static electric field as expectation value of which quantum state?

[iuval]

[[Mod. note -- Please limit your text to fit within 80 columns,
preferably around 70, so that readers don't have to scroll horizontally
to read each line. I have manually reformatted this article. -- jt]]

It is easy to see that there is no superposition of coherent photon
states that would give an expectation value of the electric field
operator that is static but not constant in space (e.g. 1/r^2 for
radial component from a point source). The reason is that the
electric field operator has an an integral of (Exp(i (k.r-wt) a-
complex conjugate) (1/2 w)d^3k. There is the (2 w) in the denominator
but that gets cancelled by the delta function normalization (2 w)
when integrating over the superposition of k's of coherent
states(Int[A_k |alpha_k> d^3k'] for whatever component of the vector
state). In the end you get <Ex>=3D2 Sqrt(hbar/2epsilonO V) Int[|alpha_k|
|A_k|^2 Sin(w t- k.r-theta_k)Sqrt[w] d^3k ]. Now w=3D|k| for photons,
so since sin(wt-theta_k-k.r)=3Dsin(wt-theta_k)
cos(k.r)+cos(wt-theta_k)sin(kr), we have the situation of how to
eliminate the t dependence. If there was only one term it would
be obvious that there are no A_k that can produce the given static
expectation value, from taking the inverse (3D, not 4D!) fourier
transform, but I think it is still true with two terms, though I
haven't proven it yet. So if a classical field emerges as an
expectation value in a quantum state, which state is it? The appeal
of coherent states is that their standard deviation goes to 0 as
1/Sqrt(volume) or for finite volume is proportional to Sqrt[h].

[Gregor Scholten]

iuval wrote:

> It is easy to see that there is no superposition of coherent photon
> states that would give an expectation value of the electric field
> operator that is static but not constant in space (e.g. 1/r^2 for
> radial component from a point source).

You mean the electrostatic field of an electric charge that goes with
1/r^2 where r is the distance from the charge? You should note that
photon states are states of the free radiation field, whereas an
electrostatic field means that the electromagnetic field is interacting.
Already in classical field theory, field configurations of the
interacting field cannot be constructed by combining configurations of
the free radiation field.

The same applies in quantum field theory: quantum states of the free
radiation field, i.e. photon states, cannot be combined to states of the
interacting field, in other words, states of the interacting field are
not contained in the Hilbert space spanned by the states of the free
radiation field.

In general, quatum states |Psi> of the electromagnetic field, no matter
if free radiation field or interacting field, can be represented as wave
functionals

Psi[A^mu(x),t]

on the space of classical field configurations A^mu(x). However,
calculating such wave functionals is a very difficual mathematical
procedure. For the free radiation field, there is a very much easier
approch: the radiation field can be decomposed into modes, where each
mode can be considered as a harmonic oscillator, allowing for apply the
algebra of the quantum mechanical harmonic oscillator with ladder
operators. The ladder operators can then be considered as creation and
destruction operators for quanta of the radiation field, i.e. for photons.

For the interacting field, however, this harmonic oscillator approach
does not work. Therefore, one invented a different approach, the
perturbation theory: one focuses on interaction processes where in the
beginning and in the end, the interaction is very weak, but becomes
strong in between, and does not consider the actual state during the
process, but transition amplitudes from initial states to end states,
which initial states and end states assumed as states of the free,
non-interacting fields. For the electrostatic field of a single electric
charge, this is not a proper approach, but for a scattering process of
two electrically charged particles, it applies quite well.

The state of the electromagnetic field during the scattering process is
out of focus in this approach, and therefore does not need to be
calculated. Only initial and final states are considered, that are
approximated as statets of the free radiation field. Take e.g. a process
that produces bremsstrahlung: the initial state is a state with no
photons, the final states a state with one photon.

Besides perturbation theory, there is another approach for the
interacting field, which is a rather numerical one: lattice gauge
theory. This approach was invented decenaries after perturbation theory
because it requires computers with sufficient performance to do the
numerical calculations. The core concept of lattic gauge theory is to
approximate space as a lattice, as a set of a finite number of discrete
points, for calculating the wave functional described above on these
points. Or, as far as vector fields like the electromagnetic field are
concerned, not on the points themselves, but on the connections between
them.

[Jos Bergervoet]

On 7/30/2015 6:01 AM, Gregor Scholten wrote:
> iuval wrote:
>
>> It is easy to see that there is no superposition of coherent photon
>> states that would give an expectation value of the electric field
>> operator that is static but not constant in space (e.g. 1/r^2 for
>> radial component from a point source).
>
> You mean the electrostatic field of an electric charge that goes with
> 1/r^2 where r is the distance from the charge? You should note that
> photon states are states of the free radiation field, whereas an
> electrostatic field means that the electromagnetic field is interacting.
> Already in classical field theory, field configurations of the
> interacting field cannot be constructed by combining configurations of
> the free radiation field.
>
> The same applies in quantum field theory: quantum states of the free
> radiation field, i.e. photon states, cannot be combined to states of the
> interacting field, in other words, states of the interacting field are
> not contained in the Hilbert space spanned by the states of the free
> radiation field.
>
> In general, quatum states |Psi> of the electromagnetic field, no matter
> if free radiation field or interacting field, can be represented as wave
> functionals
>
> Psi[A^mu(x),t]

Now that you bring it up: this would easily solve the problem if
OP is mainly interested in the wave functional with the lowest variance
around the classical configuration (which is the distinctive property
of coherent states, I would say). To do so we could just take the
wave functional of the vacuum state and re-center it around the
desired electrostatic field, giving:

Psi_OP[A^mu(x),t] = Psi[A^mu(x) - A^mu_cent(x), t]

where Psi[] would be the vacuum wave functional and A_cent^mu(x)
the classical point source field.

It would of course not solve the problem of writing the solution
as the exponentiation of some free-particle-state creation operator,
which is also a property of the coherent state (more a computational
convenience than a fundamental property, I'd say.. but anyhow, it's
OP's choice to decide whether this solves it.)

--
Jos

[iuval]

On Wednesday, July 29, 2015 at 9:01:24 PM UTC-7, Gregor Scholten wrote:
> iuval wrote:
>
>> It is easy to see that there is no superposition of coherent photon
>> states that would give an expectation value of the electric field
>> operator that is static but not constant in space (e.g. 1/r^2 for
>> radial component from a point source).
>
> You mean the electrostatic field of an electric charge that goes with
> 1/r^2 where r is the distance from the charge?

Or any static field (due to static charges)

> You should note that
> photon states are states of the free radiation field, whereas an
> electrostatic field means that the electromagnetic field is interacting.
> Already in classical field theory, field configurations of the
> interacting field cannot be constructed by combining configurations of
> the free radiation field.

Sure.

>
> The same applies in quantum field theory: quantum states of the free
> radiation field, i.e. photon states, cannot be combined to states of the
> interacting field, in other words, states of the interacting field are
> not contained in the Hilbert space spanned by the states of the free
> radiation field.
>
> In general, quatum states |Psi> of the electromagnetic field, no matter
> if free radiation field or interacting field, can be represented as wave
> functionals
>
> Psi[A^mu(x),t]

That is one way, but not the one I am interested in.

>
> on the space of classical field configurations A^mu(x). However,
> calculating such wave functionals is a very difficual mathematical
> procedure. For the free radiation field, there is a very much easier
> approch: the radiation field can be decomposed into modes, where each
> mode can be considered as a harmonic oscillator, allowing for apply the
> algebra of the quantum mechanical harmonic oscillator with ladder
> operators. The ladder operators can then be considered as creation and
> destruction operators for quanta of the radiation field, i.e. for photons.

You can fourier decompose any static field into modes as well. But you
have the constraint |3Dk|=0, so maybe this can be done with complex ks?

>
> For the interacting field, however, this harmonic oscillator approach
> does not work. Therefore, one invented a different approach, the
> perturbation theory: one focuses on interaction processes where in the
> beginning and in the end, the interaction is very weak, but becomes
> strong in between, and does not consider the actual state during the
> process, but transition amplitudes from initial states to end states,
> which initial states and end states assumed as states of the free,
> non-interacting fields. For the electrostatic field of a single electric
> charge, this is not a proper approach, but for a scattering process of
> two electrically charged particles, it applies quite well.

OK, but I am not interested in that experimental situation. If classical
mechanics emerges in some many particle or high energy limit (large
alphas in the case of coherent fields where alpha is the eigenvalue of
the annihilation operator), then we have to find what the quantum state
is for this other very common measurable, experimental situation.

>
> The state of the electromagnetic field during the scattering process is
> out of focus in this approach, and therefore does not need to be
> calculated. Only initial and final states are considered, that are
> approximated as statets of the free radiation field. Take e.g. a process
> that produces bremsstrahlung: the initial state is a state with no
> photons, the final states a state with one photon.
>
> Besides perturbation theory, there is another approach for the
> interacting field, which is a rather numerical one: lattice gauge
> theory. This approach was invented decenaries after perturbation theory
> because it requires computers with sufficient performance to do the
> numerical calculations. The core concept of lattic gauge theory is to
> approximate space as a lattice, as a set of a finite number of discrete
> points, for calculating the wave functional described above on these
> points. Or, as far as vector fields like the electromagnetic field are
> concerned, not on the points themselves, but on the connections between
> them.

It can be done either with E(x) or A(x), I don't see the difference, but
that is a tangent for now. But what is the answer in the limit of zero
lattice spacing? Is it just like the shifted ground state wave function
of the harmonic oscilator, except that it is a wave functional, with x
replaced by E(x) (1/r^2in the simplest point source example), as the
next response suggests? That would be psi(E_r(x), the other 2 E
components and 3 B components)=N Exp[-(E_r(x)-1/r^2)/(2
x0^2)]*delta(other components), where x0 is the vacuum standard
deviation/fluctuation, N is a normalization constant, and the delta
function is a functional distribution. But I thought Fock's point was
that this Hilbert space is equivalent to the many photon (but not
constant number necessarily) Hilbert space, in which case this is not
correct. There are two other possible ways I can see:

1. Use the SR Vector photon wave function, but for a many particle
state, symmetrized as is appropriate for bosons. Has anyone done this?

2. Add to the photon state Hilbert space, the state space for
electrons/positrons and allow sums of tensor products. Perhaps a much
simpler space since we are dealing with a non-relativistic case of just
one electron. Has anyone done this? I would be surprised if not, since
this is a basic test of QFT--it has to describe physically measurable
static fields.

[Jos Bergervoet]

On 7/31/2015 9:23 AM, iuval wrote:
> On Wednesday, July 29, 2015 at 9:01:24 PM UTC-7, Gregor Scholten wrote:
>> iuval wrote:
>>
>>> It is easy to see that there is no superposition of coherent photon
>>> states that would give an expectation value of the electric field
>>> operator that is static but not constant in space (e.g. 1/r^2 for
>>> radial component from a point source).

...
..
> .. Is it just like the shifted ground state wave function

> of the harmonic oscilator, except that it is a wave functional, with x
> replaced by E(x) (1/r^2in the simplest point source example), as the
> next response suggests? That would be psi(E_r(x), the other 2 E
> components and 3 B components)=N Exp[-(E_r(x)-1/r^2)/(2
> x0^2)]*delta(other components), where x0 is the vacuum standard
> deviation/fluctuation, N is a normalization constant, and the delta
> function is a functional distribution.

I don't see why you would use a delta function to restrict
anything completely here, unless you are trying to fix the
gauge.

Neglecting the gauge problem, one would expect all field
components to have vacuum standard deviaton, no more and
no less, only centered around a strong 1/r^2 field instead
of centered around zero. And if you don't neglect the gauge
problem, you'd first have some degrees of freedom not
restricted at all and then you'd have to choose your own
way of restricting them (gauge choice) or you would not
restrict them but integrate them out if your application
allows some way to do that.

> ... But I thought Fock's point was

> that this Hilbert space is equivalent to the many photon (but not
> constant number necessarily) Hilbert space,

Yes why is that a problem? You started asking for a coherent
state, which would definitely be a many photon state with no
definite photon number. (Or do you mean that re-centering
would shift it *out of* the usual Fock space?)

> .. There are two other possible ways I can see:

>
> 1. Use the SR Vector photon wave function, but for a many particle
> state, symmetrized as is appropriate for bosons. Has anyone done this?

You mean the vector spherical harmonics? They are not free
fields but all require some multipole source present. You
may then also consider plane waves that aren't divergence
free (i.e. having nonzero longitudinal E, and therefore
always an accompanying charge density wave as a source).

> 2. Add to the photon state Hilbert space, the state space for
> electrons/positrons and allow sums of tensor products. Perhaps a much
> simpler space since we are dealing with a non-relativistic case of just
> one electron. Has anyone done this? I would be surprised if not, since
> this is a basic test of QFT--it has to describe physically measurable
> static fields.

It seems here you propose to use a quantum field for the
source as well! Probably not simpler and I don't see how
you would build a static situation (OK, you can build a
kind of hydrogen atom, but its CM coordinate would still
spread out indefinitely..)

--
Jos

[iuval]

Correction on my previous shifted ground state functional (the way I
wrote it it isn't a functional, that is a map from the space of
functions on R3 to C). It needs an integral over R3, and perhaps the
delta function is wrong too, as the other components can fluctuate
about 0, so maybe they should be integrals of gaussians centered about
0. But there are then serious singularities, besides the disagreement
with Fock.

On Thursday, July 30, 2015 at 2:21:55 PM UTC-7, Jos Bergervoet wrote:
> On 7/30/2015 6:01 AM, Gregor Scholten wrote:
> > iuval wrote:
> >
> >> It is easy to see that there is no superposition of coherent photon
> >> states that would give an expectation value of the electric field
> >> operator that is static but not constant in space (e.g. 1/r^2 for
> >> radial component from a point source).
> >
> > You mean the electrostatic field of an electric charge that goes with
> > 1/r^2 where r is the distance from the charge? You should note that
> > photon states are states of the free radiation field, whereas an

> > electrostatic field means that the electromagnetic field is interacting=

.
> > Already in classical field theory, field configurations of the
> > interacting field cannot be constructed by combining configurations of
> > the free radiation field.
> >
> > The same applies in quantum field theory: quantum states of the free

> > radiation field, i.e. photon states, cannot be combined to states of th=

e
> > interacting field, in other words, states of the interacting field are
> > not contained in the Hilbert space spanned by the states of the free
> > radiation field.
> >
> > In general, quatum states |Psi> of the electromagnetic field, no matter

> > if free radiation field or interacting field, can be represented as wav=

[Gregor Scholten]

iuval wrote:

>> on the space of classical field configurations A^mu(x). However,
>> calculating such wave functionals is a very difficual mathematical
>> procedure. For the free radiation field, there is a very much easier
>> approch: the radiation field can be decomposed into modes, where each
>> mode can be considered as a harmonic oscillator, allowing for apply the
>> algebra of the quantum mechanical harmonic oscillator with ladder
>> operators. The ladder operators can then be considered as creation and
>> destruction operators for quanta of the radiation field, i.e. for photons.
>
> You can fourier decompose any static field into modes as well.

No, you can't. You could do this if you would consider the spatial part
of each mode only, i.e. exp(i k x) instead of exp[i (k x - omega t)].
Taking e.g. a 1/x^2 electrostatic field, you could write

E(x) = q / x^2 = q \int a(k) exp(i k x) dk

where a(k) would be the Fourier transform of E(x). Taking the temporal
part of each mode, exp(i omega t) with omega = c k, into account,
however, destroys this scheme. As you pointed out in your initial post:

>>> we have the situation of how to eliminate the t dependence.

Shortly summarized: there is no way to eliminate that t dependence.
Electrostatic fields cannot be composed from solutions for the free
radiation field. Neither in classical field theory nor in quantum field
theory. In quantum field theory, this means that states of the
interacting field cannot be composed from photon states.

> But you
> have the constraint |3Dk|=0

You mean, due to the relation omega = c |k|, or w = |3Dk| as you wrote
it, k has to be 0, to provide omega = 0 to eliminate the time dependency?

> so maybe this can be done with complex ks?

Imagine a mode with complex wave number k = k_re + i k_im. The field
configuration would be something like

exp[i (k x - omega t)] = exp[i (k_re x + i k_im x - omega t)]

= exp[-k_im x + i (k_re x - omega t)]

= exp(-k_im x) exp[i (k_re x - omega t)]

This would correspond to a plane wave with wave number k_re that is
spatially decaying exponentially. This could be a solution for radiation
in matter, but not for the free radiation field, and therefore cannot
correspond to a photon state.

And now imagine you implement the constraints k_re = 0 and omega = 0,
then the configuration would become simply exp(-k_im x). I don't think a
1/x^2 electrostatic field could be composed of such configurations with
different k_im.

>> For the interacting field, however, this harmonic oscillator approach
>> does not work. Therefore, one invented a different approach, the
>> perturbation theory: one focuses on interaction processes where in the
>> beginning and in the end, the interaction is very weak, but becomes
>> strong in between, and does not consider the actual state during the
>> process, but transition amplitudes from initial states to end states,
>> which initial states and end states assumed as states of the free,
>> non-interacting fields. For the electrostatic field of a single electric
>> charge, this is not a proper approach, but for a scattering process of
>> two electrically charged particles, it applies quite well.
>
> OK, but I am not interested in that experimental situation. If classical
> mechanics emerges in some many particle or high energy limit (large
> alphas in the case of coherent fields where alpha is the eigenvalue of
> the annihilation operator), then we have to find what the quantum state
> is for this other very common measurable, experimental situation.

I guess a proponent of perturbation theory would claim that the
electrostatic field of a single isolated charge is no measurable
experimental situation. He would claim that the only way to measure the
electrostatic field of a charge is to let a second charge be scattered
by the field of that charge, and that this scattering process can be
handled with perturbation theory again.

>> Besides perturbation theory, there is another approach for the
>> interacting field, which is a rather numerical one: lattice gauge
>> theory. This approach was invented decenaries after perturbation theory
>> because it requires computers with sufficient performance to do the
>> numerical calculations. The core concept of lattic gauge theory is to
>> approximate space as a lattice, as a set of a finite number of discrete
>> points, for calculating the wave functional described above on these
>> points. Or, as far as vector fields like the electromagnetic field are
>> concerned, not on the points themselves, but on the connections between
>> them.
>
> It can be done either with E(x) or A(x), I don't see the difference, but
> that is a tangent for now. But what is the answer in the limit of zero
> lattice spacing?

Lattice gauge theory has been mainly applied to QCD, the theory of
strong interaction between quarks, not that much to QED. To QED, is has
been only applied for comparison of both interactions. So, there are not
that many known results of lattice gauge theory that concern QED.
However, one known result is that the configuration of an electric field
that interacts with two charged particles looks like the dipole field
known from classical Electrodynamics, making the attractive force
between both particles decay with 1/r^2, whereas in QCD, the strong
field between two quarks takes a configuration with the field lines
confined in a thin tube, making the attractive force distance-independent.

> Is it just like the shifted ground state wave function
> of the harmonic oscilator, except that it is a wave functional, with x
> replaced by E(x)

This question refers to wrong presumptions. Namely that the situation
would be comparable to a harmonic oscillator where the minimum of V(x)
is shifted from x = 0 to some point x != 0. If you imagined the
electromagnetic field as a network of harmonic oscillators with each of
them located at a point in space, then it would be an appropriate
assumption that each oscillator could be described as having the minimum
of its potential shifted, with the shift depending on the distance
between the position of the charge and the point where the particular
oscillator is located.

However, when describing the free radiation field as a set of
oscillators, one does not consider oscillators located in position
space, but modes that correspond to oscillators in momentum space, i.e.
that are totally delocated in position space. So, the analogy of an
oscillator with shifted minimum of V(x) could only apply if that what
corresponds to the shift would be delocated in position space, too, i.e.
would be position-independent. A 1/r^2 electrostatic field, however, is
position-dependent, thefore that analogy is not applicable.

To make the analogy applicable, one has to imagine the field as a
network of oscillators located in position space rather than in momentum
space. However, this produces a new difficulty, namely that those
oscillators located in position space are coupled with each other. The
usual way to handle that coupling is to eliminate it by changing into
momentum space where one has oscillators that are de-coupled, but
exactly that cannot be done here because we must consider things in
position space to apply the described analogy.

Therefore, the known results from lattice gauge theory are rather
numerical than analytical.

> But I thought Fock's point was
> that this Hilbert space is equivalent to the many photon (but not
> constant number necessarily) Hilbert space

What do you mean with "this Hilbert space"? The Hilbert space spanned by
states of the interacting field? Fock did not consider the interacting
field, Fock space formalism is about states of the free radiation field
only. The same is true for other fields than electromagnetic field, e.g.
fermionic matter fields described by Dirac equation: Fock space
formalism deals with free fermions only, not with fermions in interaction.

The Hilbert space spanned by the states of the free radiation field is
equivalent to many photon Hilbert space, that's right, but not the
Hilbert space of the states of the interacting field.

> 1. Use the SR Vector photon wave function, but for a many particle
> state, symmetrized as is appropriate for bosons. Has anyone done this?

What do you mean with "SR Vector photon wave function"? A first
quantization N-particles wave function attributed to a set of N photons?
Since in relativistic quantum theory, particles can be created and
destroyed, especially photons because they have zero-mass, it is widely
agreed that the formalism of second quantization is required, therefore
there probably hasn't been much research about first quantization
descriptions for photons.

And since one can quite easily see that the concept of photons applies
well for the free radiation field only, I don't think that it could be
fruitfull to try to invent an approach for the interacting field based
on a first quantization consideration of N-photon states.

> 2. Add to the photon state Hilbert space, the state space for
> electrons/positrons and allow sums of tensor products.

The result is well-known: co-existence of electromagnetic field and
matter fields (like electron-positron Dirac field) without any
interaction. The tensor products in this Hilbert space are tensor
products of states of interaction-free fields: the free radiation field
in the case of electromagnetic field and matter fields without coupling
to force-carrying fields. A typical state in this Hilbert space contains
electrons, positrons and photons, but these particles do not interact:
an electron and a positron do not attract each other, there's no
repulsion between two electrons or two positrons, and photons are not
scattered by electrons or positrons.

To gather interaction, you need to add interaction terms to the
Lagrangian, and that destroys your tensor product Hilbert space. If you
want, you can apply perturbation theory: you assume asymptotic free
initial and final states that can be approximated as being located in
your tensor product Hilbert space of interaction-free fields, and
interaction processes, e.g. scattering, that imply intermediate states
that are outside this Hilber space but that are out of interest because
all you want to know is what the transition probabilities from the given
initial state to possible final states are. Take e.g.
electron-positron-annihilation: in the initial state, there is an
electron and a positron, in the final state, there are two photons, and
you can calculate to transition probability from intial state to final
state. Any intermediate states, that occur during the process and the
are outside the described Hilbert, are out of consideration in this
approach.

> Perhaps a much
> simpler space since we are dealing with a non-relativistic case of just
> one electron.

For a single electron, you do not need the Hilbert space you described
above, you come along with the very simple one-particle Hilbert space.
There are no photon states in it, though. If you are happy with a
non-relativistic description of the eletrostatic field of that single
electron, you do not need an additional quantum mechanical state for the
field, because in non-relativistic Coulomb theory of electrostatics, the
eletric field has no own dynamical degrees of freedom since it is fully
determined by the electric charge distribution. Therefore the quantum
mechanical state of the electron is all you need for both, the electron
and its electrostatic field.

If you, however, want to describe the electrostatic field of the single
electron quantum-mechanically, what implies that you describe at least
the electromagnetic field relativisticly, things become much more
complicated. At least for the electromagnetic field, you need to
consider the Hilbert space of states of the interacting field, which is
much more complicated that the Hilbert space of photon states of the
free radiation field.


> Has anyone done this?

What do you mean with "this"? You just described three different
procecuderes:

(1) Tensor product Hilbert space, made of tensor products of states of
free fields.

(2) Non-relativistic quantum mechanics with a single-electron state and
an electrostatic field according to non-relativistic Coulomb theory

(3) A single-electron state with an electrostatic field described in the
style of relativistic quantum field theory.

(1) was done by Fock himself. (2) is nothing but non-relativistic
quantum mechanics according to Schroedinger equation from 1925. (3) is
done in lattice gauge theory. In the early years of quantum field
theory, one probably considered to do this, too, but soon found it to be
too difficult to be considered analytically, and invented perturbation
theory instead.

> I would be surprised if not, since
> this is a basic test of QFT--it has to describe physically measurable
> static fields.

As I already mentioned above: a proponent of perturbation theory would
probably not agree with you that static field would be measurable.

[iuval]

On Friday, July 31, 2015 at 3:29:09 PM UTC-7, Jos Bergervoet wrote:
> On 7/31/2015 9:23 AM, iuval wrote:
> > On Wednesday, July 29, 2015 at 9:01:24 PM UTC-7, Gregor Scholten wrote:
> >> iuval wrote:
> >>
> >>> It is easy to see that there is no superposition of coherent photon
> >>> states that would give an expectation value of the electric field
> >>> operator that is static but not constant in space (e.g. 1/r^2 for
> >>> radial component from a point source).
> ...
> ..
> > .. Is it just like the shifted ground state wave function
> > of the harmonic oscilator, except that it is a wave functional, with x
> > replaced by E(x) (1/r^2in the simplest point source example), as the
> > next response suggests? That would be psi(E_r(x), the other 2 E

> > components and 3 B components)=3DN Exp[-(E_r(x)-1/r^2)/(2

> > x0^2)]*delta(other components), where x0 is the vacuum standard
> > deviation/fluctuation, N is a normalization constant, and the delta
> > function is a functional distribution.
>
> I don't see why you would use a delta function to restrict
> anything completely here, unless you are trying to fix the
> gauge.

Sorry, that was an error. I posted a correction but the moderator
didn't let it through for some reason. I should have had an integral
in the exponent first of all over the relevant volume of space since
this functional is a map from real valued functions on R3 to C. The
delta functions should be replaced by gaussian functionals centered
around the function R3->0.

>
> Neglecting the gauge problem, one would expect all field
> components to have vacuum standard deviaton, no more and
> no less, only centered around a strong 1/r^2 field instead
> of centered around zero.

Yes

> And if you don't neglect the gauge
> problem, you'd first have some degrees of freedom not
> restricted at all and then you'd have to choose your own
> way of restricting them (gauge choice) or you would not
> restrict them but integrate them out if your application
> allows some way to do that.
>
> > ... But I thought Fock's point was
> > that this Hilbert space is equivalent to the many photon (but not
> > constant number necessarily) Hilbert space,
>
> Yes why is that a problem? You started asking for a coherent
> state, which would definitely be a many photon state with no
> definite photon number. (Or do you mean that re-centering
> would shift it *out of* the usual Fock space?)

If it's equivalent than we should be able to do it (describe the
classical static field as an expectation value in some state) with
either Hilbert space, not just the functional one.

>
> > .. There are two other possible ways I can see:
> >
> > 1. Use the SR Vector photon wave function, but for a many particle
> > state, symmetrized as is appropriate for bosons. Has anyone done this?
>
> You mean the vector spherical harmonics? They are not free
> fields but all require some multipole source present. You
> may then also consider plane waves that aren't divergence
> free (i.e. having nonzero longitudinal E, and therefore
> always an accompanying charge density wave as a source).

No! I mean the Silberstein-Rieman vector, roughly (but not exactly)
E+iB (where E and B are classical!) which has all the properties
we want for a one particle wave function.

>
> > 2. Add to the photon state Hilbert space, the state space for
> > electrons/positrons and allow sums of tensor products. Perhaps a much
> > simpler space since we are dealing with a non-relativistic case of just
> > one electron. Has anyone done this? I would be surprised if not, since
> > this is a basic test of QFT--it has to describe physically measurable
> > static fields.
>
> It seems here you propose to use a quantum field for the
> source as well! Probably not simpler and I don't see how
> you would build a static situation (OK, you can build a
> kind of hydrogen atom, but its CM coordinate would still
> spread out indefinitely..)

I don't see why. There is only one charge for simplicity.
>
> --
> Jos

[Jos Bergervoet]

On 8/1/2015 3:01 PM, Gregor Scholten wrote:
...

> Electrostatic fields cannot be composed from solutions for the free
> radiation field. Neither in classical field theory nor in quantum field
> theory. In quantum field theory, this means that states of the
> interacting field cannot be composed from photon states.

Not from free-field photon states alone..

>> OK, but I am not interested in that experimental situation. If classical
>> mechanics emerges in some many particle or high energy limit (large
>> alphas in the case of coherent fields where alpha is the eigenvalue of
>> the annihilation operator), then we have to find what the quantum state
>> is for this other very common measurable, experimental situation.
>
> I guess a proponent of perturbation theory would claim that the
> electrostatic field of a single isolated charge is no measurable
> experimental situation.

Irrelevant to the question of how it is described by QFT.
Even if this proponent would want to argue that it *cannot*
be described by QFT, then the above statement alone would
not be a proof of that.

...

> However, when describing the free radiation field as a set of
> oscillators, one does not consider oscillators located in position
> space,

Of course one does. It is the starting point! Coupled oscillators.

> but modes that correspond to oscillators in momentum space, i.e.
> that are totally delocated in position space.

That is the subsequent step to decouple them. With the advantage
that the wave functional will have zero correlation between its
coordinates, which are the A^mu(k). This is computationally
convenient of course, usually. But no more than that.

> .. So, the analogy of an

> oscillator with shifted minimum of V(x) could only apply if that what
> corresponds to the shift would be delocated in position space, too, i.e.
> would be position-independent. A 1/r^2 electrostatic field, however, is
> position-dependent, thefore that analogy is not applicable.

On the contrary, it would apply in both descriptions. In position
space your wave functional gets the position-dependent shift of
its arguments, which are the A^mu(x). Why would you restrict a
shift in coordinates to be the same shift for each coordinate?!

In momentum space the static field would be Fourier transformed
to some 1/k shaped static field, which therefore would still give
unequal shifts for the wave functional's coordinates, A^mu(k).
This construction of the wave functional would of course retain
the advantage of being a product of uncorrelated Gaussians for
the modes (if you start from the vacuum state, that is..)

What makes it interesting is that the shift in coordinates by
an electrostatic field is in fact orthogonal to the directions
that are usually quantized for the free field. For that case we
do not quantize the pure-gauge directions, nor the longitudinal
field directions, only the two two transverse field directions.
Only the latter get a description of the variance around the
mean in the wave functional by the quantization. Longitudinal
configuration directions are fixed by the source and therefore
are zero without any spread in the free field vacuum.

Now we are talking about situations where there *is* a source,
so you could say that assumption of a fixed (classical) charge
distribution leads to a wave functional that is delta-function
resticted in directions of longitudinal field configurations,
but not centered around zero. And at the same time it has the
ordinary vacuum-stae variance the transversal configuration
directions, but in that case around zero.

If, however, you want to acknowledge that the source comes from
some matter field, then it will naturally also have non-zero
variance, which is on-to-one copied to the longitudinal modes
in the photon wave functional. So in that case the question:
"What wave functional does the electrostatic EM field have?"
is returned to you as:
"What wave functional does your source have?"
and you would somehow have to obtain this from a squared matter
field |phi^2|, which may require some technicalities, but in
any case requires you first to answer what state you want the
phi-field to be in! Would you use a single-particle state? If
not, then at least a definite-charge state? (Starting with a
harmonic state for phi, you would get neither, I think..)

--
Jos

[iuval]

[Moderator's note: Please post unencoded text with line breaks every 72
characters or so to avoid ugly text like that below. -P.H.]

I was in a rush during my previous post and made another error. The 3rd cri=
terion I mentioned is not a property of the candidate state. I want to make=
sure that we are talking about the same Hilbert space that is Fock space w=
ith our shifted vacuum wave functional or any other candidate such as the o=
ne JOS just proposed. But perhaps a better question to ask, one that has ex=
perimental consequences, is what are the amplitudes for finding any number =
of photons in our candidate state psi? This leads to the third criterion to=
replace my previous one: (3) <n|psi> should approach 0 in some classical l=
imit, for any n (except for n=0, the vacuum state), since experimentally =
we don't detect any photons in an electrostatic field. This presumes we kno=
w how to take inner products in the wave functional space and represent |n>=
as a wave functional (I think the latter can be done in analogy (which can=
be made rigorous by discretizing a timelike slice of spacetime or discreti=
zing 3D wavevector space as in numerical LGT) with the harmonic oscillator =
case using Hermite polynomial functionals multiplied by gaussian functional=
s. Or conversely we could represent our candidate state in Fock space, wher=
e we know how to take inner products. It seems to me we this is where we wi=
ll run into trouble with singularities (or simply won't get the inner produ=
ct to vanish, by analogy with harmonic oscilator) for both candidates propo=
sed so far and so my suggestion is to use the full interacting Hilbert spac=
e, which can be represented as the set of functionals on 4 functions (the 4=
A^u(x), forget about gauge fixing for now), AND one continuous 3D degree o=
f freedom (the position of the charge). Then impose both the gauge constrai=
nts and a constraint to localize the charge (assumed highly massive), find =
eigenvalues and eigenvectors of resulting hamiltonian (with Lagrange multip=
liers to encode constraints--use perturbation theory or numerical LGT), and=
then show that the 2 conditions I mentioned in previous post and the one i=
n this post hold for the ground state. The reason the third condition will =
hold is because the interacting eigenstates are all orthogonal, but they do=
n't look much like the non-interacting ones, or even products of field wave=
functionals and position wave function. Please see below for more on this=
.
On Sunday, August 2, 2015 at 1:56:56 AM UTC-7, Jos Bergervoet wrote:
> ...

>
> If, however, you want to acknowledge that the source comes from
> some matter field, then it will naturally also have non-zero
> variance, which is on-to-one copied to the longitudinal modes
> in the photon wave functional. So in that case the question:
> "What wave functional does the electrostatic EM field have?"
> is returned to you as:
> "What wave functional does your source have?"
> and you would somehow have to obtain this from a squared matter
> field |phi^2|, which may require some technicalities, but in
> any case requires you first to answer what state you want the
> phi-field to be in! Would you use a single-particle state? If
> not, then at least a definite-charge state? (Starting with a
> harmonic state for phi, you would get neither, I think..)

No, there will just be one wave functional for both, as I suggest above, wh=
ich is not factorizable into one for the field and one (function) for the p=
osition of the source. Maybe factorizability becomes a good approximation i=
n the limit of large source mass.
>
> --
> Jos

[iuval]

On Sunday, August 2, 2015 at 1:56:56 AM UTC-7, Jos Bergervoet wrote:

> On 8/1/2015 3:01 PM, Gregor Scholten wrote:

<snip>

> > I guess a proponent of perturbation theory would claim that the
> > electrostatic field of a single isolated charge is no measurable
> > experimental situation.

>=20

> Irrelevant to the question of how it is described by QFT.
> Even if this proponent would want to argue that it *cannot*
> be described by QFT, then the above statement alone would
> not be a proof of that.
>

Agreed. But QFT MUST be able to describe experimentally measurable
classical quantities such as electric fields in some classical
correspondence (unless it is an incomplete theory), and what that
correspondence is, is not clear here. For coherent states the
correspondence is in the limit of large eigenvalue of the annihilation
operator, which also corresponds to large mean photon number and
energy. Perhaps in the case I'm interested in, the correspondence
is a macroscopic charge, whose classical field is certainly measurable,
and which also must correspond to a large mean eigenvalue of the
interacting hamiltonian (though with vanishingly small variance).
But that might be premature, because even though we might not be
able to directly measure the electric field of a microscopic charge,
we get excellent agreement with experiment (spectroscopy in particular)
by assuming the field of a hydrogen nucleus (and more generally a
many electron atom) to be given by classical electrostatics. So I
think QFT SHOULD be able to describe even the electrostatic field
of a microscopic charge, as long as it has a large enough mass.
Note I am asking a different question here than Oppenheimer who
showed that we need not consider the wave function of the nucleus
in our quantum calculations of electron states, due to the relatively
large nuclear mass.
> ...

> > .. So, the analogy of an
> > oscillator with shifted minimum of V(x) could only apply if that what
> > corresponds to the shift would be delocated in position space, too, i.e.
> > would be position-independent. A 1/r^2 electrostatic field, however, is
> > position-dependent, thefore that analogy is not applicable.

>=20

> On the contrary, it would apply in both descriptions. In position
> space your wave functional gets the position-dependent shift of
> its arguments, which are the A^mu(x). Why would you restrict a
> shift in coordinates to be the same shift for each coordinate?!

I don't understand Gregor's statement either. The proposed state
must have two properties: 1 the expectation value of the field
operators must approach <Er>=3D1/r^2 in some classical limit, all
other components of E and B approach 0 expectation value. 2.The
variance <F^2>-<F>^2 must vanish in some classical limit, where F
is any component of E or B. Whether to use E,B or A with gauge
fixing for these computations is a matter of convenience. 3. If
one gives a wave functional of eigenvalues of the fields as the
candidate state, one must show that the operators Int(f_k |f_k><f_k|
d^3k) where f_k are eigenvalues and |f_k> are eigenstates corresponds
to the operators gotten from canonical quantization.

The first two are easy to show with proposed shifted vacuum state,
but the third is not obvious to me. It seems we switched from a
Heisenberg picture, where we couldn't easily find states satisfying
(1) and (2) (and Gregor claimed it was impossible using the free
field Hilbert space, which I am inclined to agree with), to a
Schrodinger representation where (1) and (2) are obvious for proposed
state, but (3) may not follow.

>
<snip>

> What makes it interesting is that the shift in coordinates by
> an electrostatic field is in fact orthogonal to the directions
> that are usually quantized for the free field. For that case we
> do not quantize the pure-gauge directions, nor the longitudinal
> field directions, only the two two transverse field directions.
> Only the latter get a description of the variance around the
> mean in the wave functional by the quantization. Longitudinal
> configuration directions are fixed by the source and therefore
> are zero without any spread in the free field vacuum.

>=20

Do you mean fixed by the GAUGE (you said "SOURCE")? You just said
no source for this case.

> Now we are talking about situations where there *is* a source,
> so you could say that assumption of a fixed (classical) charge
> distribution leads to a wave functional that is delta-function
> resticted in directions of longitudinal field configurations,
> but not centered around zero. And at the same time it has the
> ordinary vacuum-stae variance the transversal configuration
> directions, but in that case around zero.

OK, so this is another candidate for the state (than the one I wrote
down) that has to satisfy the 3 properties I mention above. Still
(3) is not obvious for this state. I hope (3) will enable us to
pick a unique state among all candidates satisfying (1) and (2).
This latest candidate is more singular than the previous one.

>
> If, however, you want to acknowledge that the source comes from
> some matter field, then it will naturally also have non-zero
> variance, which is on-to-one copied to the longitudinal modes
> in the photon wave functional. So in that case the question:
> "What wave functional does the electrostatic EM field have?"
> is returned to you as:
> "What wave functional does your source have?"
> and you would somehow have to obtain this from a squared matter
> field |phi^2|, which may require some technicalities, but in
> any case requires you first to answer what state you want the
> phi-field to be in! Would you use a single-particle state? If
> not, then at least a definite-charge state? (Starting with a
> harmonic state for phi, you would get neither, I think..)

I disagree with this but I have to go now. More later.
>
> --
> Jos

[Paul Colby]

On 2015-07-28 09:42:55 +0000, iuval said:

> [[Mod. note -- Please limit your text to fit within 80 columns,
> preferably around 70, so that readers don't have to scroll horizontally
> to read each line. I have manually reformatted this article. -- jt]]
>
> It is easy to see that there is no superposition of coherent photon
> states that would give an expectation value of the electric field
> operator that is static but not constant in space (e.g. 1/r^2 for
> radial component from a point source). The reason is that the
> electric field operator has an an integral of (Exp(i (k.r-wt) a-
> complex conjugate) (1/2 w)d^3k. There is the (2 w) in the denominator
> but that gets cancelled by the delta function normalization (2 w)
> when integrating over the superposition of k's of coherent
> states(Int[A_k |alpha_k> d^3k'] for whatever component of the vector

> state). In the end you get <Ex>=3D2 Sqrt(hbar/2epsilonO V) Int[|alpha=

_k|
> |A_k|^2 Sin(w t- k.r-theta_k)Sqrt[w] d^3k ]. Now w=3D|k| for photons,
> so since sin(wt-theta_k-k.r)=3Dsin(wt-theta_k)
> cos(k.r)+cos(wt-theta_k)sin(kr), we have the situation of how to
> eliminate the t dependence. If there was only one term it would
> be obvious that there are no A_k that can produce the given static
> expectation value, from taking the inverse (3D, not 4D!) fourier
> transform, but I think it is still true with two terms, though I
> haven't proven it yet. So if a classical field emerges as an
> expectation value in a quantum state, which state is it? The appeal
> of coherent states is that their standard deviation goes to 0 as
> 1/Sqrt(volume) or for finite volume is proportional to Sqrt[h].

My memory could be faulty but as I recall there is no momentum
conjugate to the scalar component of potential. Static E-fields are
classical.

[iuval]

On Wednesday, August 5, 2015 at 12:37:40 PM UTC-7, Paul Colby wrote:
> On 2015-07-28 09:42:55 +0000, iuval said:

>=20

> > [[Mod. note -- Please limit your text to fit within 80 columns,
> > preferably around 70, so that readers don't have to scroll horizontally
> > to read each line. I have manually reformatted this article. -- jt]]

> >=20

> > It is easy to see that there is no superposition of coherent photon
> > states that would give an expectation value of the electric field
> > operator that is static but not constant in space (e.g. 1/r^2 for
> > radial component from a point source). The reason is that the
> > electric field operator has an an integral of (Exp(i (k.r-wt) a-
> > complex conjugate) (1/2 w)d^3k. There is the (2 w) in the denominator
> > but that gets cancelled by the delta function normalization (2 w)
> > when integrating over the superposition of k's of coherent
> > states(Int[A_k |alpha_k> d^3k'] for whatever component of the vector

> > state). In the end you get <Ex>=3D3D2 Sqrt(hbar/2epsilonO V) Int[|alpha=
=3D
> _k|
> > |A_k|^2 Sin(w t- k.r-theta_k)Sqrt[w] d^3k ]. Now w=3D3D|k| for photons,
> > so since sin(wt-theta_k-k.r)=3D3Dsin(wt-theta_k)

> > cos(k.r)+cos(wt-theta_k)sin(kr), we have the situation of how to
> > eliminate the t dependence. If there was only one term it would
> > be obvious that there are no A_k that can produce the given static
> > expectation value, from taking the inverse (3D, not 4D!) fourier
> > transform, but I think it is still true with two terms, though I
> > haven't proven it yet. So if a classical field emerges as an
> > expectation value in a quantum state, which state is it? The appeal
> > of coherent states is that their standard deviation goes to 0 as
> > 1/Sqrt(volume) or for finite volume is proportional to Sqrt[h].

>=20

> My memory could be faulty but as I recall there is no momentum
> conjugate to the scalar component of potential. Static E-fields are
> classical.

Why is this relevant? As long as we can measure a classical static
field, we need to explain how it emerges out of quantum mechanics.
The electric field is just an operator in QM, it does not need to
have a conjugate momentum, and anyway it is given in terms of
derivatives of A^u. A^u does have a conjugate momentum, which depends
on the gauge.

[Gregor Scholten]

Jos Bergervoet wrote:

> What makes it interesting is that the shift in coordinates by
> an electrostatic field is in fact orthogonal to the directions
> that are usually quantized for the free field. For that case we
> do not quantize the pure-gauge directions, nor the longitudinal
> field directions, only the two two transverse field directions.
> Only the latter get a description of the variance around the
> mean in the wave functional by the quantization. Longitudinal
> configuration directions are fixed by the source and therefore
> are zero without any spread in the free field vacuum.

In other words: you decompose the electromagnetic field into an
electrostatic Coulomb part and a radiation part. And you implement this
in the four-potential

A^mu = (A^0, \vec A)

by using Coulomb gauge so that the scalar potential A^0 describes the
electrostatic part only and the vector potential \vec A describes the
radiation part only. And since the scalar potential is fully determined
by the charge density in this gauge, you only need to quantize the
vector potential, what is equivalent to quantize the free radiation
field only. In this approach, the answer to iuval's original question
what the quantum state is of which the electrostatic field is an
expectation value is: the electrostatic field isn't an expectation value
of a quantum state, at least not of a quantum state of the field (but
maybe of the quantum state of the charge-carrying particle).

However, there is something wrong in this approach: it ignores that
besides electrostatic part and radiation part, there is a third part of
the electromagnetic field, namely the magnetostatic part. A classical
charge the is translating or rotating does not only cause an
eletrostatic field, but also a magnetostatic one. That magnetostatic
field is described by the vector potential \vec A (since \vec B = \rot
\vec A), so the vector potential does not describe the radiation part only.

You can consider things in the rest frame of the charge, so that the
charge is not translating, but since the magnetostatic field is part of
the vector potential which is quantized, you cannot simply assume the
magnetostatic field to be fixed to be zero. And as far as an electron is
concerned, the particle has a spin and therefore a magnetic dipole
contribution to its magnetostatic field. So, the magnetostatic field
cannot be eliminated.

[Gregor Scholten]

iuval wrote:

>> My memory could be faulty but as I recall there is no momentum
>> conjugate to the scalar component of potential. Static E-fields are
>> classical.
>
> Why is this relevant? As long as we can measure a classical static
> field, we need to explain how it emerges out of quantum mechanics.
> The electric field is just an operator in QM, it does not need to
> have a conjugate momentum, and anyway it is given in terms of
> derivatives of A^u. A^u does have a conjugate momentum, which depends
> on the gauge.

Obviously, Paul is, like Jos, reffering to an approach that is based on
using Coulomb gauge. If you differentiate the Lagrange density of QED
with respect to the time derivative d(A^0)/dt of the scalar potential,
you get 0 as result, indicating that A^0 does not have a conjugate
momentum, unlike the spatial components A^i, i=1,2,3, and is therefore
no dynamical degree of freedom (it is fully determined by the eletric
charge density). For quantization, this means that A^0 is not to be
quantized, unlike the spatial components A^i. Now, using Coulomb gauge,
that defines the spatial part \vec A = A^i as being source-free:

\div \vec A = 0

one can describe electrostatic fields by the non-quantized scalar
potential A^0, and the other parts of the electromagnetic field
(radiation field and magnetostatic field) by the quantized vector
potential A^i.

There is another approach to quantize the elctromagnetic field where
alle four components of A^mu are quantized and that is therefore in
better compliance with special-relativistic Lorentz-covariance. This
approach, however, requires a change in the Lagrange density L to make

d L / d(d(A^0)/dt)

non-zero so that A^0 is equipped with a conjugate momentum. This
approach is the one that is mostly used for perturbation theory. The
trouble with this approach is that it yields four polarisation
directions for photons (besides two transveral ones a longitudinal one
and a scalar one) instead of the well-known two transversal directions.
To reduce this number of polarisation directions from four to two, there
is Gupta-Bleuler formalism in which longitudinal and scalar photon
states turn out to be non-physical. Only concerning virtual particles
that occur in perturbation theory, scalar and longitudinal photons are
relevant, e.g. for describing scattering of two electrons or an electron
and a positron.

[Paul Colby]

On 2015-08-05 22:41:02 +0000, iuval said:

> On Wednesday, August 5, 2015 at 12:37:40 PM UTC-7, Paul Colby wrote:
>> On 2015-07-28 09:42:55 +0000, iuval said:

>> =3D20

>>> [[Mod. note -- Please limit your text to fit within 80 columns,
>>> preferably around 70, so that readers don't have to scroll horizontally
>>> to read each line. I have manually reformatted this article. -- jt]]

>>> =3D20

>>> It is easy to see that there is no superposition of coherent photon
>>> states that would give an expectation value of the electric field
>>> operator that is static but not constant in space (e.g. 1/r^2 for
>>> radial component from a point source). The reason is that the
>>> electric field operator has an an integral of (Exp(i (k.r-wt) a-
>>> complex conjugate) (1/2 w)d^3k. There is the (2 w) in the denominator
>>> but that gets cancelled by the delta function normalization (2 w)
>>> when integrating over the superposition of k's of coherent
>>> states(Int[A_k |alpha_k> d^3k'] for whatever component of the vector

>>> state). In the end you get <Ex>=3D3D3D2 Sqrt(hbar/2epsilonO V) Int[|alp=
ha=3D
> =3D3D
>> _k|
>>> |A_k|^2 Sin(w t- k.r-theta_k)Sqrt[w] d^3k ]. Now w=3D3D3D|k| for photon=
s,
>>> so since sin(wt-theta_k-k.r)=3D3D3Dsin(wt-theta_k)

>>> cos(k.r)+cos(wt-theta_k)sin(kr), we have the situation of how to
>>> eliminate the t dependence. If there was only one term it would
>>> be obvious that there are no A_k that can produce the given static
>>> expectation value, from taking the inverse (3D, not 4D!) fourier
>>> transform, but I think it is still true with two terms, though I
>>> haven't proven it yet. So if a classical field emerges as an
>>> expectation value in a quantum state, which state is it? The appeal
>>> of coherent states is that their standard deviation goes to 0 as
>>> 1/Sqrt(volume) or for finite volume is proportional to Sqrt[h].

>> =3D20

>> My memory could be faulty but as I recall there is no momentum
>> conjugate to the scalar component of potential. Static E-fields are
>> classical.

>=20

> Why is this relevant? As long as we can measure a classical static
> field, we need to explain how it emerges out of quantum mechanics.
> The electric field is just an operator in QM, it does not need to
> have a conjugate momentum, and anyway it is given in terms of
> derivatives of A^u. A^u does have a conjugate momentum, which depends
> on the gauge.

It seems relevant to me because there are components of EM fields that
appear as c-numbers in the theory and are measurable. They have nothing
to do with mass shell photons.

[Jos Bergervoet]

On 8/6/2015 1:00 PM, Gregor Scholten wrote:
> iuval wrote:

>> ...
>
> .. The trouble with this approach is that it yields four polarisation

> directions for photons (besides two transveral ones a longitudinal one
> and a scalar one) instead of the well-known two transversal directions.
> To reduce this number of polarisation directions from four to two, there
> is Gupta-Bleuler formalism in which longitudinal and scalar photon
> states turn out to be non-physical.

We could of course use the Faddeev-popov approach and quantize
the gauge degrees of freedom (creating ghosts, of course). In
QED it is not needed (whether in QCD it is really unavoidable
can be debated) but one could nevertheless do it.

That would still leave one polarization untreated, but it sort
of solves the problem as follows:
1) Gauge degrees of freedom that were completely unrestricted
in the wave functional (or the in the path integral) are
now restricted by the quantized ghosts.
2) The longitudinal degrees of freedom already were delta-
function restricted to the source charge, so they're either
infinitely restricted with a classical source or "inherit"
their spread from a matter field describing the charge.


I think it is easiest to see in (1+1) dimensions with a complex
scalar matter field and an EM field. Only the ghost and the
longitudinal field exist for EM in that case. And the magnetic
field mentioned earlier is also absent in (1+1) EM.

So in that case you don't need a QFT for EM: the only thing
you can do with it is describe the (unnecessary) ghost, the
longitudinal field is directly expressed in the matter field
(which of course still does need a QFT).

--
Jos

[Roland Franzius]

Am 05.08.2015 um 21:37 schrieb Paul Colby:
>
> My memory could be faulty but as I recall there is no momentum
> conjugate to the scalar component of potential. Static E-fields are
> classical.

Generally speaking, as a theoretical professional, one should not annoy
elementary particle physicists with details of the construction of state
spaces and the representation of the algebra of observables by a
GNS-construction process in the cas of an interacting QFT.

The total nonsensical rap about static fields representing the
0-component gauge ("energy") of the four momentum vector observable
scientifically ranges on the same level as gold cooking in the times of
Galilei and Newton. Nice stuff for TV pseodo-scienctific religiosity and
free energy believers.

--

Roland Franzius

[iuval]

My philosophical objection in a reply to a previous post above is
strengthened by this observation, because one can boost to another
Lorentz frame and mix the supposedly quantum and classical components of
A^u. This is obvious nonsense. It's bad enough if part of the world is
quantum and another classical, but even worse if the classical and
quantum parts are components of a covariant vector.

>
> There is another approach to quantize the elctromagnetic field where
> alle four components of A^mu are quantized and that is therefore in
> better compliance with special-relativistic Lorentz-covariance. This
> approach, however, requires a change in the Lagrange density L to make
>
> d L / d(d(A^0)/dt)
>
> non-zero so that A^0 is equipped with a conjugate momentum. This
> approach is the one that is mostly used for perturbation theory. The
> trouble with this approach is that it yields four polarisation
> directions for photons (besides two transveral ones a longitudinal one
> and a scalar one) instead of the well-known two transversal directions.
> To reduce this number of polarisation directions from four to two, there
> is Gupta-Bleuler formalism in which longitudinal and scalar photon
> states turn out to be non-physical. Only concerning virtual particles
> that occur in perturbation theory, scalar and longitudinal photons are
> relevant, e.g. for describing scattering of two electrons or an electron
> and a positron.

So how does the GB formalism help to answer my original question, since
you can't waive away the electrostatic field, but only the longitudinal
and scalar photon states? Also, I should have mentioned this before, but
the whole concept of longitudinal and transverse only makes sense for
photons with a non-zero component of (the spatial part of) the wave
vector. For a static field (electro or magneto) that component is 0.

[iuval]

On Thursday, August 6, 2015 at 12:06:39 AM UTC-7, Gregor Scholten wrote:
> Jos Bergervoet wrote:
>
> > What makes it interesting is that the shift in coordinates by
> > an electrostatic field is in fact orthogonal to the directions
> > that are usually quantized for the free field. For that case we
> > do not quantize the pure-gauge directions, nor the longitudinal
> > field directions, only the two two transverse field directions.
> > Only the latter get a description of the variance around the
> > mean in the wave functional by the quantization. Longitudinal
> > configuration directions are fixed by the source and therefore
> > are zero without any spread in the free field vacuum.
>
> In other words: you decompose the electromagnetic field into an
> electrostatic Coulomb part and a radiation part. And you implement this
> in the four-potential
>
> A^mu = (A^0, \vec A)
>
> by using Coulomb gauge so that the scalar potential A^0 describes the
> electrostatic part only and the vector potential \vec A describes the
> radiation part only. And since the scalar potential is fully determined
> by the charge density in this gauge, you only need to quantize the
> vector potential, what is equivalent to quantize the free radiation
> field only.

This doesn't follow. Not only the electrostatic field, but any field, including
magnetostatic, or radiation follow from ("are fully determined by" the classical position and velocity of
the charges. Just because you can get the classical fields from Maxwell
equations doesn't mean we don't need to quantize the charges and fields. There
is nothing special about the electrostatic field from this point of view.

>In this approach, the answer to iuval's original question
> what the quantum state is of which the electrostatic field is an
> expectation value is: the electrostatic field isn't an expectation value
> of a quantum state, at least not of a quantum state of the field (but
> maybe of the quantum state of the charge-carrying particle).

It can't be the expectation value of the state of the charge. The charge
is not measured by an electric field operator, but only by such things as
position, momentum, energy, charge. Can you write down the electric field operator on the Hilbert space of the charge?

As to your first possibility, I have a philosphical problem with having a
part of the world be classical and a part quantum. It is either all quantum
or all classical (as in hidden variable theories). The whole motivation for
trying to quantize gravity would be gone if we could just say spacetime is
classical but everything else is quantum.

>
> However, there is something wrong in this approach: it ignores that
> besides electrostatic part and radiation part, there is a third part of
> the electromagnetic field, namely the magnetostatic part. A classical
> charge the is translating or rotating does not only cause an
> eletrostatic field, but also a magnetostatic one. That magnetostatic
> field is described by the vector potential \vec A (since \vec B = \rot
> \vec A), so the vector potential does not describe the radiation part only.
>
> You can consider things in the rest frame of the charge, so that the
> charge is not translating, but since the magnetostatic field is part of
> the vector potential which is quantized, you cannot simply assume the
> magnetostatic field to be fixed to be zero. And as far as an electron is
> concerned, the particle has a spin and therefore a magnetic dipole
> contribution to its magnetostatic field. So, the magnetostatic field
> cannot be eliminated.

Yes, even without the intrinsic spin, the magnetostatic field can't be hand-
waived away. It must emerge as the expectation value of the magnetic field
operator.

[iuval]

[[Mod. note -- Please limit your text to fit within 80 columns,
preferably around 70, so that readers don't have to scroll horizontally
to read each line. I have manually reformatted this article. -- jt]]

On Monday, August 10, 2015 at 1:11:46 AM UTC-7, Paul Colby wrote:

> >> My memory could be faulty but as I recall there is no momentum
> >> conjugate to the scalar component of potential. Static E-fields are
> >> classical.
> >
Iuval wrote:
> > Why is this relevant? As long as we can measure a classical static
> > field, we need to explain how it emerges out of quantum mechanics.
> > The electric field is just an operator in QM, it does not need to
> > have a conjugate momentum, and anyway it is given in terms of
> > derivatives of A^u. A^u does have a conjugate momentum, which depends
> > on the gauge.
>
> It seems relevant to me because there are components of EM fields that
> appear as c-numbers in the theory and are measurable. They have nothing
> to do with mass shell photons.

The question remains: describe, in whatever Hilbert space representation
you'd like, the quantum state that gives as the expectation value of the
electric field operator (whether it has c numbers as part of it or not),
the classically observed electrostatic field of a static charge
configuration, whether that charge configuration is quantum or in some
classical limit.

[iuval]

[[Mod. note -- Please limit your text to fit within 80 columns,
preferably around 70, so that readers don't have to scroll horizontally
to read each line. I have manually reformatted this article. -- jt]]

On Monday, August 10, 2015 at 1:11:58 AM UTC-7, Jos Bergervoet wrote:
> On 8/6/2015 1:00 PM, Gregor Scholten wrote:
> > iuval wrote:
> >> ...
> >
> > .. The trouble with this approach is that it yields four polarisation
> > directions for photons (besides two transveral ones a longitudinal one
> > and a scalar one) instead of the well-known two transversal directions.

> > To reduce this number of polarisation directions from four to two, ther=

e
> > is Gupta-Bleuler formalism in which longitudinal and scalar photon
> > states turn out to be non-physical.

>=20

> We could of course use the Faddeev-popov approach and quantize
> the gauge degrees of freedom (creating ghosts, of course). In
> QED it is not needed (whether in QCD it is really unavoidable
> can be debated) but one could nevertheless do it.

Sorry, I need to go back and study Ryder on ghosts and Fadeev-Popov to make=
an intelligent response but I'll try even now.
>=20

> That would still leave one polarization untreated, but it sort
> of solves the problem as follows:
> 1) Gauge degrees of freedom that were completely unrestricted
> in the wave functional (or the in the path integral) are
> now restricted by the quantized ghosts.
> 2) The longitudinal degrees of freedom already were delta-
> function restricted to the source charge, so they're either
> infinitely restricted with a classical source or "inherit"
> their spread from a matter field describing the charge.

So can you write down the wave functional from which the classical
electrostatic field emerges, with the 3 criteria I posted in a
previous post? Is it the ground state functional in the presence
of a static=20 charge, and does including the ghosts in the Hamiltonian
(rather than using the coulomb gauge or the Gupta-Bleuler formalism)
help you diagonalize it somehow? Can you show that the solution is
your 2) above (which I still hope you can write down more concretely)?
Thanks!

[[Mod. note -- 14 excessively-quoted lines snipped here. -- jt]]

[Hendrik van Hees]

On 11/08/15 09:26, iuval wrote:
>
> So how does the GB formalism help to answer my original question, since
> you can't waive away the electrostatic field, but only the longitudinal
> and scalar photon states? Also, I should have mentioned this before, but
> the whole concept of longitudinal and transverse only makes sense for
> photons with a non-zero component of (the spatial part of) the wave
> vector. For a static field (electro or magneto) that component is 0.
>

The Gupta-Bleuler formalism (for which GB seems to stand here) is just a
way to canonically quantize the electromagnetic field, which is
necessarily a gauge field, in any linear gauge and thus particularly in
the covariant Lorenz gauge. It has not too much to do with your
question, and the path-integral quantization is much more transparent.

The Coulomb field does not exist for free electromagnetic fields and
thus cannot be constructed with photon Fock states, which are always
free-photon states. You can construct coherent states, which represent
(in the limit of large average photon numbers) classical radiation
fields. As in classical electromagnetism the free electromagnetic fields
are always radiation fields. In the absence of charge-current
distributions no static electromagnetic field can exist.

The Coulomb field occurs in perturbation theory as an approximation for
the case of a very heavy charged particle when describing the
electromagnetic interaction of light particles with this charge. Then
you can work in a frame of referene, where the heavy charge is at rest
and evaluate the scattering amplitudes for collisions between this heavy
charge and the light charge. As it turns out, due to infrared and
collinear singularities of the t-channel diagrams, you have to resum an
infinite set of ladder diagrams to be consistent to a certain order in
the electromagnetic coupling constant alpha. The upshot of this
resummation is that in leading order the heavy charge can be treated as
a classical Coulomb field of this charge at rest, neglecting the recoil
in the scattering process. You can also take this approximation as the
starting point to solve the bound-state problem (e.g., the relativistic
hydrogen problem). This derivation of the usual external-field ansatz
for the bound-state problem then enables you to systematically improve
this approximation with radiative corrections, leading to a nearly
perfect desciprtion of phenomena like the Lamb shift.

--
Hendrik van Hees
Goethe University (Institute for Theoretical Physics)
D-60438 Frankfurt am Main
http://fias.uni-frankfurt.de/~hees/

[Gregor Scholten]

iuval wrote:

>> Obviously, Paul is, like Jos, reffering to an approach that is based on
>> using Coulomb gauge. If you differentiate the Lagrange density of QED
>> with respect to the time derivative d(A^0)/dt of the scalar potential,
>> you get 0 as result, indicating that A^0 does not have a conjugate
>> momentum, unlike the spatial components A^i, i=1,2,3, and is therefore
>> no dynamical degree of freedom (it is fully determined by the eletric
>> charge density). For quantization, this means that A^0 is not to be
>> quantized, unlike the spatial components A^i. Now, using Coulomb gauge,
>> that defines the spatial part \vec A = A^i as being source-free:
>>
>> \div \vec A = 0
>>
>> one can describe electrostatic fields by the non-quantized scalar
>> potential A^0, and the other parts of the electromagnetic field
>> (radiation field and magnetostatic field) by the quantized vector
>> potential A^i.
>
> My philosophical objection in a reply to a previous post above is
> strengthened by this observation, because one can boost to another
> Lorentz frame and mix the supposedly quantum and classical components of
> A^u. This is obvious nonsense. It's bad enough if part of the world is
> quantum and another classical, but even worse if the classical and
> quantum parts are components of a covariant vector.

Just to prevent misunderstandings: in the described approach, the
component A^0 *is* described quantum mechanically. However, the quantum
mechanical description is not provided by quantization of A^0 as
dynamical degree of freedom, but by quantization of the matter field
that is carrying the electric charge. So, there is no mixture of quantum
mechanical things and classical things, but rather a mixture of two
fields (matter field and electromagnetic field). This is not an
unexpected result, since both fields are interacting.

On the other hand, you're right when you point out that the different
handling of A^0 and the spatial components A^i is not in good agreement
with Lorentz covariance. However, in the situation that you considered,
a single electric charge encompassed by its electric field, it is
natural to consider things with respect to the charge's rest frame, so
that one does not need to care that much about transformations to other
frames. The situation is similar in lattice gauge theory: it is natural
to apply the inertial frame that was chosen to define the lattice.

In addition, in both situations, one can point out that Lorentz
covariance is granted by the Lagrange density being Lorentz-invariant.
Lagrange formalism implies that there are conjugate momenta that are
defined by differentiating the Lagrange density with respect to the time
derivatives of the dynamical variables of the theory, which is a
necessarily frame-dependent procedure: let L be the Lagrange density, q
a dynamical variable, qdot = dq/dt its time derivative, and p its
conjugate momentum, then applies:

p = \partial_qdot L

For qdot = dq/dt being defined, a time variable t must be defined, what
requires that a special frame of reference is used. Due to this
frame-dependency of the procedure, the resulting set of conjugate
momenta does not need to be Lorentz-covariant, although the Lagrange
density itself is Lorentz-invariant, and by this, the theory is
Lorentz-covariant.

In Electrodynamics, the Lagrange density of the electromagnetic field is

L = F^{mu,nu} F_{mu,nu} + A^mu s_mu

with

F^{mu,nu} = \partial^mu A^nu - \partial^nu A^mu

being the field strength tensor and s^mu the electric four-flux density.
Due to is asymmetric structure, the field strength tensor only contains
non-diagonal components. This especially means that there is no
\partial^0 A^0 component, i.e. no d(A^0)/dt component. Therefore, there
is no d(A^0)/dt term in the Lagrange density, and therefore, there is no
conjugate momentum existing for A^0.

Going back to the two situations considered above, single charge with
electrostatic field and lattice gauge theory, one can say in other
words: one could as well go into any other inertial frame and apply
Lagrange formalism there (maybe making calculations much more
complicated), and achieve the same results.

>> There is another approach to quantize the elctromagnetic field where
>> alle four components of A^mu are quantized and that is therefore in
>> better compliance with special-relativistic Lorentz-covariance. This
>> approach, however, requires a change in the Lagrange density L to make
>>
>> d L / d(d(A^0)/dt)
>>
>> non-zero so that A^0 is equipped with a conjugate momentum. This
>> approach is the one that is mostly used for perturbation theory. The
>> trouble with this approach is that it yields four polarisation
>> directions for photons (besides two transveral ones a longitudinal one
>> and a scalar one) instead of the well-known two transversal directions.
>> To reduce this number of polarisation directions from four to two, there
>> is Gupta-Bleuler formalism in which longitudinal and scalar photon
>> states turn out to be non-physical. Only concerning virtual particles
>> that occur in perturbation theory, scalar and longitudinal photons are
>> relevant, e.g. for describing scattering of two electrons or an electron
>> and a positron.
>
> So how does the GB formalism help to answer my original question, since
> you can't waive away the electrostatic field, but only the longitudinal
> and scalar photon states?

Gupta-Bleuler formalism doesn't help to answer your question at all. The
focus of Gupa-Bleuler formalism is Lorentz-covariant description of the
free radiation field on the one hand and perturbation theory on the
other hand, not the electrostatic field of a single charge. The approach
behind Gupta-Bleuler formalism, where all four components of A^mu are
quantized, concerns your question in that way that A^mu is no longer
fully determined by the electric charge density. But this does not help
to answer your question very much.

> Also, I should have mentioned this before, but
> the whole concept of longitudinal and transverse only makes sense for
> photons with a non-zero component of (the spatial part of) the wave
> vector. For a static field (electro or magneto) that component is 0.

No, for a static field, that component isn't even defined. A static
field cannot be described by perturbation theory, and therefore not by
virtual photons.

[Gregor Scholten]

iuval wrote:

>>> What makes it interesting is that the shift in coordinates by an
>>> electrostatic field is in fact orthogonal to the directions that
>>> are usually quantized for the free field. For that case we do not
>>> quantize the pure-gauge directions, nor the longitudinal field
>>> directions, only the two two transverse field directions. Only
>>> the latter get a description of the variance around the mean in
>>> the wave functional by the quantization. Longitudinal
>>> configuration directions are fixed by the source and therefore
>>> are zero without any spread in the free field vacuum.
>>
>> In other words: you decompose the electromagnetic field into an
>> electrostatic Coulomb part and a radiation part. And you implement
>> this in the four-potential
>>
>> A^mu = (A^0, \vec A)
>>
>> by using Coulomb gauge so that the scalar potential A^0 describes
>> the electrostatic part only and the vector potential \vec A
>> describes the radiation part only. And since the scalar potential
>> is fully determined by the charge density in this gauge, you only
>> need to quantize the vector potential, what is equivalent to
>> quantize the free radiation field only.
>
> This doesn't follow. Not only the electrostatic field, but any field,
> including magnetostatic, or radiation follow from ("are fully

> determined by") the classical position and velocity of the charges.

You're wrong. In Maxwell's theory of Electrodynamics, the free radiation
field has own dynamical degrees of freedom: you can set rho = 0 and \vec
j = 0 everywhere, and Maxwell equations have still plane wave solutions.
With charges and currents present, those can occur to be sources of
radiation, and by setting proper initial conditions, you can eliminate
all radiation that was not emitted by charges and currents, but when
quantizing Electrodynamics, you necessarily have to take into account
the dynamical degrees of freedom of the radiation field. You cannot
eliminate them by choosing proper initial conditions.

That is why it is possible to quantize the free radiation field in the
absence of charges and currents. Otherwise, this would not be possible.

So, it comes out that the radiation field is not fully determined by
charges and currents, whereas the electrostatic and magnetostatic field
are. And since the radiation field has no A^0 component, A^0 is fully
determined by the charges and currents.

> Just because you can get the classical fields from Maxwell equations
> doesn't mean we don't need to quantize the charges and fields.

You are wrong again. Only dynamical degrees of freedom are to be
quantized in quantization procedure.

>> In this approach, the answer to iuval's original question what the
>> quantum state is of which the electrostatic field is an expectation
>> value is: the electrostatic field isn't an expectation value of a
>> quantum state, at least not of a quantum state of the field (but
>> maybe of the quantum state of the charge-carrying particle).
>
> It can't be the expectation value of the state of the charge. The
> charge is not measured by an electric field operator, but only by
> such things as position, momentum, energy, charge.

Assumed, one uses the terminology that observables are "measured" by
operators, the charge of a particle is "measured" by exactly one
operator, namely the charge operator. The position operator "measures"
position, not charge, the momentum operator "measures" momentum, not
charge, an "energy operator" does not exist, except you mean the
Hamilton operator.

And like a position operator or momentum operator is defined for a
single charge, one can define a scalar potential operator for a single
charge. In first quantization, this operator can be written as

\op{A^0}(x) = q / |x - \op{x}|

with \op{x} being the position operator and q the charge (in first
quantization, the charge operator is simply the classical charge).

> Can you write down
> the electric field operator on the Hilbert space of the charge?

One can write the scalar potential operator. With the electric field
being the gradient of the scalar potential:

\vec E = - grad A^0 = -q (x - x0)^3 / |x - x0|

with x0 = position of the charge, one can write an electric field operator

\op{\vec E} = -q (x - \op{x})^3 / |x - \op{x}|

> As to your first possibility, I have a philosphical problem with
> having a part of the world be classical and a part quantum. It is
> either all quantum or all classical (as in hidden variable theories).

Refer to my parallel post: that A^0 is not quantized as dynamical degree
of freedom does not mean that wouldn't be quantum mechanical. Its
quantum mechanical description is provided by the quantization of the
charge-carrying matter field.

> The whole motivation for trying to quantize gravity would be gone if
> we could just say spacetime is classical but everything else is
> quantum.

The reason why it is considered necessary to quantize gravity is that,
similar to the electromagnetic field, there are components of the
gravitational field that are dynamical degrees of freedom. And, again
similar to the electromagnetic field, not all components are dynamical
degrees of freedom: in GR, the gravitational field can be expressed by
the metric tensor g_{mu,nu} that can be written as 4x4 matrix with 16
components, where 10 components are independent from each other.
However, only the spatial components h_ij = g_ij, i,j=1,2,3, are
dynamical degrees of freedom, the temporal component g_00 and the mixed
spatial-temporal components g_0i and g_i0 are not, those are, analog to
A^0 in Electrodynamics, fully determined by the energy-momentum tensor
T_{mu,nu}.

And analog to the two approaches described earlier to quantize
Electrodynamics, there are to approaches to quantize gravity: the
canonical one, where only the spatial components h_ij (a 3x3 matrix with
9 components where 6 are independent from each other) are quantized, and
the covariant one, where the full metric tensor g_{mu,nu} is quantized.
Results of the canonical approach are Quantum Cosmology (Wheeler-deWitt
equation) and Loop Quantum Gravity, a result of the covariant approach
is Superstring Theory.

[Roland Franzius]

[[Mod. note -- I apologise for the long delay in processing this article,
which was originally submitted on Thursday 13 Aug 2015.
-- jt]]

This is of course plainly wrong. There are no solutions of classical
Maxwell equations dF=0, d*F=0 in all of empty Minkowski space without
field singularities, given the condition of finite square integrable
|F|^2 and subject to the special smoothness condition along a Cauchy
surface as starting values for an equation of hyperbolic type over the
future of that surface in Minkowski space.

The permanent confusion about Fourier decompositon and plane wave and
other types of generalized solutions in the dual field space is
notorious for undergraduate courses in electrodynamics and late on
develops into a severe obstacle to understand quantization of the
electromagnetic field.

--

Roland Franzius

[iuval]

[[Mod. note -- I apologise for the delay in processing this article,
which was originally submitted on Saturday 15 August 2015.
-- jt]]

On Thursday, August 6, 2015 at 12:06:39 AM UTC-7, Gregor Scholten wrote:
> Jos Bergervoet wrote:
>
>

> In other words: you decompose the electromagnetic field into an
> electrostatic Coulomb part and a radiation part. And you implement this
> in the four-potential
>
> A^mu = (A^0, \vec A)
>
> by using Coulomb gauge so that the scalar potential A^0 describes the
> electrostatic part only and the vector potential \vec A describes the
> radiation part only. And since the scalar potential is fully determined
> by the charge density in this gauge, you only need to quantize the
> vector potential, what is equivalent to quantize the free radiation
> field only. In this approach, the answer to iuval's original question
> what the quantum state is of which the electrostatic field is an
> expectation value is: the electrostatic field isn't an expectation value
> of a quantum state, at least not of a quantum state of the field (but
> maybe of the quantum state of the charge-carrying particle).

Or the electrostatic field is the expectation value of a c number operator (\grad A0) in ANY state of the electromagnetic field. All states satisfy criteria (1) and (2) trivially then, but only the vacuum satisfies criterion (3).

>
> However, there is something wrong in this approach: it ignores that
> besides electrostatic part and radiation part, there is a third part of
> the electromagnetic field, namely the magnetostatic part. A classical
> charge the is translating or rotating does not only cause an
> eletrostatic field, but also a magnetostatic one. That magnetostatic
> field is described by the vector potential \vec A (since \vec B = \rot
> \vec A), so the vector potential does not describe the radiation part only.
>
> You can consider things in the rest frame of the charge, so that the
> charge is not translating, but since the magnetostatic field is part of
> the vector potential which is quantized, you cannot simply assume the
> magnetostatic field to be fixed to be zero. And as far as an electron is
> concerned, the particle has a spin and therefore a magnetic dipole
> contribution to its magnetostatic field. So, the magnetostatic field
> cannot be eliminated.

Let's not worry about intrinsic spin, since I am looking for a classical correspondence of a large field. So this is the remaining question, I am tentatively happy with the resolution of the previous question.

[Gregor Scholten]

Roland Franzius wrote:

>>> This doesn't follow. Not only the electrostatic field, but any field,
>>> including magnetostatic, or radiation follow from ("are fully
>>> determined by") the classical position and velocity of the charges.
>>
>> You're wrong. In Maxwell's theory of Electrodynamics, the free radiation
>> field has own dynamical degrees of freedom: you can set rho = 0 and \vec
>> j = 0 everywhere, and Maxwell equations have still plane wave solutions.
>
> This is of course plainly wrong. There are no solutions of classical
> Maxwell equations dF=0, d*F=0 in all of empty Minkowski space without
> field singularities, given the condition of finite square integrable
> |F|^2 and subject to the special smoothness condition along a Cauchy
> surface as starting values

As I already pointed out: by setting special initial conditions (and
border conditions, like the one you refer to here), one can eliminate
all radiation that was not emitted by charges and currents, i.e. that is
solution for j=0. But when quantizing Electrodynamics, one necessarily
has to take into account the dynamical degrees of freedom of the
radiation field.

[Jos Bergervoet]

This is why I already proposed the 1+1 dimensional case as
a better example! (No dynamical dof's, but still an E-field)

So, the same question of OP (about life, the static E-field,
the quantum state and everything) but then for one space
dimension and as a source the simplest scalar charged field.

The scalar field can be described by a wave functional with
spread and expectation value as usual, I presume. So then,
what is the description of the E-field, or the 2-potential
A^mu in that case?

In my view there is no need to introduce the potential at
all, and the single component of E can directly be expressed
in the matter field and therefore "inherits" its description
as a wave functional with spread and expectation value.

But if we *do* introduce the 2-vector A^mu (just to prepare
for the higher dimensions where it is really needed) then
how would we visualize what it *is* in this quantum state?

--
Jos

[iuval]

If the static electric field is a c-number operator (at least in acting
on the photon part of the Hilbert space), then it is that for all orders
in any perturbation theory computation, not just leading order. Also,
the c-number property does not depend on mass. First order (Lamb shift)
and higher corrections to energy eigenstates and eigenvalues come from
additional terms in the Hamiltonian which do have standard (not
multiples of unity, aka c-numbers) operators.

[iuval]

The argument about what is more fundamental, source or field is rather
like the chicken and egg argument, not very productive. Neither can
exist without the other, even though dynamic field can exist in some
region without source, but ultimately it was produced by source on some
boundary of that region.

But I think that viewing a static field in any dimension as a quantum
state described by a wave functional is wrong, given my new
understanding. There are quantum states of source (given in one
possible representation, for electrons at least, by Dirac wave
functions) and quantum states for dynamic fields (given in one possible
representation, for coulomb gauge at least, by wave functionals of the
three components of \vec A) and sums of tensor products of these.

The static field is then something which is not emerging from some
classica l limit of the dynamical dofs' quantum state. It does emerge
from the sourc e degrees of freedom quantum state. Since that state can
be described by a wave function, perhaps there is some way to describe
the static field as a wave functional, but a very different kind of
beast than the one describing a quantum state of a classical dynamic
elctromagnetic field.

On Saturday, August 22, 2015 at 11:01:17 AM UTC-7, Jos Bergervoet wrote:
> On 8/19/2015 5:22 AM, Gregor Scholten wrote:
> > Roland Franzius wrote:
> >

> >>>> This doesn't follow. Not only the electrostatic field, but any field=

,
> >>>> including magnetostatic, or radiation follow from ("are fully
> >>>> determined by") the classical position and velocity of the charges.
> >>>

> >>> You're wrong. In Maxwell's theory of Electrodynamics, the free radiat=
ion
> >>> field has own dynamical degrees of freedom: you can set rho =3D 0 and=
\vec
> >>> j =3D 0 everywhere, and Maxwell equations have still plane wave solut=

ions.
> >>
> >> This is of course plainly wrong. There are no solutions of classical

> >> Maxwell equations dF=3D0, d*F=3D0 in all of empty Minkowski space with=

out
> >> field singularities, given the condition of finite square integrable
> >> |F|^2 and subject to the special smoothness condition along a Cauchy
> >> surface as starting values
> >
> > As I already pointed out: by setting special initial conditions (and
> > border conditions, like the one you refer to here), one can eliminate

> > all radiation that was not emitted by charges and currents, i.e. that i=
s
> > solution for j=3D0. But when quantizing Electrodynamics, one necessaril=

[iuval]

The argument about what is more fundamental, source or field is rather
like the chicken and egg argument, not very productive. Neither can
exist without the other, even though dynamic field can exist in some
region without source, but ultimately it was produced by source on some
boundary of that region.

But I think that viewing a static field in any dimension as a quantum
state described by a wave functional is wrong, given my new
understanding. There are quantum states of source (given in one
possible representation, for electrons at least, by Dirac wave
functions) and quantum states for dynamic fields (given in one possible
representation, for coulomb gauge at least, by wave functionals of the
three components of \vec A) and sums of tensor products of these.

The static field is then something which is not emerging from some
classica l limit of the dynamical dofs' quantum state. It does emerge
from the sourc e degrees of freedom quantum state. Since that state can
be described by a wave function, perhaps there is some way to describe
the static field as a wave functional, but a very different kind of
beast than the one describing a quantum state of a classical dynamic
elctromagnetic field.

On Saturday, August 22, 2015 at 11:01:17 AM UTC-7, Jos Bergervoet wrote:

[Gregor Scholten]

iuval wrote:

> The argument about what is more fundamental, source or field is rather
> like the chicken and egg argument

As far as the full system, matter field and full force-carrying field
(eletromagnetic field here), is concerned, you are right. However, the
force-carrying field can be decomposed in components that are own
dynamical degrees of freedom (\vec A in case of EM field) and components
that are fully determined by the matter field (scalar potential A^0),
and when considering the latter components, you are wrong: those
components are caused by the matter field, and therfore the matter field
is more fundamental.

> not very productive. Neither can
> exist without the other, even though dynamic field can exist in some
> region without source, but ultimately it was produced by source on some
> boundary of that region.

Only if you set special boundary conditions and consider classical
theory. In quantum theory, when quantizing the force-carrying field, you
are obliged to quantize the dynamical degrees of freedom of the field as
dynamical degrees of freedom, no matter what initial conditions and
boundary conditions you apply.

Maybe you could formulate a different theory by applying the initial
conditions and boundary conditions *before* the quantization procedure,
but the resulting theory would have little to do with ordinary Quantum
Field Theorie and Quantum Electrodynamics.

In addition, when regarding the history of the universe, it is not very
appropriate to set as initial condition that there were no photons (i.e.
that there were only matter particles that then emitted the first
photons). The early universe was radiation-domiated, and there was a
high number of pair creation processes where pairs of matter particles
and anti-particles where created from pairs of photons. So, in the
initial state of the universe, there were surely more photons than
matter particles.

BTW: with your argument, you are contraditing the statement you made
just before:

> The argument about what is more fundamental, source or field is rather
> like the chicken and egg argument

because with your argument, you claim that the matter field (the source)
is more fundamental than the force-carrying (EM) field.

[iuval]

On Sunday, September 13, 2015 at 1:51:45 PM UTC-7, Gregor Scholten wrote:
> iuval wrote:

>=20

> > The argument about what is more fundamental, source or field is rather
> > like the chicken and egg argument

>=20
> As far as the full system, matter field and full force-carrying field=20
> (eletromagnetic field here), is concerned, you are right. However, the=20
> force-carrying field can be decomposed in components that are own=20
> dynamical degrees of freedom (\vec A in case of EM field) and components=
=20
> that are fully determined by the matter field (scalar potential A^0),=20
> and when considering the latter components, you are wrong: those=20
> components are caused by the matter field, and therfore the matter field=
=20
> is more fundamental.

I think this is correct and I was wrong. But the reason is more
subtle than what you say. It seems that A_0 is only a c-number when
acting on the radiation part of the fully interacting system
(described by a direct product of the matter and radiation Hilbert
spaces). When acting on the matter part it is an operator given by
A_0(x)= e \Int (j_0(x')/(x-x')) d^3 x' (with j_0(x) being an
operator=\psi^bar gamma_0 psi), which is a solution to the poisson
equation with Laplacian on the LHS and e j0(x) as source on the
RHS. That is the Euler-Lagrange equation that minimizes the fully
interacting action obtained by varying A0. It would be OK if A0 was
a c-number when acting on the matter part (not just the radiation
part) to have it minimize the action. But being a quantum operator
when acting on the matter part, it is not totally making sense to
me. So this is still an outstanding problem.

[[Mod. note -- 41 excessively-quoted lines snipped here. -- jt]]