On 28/06/2017 2:26 pm, Russell Standish wrote:
On Tue, Jun 27, 2017 at 05:09:49PM +1000, Bruce Kellett wrote:
On 27/06/2017 10:21 am, Russell Standish wrote:
No, you are just dealing with a function from whatever set the ψ and ψ_α
are drawn from to that same set. There's never been an assumption that
ψ are numbers or functions, and initialy not even vectors, as that
later follows by derivation.
psi(t) is an ensemble, psi_a is an outcome.
ψ_α is still an ensemble, just a sub-ensemble of the original. Assume
α is the result (A has the value 2.1 ± 0.15). Then all universes where
A is greater than 1.95 and 2.25 will be in the ensemble ψ_α.
That might be what you meant, but that is not what the average reader
(such as I am) is going to take from the text. You say the projections
divides the observer moment (set) into a discrete set of outcomes
(indexed by a). You then wish to calculate the probability that outcome
a is observed. Now observers observe one outcome -- even if there is
some associated measurement error, there is still one outcome -- one
observer does not see 2.0, 2.1, and 2.2. At least, that would be a very
strange way of talking about an observation. Other observer might see
that, and one observer might see a range of results on repeated
measurements, but that is not what you appeared to be talking about.
The projection produces
the outcome from the observer moment psi. It maps the ensemble to
one member of that ensemble.
No. See above. It always maps to an ensemble with an infinite number
of members.
Not necessarily. A photon polarization measurement is dichotomous -- you
see a photon downstream of the polarizer, or you do not. No uncertainty
involved.
Certainly, psi is not a number or a
function, it is a set of possible outcomes: the psi_a are single
outcomes, be they numbers, functions or vectors, but the are not
just further ensembles.
Thus, for the sum to make sense you must assume linearity.
If you are objecting that the use of the symbol '+' implies linearity
where no such thing is assumed, then feel free to replace it with the
symbol of your choice. Then once linearity is established, feel free
to replace it back again to + so that the formulae following D.8 have
a more usual notation. Fine - that is a presentational quibble. My
taste is that it is unnecessarily cumbersome, but if you find it helps
prevent confusion in your mind, please do so.
Now
linearity is at the bottom of most distinctive quantum behaviour
such as superposition, interference, and entanglement. It is not
surprising, therefore, that if you assume linearity at the start,
you can get QM with minimal further effort.
Except that I don't assume linearity from the outset.
There seems to be some confusion between outcomes of observations
and sets of possible outcomes. The \P_A*psi is actually defined as a
superposition in (D.2), ad you then seek to determine the
probability of this superposition? You define the probability of a
set of outcomes by P_psi(\P_A*psi), which is P_psi(\Sigma psi_a). I
find it hard to interpret what this might mean -- the probability of
a superposition of measurement outcomes (with equal weights, what is
more)?
The weights aren't equal. They're denoted P(ψ_α).
Again, that is not what the text says; (D.2) is a sum over outcomes with
equal weights.
I find the notation confusing again. You have A contained in S, with
probability P_psi(\P_A*psi). A is original defined as an observable,
which divides the observer moment into a set of discrete outcomes. But S
is the set of possible outcomes: a is a member of S, so A contained in S
seems to be a different A -- the operator is not a subset of the
outcomes. To make something out of this, I took the latter use of A to
be the set of possible outcomes a -- not every operator has the same set
of possible outcomes.
You then talk about this as though you were still partitioning sets,
but the probability is not defined on a set, only on a
superposition. If it is a set, then (D.2) makes no sense.
You then introduce, quite arbitrarily, multiple observers for each
observer moment. This then gives you a measure, which is then made
to be complex!! The number of observers for each observer moment,
even if there can be more than one, which is not proved, cannot be
complex.
Why? Give me one good reason - other than it doesn't match your
intuition, which is generally not a good reason.
People/observers can be complicated beings, but that does not mean that
you can have a complex number of them. Since all you are really wanting,
ISTM, is to give each observer moment a weight. The n umber of observers
observing tis moment seems a rather arbitrary source for this weight,
because it is not ever determinable. Why not just give a weight, which
can be complex if you wish, but it has nothing to do with multiple
observers.
So your introduction of a measure, or weight for each
superposition really does not make sense. You then conclude that V,
the set of all observer moment, is a vector space over the complex
numbers.
That's right. If a linear combination of observer moments is also and
observer moment, then the set of all observer moments is a vector
space. This is linear algebra 101.
Again, that is not what your text says. You say that linearity comes
from the additive property of measure, and that is really what (D.7)
appears to be about. Except that it is not the additivity of measure
that is doing the work there, it is the additivity of probabilities for
disjoint observations (when the probability measure is normalized to unity).
I think you have to do more that just asserting that a linear
combination of observer moments is also an observer moment. The notion
of an observer moment has become opaque. An observer moment is the set
of possibilities consistent with what is known at that point in time. So
it is complete in itself -- how can you add two observer moments? You
clearly cannot add them for a single observer, because adding two
moments in time is not a defined operation -- it would not be an
observer moment since no observer observes two moments in time
simultaneously. Other observers at that time? Again, if you add observer
moments for different observers, you have no guarantee that there is
another observer who has just this combination of possibilities
consistent with what they know at that point in time. That would have to
be proved, rather than just asserted. In fact, ISTM that such a result
would require that every observer knows everything at every time, and
that everything that is ever possible is part of the set of things
consistent with what is known by that observer at that point in time --
and the notion of observer moments become otiose.
Even if you widen definitions to this extent, it does not follow that
the projections for observable A are linear -- if you say that there
will always be an observable C, such that the set of outcomes for A plus
the set of outcomes for B, will be contained in the set of outcomes for
C. And even if you do want to claim this, such a result would be of no
use for quantum mechanics, since the important thing there is that the
set of outcomes for operator A in themselves form a possible basis for a
linear vector space. In QM, linear combinations of the eigenvectors of
one operator, if independent and complete, can form a set of
eigenfunctions for a different operator in the same linear space. But
that condition is not met by just any two arbitrary operators.
I understand that in your theory, an observer moment must be shown to be
a Hilbert space, possibly a product Hilbert space -- a separate
component for different classes of operator: in QM the Hilbert space for
spin measurements on an electron is not the same Hilbert space as that
for position measurements. In the product space, linearity is required
in each term of the vector product. I don't see that you have
established this.
I remain baffled. You start with an observer moment as a set of
consistent possibilities. But there is no specification of what
'consistent' might mean.
There doesn't need to be a specification. All we need to know is that
some worlds correspond to the one we see, and some don't. We don't
need a constructive procedure for determining which worlds are to be
included, and in all likelihood, no such constructive procedure will
be found anyway.
That seems as though you ultimately rely on observation. But if you
cannot make predictions about what might or might not be observed, then
you don't have a scientific theory. Because the aim of science is to
constructively predict what we might see. So you seem to be claiming
that science is impossible, and if possible, useless (we don't need such
a constructive procedure).
There is also no particular structure
imposed on this observer moment
It satisfies set axioms, otherwise you cannot apply Kolmogorov's
probability axioms. I have been criticised for this particular
assumption before, however the Nothing (the book, after all is called
"Theory of Nothing") is a set above all.
A set consisting of what?
, and you conclude, after a number of
obscure manipulations, that the set of all observer moments is a
vector space over the complex numbers. I look in vain for the magic
that converts an unstructured ensemble into a linear vector space.
The "magic" IMHO is to consider that observers are also drawn from a
distribution according to some measure, rather than just being a
single observer. This is forced onto us by the Multiverse nature of
assumed reality. An observer cannot see a∧b, where a and b are
disjoint, but two different observers in different branches of the
Multiverse can.
So what? if observations a and b are disjoint (in separate universes),
they cannot be part of a single observer moment. And it is observer
moments in one universe that must obey quantum mechanics. Even in MWI,
the separated universe do not actually play any role in our
observations. (Decoherence and all that.) Part of the problem you face,
it seems, is that the separate observers you wish to consider must all
be related -- their observer moments must essentially be branches of the
multiverse stemming from the same quantum event -- possible outcomes of
the observation at time t. If not, then there is no necessary relation
between the different observer moments and linearity (in your terms)
cannot be established.
This is surely a non-trivial restriction on the nature of observer
moments, but you do not restrict the possible generality, you only
project particular (unstructured) results from this observer moment.
What, exactly, has gone on to extract linearity?
That is why I think that you have actually built linearity in from
the start -- there is no mechanism, engine, or procedure that
extracts this linearity. And what happens to things in the ensemble
that are not linear?
This does not, to me, pass the sniff test (or, in the Australian
vernacular, the pub test). Can you wonder that I am sceptical that
you have actually proved anything?
Yes, but then you've also exhibited some fairly deep misunderstandings
of the argument, possibly because you haven't read the whole book, or
alternatively, because my presentation is not sufficiently clear.
I accept that I may have misunderstood large parts of your argument, but
I would claim that that is largely because Appendix D is far from
transparent, and far from having any clear notation. I have dipped into
the rest of the text to see if that helped, but it didn't really.
But at least we can iterate on that, modulo your patience, until we
arrive at a genuine disagreement :).
Well, I am happy to continue to try to understand your argument, but I
guess you will not be surprised if I say that I am not sympathetic to
this approach -- the leaps of faith are too great.
Bruce
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