Eugene Stefanovich:
Bilge wrote:

For example, suppose H
has an eigenvector |\Psi with angular momentum wavefunctions in the basis
|j,j_z. Then, one unitary transform is U = \exp(-iJ.\theta), with \theta
being a rotatation around any axis, say the y-axis. The matrix elements of
the new hamiltonian, U^{-1}HU in the basis |j,j_z is the same as the
matrix elements of the old hamiltonian operating on the rotated states.

I agree.
This was my case 3. of how unitary transformations are used in quantum
mechanics.

It's not ``case 3.'' I explicitly stated that and nothing I wrote
implied any such thing. What I wrote came from exacltly what you
posted:

\Psi|H'|\Psi = \Psi|U^{-1}HU|\Psi

since the above is identically equivalent to,

\Psi|U^{-1}HU|\Psi = \Psi'|H|\Psi'

which is just, \Psi|H'|\Psi = \Psi'|H|\Psi' and is exactly what I
wrote. The only question is whether or not anything is diagonal.
Your ``case 3'' is,

\Psi|H|\Psi = \Psi|UU^{-1}HUU^{-1}|\Psi

= \Psi'|U^{-1}HU|\Psi'

= \Psi'|H'|\Psi'

Note the difference in where the primes are located in that expression.
It's the same expression I've been writing.

If transformation U is applied to both the Hamiltonian
H' = U^{-1}HU and to states (you mentioned "rotated states") then matrix
elements do not change. This is a trivial change of representation that
does not have any effect on physics.

That explicitly _not_ what I've done. See above.

If you remember, we started this discussion from my assertion that
"dressing transformation" of the Hamiltonian (in spite of being unitary)
does change physics, because this transformation is applied only to
the Hamiltonian

Yes, so what?

(and to the generator of boost transformations, which is not that
relevant for our present discussion).

Well, it is, because the current discussion is due to the same
misconception you've acquired by not keeping track of what operates
on what and the conditions under which your transformation is valid.

This transformation IS NOT applied to state vectors.

I didn't say that it was. See above.

Therefore, the transformed Hamiltonian H' has matrix elements (in the
old basis) different from those of the original Hamiltonian H.

Of course. But, _matrix_ elements are not the same as the eigenvalues of
the operator represented by that matrix. If \Psi diagnolizes H, then
obviously, U^{-1}HU does not unless [H,U] = 0. But, since you aren't
transforming the wavefumctions, \Psi is not an eigenvector of your
hamiltonian, H' and therefore H' is not observable in the basis you've
chosen. Why do you think the term ``matrix elements'' is used when
referring to measurements, rather than eigenvalues? The reason is that
|f|O|i|^2 is a real probability.

The eigenvalues of the two Hamiltonians H and H' are the same, but
eigenvectors are different.

If you aren't careful, at this rate, you might actually debunk yourself
and put an end to your own charade.

The time evolutions desribed by operators exp(iHt) and exp(iH't) are
also different, although the corresponding S-matrices are exactly the
same.

I really don't see how you manage to beat around the bush without ever
actually finding it. It just isn't conceivable unless you are so hung
up on your own non-sensical interpretation that you just refuse to
comprehend the meaning of your own words. If you look real carefully,
you'll discover the reason for the \theta(t' - t), etc.

That's all I wanted to say. If you like, I can put these statements
in the form Theorem/Proof to avoid any ambiguity.

No reason to do that. The problem has nothing to do with being able
to suffle unitary matrices around. It has to do with you trying to
interpret equivalent to mean bot equivalent and not equivalent, depending
upon which best fits your argument.

this is bull****

useless, show a plot and code implementations
then we undestand everybody and we give you right

you canf do the plots becus you know your papers
is not goof

liquid mind:
this is bull****

Then why did you post it?


useless, show a plot and code implementations
then we undestand everybody and we give you right

you canf do the plots becus you know your papers
is not goof

liquid mind wrote:
this is bull****

useless, show a plot and code implementations
then we undestand everybody and we give you right

you canf do the plots becus you know your papers
is not goof

Learn some linear algebra instead of complaining all the time.

In article .com,
liquid mind wrote:
this is bull****

useless, show a plot and code implementations
then we undestand everybody and we give you right

you canf do the plots becus you know your papers
is not goof



The irony is that you couldn't create the plot and code implementations
unless you first understood what Bilge is talking about. And then
restricted it to a particular application.


--
"Outside the camp you shall have a place set aside to be used as a
latrine. You shall keep a trowel in your equipment and with it, when you
go outside to ease nature, you shall first dig a hole and afterward cover
up your excrement." -- Deuteronomy 23:13-14