say its delta(p) is infinite?
Well first off let's clear up what the "delta" means. It refers to a
statistical quantity. It has to do, not with one single measurement,
but with many measurements of a system in a given (i.e. known) state.
This is not true
so when you measure "x=a" it means that you go and measure the
position of the particle and the result of the measurement was x=a.
Now you measure the momentum of the particle at the same time and you
get P=b. you repeat this many many many times.
How do you measure momentum p = mv ?
In ONE measurement ?
and what is than delta p (if you do this many times)
Is there some one who can give an example ?
(I think this is all very very difficult
besides I do not think there IS any true uncertainty in nature
except OUR limitations to measure those quantities )
delta(x)*delta(p) >= hbar/2 (I think that's the lower limit .. I
always have to look that up).
Given the definition of dx and dp and the popstulates of quantum
mechanics you can actually derive the expression dx*dp >= hbar/2
This is true.
Pmb
On Wed, 27 Mar 2002 11:05:09 GMT, Nicolaas Vroom
"Robert Hamilton"
It is a question confuses me long time.
That is the delta(x) * delta(p) >= h / (4*pi)
If the problem is following...
There is a rest body with its mass M which its position is defined, and its
position is at X = a.
According to the hypothesis,we have delta(x) is zero. and delta(p) is
infinite!!
How do we explain this phenomenon? just said we can determine the delta(p) ?
But back to the problem, we have a rest body then its momentum is ZERO.
How can I say its delta(p) is infinite?
Well first off let's clear up what the "delta" means. It refers to a
statistical quantity. It has to do, not with one single measurement,
but with many measurements of a system in a given (i.e. known) state.
This is not true
so when you measure "x=a" it means that you go and measure the
position of the particle and the result of the measurement was x=a.
Now you measure the momentum of the particle at the same time and you
get P=b. you repeat this many many many times.
How do you measure momentum p = mv ?
In ONE measurement ?
and what is than delta p (if you do this many times)
Is there some one who can give an example ?
(I think this is all very very difficult
besides I do not think there IS any true uncertainty in nature
except OUR limitations to measure those quantities )
You calculate the
quanties delta(x) and delta(p) and you'll find that
delta(x)*delta(p) >= hbar/2 (I think that's the lower limit .. I
always have to look that up).
Given the definition of dx and dp and the popstulates of quantum
mechanics you can actually derive the expression dx*dp >= hbar/2
This is true.
Pmb
The objection: when presented this way, one is led to think of
the measurement as a coupling of the state to some external system which
performs the physical act of measurement. Moreover, that kind of
measurement is correctly described using mixed states. I think that something
as fundamental to quantum mechanics as is the uncertainty relation should
be applicable to the states themselves and not to state+measuring device,
and moreover to individual states or superpositions rather than to statistical
mixtures. That naturally leads one to look for a more fundamental statement
of the uncertainty principle in those terms. If such exists it is surely
more appropriate.
Well, in fact one does exist in the noncommutativity of the operators p
and x. Many (myself included) think of this relation, [x,p]=i as the
most fundamental statement of the uncertainty principle, and the delta p
delta x formula becomes a necessary consequence of it. The operator form
tells us, among other things, that a state cannot have both a definite
momentum and a definite position. This is a condition that has to be
satisfied by individual states of individual particles so it describes an
intrinsic property of the state itself, not just the act of measurement
of the state.
Mike Hanson
That is correct. If you know the position of a particle precisely,
you can say nothing at all about its momentum
Absolutely. However:
Roughly speaking, if you want to see with precision where a particle
is, you have to illuminate it with light which has a wavelength
smaller than the precision you wish to achieve. Such light consists
of photons which will impart a large momentum to the particle when a
photon hits it. So you have disturbed its original momentum by a
correspondingly large amount.
Absolutely not! Or rather, although this would certainly happen if
that's what you did, it has nothing whatsoever to do with the
uncertainty principle.
I would imagine that you yourself know better than to see HUP as
'unavoidable disturbance during measurement'. And I know that such an
explanation is commonly given for the sake of simplicity (even Asimov
explains it in these terms). But that does not justify giving out
misleading information which subsequently needs to be corrected once a
higher level of understanding is achieved.
If this is the way that HUP is introduced to people, then it takes an
awful lot of effort for them to unlearn this misconception and fully
appreciate the implications of uncertainty.
You are indisputably correct. I appologise for falling into this
trap.
Regrettably, that was how I was taught seventy years ago, and my
thoughts still veer in that direction if I don't keep them in check.
Franz Heymann
"franz heymann"
Mike Hanson
That is correct. If you know the position
of a particle precisely,
you can say nothing at all about its momentum
Absolutely. However:
I expect that also the reverse is true.
If you know its momentum precisely etc.
It is interesting that the sentence starts with IF
Absolutely not! Or rather, although this would certainly happen if
that's what you did, it has nothing whatsoever to do with the
uncertainty principle.
Okay - but not agreed.
If the above is not the way than how do you measure
p (and x) precisely ?
Please explain those implications.
(I have great doubts)
IMO you do not have to appologise.
If this was taught to you 70 years ago than your youth must be
at the craddle of GR and Quantum Mechanics.
Assuming you where than 15 years old than you must now be ?
I still have 25 years to go to reach that age.
"Nicolaas Vroom"
Mike Hanson
That is correct. If you know the position
of a particle precisely,
you can say nothing at all about its momentum
Absolutely.
I expect that also the reverse is true.
If you know its momentum precisely etc.
Yes, that's the crux of HUP. Or rather, it's the most extreme case.
The less you know about one of them, the more you can predict about
the other. But please note that it's not just position and momentum
that are related in this way ('conjugate attributes'). There are all
sorts of attributes one can measure of a particle, and each has its
conjugate attribute, such that one can never predict both of a
conjugate pair beyond a certain combined precision.
You can measure any attribute to any arbitrary precision, dependant
only upon how good your measuring equipment is.
Absolutely not! Or rather, although this would certainly happen if
that's what you did, it has nothing whatsoever to do with the
uncertainty principle.
Okay - but not agreed.
Well, as I went on to say:
Having had HUP introduced to you in terms of disturbance, I should
imagine you are extremely reluctant to let go of this idea. After all,
it makes such intuitive sense, doesn't it?
Unfortunately, it isn't correct. And I stress: this is not a point of
view. It is a fact. Even realist, hidden-variable models (which seek
to get rid of the random element that Einstein so disliked) do not see
HUP as a disturbance during measurement. Of course, it has everything
to do with measurement, but it is an intrinsic property of the
particle itself, rather than something that is 'done to it' from the
outside.
When an electron smacks into the phosphor screen of your monitor, you
know where it is (or where it was when it hit) because the screen
emits a flash of light at that position. The flash of light, however,
is not emitted immediately but only after a (very small) interval of
time. That interval is not the same every time round, and it cannot be
predicted beyond a certain accuracy. Yep, it's HUP.
Another way to determine position to arbitrary accuracy is to shoot
particles through a small hole that you can make smaller. The
precision of the 'sideways position' of the particle as it passes
through the hole is then constrained to be simply the width of the
hole. Making the hole smaller increases precision in this regard, but
it will result in an increased unpredictability of the particle's
sideways momentum.
Measuring momentum is easy if the particle is charged: pass it through
a magnetic field (going 'across' the path of the particle) and the
path of the particle will curve. You know how massive the particle is,
so you can work out the momentum from the curvature.
There are various clever ways of working out the particle's momentum
if it isn't charged, generally involving collisions with other
particles whose momentum you can measure.
Please explain those implications.
(I have great doubts)
Doubtless you have doubts: you have a 'disturbance picture' in your
mind as you read this. As a result, what I am about to write is going
to seem like groundless mysticism to you. But I assure you that this
is only because you cannot see the grounds for such apparent
weirdness, and it is therefore unacceptable to you. The first time you
do see them, it's quite a shock, and it does force a paradigm shift on
you.
Here goes. You will have heard it said that, prior to measurement, a
particle cannot really be said to have a position at all. This is
commonly misinterpreted as meaning "the particle's position is fuzzy".
But whilst that would certainly be weird and hard to picture, it is
still a naive oversimplification; the truth is even worse.
What is actually meant is that the attribute 'position' does not exist
in any real sense at all until, by the act of making a position
measurement, you help to create it. Likewise for all other attributes
- momentum, spin, polarisation, etc: they are all figments of the
measurement process, and have no meaningful existence outside of that
process. (N.B. this is not the same as saying the particle isn't real:
the particle *is* real, but its attributes are not, until the act of
measurement calls them into existence. Attributes are a joint creation
of the particle and of the measurement process.)
There are only two ways of escaping this conclusion. One is to adopt
the 'Many Worlds' interpretation ("If you flip a 'quantum coin', the
universe splits into two copies, identical in every respect save that
in one the coin came up heads, and in the other it came up tails.").
This gives you real attributes, but the price you have to pay for it
is a burgeoning number of almost identical copies of yourself reading
articles on USENET.
The other escape is to go for a realist, 'hidden variable' model. Very
few physicists subscribe to these, though, because they create even
greater absurdities than the ones they are trying to get rid of.
And those are your three choices. Disturbance during measurement
doesn't cut it as an explanation, I'm afraid.
Mike.
Nicolaas Vroom
"franz heymann"
Mike Hanson
That is correct. If you know the position
of a particle precisely,
you can say nothing at all about its momentum
Absolutely. However:
I expect that also the reverse is true.
If you know its momentum precisely etc.
It is interesting that the sentence starts with IF
Roughly speaking, if you want to see with precision
where a particle is, you
have to illuminate it with light which has a wavelength
smaller than the precision you wish to achieve.
Such light consists of photons which will
impart a large momentum to the particle when a
photon hits it. So you have disturbed its original
momentum by a correspondingly large amount.
Absolutely not! Or rather, although this would certainly happen if
that's what you did, it has nothing whatsoever to do with the
uncertainty principle.
Okay - but not agreed.
I would imagine that you yourself know better than to see
HUP as 'unavoidable disturbance during measurement'.
And I know that such an explanation is commonly
given for the sake of simplicity (even Asimov explains it in
these terms). But that does not justify giving out
misleading information which subsequently needs to
be corrected once a
higher level of understanding is achieved.
If the above is not the way than how do you measure
p (and x) precisely ?
If this is the way that HUP is introduced to people, then
it takes an awful lot of
effort for them to unlearn this misconception and fully
appreciate the implications of uncertainty.
Please explain those implications.
(I have great doubts)
You are indisputably correct.
I appologise for falling into this trap.
IMO you do not have to appologise.
Yes, It was correct of me to apologise, since the example I gave was a
bad one. The real background to the HUP may be illustrated by the
following example:
A particle with a perfectly well known momentum has a wave function of
the form exp(ikx). This wave is smeared out over all x. To make a
wave which has an amplitude which is large only in a restricted region
of x, you have to add together (integrate) a range of waves with a
range of values of k. (Consider the Fourier spectrum of a pulse
extending over a short range of x only). If you look at the
relationship between a pulse and its Fourier transform, you will find
that (delta x)/(delta k) just gives the usual HUP relationship between
position and momentum.
Regrettably, that was how I was taught seventy years ago,
and my thoughts still veer in that direction if I don't keep
them in check.
If this was taught to you 70 years ago than your youth must be
at the craddle of GR and Quantum Mechanics.
Assuming you where than 15 years old than you must now be ?
I still have 25 years to go to reach that age.
Slip of the tongue! 60 years ago, not 70. (Actually 58 years ago, to
be precise). It just shows what creeping senility is doing to me.
:-)
Franz Heymann
"Mike Hanson"
I expect that also the reverse is true.
If you know its momentum precisely etc.
Yes, that's the crux of HUP. Or rather, it's the most extreme case.
The less you know about one of them, the more you can predict about
the other. But please note that it's not just position and momentum
that are related in this way ('conjugate attributes'). There are all
sorts of attributes one can measure of a particle, and each has its
conjugate attribute, such that one can never predict both of a
conjugate pair beyond a certain combined precision.
It is interesting that the sentence starts with IF
You can measure any attribute to any arbitrary precision, dependant
only upon how good your measuring equipment is.
If this is the way that HUP is introduced to people, then
it takes an awful lot of
effort for them to unlearn this misconception and fully
appreciate the implications of uncertainty.
Having had HUP introduced to you in terms of disturbance, I should
imagine you are extremely reluctant to let go of this idea. After all,
it makes such intuitive sense, doesn't it?
Unfortunately, it isn't correct. And I stress: this is not a point of
view. It is a fact. Even realist, hidden-variable models (which seek
to get rid of the random element that Einstein so disliked) do not see
HUP as a disturbance during measurement. Of course, it has everything
to do with measurement, but it is an intrinsic property of the
particle itself, rather than something that is 'done to it' from the
outside.
Starting point is first: measurements
Those measurements IMO will disturb the particle involved.
General speaking the more accurate you want to measure
the more you will disturb the particle.
Using those measurements you can come up
with a model of the particle and or with
intrinsic properties of the particle.
When an electron smacks into the phosphor screen of your monitor, you
know where it is (or where it was when it hit) because the screen
emits a flash of light at that position. The flash of light, however,
is not emitted immediately but only after a (very small) interval of
time. That interval is not the same every time round, and it cannot be
predicted beyond a certain accuracy. Yep, it's HUP.
Another way to determine position to arbitrary accuracy is to shoot
particles through a small hole that you can make smaller. The
precision of the 'sideways position' of the particle as it passes
through the hole is then constrained to be simply the width of the
hole. Making the hole smaller increases precision in this regard, but
it will result in an increased unpredictability of the particle's
sideways momentum.
If the hole is larger you will not observe any change ie no x
and you will not measure x
If the hole is small and the particle can go through the whole
what you claim is that than you can measure x ie
is a function of the size of the hole.
If the hole is very small the particle will not pass
and again you will not measure x
ie. you can not measure x as precise as you want.
Do you call this a precise or a not precise measurement ?
If it is precise
does this mean that you can measure p of such a particle
twice precisely ?
I think this is very very tricky and the uncertainty principle
prevents you from doing this.
Suppose I start with particle_b and particle_c and
I let the two collide.
After the collision I must measure the
p of particle_b and the p of particle_c precisily
in order to calculate the original p of particle_c
But how do I do that ?
I need collisions with other particles with
other particles you can measure "precisely?"
IMO you will not succeed.
Please explain those implications.
(I have great doubts)
Doubtless you have doubts: you have a 'disturbance picture' in your
mind as you read this. As a result, what I am about to write is going
to seem like groundless mysticism to you. But I assure you that this
is only because you cannot see the grounds for such apparent
weirdness, and it is therefore unacceptable to you.
Here goes.
What is actually meant is that the attribute 'position' does not exist
in any real sense at all until, by the act of making a position
measurement, you help to create it.
Each measurement is a small process an action by using
a measurement tool.
As a result of this process or action both the tool and the object
will change. The amount of change of both give you information
ie you get more information about the object.
You learn something from this experiment.
It is difficult to accept that as such that you create something.
(IMO parameters of objects do not exist and they cannot
be created. Only the particle ansich exists)
IMO general speaking it is for me difficult to accept that
a measurement creates the spin of a particle and that before
the measurement it did not have a spin.
You can also repeat this measurement n times.
Does that mean that each time you create this spin ?
(or position or momentum ?)
Mike.
Nick
Back to my home page Contents of This Document
>
>
ZERO.
If the momentum is truly zero, then the position of the body cannot be
determined at all.
>
just said we can determine the delta(p) ?
But back to the problem, we have a rest body then its momentum is
>
How can I say its delta(p) is infinite?
3 A Question about Uncertianty Principle
Van: "Pmb"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: dinsdag 26 maart 2002 19:40
>
It is a question confuses me long time.
4 A Question about Uncertianty Principle
Van: "Robert Hamilton"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: woensdag 27 maart 2002 5:25
>
"Akira"
>>
>
>
so when you measure "x=a" it means that you go and measure the
position of the particle and the result of the measurement was x=a.
Now you measure the momentum of the particle at the same time and you
get P=b. you repeat this many many many times. You calculate the
quanties delta(x) and delta(p) and you'll find that
>
5 A Question about Uncertianty Principle
Van: "Mike Hanson"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: woensdag 27 maart 2002 11:49
>
That is correct. If you know the position of a particle precisely,
you can say nothing at all about its momentum
>
Roughly speaking, if you want to see with precision where a particle
is, you have to illuminate it with light which has a wavelength
smaller than the precision you wish to achieve. Such light consists
of photons which will impart a large momentum to the particle when a
photon hits it. So you have disturbed its original momentum by a
correspondingly large amount.
6 A Question about Uncertianty Principle
Van: "Nicolaas Vroom"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: woensdag 27 maart 2002 13:05
>
On 26 Mar 2002 09:40:57 -0800, Pmb
> >
"Akira"
> >>
> >
>
> >
> >
You calculate the
quanties delta(x) and delta(p) and you'll find that
>
> >
>
7 A Question about Uncertianty Principle
Van: "Robert Hamilton"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: woensdag 27 maart 2002 15:09
>
>>
On 26 Mar 2002 09:40:57 -0800, Pmb
>> >
"Akira"
>> >>
>> >
>>
>> >
>
>> >
>>
>> >
>>
Not objecting to the measurment theorem. The theorem itself is correct as
far as it goes. But it's a mistake to think of the uncertainty principle
in only in terms of a statistical sampling, because it is more fundamental
than that.
>
8 A Question about Uncertianty Principle
Van: "franz heymann"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: woensdag 27 maart 2002 23:47
>
"franz heymann"
> >
>
> >
>
9 A Question about Uncertianty Principle
Van: "Nicolaas Vroom"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: donderdag 28 maart 2002 19:02
>
> >
"franz heymann"
> > >
> >
> > >
Roughly speaking, if you want to see with precision
where a particle is, you
have to illuminate it with light which has a wavelength
smaller than the precision you wish to achieve.
Such light consists of photons which will
impart a large momentum to the particle when a
photon hits it. So you have disturbed its original
momentum by a correspondingly large amount.
> >
> >
I would imagine that you yourself know better than to see
HUP as 'unavoidable disturbance during measurement'.
And I know that such an explanation is commonly
given for the sake of simplicity (even Asimov explains it in
these terms). But that does not justify giving out
misleading information which subsequently needs to
be corrected once a
higher level of understanding is achieved.
Do you use light ?
I think it is not so easy.
> >
If this is the way that HUP is introduced to people, then
it takes an awful lot of
effort for them to unlearn this misconception and fully
appreciate the implications of uncertainty.
>
You are indisputably correct.
I appologise for falling into this trap.
>
Regrettably, that was how I was taught seventy years ago,
and my thoughts still veer in that direction if I don't keep
them in check.
Nick
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Franz Heymann
https://www.nicvroom.be/
10 A Question about Uncertianty Principle
Van: "Mike Hanson"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: vrijdag 29 maart 2002 18:13
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"franz heymann"
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"franz heymann"
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It is interesting that the sentence starts with IF
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Roughly speaking, if you want to see with precision
where a particle is, you
have to illuminate it with light which has a wavelength
smaller than the precision you wish to achieve.
Such light consists of photons which will
impart a large momentum to the particle when a
photon hits it. So you have disturbed its original
momentum by a correspondingly large amount.
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If this is the way that HUP is introduced to people, then
it takes an awful lot of
effort for them to unlearn this misconception and fully
appreciate the implications of uncertainty.
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If the above is not the way than how do you measure
p (and x) precisely ?
Do you use light ?
I think it is not so easy.
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the implications of uncertainty.
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11 A Question about Uncertianty Principle
Van: "franz heymann"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: vrijdag 29 maart 2002 20:47
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"franz heymann"
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Do you use light ?
I think it is not so easy.
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12 A Question about Uncertianty Principle
Van: "Nicolaas Vroom"
Onderwerp: Re: A Question about Uncertianty Principle
Datum: maandag 1 april 2002 10:49
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"Nicolaas Vroom"
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"franz heymann"
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If the above is not the way than how do you measure
p (and x) precisely ?
Do you use light ?
I think it is not so easy.
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Measuring momentum is easy if the particle is charged: pass it through
a magnetic field (going 'across' the path of the particle) and the
path of the particle will curve. You know how massive the particle is,
so you can work out the momentum from the curvature.
It seems that for such particles HUP does not apply.
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There are various clever ways of working out the particle's momentum
if it isn't charged, generally involving collisions with other
particles whose momentum you can measure.
IMO if you want to measure p precisely of particle_b you need
a particle_c with p precisely but than you do not know the x of
that particle.
Of particle_b I know p precise before the collision.
The original p of particle_c is unknown.
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the implications of uncertainty.
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IMO as far as I can guess nothing new is unacceptable to me
but it should be clear.
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Thanks for your effort
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The first time you
do see them, it's quite a shock, and it does force a paradigm shift on
you.
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You will have heard it said that, prior to measurement, a
particle cannot really be said to have a position at all. This is
commonly misinterpreted as meaning "the particle's position is fuzzy".
But whilst that would certainly be weird and hard to picture, it is
still a naive oversimplification; the truth is even worse.
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Likewise for all other attributes
- momentum, spin, polarisation, etc: they are all figments of the
measurement process, and have no meaningful existence outside of that
process. (N.B. this is not the same as saying the particle isn't real:
the particle *is* real, but its attributes are not, until the act of
measurement calls them into existence. Attributes are a joint creation
of the particle and of the measurement process.)
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Created: 15 April 2002