On 1/07/2017 11:18 am, Russell Standish wrote:
On Fri, Jun 30, 2017 at 04:57:00PM +1000, Bruce Kellett wrote:
In other words, observer moments are not vectors (or rays) in a
linear vector space because a linear superposition of observer
moments is not another observer moment: your derivation of Hilbert
space fails.
Bruce
To summarise, you are happy with everything up to (D.6), but think the
move to the linear superposition (D.7) is not justified, because you
say it cannot be an observer moment.
I would not say that I am happy with everything up to (D.7)! I find the
notation confusing, and some of the manipulations do not seem to make
operational sense. Your comments through these exchanges have certainly
helped me to see what is going on but, as you know, I am fairly out of
sympathy with the general approach anyway....
Maybe you are leaning on the fact that if A and B are projections,
then aA+bB is not in general a projection, since idempotency is not preserved?
That could be an issue, but it is not the main issue here. I think the
concentration on projections and events/observations/outcomes is
probably a mistake. You start with the concept of an observer moment as
a set of possibilities consistent with what is known at time t. The
elements of this set are infinite strings of bits encoding the
information and possible continuations. This is fair enough, I suppose.
But if you want to make contact with ordinary QM, you have to see this
psi(t) essentially as a wave function (or equivalent). So it is this OM
that is to be interpreted as a vector or ray in some space, and you have
to establish that this is a linear space, with an defined inner product,
so it is a Hilbert space.
In this endeavour, concentration on projections as measurements is not
actually helpful. These projections would correspond to operators on the
vector space of OMs, and the existence and/or actions of operators is
not actually going to help with establishing linearity. The point here
is that the action of an operator on a wave function does not actually
change the wave function, it just gives a set of eigenfunctions in terms
of which the original wave function can be expanded. Particular results
can then be projected out of the expansion by saying that an observation
corresponds to a particular eigenfunction, and the measurement result is
the corresponding eigenvalue. In your notation, the action of the
operator seems to be \P_A*psi = \Sigma_a \P_{a}*psi, where \P_{a} is the
projection of a particular result. Since applying the operator is
\P_A*psi, the result is still an observer moment -- in fact, the
original observer moment in some sense. The projections of individual
results, \P_{a}*psi, take one branch of the total wave function, but
this is still a wave function, and also still a possible observer
moment. So taking projections of OMs leads to further OMs, and it is the
linearity of these OMs that is the issue. If they are vectors in a
linear space, the the operators or projections act in a linear space,
and their linearity is a separate issue.
In order to establish that OMs are vectors in a linear space, you have
to show that linear combinations of OMs are just further OMs.
However, I don't think that's what I'm relying on. Given a starting
vector ψ, the vector (αA+βB)ψ is going the the result of some
projection C, where Cψ=(αA+βB)ψ, modulo an arbitrary complex
factor, and so (αA+βB)ψ is still an observer moment.
Well, that might be the case for some linear operators, but it is not
generally the case. One of the significant ways in which QM differs from
classical mechanics is that some observations are mutually exclusive -
the corresponding operators do not commute, so adding them does not
result in another possible operator: (x + p) is not an operator in
either x-space or p-space. In formal developments of QM, such as Landau
and Lifschitz, or von Neumann, a lot of care is taken to distinguish
between compatible and incompatible observations (commuting and
non-commuting operators). I do not see how this could be incorporated in
your approach -- just like the question of tensor product Hilbert spaces
for different (commuting) operators.
Thinking along those lines some more, I'm incorrect to say that the
vector space V is the set of all observer moments. It must be the set
of all successor observer moments to ψ, or all continuations as I
think you put it earlier. The clue lies in the linear span (D.8). OMs
that can't be reached from ψ just simply cannot be put in the linear
span of outcomes from an observation on ψ.
Any observer moment is the successor of some previous observer moment,
so concentrating on successor OMs of psi achieves nothing.
I need to think some more about what a linear superposition of
observers actually means in terms of selecting out subsets of infinite
strings, which is the original model I proposed. It may help resolve
the observer measure issue.
Would it be fair to say that the remainder of the appendix follows
through once linearity is demonstrated? It seems fairly
uncontroversial to me (aside from the assumption of continuous time,
which for me is a necessary ad-hoc assumption to make contact with
regular QM).
Well, as I said, I am out of sympathy with the basic approach, but that
aside, I am not convinced by your definition of the inner product in
terms of the probabilities of outcomes -- that seems to beg the
question. In the superposition in terms of the eigenfunctions of some
observable (operator), the coefficients are used in the definition of
the inner product -- and it is those coefficients that have then to be
interpreted as probabilities via the Born rule. You can't start by
assuming that they represent probabilities.
The derivation of the Schrödinger equation depends on the assumption of
unitarity. OK, that is your assumption of heritability = conservation of
information. But you would have to prove that your OMs actually had this
property. When you start with unstructured infinite bit strings, you
have to have some independent reason to impose these requirements -- it
might turn out that none of your OM bit strings have this property!
Bruce
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