On 10/07/2017 11:40 am, Russell Standish wrote:
On Fri, Jul 07, 2017 at 10:56:27AM +1000, Bruce Kellett wrote:
On 7/07/2017 10:40 am, Russell Standish wrote:
On Thu, Jul 06, 2017 at 10:22:40PM +1000, Bruce Kellett wrote:
No, position and momentum are dual in the sense I defined. The
observables are not compatible -- position and momentum are not
simultaneously observable.
I know what you mean by "dual", although the conventional term is
"complementary".
Yes, I was avoiding that term because it is sometimes controversial.
But complementarity is a feature of conjugate variables.
I didn't know the terminology was controversial. Why?
Well, the term originated with Bohr, I think, and there are some who
reject more or less everything Bohr did as part of their rejection of
the Copenhagen Interpretation. More recently, the term "Black Hole
Complementarity" has been invented by Susskind and others. They use this
as a way to escape from the contradictions inherent in their solution of
the BH information paradox. Their use of the term there has really
nothing to do with Bohr's notion of complementarity, and I think it
merely confusing.
Observing S=X+P does not imply simultaneously observing X and P.
As far as I can see, it does. It is not just something you construct
by measuring X then P (or vice versa).
Exactly, the combined error on S will be the sum of errors on X & P,
so there's no point in doing that.
X and P are operators in
different spaces, related by a Fourier transform. Unless you mean
measuring the conjugate variables on different members of an
ensemble of identically prepared states?
Prove that I can't observe S, or provide a reference to someone doing
so. It appears rather crucial to your critique.
Again, I do not accept the reversal of the burden of proof. X and P
are conjugate variables so they are not simultaneously measurable.
Yes - but failure of imagination is not proof of failure. You were
quite clear that X+P could not be measured independently of X or
P. Why? Might there not be something like a squeezed light scenario
that could do it?
Squeezed states are simply states that reach the limits of the spread in
conjugate variables consistent with the Uncertainly Principle -- you can
squeeze in one dimension at the expense of an increase in uncertainty in
the other dimension. I think it is up to you to show how this can be
used to give meaning to the operator X+P. My argument against this
possibility relies ultimately on complementarity (or the dual nature of
position and momentum spaces) -- the spaces are alternatives, not parts
of the same thing.
On your last point, this is not crucial to my critique of your
theory. Well, not unless you are going to rely on this in your proof
of linearity.
I don't, but clearly if not all linear combinations of observer
moments are observer moments in themselves, then that is an important
critique. If (X+P) is not a physical observable, then (\P_X+\P_P)ψ is
probably not an OM, and it makes it just that much more difficult for
me to demonstrate the emergence of QM. ITOH, if X+P is a physical
observable, then clearly (\P_X+\P_P)ψ is an observer moment.
There is a point about operators on observer moments that I mentioned
briefly before, but would like to raise in more detail again. Your
projection operator \P_A produces the set of possible outcomes of the
observation A, \P{a}*psi = psi_a, the set of possible outcomes of the
observation. My question was, do you consider it possible for these
alternative outcomes to interfere, or not? If they can interfere, then
you have assumed a pure quantum state. If they cannot interfere, then
you have a mixed state, which is just a range of disjoint possible
classical outcomes. In which case, the result of the observation is not
a single vector in some space.
It seems that you take these projection operators to produce pure
quantum states, since later (in the derivation of the Schrödinger
equation, for instance) you assume unitarity. It is only pure quantum
states that preserve unitarity -- mixed states are not reversible, hence
not the product of unitary evolution from a single predecessor state.
But pure states are uniquely quantum, so if your operators produce pure
states, you have already built a large amount of quantum mechanics into
your derivation.
Bruce
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