From: *Brent Meeker* <[email protected] <mailto:[email protected]>>
On 6/13/2018 9:18 PM, Bruce Kellett wrote:
From: *Brent Meeker* <[email protected] <mailto:[email protected]>>
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.
How do you know that this gives the same physical results? The position
operator is different, so the interaction Hamiltonian is no longer given
by interactions between point particles, obeying a separation force law
depending on the distance. If you are right and the physics is unchanged
by a basis change, then there is no preferred basis problem because all
bases would give the same physics. But that is manifestly false. Just
consider the expansion of some state in two different bases for the same
measurement space (different operators, mind):
}psi> = Sum_i c_i |a_i> = Sum_j d_j |b_j>.
the c_i =/= d_j in general. So the worlds are split into different
branches, with different weights, depending on which basis is chosen.
This does not seem like the same physics to me.
Sure, it is the same initial vector, so we can related the bases by a
simple linear transformation. But decoherence will operate on different
sets of states in the two cases; the branching worlds will be different.
So there is no way this could be said to represent the same physics in
general.
Bruce.
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