On 6/07/2017 2:33 pm, Russell Standish wrote:
Sorry for having gone dark, although maybe you relished the
respite. I've been travelling, and its not been all that convenient to
check and respond to emails.
On Sat, Jul 01, 2017 at 02:56:58PM +1000, Bruce Kellett wrote:
On 1/07/2017 11:18 am, Russell Standish wrote:
I'm not asking for your sympathy. Actually, it is helpful if you
aren't sympathetic, as you're more likely to find a critical flaw.
Obviously, you need to be motivated enough to poke around in it, though.
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.
Establishing linearity is key.
Yes, and you haven't made progress with that.
Establishing the resultant vector space
is a Hilbert space does follow fairly easily from Kolmogorov's axioms
(although its possible you have beef with those :).
I think the issue I would have here is that you assume that a projection
can give a range of results. I don't think that is necessarily true even
in a multiverse. Since projections are as undefined as everything else
here, it could be the case that a projection gave a single value -- the
same value in every world of the multiverse, so that you had a version
of classical mechanics. Assuming a range of different values, assigning
probabilities to individual outcomes makes some sort of sense, but that
is assuming a lot of quantum mechanics at the start. Hence my worries
about circularity.
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.
Observations are projections (in the general sense, not necessarily
linear space sense). This is why I focus on the projection operator
form of QM.
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.
That is the case of a traditional observable - a full rank Hermitian
operator.
Going from a projection operator formalism to the traditional observer
formulation is mathematically quite trivial, however.
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.
S=X+P is definitely an observable, and corresponds to measuring the sum
of position and momentum crisply. You can work out the equivalent
Heisenberg uncertainty relations too
ΔsΔx ≥ ℏ and ΔsΔp ≥ ℏ
So if you measure with S, you will have uncertainty in both position
and momentum.
So it is not a quantum operator or quantum observable. These are defined
to have precise values: eigenvalues corresponding to the eigenfuntions
of the Hermitian operator. And the eigenfunctions span the corresponding
Hilbert space. My point here is that position space, spanned by the
eigenfunctions of the position operator, is 'dual' to momentum space,
spanned by the eigenfunctions of the momentum operator. These spaces are
related by the Fourier transform and each serves to give a distinct
complete representation of the underlying quantum state. (That is the
sense in which I refer to these operators and spaces as 'dual' -- it
makes no quantum sense to add them,)
In general, any bounded hermitian operator is an observable of some
sort (one can create a machine, albeit weirdly wonderful in a Heath
Robisonesque way, that will measure that particular quantity).
Certainly. If one has the Hilbert space for some quantum operator,
linear combinations of the eigenvector basis vectors that also span the
space will be the eigenvalues of some other hermitian operator in the
space. As you say, not all of the resultant operators correspond to
natural observables.
The sum of two bounded hermitian operators is also a bounded hermition
operator. In fact a linear combination with real coefficients will
too, but not necessarily complex coeficients (since the hermitian
property may not be preserved).
Yes, but my concern here is that this generality does not extend to
tensor products of Hilbert spaces, and the product space is what one has
in quantum mechanics -- projections in one space are not projections in
another of the component spaces. You have also to distinguish carefully
between compatible and incompatible observations -- commuting and
non-commuting observables (projections). I don't see how you could
establish this distinction in your approach without just assuming that
it exists.
All bets are off with unbounded operators, of course, but my attitude
is that unbounded operators are stictly unphysical, albeit sometimes
convenient for computational purposes.
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 don't believe so. What is the observer moment you experienced prior
to being born? It is not necessary for all OMs to have a predecessor.
Maybe your mother had some observer moments?
But this just highlights one of the main reasons that I am out of
sympathy with the approach based on observer moments. The history of
quantum mechanics over the last hundred or so years has been an attempt
to find an account of the theory that removes the observer from any
central role. One of the concerns here, of course, is that the universe
showed quantum behaviour long before there were any observers around --
long before any consciousness at all, in fact. So physics must
ultimately be independent of OMs. OMs might not even be quantum in
general -- we could have lived in a classical world -- we don't, but can
you swear that consciousness is irreducibly quantum? What about
conscious digital machines that could be entirely classical in
construction and operation?
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 is quite the inverse of that. You start with a
probability measure over the outcomes of an observation, and in order
to recover those probabilities from the projections, you construct an
inner product space such that the probabilities.
This might seem trivial and arbitrary, but there is a lemma related to
the Riesz representation theorem that gurantees the equivalence of
inner products provided certain properties are satisfied (which they
are in this case). This means that the inner product so generated is
the same as the inner product generated by a different observable. I
don't mention this in my presentations, but it seems a reasonable
thing that should be in the exposition. I can dig into that when I get
back to my functional analysis textbook next week.
Assuming that projections give a range of outcomes so that probabilities
are required -- that appears to beg the question.
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!
It seems pretty obvious to me that when you partition a set, and you
consider the union of all of those partitions, you get the original
set. Did you have a scenario in mind where this mightn't be the case?
The union of the partitions is not another observer moment -- an
observation of a result is not the whole set. No observer sees all
possibilities simultaneously. Unitarity is not generally preserved for
observer moments because they change discontinuously.
Bruce
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