Re: Bras, kets etc.


"Timo A. Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote in message
news:Pine.WNT.4.64.0609020622380.1588@xxxxxxxxxxxx
On Fri, 1 Sep 2006, Ron Baker, Pluralitas! wrote:

"Timo Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote:
On Thu, 31 Aug 2006, Ron Baker, Pluralitas! wrote:
"Timo A. Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote:

If the basis
vectors correspond to measurements you might make, the probability of
outcomes of the measurements.

How does a measurement translate to a basis?

If the measurements correspond to a mutually exclusive complete set of
properties, then the possible measurement results _are_ an orthogonal
basis for the description of the possible states of the system.

I get the concept but I'm not ready to buy in yet.
Seems to me we might be getting to the crux of the
matter.
I'm thinking of Bell's inequality and the Aspect experiment.
I've read several descriptions of Bell's inequality. They
all describe a logical tautology based on something
that seems equivalent to your "orthogonal 'measurement'
basis". Then they say that QM predicts something different
and totally gloss over the QM prediction.

"Measurement" is the crux.

I've listened to the Feynman lectures. What he describes
is totally (forgive my California accent) classical waves
except for "measurement".

I am definitely not ready to accept that one can
simply posit an orthogonal basis and say it is
a 'measurement'.

"Classical waves except for measurement" is a pretty good description of
QM. "Classical waves + annihilation/creation operators" for QED.

The relationship between basis and measurement is more the other way
around:

Even so.

a "good" measurement - with complete and mutually exclusive results -
tells you a useful basis.

Yes, even so. When I typed in the previous reply
I was also thinking:
"Conversely I am not ready to accept that
every measurement establishes a valid underlying
orthogonal basis."



It's very natural for us to think about specifying a position as (x,y,z)
coordinates because this is what we _see_. Years of measurement (ie visual
observation) tell us this is a useful basis (although untrained, we're
more likely to express it as "r far away, in that direction (pointing)").

This is the same whether you are talking about x y z Euclidean basis
vectors, a j_n Y_nm basis for the Helmholtz equation, an |H> |V> basis
for
polarisation, or even the behaviour of children with autism (eg see my
consistency in behaviour classification behaviour).

(Is that a paper you wrote? Can you post a link?)

Conference paper. Html and pdf versions at:
http://www.aare.edu.au/02pap/cho02101.htm
http://eprint.uq.edu.au/archive/00000917/

:) Interesting. Well you are a renaisance man. :)
But I didn't see orthogonal basis in that. But
I did see set partitioning which is similar.


The point is: suppose your measurement can distinguish between 3 discrete
states, |1>, |2>, |3>, and the system is always found to be in one of
these states when measured, and only one of these states. That means that
they form a complete orthogonal basis. As far as your measurement is
concerned, you can describe a state as S = a|1> + b|2> + c|3>. Yes, there
might be other degrees of freedom, but since they're inaccessible to your
measurement, they're irrelevant if you want to know the probabilities of
the possible outcomes of your measurement.

A QM example. Consider an electron. Consider detecting its position.
S(x,y,z) is sufficient to tell you the probabilities. Alas, I have to
write S as a function of (x,y,z) rather than a sum over basis vectors,
because the position basis is continuous. Anyway, this tells you _nothing_
about S(px,py,pz), the momentum state of the electron. It also tells you
nothing about its spin. But if you're only measuring the position, so
what?

But I don't yet see a solid link between a particular
PMT click and the spherical basis to the HE.


If the measurement describes the system,

But measurements can't completely describe a system.
HUP.

then it can be used as a basis.

No. ...? It can't be complete. HUP.

(Are we getting into "a priori" vs "a posteriori" here?)

No. Just extra degrees of freedom that we ignore. If we measure position
(eg of an electron) exactly, we have the exact description of the
position. We have no information about the momentum, but if we wanted
that, we wouldn't measure the position exactly.

So a measurement of one property cannot be a
complete description of a system.

A measurement might have a basis but the measurement
basis is not equivalent to the system basis.
The measurement basis is at best be a lower dimensional projection
of the original system basis, isn't it?



What if you want information about both? Can we write a sensible basis?
Yes. Consider a laser beam. We want information about both position as the
photon crosses the focal plane (yes, this might be commiting a
"particle-photon" sin, but ...)

:)

and the momentum. Answer: Hermite-Gauss laser beam modes. Each basis mode
has a certain intensity distribution in the focal plane, and a certain
far-field intensity distribution (the relationship between these is
related to the momentum).

Compare with the <a|B|c> triple integral.

Actually I'd say I'm having trouble reading your
notation there. That is an integral? What are
the commas?

If a and b are basis functions, then the integral is something like

<a|B|c> = integral a*(x2) B(x2,x1) c(x1) dx2.

B is a transformation from c(x1) to (Bc)(x2). [Warning! This is
off-the-cuff without looking anything up. Perhaps wrong!]

Yeah, I think I can see an inconsistancy.
The result should be a scalar, right? If you only integrate
with dx2 you get a function of x1.
Shouldn't it be something like:
<a|B|c> = integral( a*(u) B(u) c(u) )du

Try:
<a|B|c> = integral { a*(x2) integral [ B(x2,x1) c(x1) ] dx1 } dx2.

Recall that if B and c are a transformation matrix and a vector, then the
product Bc is equivalent to the integral above. B(x2,x1) could, for
example, be a Green function.

But this depends on whether you have B(x2,x1) or B(u); in the latter case,
yes, <a|B|c> = integral( a*(u) B(u) c(u) )du.

Which of these (either/both/neither) correspond to the usual QM cases?
Classically, both kinds of integrals turn up. As you say, the scalar
product must be a scalar, so if you have B(x2,x1), you get the former. The
matrix-vector equivalent is (row vector)(matrix)(column vector). What is
the same for the latter integral? (row)(scalar)(column)?

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html


--
rb


.

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