Pmb wrote:

I'm looking for examples of closed form solutions, and their derivations, to
the Dirac equation. Does anyone know of any online resources

Try

http://membres.lycos.fr/pvarni/dirac/rapport.html

I entered "Solution Direac Equation" to Google and got many hits. Check
them out.

Bob Kolker

"Robert J. Kolker" wrote in message
...


Pmb wrote:

I'm looking for examples of closed form solutions, and their
derivations, to
the Dirac equation. Does anyone know of any online resources

Try

http://membres.lycos.fr/pvarni/dirac/rapport.html

I entered "Solution Direac Equation" to Google and got many hits. Check
them out.

Bob Kolker

Thanks Bob.

I'm also curious as to something - The Hamiltonian for a relativistic
charged particle in an electromagnetic field is given by

H = sqrt[(cP - eA)^2 + m^2 c^4] + e*Phi

where

P = canonical momentum
e = charge
A = vector potential
m = rest mass
Phi = Coulomb potential

I've seen two discussions in the texts that I have on this. One is
Cohen-Tannoudji. They do the special case of the Hydrogen atom. However they
do it using perturbation theory. The rule for a function of an operator F(A)
is to expand the function F(z) in a power series and then substitute A -
F(A). In this case Hamiltonion the operator for the canonical momentum is
inserted into the Hamiltonian above. It's been close to a decade since I've
done this so please bear with me. I've only seen the solution approximated,
and then only by using the above Hamiltonitan in Schrodinger's equation and
then using perturbation theory. This is what Cohen-Tannoudji do. Since the
series is an infinite power series it appears there's no closed form
solution. Is that true?


I have another QM text that states that when the operators are placed in
the above Hamiltonian the result is not symetric in space and time
derivatives. But it doesn't say that it can't be done. Is this what
Cohen-Tannoudji is doing?? (see page 1213)

I'm also interested in finding an example of an eigenvalue problem F(A)Psi =
F(a)Psi where there's no closed form expression for Psi

Does anyone know of an example?

Pmb

Subject: Relativistic Quantum Mechanics
From: "Pmb"
Date: 8/16/03 2:35 PM US Mountain Standard Time
Message-id:


"Robert J. Kolker" wrote in message
...


Pmb wrote:

I'm looking for examples of closed form solutions, and their
derivations, to
the Dirac equation. Does anyone know of any online resources

Try

http://membres.lycos.fr/pvarni/dirac/rapport.html

I entered "Solution Direac Equation" to Google and got many hits. Check
them out.

Bob Kolker

Thanks Bob.

I'm also curious as to something - The Hamiltonian for a relativistic
charged particle in an electromagnetic field is given by

H = sqrt[(cP - eA)^2 + m^2 c^4] + e*Phi

where

P = canonical momentum
e = charge
A = vector potential
m = rest mass
Phi = Coulomb potential

I've seen two discussions in the texts that I have on this. One is
Cohen-Tannoudji. They do the special case of the Hydrogen atom. However they
do it using perturbation theory. The rule for a function of an operator F(A)
is to expand the function F(z) in a power series and then substitute A -
F(A). In this case Hamiltonion the operator for the canonical momentum is
inserted into the Hamiltonian above. It's been close to a decade since I've
done this so please bear with me. I've only seen the solution approximated,
and then only by using the above Hamiltonitan in Schrodinger's equation and
then using perturbation theory. This is what Cohen-Tannoudji do. Since the
series is an infinite power series it appears there's no closed form
solution. Is that true?


I have another QM text that states that when the operators are placed in
the above Hamiltonian the result is not symetric in space and time
derivatives. But it doesn't say that it can't be done. Is this what
Cohen-Tannoudji is doing?? (see page 1213)

I'm also interested in finding an example of an eigenvalue problem F(A)Psi =
F(a)Psi where there's no closed form expression for Psi

Does anyone know of an example?


Try to find sqrt[a(d^2/dx^2) +b(d/dx) +c] which is what your hamiltonian
contains so that you'll find why the correct exactsolution for the Hamiltonian
from the energy expression before you did the square root is Diracs/the one at
my site. Finally got tired of all the zero subscripts huh.

Pmb:

I've seen two discussions in the texts that I have on this. One is
Cohen-Tannoudji. They do the special case of the Hydrogen atom. However they
do it using perturbation theory. The rule for a function of an operator F(A)
is to expand the function F(z) in a power series and then substitute A -
F(A).

Which is why dirac didn't do it that way. Expanding the square root
leads to an intractable problem.

In this case Hamiltonion the operator for the canonical momentum is
inserted into the Hamiltonian above. It's been close to a decade since I've
done this so please bear with me. I've only seen the solution approximated,
and then only by using the above Hamiltonitan in Schrodinger's equation and
then using perturbation theory. This is what Cohen-Tannoudji do.

Find a copy of bjorken and drell, vol I. They address all of the above
and more in the first 6 pages and without the verbose rambling found
in cohen-tannoudji. You can also look in schiff, or just about any
other decent quantum mechanics, particle or introductory field theory
text.

Since the series is an infinite power series it appears there's no
closed form solution. Is that true?

No, it obviously isn't true, since the dirac equation is a solution
in closed form. Choose coefficients for p and m and rewrite the
hamiltonian as:

E^2 = [(a.p) + (bm)] ^2

Require that:

[(a.p) + (bm)] ^2 = p^2 + m^2

Solve for the coefficients a_i and b from the anti-commutation
relations that have to be satisfied for the cross terms to
disappear. You can also otain different solutions if you factor
it directly:

E^2\Psi = (p + im)(p - im)\Psi

But interpreting the solution to those two differential equations
is up to you.

I have another QM text that states that when the operators are placed in
the above Hamiltonian the result is not symetric in space and time
derivatives. But it doesn't say that it can't be done. Is this what
Cohen-Tannoudji is doing?? (see page 1213)

From what you've written, cohen-tannoudji appears to have given
only a low energy approximation to a relativistic hamiltonian.


I'm also interested in finding an example of an eigenvalue problem
F(A)Psi = F(a)Psi where there's no closed form expression for Psi

Does anyone know of an example?

Pick one. Just about any operator will do. Very few eigenvalue problems
give solutions which are expressable in closed form. That's why everyone
does perturbation theory and manipulates operators assuming plane waves
rather than solves differential equations for the exact wavefunctions.

First off - Let me state right here and now so that nobody gets the
wrong idea. I should have mentioned this the other day but didn't
think of it.

I've only done the rudimentary stuff when it comes to relativistic
quantum mechanics. I.e. simple stuff that can be found in, say
Liboff's "Relativistic Quantum Mechanics" section. And it's been over
10 years since I've done perturbation theory. So I make no claims at
being anything but just beginning to learn/brush up/re-learn this
stuff, i.e. Klein-Gordon Eq. etc.

That said.........


(Bilge) wrote in message ...
Pmb:

I've seen two discussions in the texts that I have on this. One is
Cohen-Tannoudji. They do the special case of the Hydrogen atom. However they
do it using perturbation theory. The rule for a function of an operator F(A)
is to expand the function F(z) in a power series and then substitute A -
F(A).

Which is why dirac didn't do it that way. Expanding the square root
leads to an intractable problem.

Why? Seems to me that this is what is used in perturbation theory to
examine the fine-structure and hyperfine-struction. Cohen-Tannoudji
expand the momentum in terms of a power series.In fact this is what
they seem to be doing in the advanced quantum mechanice course at
CalTech.

http://minty.caltech.edu/Ph195/entrypage.htm


In fact they refer to Cohen-Tannoudji as well
http://minty.caltech.edu/Ph195/wednesday1c.pdf


Find a copy of bjorken and drell, vol I. They address all of the above
and more in the first 6 pages and without the verbose rambling found
in cohen-tannoudji. You can also look in schiff, or just about any
other decent quantum mechanics, particle or introductory field theory
text.

Thank you.


Since the series is an infinite power series it appears there's no
closed form solution. Is that true?

No, it obviously isn't true, since the dirac equation is a solution
in closed form.

If it's not true then why doesn't Cohen-Tannoudji given the energy
values of the hydrogen atom in terms of a closed form?

Choose coefficients for p and m and rewrite the
hamiltonian as:

E^2 = [(a.p) + (bm)] ^2

Require that:

[(a.p) + (bm)] ^2 = p^2 + m^2

Solve for the coefficients a_i and b from the anti-commutation
relations that have to be satisfied for the cross terms to
disappear.

While I understand that in principle that may be true, I see no reason
to assume that it can be done without numerical methods or using
perturbation theory etc. Do those texts you mention derive a closed
form solution?


From what you've written, cohen-tannoudji appears to have given
only a low energy approximation to a relativistic hamiltonian.

They refer to it as the "weakly relativistic domain." Even so it seems
that they're using that Hamiltonian, even if its just only low energy
it seems like a valid approach - even if your just interested in
things like the fine-structure.

My intent here is to investigate claims made regarding this
Hamiltonian,

H = sqrt[ (cP - eA)^2 + m^2 c^4] + e*Phi

I never said anything about this with respect to rel-QM. The claim was
that the square root of an operator is meaningless. While I don't know
exacatly if and how



I'm also interested in finding an example of an eigenvalue problem
F(A)Psi = F(a)Psi where there's no closed form expression for Psi

Does anyone know of an example?

Pick one. Just about any operator will do. Very few eigenvalue problems
give solutions which are expressable in closed form.

Damn! I missed the most obvious one. And it was staring me right in
the face. If Psi is an energy eigenfunction, then

exp[-itH/hbar]Psi = exp[-itE/hbar]Psi

Thank you bilge.

This arose from a claim that such an equation as this is a
differential equation of infinite power and the person wsuggested that
it was nonsense or meaningless or whatever.

Pmb

(Gauge) wrote in message . com...

I hope I'm not too tangential to the developement of this thread,
but I'm still stuck on Pmb's 1st question....

For example: Consider a frame, S, where there there is a uniform magnetic
field. Has anyone every solved such a problem in closed form soution for a
relativistic charged particle moving in such a field?
Pmb

You specified a "uniform magnetic field" and presuming the Electric
field (F_0i) is zero then, Lorentz force is,

f_i = qF_ij U^j , i,j = 1,2,3

U^j is 3-velocity, F_ij is the field tensor.

Power = force x velocity = f_i U^i = 0

= qF_ij U^j U^i =0 (summing on an antisymmetrical tensor =0).

Integrating ($) Power provides

$ f_i dx^i = f_i x^i = $(qF_ij U^i U^j) ds
= constant of integration = force x distance = energy.

Radiation is usually described by Maxwell's

F_uv;w + F_vw;u + F_wu;v = 0

";" may be considered covariant or partial differentials in
this case, and these require a changing field. In Pmb's
example all field tensor components (F_uv) are constant,
so there is no source of radiation, hence the charged
particle is unable to alter it's energy or velocity.

Here's where I maybe screwed up, I can't see how a charged
particle can accelerate in a uniform magnetic field!

Pmb's question really highlights the difference between
electromagnetism and gravitation. A material object
falling in a constant g-field can acquire relative power
and energy, described in the Newtonian notation,

Power = Force x velocity

Energy = Force x distance

However - in electromagnetism - to generate power and
energy, one must vary the relative field strengths as is done
in a generator. Now at this point, you might be thinking
about Coulombs force law that acts similiar to gravitation.
But I ask, how can a constant electric field produce
radiation?

OK, flames welcome, and thanks Pete for the neat question!

Regards
Ken S. Tucker

Gauge:


Which is why dirac didn't do it that way. Expanding the square root
leads to an intractable problem.

Why?

Because you have an infinite number of terms. If you expand the
square root, you get:

E = m sqrt(1 + (p/m)^2) = m ( 1 + (1/2)(p/m)^2 - (1/4)(1/2!)(p/m)^4 + ...)

E - m = (p/m)^2 - (1/8)(p/m)^4 + ...

The first term is the schroedinger equation (after dividing everything
by m). What do you want to d about the terms p^4, p^6, p^8, p^10, etc?

Moreover, the hamiltonian is non-local, because it contains derivatives
to all orders, so the space and time variables are not on equal footing.


Seems to me that this is what is used in perturbation theory to
examine the fine-structure and hyperfine-struction. Cohen-Tannoudji
expand the momentum in terms of a power series.In fact this is what
they seem to be doing in the advanced quantum mechanice course at
CalTech.


That's rather illusory. You are not getting the true significance of
relativity and you aren't really getting the fine and hyperfine structure
from the theory. What you are getting are only the terms which contribute
to the kinetic energy from the v/c dependence. The spin is put in by hand.
However, the spin is the only really interesting feature that is novel
to quantum mechanics. You'll notice that your reference merely asserts
the spin-orbit term and then uses the relation for total J = L+S to
rewrite L^2 as J^2 - S^2 - 2L.S. While that is useful to know, since
it's a simple way to obtain the (mostly) correct energy levels, it
really has little to do with the relativistic aspect of quantum mechanics.

The proper way to do that is to start with the dirac equation:

H^2\Psi = E^2\Psi = [(a.p) + (bm)]^2\Psi = [p^2 + m^2]\Psi

E^2 = (1/2)[a_i a_j + a_j a_i] p_i p_j + [a_i b + b a_i]m p_i + (bm)^2

which immediately gives:

a_i b + b a_i = 0,

(a_i)(a_i) = b^2 = 1

[a_i, a_j]_+ = 2\delta_jk ([ ]_+ means anti-commutator)

Solving for the a_i gives you (at the least) 4x4 matricies with the
pauli matricies \delta_i on the off-diagonals of a 2x2 matrix and
the matrix b as I and -I (2x2 identity matrix) on the diagonals
of a 2x2 matrix:

[ 0 \delta_i ] [ I 0 ]
a_i = [ ] b = [ ]
[ \delta_i 0 ] [ 0 -I ]


So, first of all, from obtaining the exact relativistic hamiltonian,
you get the spin automatically. Since the matrices are 4 x 4, the
soultion is a set of 4 coupled first order differential equations.

Now if you you substitute the potentials into the dirac hamiltonian,
rather than the equation you expanded, you get:

E\Psi = (a.(p - eA) + bm - e\phi)\Psi


Using the representation for the a_i and b above and writing \Psi as
a two-component column matrix (spinor), i.e.,

[Q]
\Psi = [ ] Where Q and X are each two component wavefunctions,
[X] (Q is called the "large component" and X is called
the "small component" of the wavefunction, \Psi)

which after some algebra I'm not going to reproduce, gives in the
limit for E mc^2:

X = (1/2mc)\delta.(p-eA)Q, which gives an equation for the
spinor Q that uncouples the two equations for each spinor:

X = (a.(p-eA)/2m) Q

Which plugging back into the equation for Q, gives:

E Q = [(a.(p-eA))^2/2m + e\phi] Q

Write P == p - eA and you have the (a.P)^2 which since the a_i are
just the pauli matrices, \delta_i, can be reduced using
the identity for two vectors u and v:


\delta.u \delta .v = u.v + i\delta . (u x v)

In this case both u and v are P so write out P as (p - eA) and you get:


(\delta.P)^2 = (p - eA)^2 - i\delta .(p x A)

The hamiltonian given in cohen-tannoudji contains _only_ the term
(p-eA)^2. It does not contain the second term above. That was
partially handwaved in by asserting the existence of the spin.
From this, you can multiply out the terms and use the fact that
p = -i\hbar\grad to find that the electron has a magnetic moment,

i\delta.(i\hbar\grad x A) = -\hbar \delta . B

Bjorken and Drell solve the real hydrogen atom and give the additional
splitting from the lamb shift (vacuum polarization).

[...]

No, it obviously isn't true, since the dirac equation is a solution
in closed form.

If it's not true then why doesn't Cohen-Tannoudji given the energy
values of the hydrogen atom in terms of a closed form?

The energy eigenvaluse are not the same thing as writing the
eigenvalue equation in closed form. Cohen-Tannoudji writes an
approximation to the hamiltonian and then inserts some stuff
obtained from the dirac equation. The point is that you can get
the exact equation and solve that instead. How you solve the
exact hamiltonian depends upon whether or not you can solve that
equation in closed form. In general, you can't so what you do
is perform what's called a "foldy-wouthuysen transformation",
which decouples the large and small components by assuming a
unitary transform of the form:

H' = \exp(iS) H \exp(-iS)

and then solving for S using a power series expansion of the exponential
and rewriting the terms as an infinite series of iterated commutators.


[(a.p) + (bm)] ^2 = p^2 + m^2

Solve for the coefficients a_i and b from the anti-commutation
relations that have to be satisfied for the cross terms to
disappear.

While I understand that in principle that may be true, I see no reason
to assume that it can be done without numerical methods or using
perturbation theory etc. Do those texts you mention derive a closed
form solution?

You need to specify what you mean by "closed form solution". In
one place it appears you mean an expression for the hamiltonian and
in others it appears to mean an expression for the eigenvalues.



From what you've written, cohen-tannoudji appears to have given
only a low energy approximation to a relativistic hamiltonian.

They refer to it as the "weakly relativistic domain." Even so it seems
that they're using that Hamiltonian, even if its just only low energy
it seems like a valid approach - even if your just interested in
things like the fine-structure.

It's fine for that, but in doing so, it's very misleading to imply that
it represents a relativistic equation, since it merely asserts the parts
that come from relativity (apart from the expansion of the sqaure root).
You could just as easily assert those terms and tack them on to the
schroedinger equation. This is one of the reasons that I object to that
book. What they are doing is essentially showing you how to do
perturbation theory and obscuring the point by introducing it as
relativity, but then assert all of the stuff that really comes from the
relativistic equations. So, the introduce pages of material which are
extraneous to the point they are making leaving one to wonder what the
example was supposed to address. Since that book has the absolute worst
treatment of addition of angular momentum that I have ever encountered,
they would have been better off explaining addition of angular momentum,
which happens to be important for the example at hand.



My intent here is to investigate claims made regarding this
Hamiltonian,

H = sqrt[ (cP - eA)^2 + m^2 c^4] + e*Phi

I never said anything about this with respect to rel-QM.

The Subject header you wrote plainly says:

"Subject: Relativistic Quantum Mechanics"


The claim was that the square root of an operator is meaningless.

It's obviously not meaningless, as dirac demonstrated by showing that
it gives you the electron spin and antimatter. Solving it by expanding
the square root just happens to be an extremely cumbersome way to
go about it and doesn't give any physical insight as to what the
terms mean.


This arose from a claim that such an equation as this is a
differential equation of infinite power and the person wsuggested that
it was nonsense or meaningless or whatever.

It's not meaningless. It took someone like dirac (if there actually
could be anyone like dirac), to find a way to make that expression
meaningful and interpret the terms. The klein-gordon equation was
proposed very early - early enough that the schroedinger equation
might have been deemed irrelevant or at least given less importance -
but it couldn't be interpreted because of the second order time
derivative. Dirac believed that his solution to obtaining a first order
theory, was simply too elegant to not be true and ended up prediction
antimatter and the spin of fermions.

(Ken S. Tucker) wrote in message om...
(Gauge) wrote in message . com...

I hope I'm not too tangential to the developement of this thread,
but I'm still stuck on Pmb's 1st question....

Nah. I left this topic wide open with the broad label "Relativistic
Quantum Mechanics" so feel free to indulge. I'm rather bored with the
same-ole same-ole. Time for something new! :-)

For example: Consider a frame, S, where there there is a uniform magnetic
field. Has anyone every solved such a problem in closed form soution for a
relativistic charged particle moving in such a field?
Pmb

You specified a "uniform magnetic field" and presuming the Electric
field (F_0i) is zero

Yes. The E-field is zero.

... then, Lorentz force is,

f_i = qF_ij U^j , i,j = 1,2,3

U^j is 3-velocity, F_ij is the field tensor.

Power = force x velocity = f_i U^i = 0

= qF_ij U^j U^i =0 (summing on an antisymmetrical tensor =0).

Hold on there. Becareful with what you're calling "3-velocity." The
quantinty "v" defined as

v_x = dx/dt
v_y = dy/dt
v_z = dz/dt

is what is normally called a 3-vector. The spatial part of the
4-velocity is not a 3-vector. They are related buy

U^k = gamma*v_k


Integrating ($) Power provides

$ f_i dx^i = f_i x^i = $(qF_ij U^i U^j) ds
= constant of integration = force x distance = energy.

Radiation is usually described by Maxwell's

F_uv;w + F_vw;u + F_wu;v = 0

It is? How so?

";" may be considered covariant or partial differentials in
this case, and these require a changing field. In Pmb's
example all field tensor components (F_uv) are constant,
so there is no source of radiation, hence the charged
particle is unable to alter it's energy or velocity.

Hold on - There is the field the particle is moving in and there is
the field of the particle. The field can basically be constant while
it interacts with the particle. In the inertial frame S (the one in
which there is only a uniform B field) the field does no work and
therefore the energy of the particle is a constant. The velocity may
change but the magnitude doesn't.

Result - In S energy is conserved

In another inertial frame, S' moving with with respect to S, there may
be an electric field. If so then the field does work on the particle.

Result - In S' energy is not conserved


Conservation of energy is frame dependant for systems which are not
closed. Energy can be conserved in one frame and not another. This is
within a certain approximation of course since the particle looses
radiation due to its acceleration. And of course this ignores the
effect the particle has on the source etc.

Here's where I maybe screwed up, I can't see how a charged
particle can accelerate in a uniform magnetic field!

It moves perpedicular to it.

However - in electromagnetism - to generate power and
energy, one must vary the relative field strengths as is done
in a generator.

That's not true in general. E.g. consider a uniform Electric fiedl.
The force on a charge is F= qE. The power being delivered to the
charge is P = Fv

Now at this point, you might be thinking
about Coulombs force law that acts similiar to gravitation.
But I ask, how can a constant electric field produce
radiation?

It doesn't. Why should it? It's not field that's producing radiation.
Its the particle which is accelerating. It gets its energy from
whatever is generating the field.

Pete

"Gauge" wrote in message
om...
(Ken S. Tucker) wrote in message
om...
(Gauge) wrote in message
. com...

I hope I'm not too tangential to the developement of this thread,
but I'm still stuck on Pmb's 1st question....

Nah. I left this topic wide open with the broad label "Relativistic
Quantum Mechanics" so feel free to indulge. I'm rather bored with the
same-ole same-ole. Time for something new! :-)

For example: Consider a frame, S, where there there is a uniform
magnetic
field. Has anyone every solved such a problem in closed form soution
for a
relativistic charged particle moving in such a field?
Pmb

You specified a "uniform magnetic field" and presuming the Electric
field (F_0i) is zero

Yes. The E-field is zero.

... then, Lorentz force is,

f_i = qF_ij U^j , i,j = 1,2,3

U^j is 3-velocity, F_ij is the field tensor.

Power = force x velocity = f_i U^i = 0

= qF_ij U^j U^i =0 (summing on an antisymmetrical tensor =0).

Hold on there. Becareful with what you're calling "3-velocity." The
quantinty "v" defined as

v_x = dx/dt
v_y = dy/dt
v_z = dz/dt

is what is normally called a 3-vector. The spatial part of the
4-velocity is not a 3-vector. They are related buy

U^k = gamma*v_k


Integrating ($) Power provides

$ f_i dx^i = f_i x^i = $(qF_ij U^i U^j) ds
= constant of integration = force x distance = energy.

Radiation is usually described by Maxwell's

F_uv;w + F_vw;u + F_wu;v = 0

It is? How so?

";" may be considered covariant or partial differentials in
this case, and these require a changing field. In Pmb's
example all field tensor components (F_uv) are constant,
so there is no source of radiation, hence the charged
particle is unable to alter it's energy or velocity.

Hold on - There is the field the particle is moving in and there is
the field of the particle. The field can basically be constant while
it interacts with the particle. In the inertial frame S (the one in
which there is only a uniform B field) the field does no work and
therefore the energy of the particle is a constant. The velocity may
change but the magnitude doesn't.

Result - In S energy is conserved

In another inertial frame, S' moving with with respect to S, there may
be an electric field. If so then the field does work on the particle.

Result - In S' energy is not conserved


Conservation of energy is frame dependant for systems which are not
closed. Energy can be conserved in one frame and not another. This is
within a certain approximation of course since the particle looses
radiation due to its acceleration. And of course this ignores the
effect the particle has on the source etc.

Hold on. Let me put that notion on hold Ken. While its' true that work is
done in S' but not in S the systems may or may not be conservative

Regarding frame dependance on conservation of energy - remember - this is
*not* a closed system now.

Consider the classical case from Newtonian mechancis - A completely elastic
collision of a ball and a fixe wall. In the inertial frame moving with the
ball after it bounces the ball is at rest in that frame. The energy of the
ball, all of which is kinetic, is then zero. But the kinetic energy of the
ball didn't become zero until the ball hit the wall. So it had energy before
the bounce but not after. Energy is not conserved in that frame. In the
frame rest frame of the wall the ball bounces and has the same kinetic
energy before as after the collision

What happened? Simple - In one frame the wall did no net work. In another
frame the wall did net work.I.e. in one frame the dx in

dW = F*dx

is zero (wall frame) and in another frame dz is not zero and therefore does
work in that frame.

With systems that are not closed this things have to done carefully

Pete

Gauge:
(Bilge) wrote in message
e-al.net...

Moreover, the hamiltonian is non-local, because it contains derivatives
to all orders, so the space and time variables are not on equal footing.

Yeah. That part I was aware of. So it's possible to use the
Hamiltonian by taking the first few terms.

However if there are an equal number of terms I don't see how that's a
problem. I take it its a practical problem and not a theoretical.
Please elaborate.

To me this seems normal. Consider, in principle, the corresponding
problem in non-relativistic quantum mechanics. In the postion
representation there's no problem of course since there's only one
second derivative

There is not just a second order derivative. Each term containing p
contains i\hbar\grad. So there are terms like \grad^2, \grad^4, etc.

and with an abitrary potential its not a problem
since the position operator has no derivatices in this representation.
However consider the exact Hamiltonian but now in the momentum
representation. As such the potential can, in principle, have a
corresponding power series expansion which my have an infinite number
of terms. In the momentum representation its now the postion operator
which has derivatives and as such we're back to the same situation - A
power series with an infinite number of terms. While not practical
it's still meaningful and well defined.

The meaning of the terms are not obvious. For example, you don't
get any idea that there is something called "spin" just from the
series expansion. As dirac factored it, the spin falls just out.
In addition, when writing the dirac equation in its "natural"
form, the operators (matrices) a and b are redefined by multiplying
the entire equation through by b to get:

(bE)\Psi = (ba + m)\Psi

and one defines the \gamma (dirac) matrices by:

b = \gamma^{0}

ab = \gamma^{i}, i = 1,2,3

which then makes the connection of the spin as a relativistic feature
completely transparent, since the \gamma matrices satisfy the
following commutation and anti-commutation relations:

[\gamma^{u}, \gamma^{v}]_{+} = 2g^{uv}

[\gamma^{u}, \gamma^{v}] = 2i\sigma^{uv}

where \sigma^{uv} is the spin tensor. From that, one can derive
the allowable forms for a relativistic potential (interaction,
that is), of which there are exactly five. For example, the
electromagnetic current, j^{u} is ie\Psibar\gamma^{u}\Psi
and the interaction with the four-vector potential is just
j^{u}A_{u} = ie\Psibar\gamma^{u}\Psi A_{u} ==\Psibar A/\Psi
(A/ == A-slash, the fourvector with a diagonal slash through it).

One also gets the the four-vector identity for any two four-vectors,
A_{u}, B_{u} contracted with \gamma matrices:

A/ B/ = A^{u} B_{u} - 2i\sigma^{uv}A_{u}B_{v}


from which the pauli identity can be obtained as a non-relativistic
reduction.


[...]
You need to specify what you mean by "closed form solution".

I've always seen and heard the term "closed form solution" to mean
that the result is represented as a finite number of elementary
functions.

It does. What wasn';t clear was what solution you wanted in closed form.
The dirac equation is the square root of p^2 + m^2 in closed form.
That doesn't mean the solutions to the dirac equation will be in
closed form.

[...]

The authors do say something to the effect that this problem is
appropriately handled with the Dirac equation and indicate that what
all falls out is that series. At least that's what I got from that
section.

Do you have this text?

I owned both volumes at one time. I was going to throw both volumes
into the trash, but someone offered to exchange both volumes for a copy
of gottfried, so I took her up on the offer. I truly loathe that text,
more than I can possibly describe. It spends 2000+ pages on the same
material that is essentially covered in the first 200 pages of schiff
and it does so with less clarity and less physical intuition. Both saxon
and gasciorowicz (sp?) are much better textbooks, too.

[...]

If the particle is something like a pion do you still use the Dirac
equation or is that strictly for particles with spin? Do you use the
Klein-Gordon equation in that case? This is what my text seems to
imply.

The dirac equation only describes 1/2 integer spin particles.
For spin 1/2, you you have the 4x4 dirac matrices. For spin 3/2
you have 8x8 matrices, and so on. The klein-gordon equation
describes integer spin particles as well as 1/2 integer spin
particles. Since the klein gordon hamiltonian is just the square
of the dirac hamiltonian and it's also just the energy-momentum
relations from relativity, the klein-gordon equation has to
be satisfied by any particle in a relativistic theory.