mathur@pollux.usc.edu (Samir Kumar Mathur) (08/03/87)
Here is a thought experiment which seems to violate Uncertainity Principle (UP)
that came up at a casual discussion with a friend.
Scenario :
There are two particles A and B and I am trying to measure their
momentum and positions accurately.
Experiment :
(1) At time t:
(a) I measure the momentums p(A,t) & p(B,t) and hence
p(A&B,t) = p(A,t) + p(B,t) as accurately as I wish.
(2) At time t+dt:
(a) I measure the momentum of A, p(A,t+dt) as accurately
as I wish.
(b) I measure the position of B, q(B,t+dt) as accurately
as I wish.
(c) I calculate p(B,t+dt) = p(A&B,t) - p(A,t+dt)
{conservation of momentum} as accurately as I wish.
Conclusion:
From steps 2(b),(c) I conclude that I can measure p(B,t+dt),q(B,t+dt)
as accurately as I wish. This obviously violates Heisenberg's UP.
-----------------------------------------------------------------------------
I could come up with the following possible explanations for the above
paradox(??) :
1. In the act of observing the two momentums p(A,t),p(B,t) accurately,
I will always disturb the total momentum of the system and thus
I cannot calculate 2(c) from the principle of conservation of
momentum.
2. The two events 2(b) & 2(c) are not independent and doing one of them
*instantaneously* effects the other, ie. the two cannot be
carried out simultaneously with arbitrary accuracy.
3. My step 2(c) is incorrect because I am assuming some kind of *objectivity*
(is it the right word in this context ?) in this experiment which
leads to a wrong conclusion. ie. unless I actually observe p(B,t+dt),
I can never say that p(B,t+dt) = p(A&B,t) - p(A,t+dt) and that I
cannot claim that conservation of momentum holds unless I directly
measure all the component momentums. But of course, if I do that,
step 2(b) will not be possible.
------------------------------------------------------------------------------
I would welcome any knowledgeble/reasonable comments on the subject.
Cheers,
Samir.
ps. I am relatively new on this net and in case an isomorph of this question
has already been discussed, I apologise for reposting it.
*****************************************************************************
Thought for the day : Mu.
Samir K. Mathur (213)-743-2746 (off)
CS dept., USC. (213)-748-2524 (Res)
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platt@emory.uucp (Dan Platt) (08/04/87)
In article <3794@oberon.USC.EDU> mathur@pollux.usc.edu (Samir Kumar Mathur) writes: >Here is a thought experiment which seems to violate Uncertainity Principle (UP) >that came up at a casual discussion with a friend. > >Scenario : > There are two particles A and B and I am trying to measure their > momentum and positions accurately. > >Experiment : > (1) At time t: > (a) I measure the momentums p(A,t) & p(B,t) and hence > p(A&B,t) = p(A,t) + p(B,t) as accurately as I wish. > > (2) At time t+dt: > (a) I measure the momentum of A, p(A,t+dt) as accurately > as I wish. > (b) I measure the position of B, q(B,t+dt) as accurately > as I wish. > (c) I calculate p(B,t+dt) = p(A&B,t) - p(A,t+dt) > {conservation of momentum} as accurately as I wish. > >Conclusion: > From steps 2(b),(c) I conclude that I can measure p(B,t+dt),q(B,t+dt) > as accurately as I wish. This obviously violates Heisenberg's UP. > >----------------------------------------------------------------------------- A simple answer to this paradox could center around several points: 1) How repeatable are your results? 2) Exactly how is your system configured? In answer to 1), you may be able to measure q(B,t+dt), but if you run several experiments, you'll find that the value of q is not predictable to within a factor of hbar/2/deltap where deltap is the uncertainty in p(A) and p(B). If you adjust the experiment to improve the measurement of q, then the values of the p's will deteriorate (assuming your experiment is capable of measureing to the accuracies required to observe these). For point 2) it looks like you have two particles A and B which are interacting only with each other, and that they may exchange momentum between t and t+dt. If this is so, how did you measure the momentum at t so that you knew what the momentum was after the measurement? (I won't suggest that the act of measurement produces effects which can't be determined; limitations in measurements are inherent in the system, and should be determinable, but it is important to consider that your measurement techniques should be reflected in the conservation of momentum equation (have you added or changed momentum?)). Lastly, I'd like to point out that the Heisenberg Uncertainty Principle is mathematically equivalent to the restriction of the rate of transmission of information via radio. If you send a signal with amplitude modulation (AM) or frequency modulation (FM -- though this is a little more complicated) a pure tone (the carrier) contains no information. When you modulate the carrier, you induce new frequencies. If you want to localize a signal (information) in a smaller and smaller dt, then you need a wider and wider dfrequency (bandwidth of induced frequencies) to handle the information. This is equivalent to trying to localize the location of a particle to dq; it requires a larger and larger bandwidth of dp's to handle the information required to define dq. The reason for this is that in quantum mechanics, the p,q pair are 'conjugate' to each other in the same way that frequency,t are 'conjugate' to each other in radio transmission. If you accept the DeBrogie hypothesis (which has been experimentally confirmed in any experiment dependant on quantum effects), then the Heisenberg principle follows. It's built into the structure of QM. Hope this is a help. Dan
ronse@prlb2.UUCP (Christian Ronse) (08/05/87)
In article <3794@oberon.USC.EDU>, mathur@pollux.usc.edu (Samir Kumar Mathur) writes: } Here is a thought experiment which seems to violate Uncertainity Principle (UP) } that came up at a casual discussion with a friend. } } Scenario : } There are two particles A and B and I am trying to measure their } momentum and positions accurately. } } Experiment : } (1) At time t: } (a) I measure the momentums p(A,t) & p(B,t) and hence } p(A&B,t) = p(A,t) + p(B,t) as accurately as I wish. } } (2) At time t+dt: } (a) I measure the momentum of A, p(A,t+dt) as accurately } as I wish. } (b) I measure the position of B, q(B,t+dt) as accurately } as I wish. } (c) I calculate p(B,t+dt) = p(A&B,t) - p(A,t+dt) } {conservation of momentum} as accurately as I wish. } } Conclusion: } From steps 2(b),(c) I conclude that I can measure p(B,t+dt),q(B,t+dt) } as accurately as I wish. This obviously violates Heisenberg's UP. This is very close to the original statement of the Einstein-Podolsky-Rosen paradox. They assume two particles A and B having interacted at time t. Then their respective position and momentum are related. At time t one measures the position of A and the momentum of B, which allows one to deduce the position of B and the momentum of A, and so violate the uncertainty principle. This has been tested in various experiments with the X, Y, and Z components of spin instead of position and momentum, especially in the Aspect experiment. That topic has been discussed at length in sci.physics a few months ago. -------------------- Christian Ronse maldoror@prlb2.UUCP {seismo|philabs|mcvax|...}!prlb2!{maldoror|ronse} ``So goodnight, kids, and remember...everybody has a pointer in their heart, set to that special someone...don't keep yours set to nil...'' Basil Hosmer Kaptain Kloodge Saga -- Christian Ronse maldoror@prlb2.UUCP {seismo|philabs|mcvax|...}!prlb2!{maldoror|ronse}
cpf@batcomputer.tn.cornell.edu (Courtenay Footman) (08/09/87)
[All >'s have been replaced by )'s to confuse Pnews]
I have seen several responses to this article; none seemed to give
a clear answer to simple problem.
In article <3794@oberon.USC.EDU> mathur@pollux.usc.edu (Samir Kumar Mathur) writes:
)Here is a thought experiment which seems to violate Uncertainity Principle (UP)
)that came up at a casual discussion with a friend.
)
)Scenario :
) There are two particles A and B and I am trying to measure their
) momentum and positions accurately.
)
)Experiment :
) (1) At time t:
) (a) I measure the momentums p(A,t) & p(B,t) and hence
) p(A&B,t) = p(A,t) + p(B,t) as accurately as I wish.
Possible
)
) (2) At time t+dt:
) (a) I measure the momentum of A, p(A,t+dt) as accurately
) as I wish.
Possible
) (b) I measure the position of B, q(B,t+dt) as accurately
) as I wish.
Possible. However, when dealing with possible violations of the
uncertainty principle, it is best to use the technique Bohr used
in his debates with Einstein: be as concrete as possible. Do not
just say "measure the position"; say how one is doing it. In this
case, say that the measurement is made visually, by shining a light
on it. (Which can be done as accurately as one wishes.)
) (c) I calculate p(B,t+dt) = p(A&B,t) - p(A,t+dt)
) {conservation of momentum} as accurately as I wish.
Oh. At this point, realize that "shining a light on it" is not
the phrase a physicist would use. A physicist would say "scattering
light off it". This practically shouts out what is going on in this
problem. What does "scattering light off it" mean? It means that
one is shining a light on the, and seeing which photon bounce off
particle B. Bounce off? What does this do to the momentum of particle
B? It changes it by an unknown amount; the uncertainty of your
knowledge of the momentum of particle B will be greater than or equal
to h divided by the uncertainty with which you know its position. At
this point, one can dispense with the concrete example, and realize
that this will be the case no matter what method was used to measure
the position of particle B; any method would have change the particle's
momentum. One does know what the momentum of particle B was before you
measured its position, p(B), but this information is of no use in predicting
the particle's motion after the position measurement.
This is not one of the possible explanations that you gave -- one can
measure momentum and not disturb the momentum -- it is measuring position
that disturbs the momentum. Conservation of momentum is not a problem
because the system is not just particle A and particle B -- there is, in
addition, whatever measures the position of particle B.
I hope that this answer is reasonably clear; please let me know if it
was not.
--
--------------------------------------------------------------------------------
Courtenay Footman ARPA: cpf@lnssun9.tn.cornell.edu
Lab. of Nuclear Studies Usenet: Not currently available.
Cornell University Bitnet: cpf@CRNLNUC.BITNETplatt@emory.uucp (Dan Platt) (08/12/87)
In article <1938@batcomputer.tn.cornell.edu> cpf@tcgould.tn.cornell.edu (Courtenay Footman) writes: ... >In article <3794@oberon.USC.EDU> mathur@pollux.usc.edu (Samir Kumar Mathur) writes: >)Here is a thought experiment which seems to violate Uncertainity Principle (UP) >)that came up at a casual discussion with a friend. >) >)Scenario : >) There are two particles A and B and I am trying to measure their >) momentum and positions accurately. >) >)Experiment : >) (1) At time t: >) (a) I measure the momentums p(A,t) & p(B,t) and hence >) p(A&B,t) = p(A,t) + p(B,t) as accurately as I wish. >Possible >) >) (2) At time t+dt: >) (a) I measure the momentum of A, p(A,t+dt) as accurately >) as I wish. >Possible >) (b) I measure the position of B, q(B,t+dt) as accurately >) as I wish. >Possible. However, when dealing with possible violations of the >uncertainty principle, it is best to use the technique Bohr used >in his debates with Einstein: be as concrete as possible. Do not >just say "measure the position"; say how one is doing it. In this >case, say that the measurement is made visually, by shining a light >on it. (Which can be done as accurately as one wishes.) >) (c) I calculate p(B,t+dt) = p(A&B,t) - p(A,t+dt) >) {conservation of momentum} as accurately as I wish. >Oh. At this point, realize that "shining a light on it" is not >the phrase a physicist would use. A physicist would say "scattering >light off it"... >...Bounce off? What does this do to the momentum of particle >B? It changes it by an unknown amount; the uncertainty of your >knowledge of the momentum of particle B will be greater than or equal >to h divided by the uncertainty with which you know its position. This explanation has always left me unsatisfied for the following reason: The Heisenberg Uncertainty Principle must hold for the particles regardless of whether they are being 'measured' or not. The quantum mechanical description implies limitations in the simultaneous resolution of location and momentum even before measurement. Furthermore, QM makes definite statements of how a measurement may affect the momentum of a particle due to measurement (such as scattering amplitudes from S matrices, or linear responses from Kubo formulas, or any other appropriate response of the system to measurement). It is true that the knowlege isn't deterministic in that you get an amplitude as a function of momentum transfer, but you do get a lot more information than just 'measurement of location at resolutions (wavelengths) as small as we want implies a larger and larger momentum transfer which isn't precisely known.' Also, the limitation on the knowlege of location and momentum via scattering with photons is itself based on the wave/particle duality embodied in the relationship between a function and its Fourier transform. In this sense the measurement argument is redundant once you specify the deBroglie relationships; the quantum formalism has the Heisenberg Uncertaity principle built into it one way or the other (even when you do try to determine the effect a measurement would have on the system while you are measuring it - as in the Kubo formalism). It seems to me that the reason for the extreme operationalism of Bohr (there isn't really a particle there unless you measure it) is that it is one of the easiest positions to defend, and therefore to accept, when confronted with the idea that things aren't deterministic: it's easier to say that the act of measuring something messes up the possibility of simultaneous measurement of conjugate measureables, than it is to say that conjugate measureables really aren't simultaneously measureable or even definable. The operational definitions free one from having to say that quantum mechanics is 'really' what 'reality' is doing, it's just a description. However, this means that a particle's location isn't defined except in terms of a measurement (and there isn't really a location unless you measure it) which I find to be a bit extreme. Therefore, I'd suggest that 1) the deBroglie relationships have been supported in all sorts of experiments from scattering, to tunneling, to coherence; 2) The deBroglie relationships have built into it a limitation on the definition of two conjugate measureables. I don't feel I have to know 'why' in the deterministic Newtonian sense any more than an Aristotilian would need to know 'why' a body could keep moving without the application of force... Dan