From: *Brent Meeker* <[email protected] <mailto:[email protected]>>
On 6/13/2018 3:53 PM, Bruce Kellett wrote:
From: *Brent Meeker* <[email protected]
<mailto:[email protected]>>
On 6/12/2018 10:26 PM, Bruce Kellett wrote:
From: *Brent Meeker* <[email protected]
<mailto:[email protected]>>
On 6/12/2018 8:25 PM, Bruce Kellett wrote:
From: *Brent Meeker* <[email protected]
<mailto:[email protected]>>
An isolated system has energy eigenvalues. But any realistic
macroscopic system is only going to conserve energy
approximately. I think energy eigenvalues are found in atoms
and maybe molecules. But larger systems (C60 Bucky balls?) tend
to emit and absorb photons that localize them in a position basis.
I am glad you said "a position basis" and not "the position
basis" -- a mistake that is frequently made. Position is an
operator in a high dimensional Hilbert space, and there are an
infinite number of possible bases for this space, each
corresponding to a different operator in the space. Which one of
these operators (and bases) is "the" position basis? The answer
from decoherence theory is that it is the basis that is stable
against environmental decoherence. But, as I pointed out in a
post on the 'Entanglement' thread, this is defined by the
operator that commutes with the interaction Hamiltonian. However,
the interaction Hamiltonian is usually defined in terms of point
particle interactions, so commutes with the position operator
because it contains that operator itself. So that particular
definition of the stable basis is circular -- any chosen operator
in the position Hilbert space would fit the bill provided it was
used for both the position measurement and the interaction
Hamiltonian.
But is it a vicious circle? Aren't all the position bases going to
be physically equivalent?
Well, yes. Insofar as you can describe any vector in a linear space
in terms of any of the possible bases. But no. Not all of these
descriptions are the same -- what is given by the eigenvalues of
one operator will be a superposition of the eigenvalues of another
operator. In terms of position measurements, we get single dots on
the screen in the basis consisting of delta functions for positions
along the line.
I don't see that. Suppose I did a Fourier transform of the basis
consisting little bins across the screen. The indeed each spot on
the screen will be represented by a superposition of Fourier
components, but it will still be a spot in that representation. And
the Schroedinger eqn solution for the interference pattern on the
screen will also be a superposition of Fourier components.
So you are saying that there is no preferred basis problem? What do
you think the problem is?
No. There's a preferred basis in which this "world" and it's spots on
the screen, is spanned by basis vectors which are orthogonal to the
basis vectors of the "worlds" in which the spots are in different
places on the screen. But in each world there are different (not
necessarily position) bases, but they describe the same physics.
I don't think that is correct. The preferred basis is selected as the
eigenvectors of the operator that commutes with the interaction
Hamiltonian. If you choose a different basis for the Hilbert space, even
by a simple rotation of your present basis, you are going to get
eigenvectors (and eigenvalues) of a different operator. Since this
operator must also be dominant in the interaction Hamiltonian, the
physics is necessarily going to be different. A different position basis
is going to result in more than different places on the screen for the
spots.
Bruce
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