Dear Friends:

In some earlier threads, I have stated that I was seeking to draw a
connection between the Heisenberg uncertainty relation and the magnetic
moment anomaly first predicted with precision by Schwinger.

In a brand new section three of a draft paper posted at:

http://jayryablon.files.wordpress.co...-schwinger.pdf

I have written up the details of how to draw this connection. Sections
1 and 2 are substantially the same as in previous posts regarding the
operator decomposition of the intrinsic spin (helicity) operators.
Section 3 resoundingly answers the question "OK, so the math is neat and
looks correct, what new physics can you do with this decomposition?"

I have also posted a new entry about this at my weblog, linked below.

Looking forward to your comments, and also, to an answer on the question
I raised in the new thread "Trying to straighten out factors of i in
Heisenberg relationships," which is the one aspect of this calculation
in the new section 3 that I want to make sure is straightened out.

Best regards,

Jay.

PS: I had some trouble opening this file with the link above, but it
opens fine if you go to the Weblog link below, and then click the link
for this file. So, if the link above does not work, try that.
____________________________
Jay R. Yablon
Email:
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm

On Apr 24, 7:48*am, "Jay R. Yablon" wrote:
Dear Friends:

In some earlier threads, I have stated that I was seeking to draw a
connection between the Heisenberg uncertainty relation and the magnetic
moment anomaly first predicted with precision by Schwinger.

In a brand new section three of a draft paper posted at:

http://jayryablon.files.wordpress.co...g-and-schwinge...

I have written up the details of how to draw this connection. *Sections
1 and 2 are substantially the same as in previous posts regarding the
operator decomposition of the intrinsic spin (helicity) operators.
Section 3 resoundingly answers the question "OK, so the math is neat and
looks correct, what new physics can you do with this decomposition?"

I have also posted a new entry about this at my weblog, linked below.

Looking forward to your comments, and also, to an answer on the question
I raised in the new thread "Trying to straighten out factors of i in
Heisenberg relationships," which is the one aspect of this calculation
in the new section 3 that I want to make sure is straightened out.

Best regards,

Jay.

PS: *I had some trouble opening this file with the link above, but it
opens fine if you go to the Weblog link below, and then click the link
for this file. *So, if the link above does not work, try that.
____________________________
Jay R. Yablon
Email:
co-moderator: sci.physics.foundations
Weblog:http://jayryablon.wordpress.com/
Web Site:http://home.nycap.rr.com/jry/FermionMass.htm

You can not pull the gamma_0 and \sigma matrices out of the
expectation value as you did Eq. 3.6. They are operators on the wave
function.

kp

"kp" wrote in message
...
On Apr 24, 7:48 am, "Jay R. Yablon" wrote:
Dear Friends:

In some earlier threads, I have stated that I was seeking to draw a
connection between the Heisenberg uncertainty relation and the
magnetic
moment anomaly first predicted with precision by Schwinger.

In a brand new section three of a draft paper posted at:

http://jayryablon.files.wordpress.co...g-and-schwinge...

I have written up the details of how to draw this connection. Sections
1 and 2 are substantially the same as in previous posts regarding the
operator decomposition of the intrinsic spin (helicity) operators.
Section 3 resoundingly answers the question "OK, so the math is neat
and
looks correct, what new physics can you do with this decomposition?"

I have also posted a new entry about this at my weblog, linked below.

Looking forward to your comments, and also, to an answer on the
question
I raised in the new thread "Trying to straighten out factors of i in
Heisenberg relationships," which is the one aspect of this calculation
in the new section 3 that I want to make sure is straightened out.

Best regards,

Jay.

PS: I had some trouble opening this file with the link above, but it
opens fine if you go to the Weblog link below, and then click the link
for this file. So, if the link above does not work, try that.
____________________________
Jay R. Yablon
Email:
co-moderator: sci.physics.foundations
Weblog:http://jayryablon.wordpress.com/
Web Site:http://home.nycap.rr.com/jry/FermionMass.htm

You can not pull the gamma_0 and \sigma matrices out of the
expectation value as you did Eq. 3.6. They are operators on the wave
function.

[Yablon]
I won't say your are right, and I won't say you are wrong. I'll ask:
why?

Specifically, in 3.3 we are certainly operating on a wavefunction. But
in (3.4), we pull the "meat" out of the "sandwich" and now obtain an
equation solely among "non-operating" operators. Going back to Ohanian
at
http://jayryablon.wordpress.com/file...at-is-spin.pdf,
see equation (18) and how he then gets to (19).

My (3.4) then goes to (3.5), which is still an operator equation where
the operators have been removed from any operation on a wavefunction and
now assume their "expected values." So, if these are constant
operators, why can I not take Kx = Kx to get to (3.6). Isn't this
essentially the same as what Ohanian does going from(18) to (19)?

Thanks,

Jay.

kp

On Apr 24, 6:25*pm, "Jay R. Yablon" wrote:
"kp" wrote in message

...
On Apr 24, 7:48 am, "Jay R. Yablon" wrote:



Dear Friends:

In some earlier threads, I have stated that I was seeking to draw a
connection between the Heisenberg uncertainty relation and the
magnetic
moment anomaly first predicted with precision by Schwinger.

In a brand new section three of a draft paper posted at:

http://jayryablon.files.wordpress.co...g-and-schwinge...

I have written up the details of how to draw this connection. Sections
1 and 2 are substantially the same as in previous posts regarding the
operator decomposition of the intrinsic spin (helicity) operators.
Section 3 resoundingly answers the question "OK, so the math is neat
and
looks correct, what new physics can you do with this decomposition?"

I have also posted a new entry about this at my weblog, linked below.

Looking forward to your comments, and also, to an answer on the
question
I raised in the new thread "Trying to straighten out factors of i in
Heisenberg relationships," which is the one aspect of this calculation
in the new section 3 that I want to make sure is straightened out.

Best regards,

Jay.

PS: I had some trouble opening this file with the link above, but it
opens fine if you go to the Weblog link below, and then click the link
for this file. So, if the link above does not work, try that.
____________________________
Jay R. Yablon
Email:
co-moderator: sci.physics.foundations
Weblog:http://jayryablon.wordpress.com/
Web Site:http://home.nycap.rr.com/jry/FermionMass.htm

You can not pull the gamma_0 and \sigma matrices out of the
expectation value as you did Eq. 3.6. They are operators on the wave
function.

[Yablon]
I won't say your are right, and I won't say you are wrong. *I'll ask:
why?

Here is a simple example. It seems you want to write the expectation
value of sigma_z as

\sigma_z=\sigma_z1=\sigma_z

where the 1 is the identity. This is obviously wrong the expectation
value is a single real number and we have that equal to a matrix
(\sigma_z) now.

When one is dealing with spin the wave functions are no longer simple
scaler functions they are spinors, i.e. vectors with complex
componets. Now to take the expectation value of \sigma_z one has to
perform matrix/vector multiplication.

kp

"kp" wrote in message
...

Take the 2x2 matrix:

/ 5i 3 \
M = | |
\ -2 7 /

Isn't M = M? We are not asking about eigenvalues, we are asking
about
expected values.


This is covered in any standard quantum mechanics text so I'll just
touch on it. Given an arbitrary state \psi and observable O, with a
set of eigenvalues. Any single measurement of O with respect to \psi
will give some eigenvalue of O. Unless \psi happens to be an
eigenvector of O another measurement of an identically prepared \psi
with give (likely) some other eigenvalue of O. The expectation value
of O with respect to \psi, given as \psi|O|\psi gives a weighted
average of all eigenvalues or in other words the most likely
measurement of O given a state \psi. Thus the expectation value of an
operator has to be a single real number.

kp

[Yablon]

Please take a look at:

http://en.wikipedia.org/wiki/Expecte...on_of_matrices

"the expected value of the matrix is defined as the matrix of expected
values."

Jay.

[Yablon]

Please take a look at:

http://en.wikipedia.org/wiki/Expecte...on_of_matrices

"the expected value of the matrix is defined as the matrix of expected
values."

Jay.

As far as proprobiltiy theory goes you can define something like this,
but it has nothing to do with quantum mechanics.

I'm not a huge fan of wikipedia but look at the quantum mechanics page
on this

http://en.wikipedia.org/wiki/Expecta...ntum_mechanics)

the very last sentence states

"Not all operators in general provide a measureable value. An operator
that has a pure real expectation value is called an observable and its
value can be directly measured in experiment."

if this wasn't true you could never measure spin.

kp