On 29/06/2017 5:36 pm, Russell Standish wrote:
On Thu, Jun 29, 2017 at 03:19:40PM +1000, Bruce Kellett wrote:
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.
Any measurement on a continuum will have an uncertainty. A measurement
of 2.0 ± 0.1 will be compatible with an infinite variety of observer
moments, observing the value in the range 1.9 to 2.1.
I think, in the light of your response to the dichotomous measurement
below, that this is just a red herring. If the outcome psi_a is an
ensemble, it has nothing to do with measurement errors. Besides, one
does not measure errors with a single measurement. A measurement of a
continuous variable may be subject to error, but a single measurement
gives a single value: errors can be estimated by doing statistics over
repeated independent measurements of the same variable on similar
systems, but that does not seem to be what your projections operators
are about.
The average reader should remember that, at least if they're scientist.
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.
A discrete partitioning of the "Nothing" will still involve infinite
sized ensembles. Just in this case, there is not way of disciminating
them - with this observable at least.
I think this is revealing. What you are really saying is that a
measurement does not project out subsets from the observer moment, it
produces a complete set of new observer moments, \P_{a} maps the
ensemble of observer moments onto itself, with one new observer moment
for each possible measurement outcome a. That could have been made
clearer. But it also needs justification because it is not implicit in
the notion of an observation, unless you assume QM and MWI from the outset.
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.
It is the passage 1p -> 1p plural -> 3p, where 3p is all observer
moments, not any single one. That could be expressed more clearly.
That is where I must object. 3p, or 0p as I would prefer to refer to the
bird view, is not an observer moment because there is no such observer.
If we take a simple example, where observer moments are, say, the sets
{1,2,3}, {4,5,6}, and {7,8,9}, then the set of all observer moments, a
set of sets, each being one of these sets, does not contain the union of
the first two observer moments, which is the set {1,2,3,4,5,6}. So the
sum or union of two observer moments is not necessarily another observer
moment, in fact, I doubt that it ever could be; 3p is not an observer
moment.
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.
The set of outcomes of any observable X will be the "certain
event". The projections over these sets of outcomes will just be the
identity. Adding the projection over the certain event for two
observable A & B will just be twice the projection over the certain
event for observable C.
However, I suspect this is not what you're alluding to. "But that
condition is not met by just any two arbitrary operators." My
knowledge of functional analysis is some 3 decades old, but what I
remember is that this is only possible for unbounded linear
operators. All bounded operators have a set of eigenfunctions that span
the whole linear space, ie is complete (and independent, obviously).
Now it is true that classical quantum theory uses unbounded operators
with gay abandon, much to the horror of mathematicians, who tend to be
rather more careful about such things. x and ∂/∂x being classic examples.
My own attitude is the x and ∂/∂x are not actually physical
observables, as they correspond to observing position and momentum to
infinite precision. When you modify the operators to take finite
measurement resolution into account, you end up back with a bounded operator.
But maybe I'm missing your point.
I agree that I did not express myself very clearly in the above. I was
just thinking in terms of what happens in standard QM, where an
observable is a Hermitian operator in a linear vector (Hilbert) space
(or "rigged Hilbert space" for unbounded operators like x and p). Then,
you have explicitly assumed linearity, so linear combinations of a set
of vectors in this space is also a vector in the space (think vectors in
Euclidean space). But that assumes linearity -- and you have not
established linearity by summing observer moments.
I understand that in your theory, an observer moment must be shown
to be a Hilbert space,
^ a vector in a Hilbert space
Right.
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'm guessing you mean tensor product, or cartesian product here. The
vector product a × b is a 3D vector at right angles to a and b, with
magnitude |a||b|sin(θ).
Yes, tensor product (or outer product). The correct term had escaped me
for the moment!
I haven't established this, because it is not needed to make contact
with the regular set of axioms assumed in quantum theory. But to the
extent it can be shown within regular QM, it will be shown within my theory.
But it is not a derived result in standard QM -- it is assumed as part
of the postulates. The linear spaces for each quantum operator are
disjoint -- spin operators act in a different space from that for
position operators -- so the combined Hilbert space must be a tensor
product of distinct Hilbert spaces. You would have to prove this from
your starting point if you want to make contact with quantum mechanics.
Bruce
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to everything-list+unsubscr...@googlegroups.com.
To post to this group, send email to everything-list@googlegroups.com.
Visit this group at https://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/d/optout.
