Re: Hydrogen in Relativistic QM
- From: Igor Khavkine <igor.kh@xxxxxxxxx>
- Date: Tue, 15 Nov 2005 23:59:53 +0000 (UTC)
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
.
- References:
- Hydrogen in Relativistic QM
- From: Chris H. Fleming
- Hydrogen in Relativistic QM
- Prev by Date: Re: Is general relativity incompatible with the Newtonian limit?
- Next by Date: Re: Karl Hess (hidden variables)
- Previous by thread: Re: Hydrogen in Relativistic QM
- Next by thread: Re: Hydrogen in Relativistic QM
- Index(es):
Relevant Pages
|
|
