Re: Bras, kets etc.


"Timo Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote in message
news:Pine.LNX.4.50.0608311641260.7790-100000@xxxxxxxxxxxx
On Thu, 31 Aug 2006, Ron Baker, Pluralitas! wrote:

"Timo A. Nieminen" <timo@xxxxxxxxxxxxxxxxx> wrote:
On Sun, 27 Aug 2006, Ron Baker, Pluralitas! wrote:

Consider the case when the <foo| and |bar> aren't the same basis sets.
For
example, when rotating. We have |x1>, |y1>, |z1> in the original axes,
and
|x2> etc in the new axes. In this case, what are the B_ab etc in terms
of
scalar products <x2|x1> etc?

In my previous work B was the 'direction cosine' matrix.
That would be when x1, y1, z1 are orthonormal and x2, y2, and
z2 are orthonormal and the difference is consequently
purely a rotation. (Euler and all, ya know.)

Yes, and for the connection to the present discussion, recall that
a.b = cos(theta) when |a| = |b| = 1.

Is that all there is to a QM operator? What does <a| _mean_?

Wouldn't <a| resolve, extract, or decompose the
the |a> component out of whatever is to the right of <a| ?

Pretty much. Basically, projections onto a basis vector. 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'.


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?)


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?)

Isn't this the origin of using x y z as a basis in Euclidean space? We
observe we something is, and we can measure these x y z values, and can
write the position as r = x|x> + y|y> + z|z> in bra-ket-ese.

But that is macroscopic and not QM.

(And I thank you for clarifying bra-ket notation for me.
A little while back I think I raised a midterm B to a final A by giving
a distilled description of Euler rotations.
If I could have tossed in bra-kets I might have gotten
an A+.)


Likewise, if what you measure is momentum (eg direction of travel) or
polarisation, then either can be a perfectly fine basis for the purposes
of QM, which is all about the results of measurements.

Forgive me for being a slow learner but I need some
further explaination.


What does this mean? Well, consider that a.b = a^T b is really only
true
for a special case: Euclidean space. In general, you have a metric
tensor
g, and a.b = a^T g b.

What is a metric tensor?

I'll email you some stuff. Be warned: it's beyond what most graduate
students are given (it isn't that hard, but it's usually neglected).

I received it.
(I wish I had nothing to do but study physics.)


[cut]

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


--
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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