Eugene Stefanovich:
Bilge wrote:

That isn't a refutation. The qed lagrangian is,
_
L = u(p/ - A/ - m)u + (1/4)F^uvF_uv

If I have a static potential, such as an electron in a hydrogen atom,
I don't need the F_uvF^uv term and,
.
H = \integral (pu - L) d3x

gives me the hamiltonian in bjorken & drell, for which a dervation of
the darwin term id given and which you seem to think involves something
that can't be done starting with qed.

You did a lot of cheating along the way. Just a few examples:

I used physics that is well over century old and can be found
in goldstein.

Bilge wrote:
Eugene Stefanovich:
Bilge wrote:

present. For example

dp_1(t)/dt = i [H, p_1(t)] /= 0

which means that there is a non-zero force acting on particle 1.

That doesn't mean it's due to a particle at the same time. You
``prove'' below that it isn't.


The potential is non-local and can't conserve momentum

This potential commutes with the total momentum operator, so
momentum is conserved. For example, the Coulomb potential commutes with
P

[P, 1/|r_1 - r_2|] = 0

Then, if you are correct, (which you are not), obviously there are
no forces due to that potential:

[p, V]\Psi = -i\hbar [(\grad V)\Psi + V\grad\Psi) - V\grad\Psi]

[p, V] = -i\hbar\grad V

So, if [p, V] = 0, then \grad V is zero and you have no interaction.
Classically, we would call that F = -\grad V = q\grad\phi = qE, but
since you don't believe in electric fields, there's not much more
I can do but say your theory has no interaction, as you've just
``proved.''

0 = [P, V] = [P, H_0 + V]

therefore P commutes with H and the total momentum is conserved.
You may say that "interaction does not act on the total momentum".
However, interaction does act on momenta of individual particles
(P = p_1 + p_2), for example take the Coulomb interaction above

[p_1, H] = [p_1, V] = [p_1, 1/|r_1 - r_2|] = - d/dr_1 1/|r_1 - r_2|

= (r_1 - r_2)/|r_1 - r_2|^3



your theory formally doesn't exist, since there are no eigenstates
of your full hamiltonian, even in principle.

There are eigenstates. Again, in the simplest case of Coulomb potential
V, the eigenstates and eigenvectors are given by well-known solutions
for the hydrogen atom.

Then obviously, there isn't an interaction. You just said so above.
I'll give you a hint. The first order perturbation is calculated with
the groundstate wavefunctions. The second order perturbation is not.
So, the eigenvectors you need to calculate the second order perturbation
are the first order corrections to the ground state wavefunctions.
Figure out what that means physically.

Why did you bring up perturbations? The Coulomb problem for the hydrogen
atom can be solved analytically.

Eugene.


The remaining stuff was the same old crap. Learn what a measurement
is and go spend some time in a real physics lab.

Eugene Stefanovich:
Bilge wrote:
Eugene Stefanovich:
This potential commutes with the total momentum operator, so
momentum is conserved. For example, the Coulomb potential commutes with
P

[P, 1/|r_1 - r_2|] = 0

Then, if you are correct, (which you are not), obviously there are
no forces due to that potential:

[p, V]\Psi = -i\hbar [(\grad V)\Psi + V\grad\Psi) - V\grad\Psi]

[p, V] = -i\hbar\grad V

So, if [p, V] = 0, then \grad V is zero and you have no interaction.
Classically, we would call that F = -\grad V = q\grad\phi = qE, but
since you don't believe in electric fields, there's not much more
I can do but say your theory has no interaction, as you've just
``proved.''

0 = [P, V] = [P, H_0 + V]

therefore P commutes with H and the total momentum is conserved.

Them you have no force.

You may say that "interaction does not act on the total momentum".
However, interaction does act on momenta of individual particles
(P = p_1 + p_2), for example take the Coulomb interaction above

[p_1, H] = [p_1, V] = [p_1, 1/|r_1 - r_2|] = - d/dr_1 1/|r_1 - r_2|

= (r_1 - r_2)/|r_1 - r_2|^3

That is what I wrote above in response to your previous post
where you tried to tell me that commutator was zero. Make up your
mind. If you believe what you just wrote down, it shoud be
obvious why the speed of the interaction must be `c'.

Your an idiot, eugene. Go pester someone else with this ****.


Then obviously, there isn't an interaction. You just said so above.
I'll give you a hint. The first order perturbation is calculated with
the groundstate wavefunctions. The second order perturbation is not.
So, the eigenvectors you need to calculate the second order perturbation
are the first order corrections to the ground state wavefunctions.
Figure out what that means physically.

Why did you bring up perturbations? The Coulomb problem for the hydrogen
atom can be solved analytically.

Because you insist on doing perturbation theory.

Bilge wrote:
Eugene Stefanovich:
Bilge wrote:
Eugene Stefanovich:
This potential commutes with the total momentum operator, so
momentum is conserved. For example, the Coulomb potential commutes with
P

[P, 1/|r_1 - r_2|] = 0

Then, if you are correct, (which you are not), obviously there are
no forces due to that potential:

[p, V]\Psi = -i\hbar [(\grad V)\Psi + V\grad\Psi) - V\grad\Psi]

[p, V] = -i\hbar\grad V

So, if [p, V] = 0, then \grad V is zero and you have no interaction.
Classically, we would call that F = -\grad V = q\grad\phi = qE, but
since you don't believe in electric fields, there's not much more
I can do but say your theory has no interaction, as you've just
``proved.''

0 = [P, V] = [P, H_0 + V]

therefore P commutes with H and the total momentum is conserved.

Them you have no force.

You may say that "interaction does not act on the total momentum".
However, interaction does act on momenta of individual particles
(P = p_1 + p_2), for example take the Coulomb interaction above

[p_1, H] = [p_1, V] = [p_1, 1/|r_1 - r_2|] = - d/dr_1 1/|r_1 - r_2|

= (r_1 - r_2)/|r_1 - r_2|^3

That is what I wrote above in response to your previous post
where you tried to tell me that commutator was zero. Make up your
mind. If you believe what you just wrote down, it shoud be
obvious why the speed of the interaction must be `c'.

Your an idiot, eugene. Go pester someone else with this ****.



You should understand the difference between momenta of individual
particles p_1 and p_2 and the total momentum P = p_1 + p_2.
Individual momenta do not commute with interaction V, total momentum P
does commute with interaction V.

Eugene.



Then obviously, there isn't an interaction. You just said so above.
I'll give you a hint. The first order perturbation is calculated with
the groundstate wavefunctions. The second order perturbation is not.
So, the eigenvectors you need to calculate the second order perturbation
are the first order corrections to the ground state wavefunctions.
Figure out what that means physically.

Why did you bring up perturbations? The Coulomb problem for the hydrogen
atom can be solved analytically.

Because you insist on doing perturbation theory.

Eugene Stefanovich:

You should understand the difference between momenta of individual
particles p_1 and p_2 and the total momentum P = p_1 + p_2.
Individual momenta do not commute with interaction V, total momentum P
does commute with interaction V.

You're an idiot, eugene. Even your classical picture of the
world is wrong. plonk