Re: a question on incompatibility of properties in a one particle system

From: Bilge (dubious_at_radioactivex.lebesque-al.net)
Date: 11/02/04 

Date: Tue, 02 Nov 2004 07:27:38 -0000

 bernard.chaverondier:
>"Bilge" <dubious@radioactivex.lebesque-al.net> a écrit dans le message de
>news:slrnco9d2h.ibg.dubious@radioactivex.lebesque-al.net...
>
>Bilge
>>>> I have no idea what you are talking about. There is no inherent
>>>> ``uncertainty'' in the measurement process. The inherent
>>>> uncertainty is in the indeterminacy of what is being measured.
>
>Chaverondier
>>>Quantum indeterminacy is caused by the absence of knowledge
>>>of the quantum state of the measuring apparatus S2 and that
>>>of the environment S3 that interact with it.
>
>Bilge
>> Not in the least. The indeterminacy is stated explicitly
>> by the theory and the phenomena which exhibit that
>> indeterminacy are the pure states.
>
>Chaverondier
>There is no indeterminacy in the quantum evolution of pure
>states. Indeterminacy shows up only when the system in a
>known pure state interacts with a measuring apparatus
>which quantum state is not known (or incompletely known).
>
>Bilge
>> An interference pattern occurs because the phases are
>> completely indeterminate, i.e., there is a non-zero commutator
>> between the phase and the photon number operator:
>> [N, \phi] = -i
>
>Chaverondier
>This indeterminacy shows up only if you perform a measurement
>of photon number (by a photographic plate for instance).
 
   Try writing a wavefunction down that provides well-defined
values for both caonical conjugates.

>This implies that your wave interacts with a measuring apparatus
>which quantum state is unknown (or intractable deterministically
>even if known). Now, you have a very good statistical model
>of the indeterminacy that stems from your lack of knowledge.
 
  I didn't assume anything was unknown. I assumed that you could write
a wavefunction down that include any and all of the apparatus. You
can certainly do that in principle, just by calling it \Psi. What's
the commutator of p and x applied to this wavefunction that now
includes the environment?

>That's not that different from the case when I play head and tail by
>flipping a coin.
 
  Yes, it is fundanentally different and the purpose of the epr
experiment was to demonstrate precisely the difference in a quantum
mechanical result and a purely statistical result.
 
>I can provide a mathematical statistical model which
>will provide me with an excellent agreement with observed statistics.
 
  But not a quantum mechanical model. If you think otherwise, write
the appropriate quantum mechanical wavefunction for two identical
coins in a singlet state, |heads> - |tails> and see if what you're
flipping looks like that wavefunction.
 
>Nevertheless, this mathematical model doesn't oppose the interpretation
>that this indeterminacy stems from my lack of knowledge of the little
>details that cause the coin to fall on one face or the other one.
>
>Bilge
>> That occurs for a state of _maximum_ knowledge,
>
>Chaverondier
>Of maximum knowledge of _a part_ of the quantum whole
>comprising the observed system, the measuring apparatus
>+the environment that interact with them. The part comprising
>only the observed system.
 
  In the epr experiment, both photons are measured. There is no unobserved
part of the system.

>
>The maximum knowledge of the observed system is not enough
>to predict deterministically what happens to the inseparable
>quantum whole comprising the observed system, the measuring
>apparatus + the environment that interact with them
>
>Bilge
>> i.e., a pure state in which the paths of the interfering photons
>> are indeterminate, not unknown or uncertain by virtue of
>> an interaction.
>
>Chaverondier
>There are no paths for the photon whatever the experiment you
>consider. That's a classical concept which is inappropriate.
 
  Now you're being ridiculous. Feynman pah integrals are a perfectly
good way to formulate quantum mechanics, so I can discuss paths so
long as I don't treat the photons as traversing any particular paths.

>Now, when you measure the position of a photon you cannot
>say that the photon was where you have measured it to be before
>you performed the quantum measurement. The measuring apparatus
>(hence its exact quantum sate) participates strongly to the creation
>of this interaction event.
>
 
  I'm sorry, but this is like reading ``In Search of Schroedinger's Cat''.
You've made quantum mechanics out to be some mystical theory. It might
be weird, but it's not nearly as weird as you are trying to make it
out to be.

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