On 6/13/2018 9:18 PM, Bruce Kellett wrote:
From: *Brent Meeker* <[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]>
On 6/12/2018 8:25 PM, Bruce Kellett wrote:
From: *Brent Meeker* <[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.
But what about my example of taking the Hilbert space of Fourier
components of the distribution of spots on the screen? It doesn't have
delta functions as the basis vectors, but it spans the same physical
results.
Brent
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