In nonrelativistic classical mechanics and electrostatics the two body
bound state is an eliptical orbit.

In relativistic classical mechanics and electrodynamics the electron
will radiate as it accelerates and spiral into the nucleus.

In nonrelativistic quantum mechanics the two body bound state is fine.

My question is this: how do I know that the electron will not spiral
into the nucleus in relativistic quantum mechanics?

In the only QFT book I have that deals with bound states thoroughly,
the relativistic Dirac equation is used, but then a static field used.
Everything else is just a perturbation off of that. That would be like
doing relativistic classical mechanics but without the electrodynamics.
But there is no problem there to be expected.

Is Hydrogen stable in relativistic quantum mechanics when you do not
approximate the field as static? I am extra confused because
positronium is not stable. The spiral occurs there. What is special
about Hydrogen that stops the spiral?

Chris H. Fleming wrote:
In nonrelativistic classical mechanics and electrostatics the two body
bound state is an eliptical orbit.

In relativistic classical mechanics and electrodynamics the electron
will radiate as it accelerates and spiral into the nucleus.

In nonrelativistic quantum mechanics the two body bound state is fine.

My question is this: how do I know that the electron will not spiral
into the nucleus in relativistic quantum mechanics?

In the only QFT book I have that deals with bound states thoroughly,
the relativistic Dirac equation is used, but then a static field used.
Everything else is just a perturbation off of that. That would be like
doing relativistic classical mechanics but without the electrodynamics.
But there is no problem there to be expected.

There is nothing wrong in using static field. You are trying to find
the stationary lowest energy state of the hydrogen atom. This problem
is time-independent.


Is Hydrogen stable in relativistic quantum mechanics when you do not
approximate the field as static? I am extra confused because
positronium is not stable. The spiral occurs there. What is special
about Hydrogen that stops the spiral?

There is a sort of "spiraling" when excited states of the hydrogen atom
are considered in quantum mechanics. The atom jumps to a lower
energy state and the excess
energy is released as a quantum of radiation - the photon.
However, at the ground state, nothing can happen, because this is the
lowest energy state (by definition). This state remains stable forever.
Conservation laws (e.g., the conservation of the baryonic and leptonic
charges) prevents its decay into lighter particles. Note that these
decay laws do not apply to the positronium (all charges are zero),
so it is allowed to decay into photons, as it does.

Eugene.

On Mon, 14 Nov 2005, Chris H. Fleming wrote:

Is Hydrogen stable in relativistic quantum mechanics when you do not
approximate the field as static? I am extra confused because
positronium is not stable. The spiral occurs there. What is special
about Hydrogen that stops the spiral?

Positronium isn't unstable because the electron "spirals" into the
positron. It is unstable because the electron and positron are
antiparticles, and hence the pair can annihilate into two photons. An
electron and a proton aren't mutual antiparticle, and hence cannot do
that.

In relativistic QM, just as in classical QM, the hydrogen problem has a
discrete spectrum of bound states. There is simply no way for an electron
to go into a lower state than the ground state (because there is no such
state), and hence the ground state is stable.

You seem to think in terms of classical orbits ("spiral"), but these don't
really exist in QM (whether classical or relativistic). An electron can
transition from a state of higher energy into a state of lower energy by
emitting a photon that carries away the energy difference. This is
sometimes referred to as a "quantum jump", and is the reason for the
spectra of discrete lines that you get from e.g. neon lamps.

Best wishes,

Georg

--
************************************************** *************************
Dr. Georg M. von Hippel
Postdoctoral Research Fellow, Subatomic Theory Group

Department of Physics Phone: +1 (306) 337-2359
University of Regina Fax: +1 (306) 585-5659
Regina, Saskatchewan E-Mail:
S4S 0A2
Canada Web: http://uregina.ca/~vonhippg/
************************************************** *************************

Chris H. Fleming wrote:
In nonrelativistic classical mechanics and electrostatics the two
body bound state is an eliptical orbit.

In relativistic classical mechanics and electrodynamics the electron
will radiate as it accelerates and spiral into the nucleus.

In nonrelativistic quantum mechanics the two body bound state is
fine.

In either relativistic or non-relativistic classical mechanics, the
orbiting electron will radiate only when its interaction with the
electromagnetic field is included. Once that is don, in both the
relativistic and non-relativistic contexts, the electron orbit is
unstable as it will lose energy through radiation.

My question is this: how do I know that the electron will not spiral
into the nucleus in relativistic quantum mechanics?

Here's a very heuristic explanation. In classical mechanics, when
looked at as a central force problem, the reason the electron orbit is
stable is the contention between the attractive Coulomb force and the
repulsive centrifugal force (as felt in the frame of the electron). If
the electron is allowed to radiate, then the centrifugal force will
become smaller and smaller until the Coulomb attraction makes the
electron crash into the nucleus. In quantum mechanics, relativistic or
not, due to the Heisenbert uncertainty relation, if the electron gets
confined to a smaller and smaller region (say the orbit radius steadily
decreases), then it's momentum and hence energy will become larger and
larger. Hence, there will be an effective force pushing the electron
outward preventing it from being confined to a progressively smaller
region. Hence, equilibrium can be established even if the electron is
allowed to radiate. Once equilibrium is reached (pretty much by
definition), the electron will no longer radiate. And this is exactly
what we see in the ground state of the Hydrogen atom.

In the only QFT book I have that deals with bound states thoroughly,
the relativistic Dirac equation is used, but then a static field
used. Everything else is just a perturbation off of that. That would
be like doing relativistic classical mechanics but without the
electrodynamics. But there is no problem there to be expected.

There is no loss of generality in writing the electromagnetic potential
as A(x) - A_0(x) + a(x), where A_0(x) is a classical background and
the correction a(x) is allowed to fluctuate classically or quantum
mechanically. This is no more than field redefinition. In principle it
is not necessary, but it does make things easier in practice if A_0(x)
is chosen such that the expectation value a(x) vanishes in the ground
state. The reason that no problem with stability is expected is the
same as in my last paragraph. And the fluctuating a(x) field is
important. It's dynamics allow one to calculate energy shifts in the
states of the Hydrogen atom, e.g. the Lamb shift.

Is Hydrogen stable in relativistic quantum mechanics when you do not
approximate the field as static? I am extra confused because
positronium is not stable. The spiral occurs there. What is special
about Hydrogen that stops the spiral?

As pointed out in other replies. The instability of positronium is not
related to radiative losses, but rather to the possibility of
annihilation between the two particles.

Hope this helps.

Igor

Eugene Stefanovich wrote:
Chris H. Fleming wrote:
In nonrelativistic classical mechanics and electrostatics the two body
bound state is an eliptical orbit.

In relativistic classical mechanics and electrodynamics the electron
will radiate as it accelerates and spiral into the nucleus.

In nonrelativistic quantum mechanics the two body bound state is fine.

My question is this: how do I know that the electron will not spiral
into the nucleus in relativistic quantum mechanics?

In the only QFT book I have that deals with bound states thoroughly,
the relativistic Dirac equation is used, but then a static field used.
Everything else is just a perturbation off of that. That would be like
doing relativistic classical mechanics but without the electrodynamics.
But there is no problem there to be expected.

There is nothing wrong in using static field. You are trying to find
the stationary lowest energy state of the hydrogen atom. This problem
is time-independent.


Is Hydrogen stable in relativistic quantum mechanics when you do not
approximate the field as static? I am extra confused because
positronium is not stable. The spiral occurs there. What is special
about Hydrogen that stops the spiral?

There is a sort of "spiraling" when excited states of the hydrogen atom
are considered in quantum mechanics. The atom jumps to a lower
energy state and the excess
energy is released as a quantum of radiation - the photon.
However, at the ground state, nothing can happen, because this is the
lowest energy state (by definition). This state remains stable forever.

How do I know there is a ground state in the fully relativistic and
electrodynamic, quantum problem. I only know there is a bound state in
the nonrelativistic problem, or the relativistic problem with an
electrostatic field. But neither of these cases had pathology in the
classical regime. I was never in doubt of these cases.

Conservation laws (e.g., the conservation of the baryonic and leptonic
charges) prevents its decay into lighter particles. Note that these
decay laws do not apply to the positronium (all charges are zero),
so it is allowed to decay into photons, as it does.

That is more in line with what I was thinking.

So is it true that if you modeled hydrogen simply with the electron and
proton as different mass, spin-1/2 particles interacting via photon
exchange, and ignored all other forces, then hydrogen would not be
stable. Is it true that forces other than electrodynamics keep hydrogen
and all other elements electronically stable.

Igor Khavkine wrote:
Chris H. Fleming wrote:
In nonrelativistic classical mechanics and electrostatics the two
body bound state is an eliptical orbit.

In relativistic classical mechanics and electrodynamics the electron
will radiate as it accelerates and spiral into the nucleus.

In nonrelativistic quantum mechanics the two body bound state is
fine.

In either relativistic or non-relativistic classical mechanics, the
orbiting electron will radiate only when its interaction with the
electromagnetic field is included. Once that is don, in both the
relativistic and non-relativistic contexts, the electron orbit is
unstable as it will lose energy through radiation.

I know but nonrelativistic electrodynamics is inconsistent so I didn't
consider it.

My question is this: how do I know that the electron will not spiral
into the nucleus in relativistic quantum mechanics?

Here's a very heuristic explanation. In classical mechanics, when
looked at as a central force problem, the reason the electron orbit is
stable is the contention between the attractive Coulomb force and the
repulsive centrifugal force (as felt in the frame of the electron). If
the electron is allowed to radiate, then the centrifugal force will
become smaller and smaller until the Coulomb attraction makes the
electron crash into the nucleus. In quantum mechanics, relativistic or
not, due to the Heisenbert uncertainty relation, if the electron gets
confined to a smaller and smaller region (say the orbit radius steadily
decreases), then it's momentum and hence energy will become larger and
larger. Hence, there will be an effective force pushing the electron
outward preventing it from being confined to a progressively smaller
region. Hence, equilibrium can be established even if the electron is
allowed to radiate. Once equilibrium is reached (pretty much by
definition), the electron will no longer radiate. And this is exactly
what we see in the ground state of the Hydrogen atom.

So there is a pressure associated with HU? I have never heard of such a
thing. I know of regular pressure due to momentum, and other pressures
due to things like PEP.

How can I derive this? Do you have any references?

In the only QFT book I have that deals with bound states thoroughly,
the relativistic Dirac equation is used, but then a static field
used. Everything else is just a perturbation off of that. That would
be like doing relativistic classical mechanics but without the
electrodynamics. But there is no problem there to be expected.

There is no loss of generality in writing the electromagnetic potential
as A(x) - A_0(x) + a(x), where A_0(x) is a classical background and
the correction a(x) is allowed to fluctuate classically or quantum
mechanically.

Yes, but there is a loss of generality when a(x) is taken to be a small
perturbation with higher order terms thrown away. Especially if I am
trying to look for a large difference.

This is no more than field redefinition. In principle it
is not necessary, but it does make things easier in practice if A_0(x)
is chosen such that the expectation value a(x) vanishes in the ground
state. The reason that no problem with stability is expected is the
same as in my last paragraph. And the fluctuating a(x) field is
important. It's dynamics allow one to calculate energy shifts in the
states of the Hydrogen atom, e.g. the Lamb shift.

Is Hydrogen stable in relativistic quantum mechanics when you do not
approximate the field as static? I am extra confused because
positronium is not stable. The spiral occurs there. What is special
about Hydrogen that stops the spiral?

As pointed out in other replies. The instability of positronium is not
related to radiative losses, but rather to the possibility of
annihilation between the two particles.

I know it's not the same thing, but the classical spiral of death seems
somewhat analagous to positronum decay into photons. It's just that in
the classical case, there can be no particle production and infinite
radiation comes out for point particles. I could imagine fixing the
situation with extended particles and looking for some kind of
collision.

Chris H. Fleming wrote:

How do I know there is a ground state in the fully relativistic and
electrodynamic, quantum problem. I only know there is a bound state in
the nonrelativistic problem, or the relativistic problem with an
electrostatic field. But neither of these cases had pathology in the
classical regime. I was never in doubt of these cases.

That's a good question. Relativistic QED has a trouble in describing
the bound state of the hydrogen atom. This trouble is explained by
Weinberg in the beginning of chapter 14 of his "The quantum theory of
fields" vol. 1. All we can do in QED is to calculate the S-matrix.
It is known that the bound states of the hydrogen atom must show up
as poles of the electron-proton scattering amplitudes. So, in principle,
the renormalized QED has a chance to give us the energies of
the stable bound states which correspond to the positions of the poles
(note that getting the corresponding wave functions is a much more
difficult task).
However, in order to reproduce these poles one needs to sum up a very
large (infinite?) number of Feynman diagrams. Weinberg proposes a
solution that does not look elegant: approximate the contribution of
some (infinite number) of these diagrams by a solution of the
Dirac equation with fixed 1/r potential and treat other diagrams as
perturbations leading to the Lamb shifts. This seems to be working fine,
but it doesn't answer directly your question, whether the bound state
exist in the full relativistic renormalized theory. We are assuming
their existence instead of deriving it ab initio.

Fortunately, there is another approach to the bound state problem
in QED that moves us a bit closer to the answer. This is the
"dressed particle" approach described (among other places) in chapter
12 of my book http://arxiv.org/abs/physics/0504062 .
In this approach, the Hamiltonian of QED is rewritten in terms of
physical particles (electrons and protons). In the position
representation, this Hamiltonian has explicit 1/r interaction
term (plus relativistic correction) already in the 2nd perturbation
order. So, by simply diagonalizing this 2nd order Hamiltonian
in the electron-proton sector of the Fock space you can get 99%
complete description of the bound state of the hydrogen atom. Additional
corrections (line widths or Lamb shifts) appear from higher order
contributions. So, in the "dressed particle" approach, the existence
and properties (including the wave functions) of the bound states
follow directly from the Hamiltonian.

Conservation laws (e.g., the conservation of the baryonic and leptonic
charges) prevents its decay into lighter particles. Note that these
decay laws do not apply to the positronium (all charges are zero),
so it is allowed to decay into photons, as it does.


That is more in line with what I was thinking.

So is it true that if you modeled hydrogen simply with the electron and
proton as different mass, spin-1/2 particles interacting via photon
exchange, and ignored all other forces, then hydrogen would not be
stable. Is it true that forces other than electrodynamics keep hydrogen
and all other elements electronically stable.

I think you misunderstood. The situation is quite opposite, in my view.
Electromagnetic interactions obey the laws
of conservation of the baryonic and leptonic charges, so the hydrogen
atom cannot decay if only EM interactions are at play.
There are hypotheses about additional forces of
nature that do not obey the law of conservation of the baryonic
charge. If these forces existed, then the proton would be unstable
and the hydrogen atom would decay as well. Needless to say, that
such interactions have not been observed yet.

Eugene.

Chris H. Fleming wrote:
Igor Khavkine wrote:

In either relativistic or non-relativistic classical mechanics, the
orbiting electron will radiate only when its interaction with the
electromagnetic field is included. Once that is done, in both the
relativistic and non-relativistic contexts, the electron orbit is
unstable as it will lose energy through radiation.

I know but nonrelativistic electrodynamics is inconsistent so I didn't
consider it.

It is no more inconcistent than any other approximation. You have to
realize that it does not apply in all situations. In particular, it
only applies if the electron's speed with respect to the nucleus is
much smaller than c.

My question is this: how do I know that the electron will not spiral
into the nucleus in relativistic quantum mechanics?

Here's a very heuristic explanation. In classical mechanics, when
looked at as a central force problem, the reason the electron orbit is
stable is the contention between the attractive Coulomb force and the
repulsive centrifugal force (as felt in the frame of the electron). If
the electron is allowed to radiate, then the centrifugal force will
become smaller and smaller until the Coulomb attraction makes the
electron crash into the nucleus. In quantum mechanics, relativistic or
not, due to the Heisenbert uncertainty relation, if the electron gets
confined to a smaller and smaller region (say the orbit radius steadily
decreases), then it's momentum and hence energy will become larger and
larger. Hence, there will be an effective force pushing the electron
outward preventing it from being confined to a progressively smaller
region. Hence, equilibrium can be established even if the electron is
allowed to radiate. Once equilibrium is reached (pretty much by
definition), the electron will no longer radiate. And this is exactly
what we see in the ground state of the Hydrogen atom.

So there is a pressure associated with HU? I have never heard of such a
thing. I know of regular pressure due to momentum, and other pressures
due to things like PEP.

How can I derive this? Do you have any references?

As I mentioned, the explanation I just gave is very heuristic. You most
likely won't find a direct interpretation of the Heisenberg uncertainty
as a kind of pressure. That is because in quantum mechanics, we don't
solve equations of the form d^2 x/dt = F_1 + F_2 + ..., where each
term on the right hand side can be given an interpretation (such as
pressure or Coulomb attraction for example). Instead we solve problems
of the form: given

E[psi] = int psi*(x) [ p^2/2m + V(x) ] psi(x) dx

minimize E as a function of psi(x) subject to the normalization
constraint

int |psi(x)|^2 dx = 1 .

The solution phi(x) to this problem is the wave function of the ground
state of a particle in potential V(x), and E_0 = E[phi] is the ground
state energy. The terms p^2/2m and V(x) in the energy functional are
the kinetic and potential energy contributions, respectively. In
quantum mechanics, the momentum observable is represented by a
derivative p = -i d/dx. So minimizing the kinetic energy tends to make
the wave function as smooth and as flat as possible. On the other hand,
minimizing the potential energy, tends to assign to psi(x) large values
where V(x) is small (or more negative) and small values where V(x) is
large (or more positive). However, if a wave amplitude is concentrated
in a small region, it will necessarily have large slope there, while if
the wave amplitude is very flat and uniform, the potential energy term
will be far from minimized. Hence, there is contention between the two
terms in the minimization problem. This contention can be traced to the
same root as the Heisenberg uncertainty principle, simply that the
position and momentum observables do not commute, xp != px.

For the case where V(x) is the Coulomb potential, the classical
instability of the Hydrogen atom corresponds to the tendency of the
wave function to become a progressively narrow and tall peak at x=0.
However, the true minimum of the quantum energy functional E[psi], i.e.
the ground state, strikes a balance between this tendency and the
tendency of the wave packet to diffuse. That is how stability is
achieved.

As you can see, it is hard to assign a direct pressure-like
interpretation to the tendency of a wave packet to diffuse due to
Heisenberg uncertainty. But if you really want to, you could do
something like the following. Setup a trial wave function, say a
Gaussian of a certain width R, psi(x,R) = N(R) exp(-x^2/R). The
normalization factor N(R) is needed to keep the wave functio of unit
norm. Then you can calculate the energy functional with this trial wave
function, U(R) = E[psi(x,R)]. Then the derivative dU(R)/dR will give
you an idea of the magnitude of the force resisting the change of the
width of the wave packet by an amount dR.

There is no loss of generality in writing the electromagnetic potential
as A(x) - A_0(x) + a(x), where A_0(x) is a classical background and
the correction a(x) is allowed to fluctuate classically or quantum
mechanically.

Yes, but there is a loss of generality when a(x) is taken to be a small
perturbation with higher order terms thrown away. Especially if I am
trying to look for a large difference.

I don't see how higher order terms in a(x) are thrown away. The usual
perturbation expansion in QED is done with respect to the coupling
constant, alpha - the fine structure constant, not with respect to
field magnitude. In any case, there is a simple self-consistency check
to see if the split into A_0(x) and a(x) was done "correctly". Once the
split is done, evaluate the expectation value a(x). If a(x) is zero
or remains small, then everything is fine. If not, you need to find a
better A_0(x). For example, if you start with A_0(x) = 0, then you'll
find that a(x) does not vanish and that the leading contribution is
exactly the same as the classical Coulomb potential. However, this
problem is far from trivial, so this step is skipped in elementary
treatments.

As pointed out in other replies. The instability of positronium is not
related to radiative losses, but rather to the possibility of
annihilation between the two particles.

I know it's not the same thing, but the classical spiral of death seems
somewhat analagous to positronum decay into photons. It's just that in
the classical case, there can be no particle production and infinite
radiation comes out for point particles. I could imagine fixing the
situation with extended particles and looking for some kind of
collision.

There is no need for a spiral or a collision. If you examine the wave
functions for the electron and the positron in positronium, you'll find
that they have large overlap. Thus they are in a position to annihilate
at any time. What regulates the decay time, though, is the weakness of
the electromagnetic interaction.

Hope this helps.

Igor

Eugene Stefanovich wrote:

That's a good question. Relativistic QED has a trouble in describing
the bound state of the hydrogen atom. This trouble is explained by
Weinberg in the beginning of chapter 14 of his "The quantum theory of
fields" vol. 1. All we can do in QED is to calculate the S-matrix.

Nonsense. Almost everything of interest can be calculated in QED.
The first and still the most conspicuous nontrivial calculation
in QED was that of the Lamb shift, which is a statement about a
bound state of hydrogen. One can also calculate the spectrum
of positronium to high accuracy.


It is time you give up your narrow views about QED!


Arnold Neumaier

Arnold Neumaier wrote:
Eugene Stefanovich wrote:

That's a good question. Relativistic QED has a trouble in describing
the bound state of the hydrogen atom. This trouble is explained by
Weinberg in the beginning of chapter 14 of his "The quantum theory of
fields" vol. 1. All we can do in QED is to calculate the S-matrix.


Nonsense. Almost everything of interest can be calculated in QED.
The first and still the most conspicuous nontrivial calculation
in QED was that of the Lamb shift, which is a statement about a
bound state of hydrogen. One can also calculate the spectrum
of positronium to high accuracy.

I agree that you can calculate ENERGIES of bound states to high
accuracy (e.g., Lamb shifts) in the traditional QED based on the
bare particle picture. This is because these energies are directly
reflected in the S-matrix as positions of poles.
Everything related to the S-matrix can
be calculated in the traditional QED. Here we fully agree.

My point is that in the bare-particle QED you cannot calculate
WAVE FUNCTIONS of the bound states (unless you drastically
mutilate the theory, e.g., introduce the Dirac-Fock Hamiltonian).
The reason is that the wave functions of bound states are not
reflected in the S-matrix.

Suppose that you found a way to calculate both energies and wave
functions of the bound states with high-order radiative corrections.
This would mean that you know precisely the Hamiltonian of
interacting physical particles (at least in the range of bonding
energies): any Hermitian operator can be uniquely restored from its
eigenvalues and eigenvectors.
This would mean that you are working in the
"dressed particle" picture. I haven't seen anybody doing that
except in Shebeko-Shirokov works. Did I miss something?
Then I would appreciate a reference.

If your statement is "Almost everything of interest can be calculated
in _dressed particle_ QED", then I totally agree.

Eugene.