Subject: Re: extra baggage?
On Sat, 18 Mar 2000, Stanley Klein wrote:
> [Henry]
> But if one does not know which outcome occurs,
> or wants to determine WHAT QUANTUM THEORY SAYS
> about what will happen, then the correct description
> of the state is PSP + (1-P)S(1-P).
>
> [Stan]
> Maybe I'm being dense here but I still don't get it.
> I fully appreciate that the critical step is the
> von Neumann choice where the interference terms are
> eliminated. That is what converts the cat from being
> dead AND alive to dead OR alive. And that is the step
> that does the QZE.
> That is well understood.
>
> What I don't get is why don't you do the trimming
> right away so that you
> avoid the splitting universe extra baggage that
> is reminiscent of Everrett.
> I am aware that having PSP OR (1-P)S(1-P) isn't
> that bad since that is what we have classically.
> But
> it still seems cleaner to me to do the Dirac choice at
> the same time as the von Neumann choice. You even said
> that you think of it happening right away.
>
First, let me mention the historical situation.
There the experimenter sets up the experiment that would
in, due course, cause the apparatus to separate into
the two superposed state, one of which would be experienced
by the human observer. Exactly what happens in the brain that
corresponds to this is not known. But I have for mathematical
neatness collapsed the two choices so that they occur in
tandem.
von Neumann did not speak of the second choice, and Wigner
did not go into this detail.
But the main point is that quantum theory does not specify
the individual outcomes. So if on wants to compute only
that which quantum theory specifies, then one will not include
the final Dirac choice. The full content of what quantum theory
predicts is given by the transformation
S-->PSP + (1-P)S(1-P). This contains all the information
that quantum theory provides.
Let me elaborate upon this.
Suppose one says that we also know that one OR the other
outcome will occur. The we could say that the state is
either PSP or (1-P)S(1-P). But we do not know which.
But we do know something else! We DO know the probability
of each of the two possibilities. So if we include that
information we will be adding something else that quantum
theory tells us. To incorporate this extra knowledge
provided by quantum theory we should use for our computation
the properly weighted sum of computations for the two
individual alternative possibilities. This we should use
for computations that incorporate ALL that quantun theory
can tell us is the normalized state
S/TrS =(PSP/TrPSP)(TrPSP/Tr S)
+ ((1-P)S(1-P)/Tr(1-P)S(1-P))(Tr(1-P)S(1-P)/TrS).
The first factor, (PSP/TrPSP), is the normalized state that
would be used if we knew that the state was PSP, and the
second factor, (TrPSP/TrS), is the probability that the
outcome of question P is Yes. The second term is analogous.
But then cancellations yield
S= PSP + (1-P)S(1-P).
So to get ALL the information that quantum theory supplies
one should simply replace S by PSP+(1-P)S(1-P) each time
the question P is posed and answered. Quantum theory
does not tell us whether the answer is P=1 or P=0, so that
information is not included.
This is the reason that von Neumann did not even mention
the final Dirac choice.
> So let me ask it again. Is there any problem (other than
> possible
> mathematical inelegance) with having the Dirac choice
> happen simultaneous
> with the vN choice? I don't think you answered that.
>
I do assume that the Dirac choice happens immediately
after the vN choice: I assume that the two always appear
in tandem. But what enters into the computation is
only the vN process S-->PSP +(1-P)S(1-P).
This process is determined by, and determines, the
choice of P. This determination is not dependent upon
whether the Dirac choice is Yes or No.
I suppose one could say that the Dirac choice
chooses both the question and the answer. But I believe
that that is not the right way to look at it. The reason
is that the von Neumann process S --> PSP + (1-P)S(1-P)
is specified by just fixing question, not the
outcome. But fixing the question is a local operation:
it corresponds to the experimenter in a region
fixing his local apparatus in some particular way.
But the choice of the answer has global effects.
It is central to my entire endeavor to distinguish
what is under the control of the local processor and
what is global. I do this by associating the von Neumann
process with the local processor, but the Dirac choice
with some global process.
> I have another quibble over language. You call the vN
> choice the 'posing of
> the question'. That seems pretty benign. But isn't the
> vN choice the
> critical one in actually doing the measurement. Isn't the
> most important
> and critical aspect of measurement the erasing of the
> cross terms. The
> words 'posing the question' seems too mild for what is
> happening. But that
> is just a semantic problem and not a fundamental one at all.
>
The von Neumann process is indeed critical. It corresponds
to the choosing by the experimenter of the experimental
set up. This is often described as choosing what aspect
of the probed system is going to be probed: choosing what
question is going to be asked. This is normal terminology,
and it accurately captures what is going on,
since the two possible outcomes, P=1 or P=0 do correspond
to the two possible outcomes to the query: Does the system
have the property charactized by P? Choosing P chooses
the question that is being posed.
So this is not just a matter of arbitrary choice of
terminology: the chosen terminology is both the one
normally used, and the one that accurately portrays
what is going on.
Henry
From stapp@thsrv.lbl.gov Sun Mar 19 14:23:19 2000
Date: Sun, 19 Mar 2000 14:12:56 -0800 (PST)
From: stapp@thsrv.lbl.gov
Subject: Re: Epistomology vs ontology
On Sat, 18 Mar 2000, Stanley Klein wrote:
> At 5:53 PM -0800 3/18/00, Stanley Klein wrote:
> >[Henry]
> >But if one does not know which outcome occurs,
> >or wants to determine WHAT QUANTUM THEORY SAYS
> >about what will happen, then the correct description
> >of the state is PSP + (1-P)S(1-P).
> >
>
> Let me try again with a little fancier language.
> It seems to me that the measurement process (or the conscious awareness
> process) reduces the ontological state to one of the two possibilities. I
> would think this is the reduction of S to either PSP or to (1-P)S(1-P).
> However, from an epistomological point of view the observer might not know
> which state was selected so it makes sense to keep the density matrix with
> both possibilities: PSP + (1-P)S(1-P).
>
> Isn't the following a possible interpretation of what's going on?
>
> Ontology: PSP OR (1-P)S(1-P)
> Epistomology: PSP + (1-P)S(1-P)
>
> Stan
>
Yes. But what is "known" to one processor, because
its "own" knowing has registered a P=1, may be unknown to
another processor. So the formalism accomodates nicely
the distinction between subjective knowings and objective
knowledge, the later being the total knowledge of all the
processor, up to the present. This is what the state S(t)
represents.
Predictions are computed from the formula
= Tr PS(t)/Tr S(t),
where
is the absolute or objective probability
that the answer to the question P asked at time t will be Yes.
But no individual processor "knows" the absolute or objective
state S(t), which could be written
S(t)=P(t_n)U(t_n,t_(n-1))P(t_(n-1)...P(t_1)S(0)P(t_1)...P(t_n).
where P(t_i) is the question answered Yes at time t_i,
and
U(t_i,t_j) = exp(-iH(t_i - t_j)).
If some processor knows only part of the information
needed to construct S(t) then he must in principle
ADD the contributions corresponding to the various
possibilities compatible with what he knows. Thus if
he knows that question P(t_i) was posed at time t_i,
but does not know that the answer was Yes, then he must
add to the actual S(t) the similar expression with
P(t_i) replaced by (1-P(t_i)). This gives a personal
S(t) that represents his own knowledge. It can be used
in place of the objective S(t) in the formula for
to give the probability that is entailed by
his knowledge.
Of course, no processor knows even all the questions
that have been posed over the course of history.
But one can make prediction about small subsystems
that have been prepared in a known way, and are known to
be isolated from interacions with the rest of nature
between preparation and subsequent measurement. In
such a case the space can be divided into the part built
on the degrees of freedom of the isolated system and
the rest, so that the matrix elements of S(t) are
. After preparation the system and its
environment evolve independently up until the measurements.
von Neumann's theory explains how information about
the preparation and the measurements are conveyed into the
brain of the observer. If P_0 is the image in the space of the
system of the knowledge of the preparation, and P is the image
in the space of the system of the knowledge corresponding to
a possible outcome then the probability of P given P_0 and
a state S(t) at the time of the preparation is
= Tr PU(t',t)P_0 S(t) P_0 U(t,t')P/Tr P_0 S(t),
where U(t',t) =exp -iH(t'-t) is the inverse of U(t,t').
The definition of Tr X gives
Tr X = Sum over all s, Sum over all r of .
The U(t',t) and U(t,t') break into a product of
two independent factors,
= ,
due to the isolation of the system during the interval
in question.
Then the fact that P and P_0 act as unit operators
in the r variables,
entails that the r-dependent factors separate out,
and cancel out of the quotient,
leaving
=
Tr_(s) PU'(t',t)P_0 S_s P_0 U'(t,t') P/Tr_(s) P_0 S_s
where Tr_(s) means trace over the s variables, and
S_s = Tr_(r) S = Tr_s S,
[according to our notation, which makes Tr_s the trace
over all variables "other" than those associated with
the system, which, in an abuse of language I call s:
I use s also to denote the indices that label the basis
vectors of the system s).
In an ideal preparation the initial state of s is completely
fixed by P_0: S_s(t) = a multiple of the identity. Then
= Eq. 1
Tr_s P U'(t',t) P_0 U(t,t') /Tr_s P_0,
where I have repeatedly used PP=P, U(t',t)U(t,t')=1, and
Tr_(s) AB = Tr_(s) BA.
Eq. 1 is the Copenhagen formula.
Suppose P_1 is measured in region 1, and P_2 is
measure at the same time in the spacelike separated
region 2.
This means P=(P_1)(P_2), and (P_1)(P_2)= (P_2)(P_1).
Then the general formula gives
= Tr_(s) P_1 P_2 U(t',t) P_0 U(t,t')/ Tr_(s) P_0.
Similarly,
= Tr_(s) P_1 (1-P_2) U(t',t) P_0 U(t,t')/ Tr_(s) P_0.
If it is not known whether the outcome in region 2
is P_2 or (1-P_2) then one must add the two probabilities
to get the probability that P_1 is 1. This gives
= Tr_(s) P_1 U(t', t)P_0 U(t',t)/ Tr_(s) P_0,
which is just the formula for the case in which no
measurement is made in region 2.
It also shows that the probability for P_1=1
does not depend on which measurement, (P_2) or
some other (P_2)', is performed in region 2,
if one does not know the outcome.
This is the basis for saying that no SIGNAL
can be transmitted faster than light: the probability
for P_1 =1 (or for P_1=0) does not depend on which
measurment is performed in the other region, or on whether
any measurement at all is performed there.
I have included here these more technical details so that
the interested readers can actually see how the theory works.
One sees, in particular, the close connection between
ontology and epistemology: the ontological structure
really seems to be, insofar is it is captured by quantum
theory, a theory of knowledge: the objective "physical"
structure, S(t), represent objective knowledge, and
represents it in a way that naturally accomodates also
subjective knowledge.
Henry
From stapp@thsrv.lbl.gov Sun Mar 19 16:23:53 2000
Date: Sun, 19 Mar 2000 16:09:09 -0800 (PST)
From: stapp@thsrv.lbl.gov
Reply-To: hpstapp@lbl.gov
Subject: Re: epistemology vs ontology
On Sun, 19 Mar 2000, Stanley Klein wrote:
> [Henry]
> First, let me mention the historical situation.
> There the experimenter set up the experiment that would
> in, due course, cause the apparatus to separate into
> the two superposed state, one of which would be experienced by the human
> observer.
>
> [Stan]
> 1) Ahh, I think we are getting close to clarifying my
> confusion. Your use
> of the words 'in due course' is critical. Are you
> saying that setting up
> the orientation of the polarizers ('posing the question')
> is really
> eliminating the cross terms, way before the particle ever
> comes near the
> polarizers?
No! I have specified that the posing and answering
are in tandem as different limits to a common time t.
I mentioned the historical situation only to bring in
the idea that the choice as to which question will be
asked is a local question: it is basically the
experimenter/processor's choice as to which question
will be asked: which aspect will be probed. We are
considering now going over to the limit in which
the processor is both the apparatus, the system being
probed, and the memory device that stores the record
of the outcome given to it by nature. But I retain the
idea from the Copenhagen interpretation that the
experimenter (Processor) decides what question will be
asked, and stores the answer that then comes immediately.
>
> 2) I'm looking forward to your comments on my distinction
> between ontology
> and epistemology. That is, the combined vN/Dirac choice
> is what is actually
> happening in that nature chooses one or the other
> (ontology). But we might
> not know which choice it was (epistemology). In terms of
> normalized states:
>
> Epistemology: S -> PSP + (1-P)S(1-P)
>
> Ontology: S* -> PSP* with probability TrPSP/Tr S
> or S* -> (1-P)S(1-P)* with probability Tr(1-P)S(1-P)/TrS
>
> where S* = S/Tr S.
>
>
>
> 3) You say that eliminating the cross terms (posing
> the question) is a
> local operation. I agree that the experiment itself is
> local. But is the
> outcome of the vN choice really local? Doesn't the
> elimination of the cross
> terms have global impact? If eliminating the cross terms
> has no global
> implication then I have indeed learned something quite
> i0mportant and this
> has been very educational. If the Dirac choice is the
> only step that does
> the nonlocal transfer of information then my point 2 about
> epistemology and
> ontology is plain wrong.
>
In my answer to "epistemology and ontology"
I showed that just posing the question
(i.e., S--> PSP + (1-P)S(1-P))
does not affect expectation values in regions
spacelike separated from the region where the
question was posed. (Actually, I showed it only
for the case in which the two measurements were
simultaneous, but in the relativistic version
that I am considering that conclusion holds
insofar as the two are spacelike separated.)
It is only when the answer is taken into
account that the far-away expectation
value is affected.
This is the no-faster-than-light signalling
result. It is, of course, completely to be expected
that what the far-away experimenter freely chooses to
measure, and then does measure, cannot affect what
the theory predicts about what happens here, if no
account is taken of the outcome of that faraway
measurement. This non-dependence upon the
vN process itself is what makes reasonable the notion
that the vN process is a local process. And this process
both determines what the question P is, and is fixed
by the fixing of the question P.
Henry
From stapp@thsrv.lbl.gov Mon Mar 20 10:04:56 2000
Date: Sun, 19 Mar 2000 18:02:50 -0800 (PST)
From: stapp@thsrv.lbl.gov
Reply-To: hpstapp@lbl.gov
Subject: Re: epistemology vs ontology (fwd)
Dear Stan,
Let me get right to the point.
I break the jump process into two steps:
1) The von Neumann process I,
S-->PSP +(1-P)S(1-P),
which I call "posing the question",
followed immediately by
2) Picking the answer PSP or (1-P)S(1-P).
The first step entails that there will immediately be a jump
to either PSP or (1-P)S(1-P), but it does not specify which.
The first is a local process, for if P_1 and P_2
act locally in two spatially separated regions
then the expectation value of P_1 is unaffected by
the process S--> [P_2 S P_2 + (1-P_2) S (1-P_2)]:
= Tr P_1 [(P_2)S(P_2) + (1-P_2)S(1-P_2)]
/Tr [(P_2)S(P_2) + (1-P_2)S(1-P_2)]
= Tr P_1/Tr S
= .
The middle step is proved immediately by just using, repeatedly,
PP=P, and Tr XY =Tr YX , for any (bounded) X and Y,
and P_1 P_2 = P_2 P_1, which is a consequence of the
space-like separation.
I do not want this simple basic point to become
obscured by the discussion.
Henry
From stapp@thsrv.lbl.gov Tue Mar 21 07:56:19 2000
Date: Tue, 21 Mar 2000 06:15:21 -0800 (PST)
From: stapp@thsrv.lbl.gov
Reply-To: hpstapp@lbl.gov
Subject: vN process summary
I put down in writing many things during this
discussion with Stan, and I will not try to
mention all of them. But let me briefly summarize
the situation as regards the vN process I (the processors
choice)and the Dirac process (nature's choice).
In a sense all that happens ontologically is the
sequence of quantum jumps S(t+0)=P(t)S(t-O)P(t) at
a sequence of times. But vN never mentioned these
jumps. He mentioned only the jumps S-->PSP + (1-P)S(1-P),
which he called process I, and which I call vN process I,
and process II, which is the unitary evolution between jumps.
Indeed, it is not wholly clear that vN really was making
an ontology: he may have gone along, at least in part,
with the Copenhagen philosophy that the objective was to
provide a theory of computation. But he also made a lot
of statements that can be construed as supporting an
ontological stance. Wigner seemed a little more clearly
ontological.
But since vN talked only about the vN process, not the
Dirac process I felt I should, to be true to him, bring
it in. Wigner added to the von Neumann approach the
further selection of one alternative or the other .
I already mentioned that von Neumann's likely reason for
bringing in only the vN jump, and not the final selection,
was that he considered the theory to be a statistical
theory that makes only statistical prediction: it makes
no prediction about individual outcomes. (Well, sometimes
the statistical prediction happens to have probability one,
but this is still a statistical prediction, strictly speaking).
I made good use of von Neumann's process I.
Because this process is local, in the sense that
it does not affect the predicted probabilities
for outcomes of far away experiments, I could
associate it with the processor, and allow just
the final "Dirac" choice "on the part of nature"
to contain the global effect.
Since S--> PSP + (1-P)S(1-P) does define P, and P
is associated with "the question" that has two possible
answers, P=1 and P=0, I called this vN process
"posing the question" or "asking the question"
or "choosing the question". It can be construed as the
image in the physical world of a choice made by the
processor. Stan wanted to say that the choice of P
could be considered as different from the associated
vN process I, which is, ontologically, a change in the
physical state, not just a mere choosing of a question
that would be put to nature. That is a possible point
of view. But I am trying to keep things as physical
as possible, and so preferred to give a physical
representation of this choice, not a more nebulous
choice that has no effect until the final jump to
PSP or (1-P)S(1-P). Representing the choice in the
physical way, as the vN process I allowed me to keep
things physical, stay true to von Neumann, and identify
a process that could be attributed to the processor.
This step S-->PSP + (1-P)S(1-P) poses a definite
yes-no question: it allows nature to choose between
two alternative possibilities, where these alternative
possibilities have been specified by a local processor.
Henry