The stability of the hydrogen atom

[eoghan]

Hi there!
If I want to find the eigenstates of the hydrogen atom in QM I start with the hamiltonian of a free particle in a Coulomb potential. But an electron in a coulomb potential is stable in classical mechanics too! The instability of the system comes from the fact that the electron accelerated radiates energy away. So, in the quantum hamiltonian shouldn't I consider also the energy radiated away by the accelerated charge?
I mean, if I consider only the hamiltonian:

$$H=\frac{P^2}{2m}-\frac{e^2}{4\pi \epsilon_0 r}$$

it is as I am considering in classical mechanics only the force:

$$m\ddotx=-\frac{e^2}{2\pi\psilon_0 r^2}$$
and I don't consider the energy radiated away, so if I solve this equation I get stable orbits!

 

[jfy4]

what acceleration is taking place in the hydrogen atom using QM?

 

[alxm]

Well, sure - if you neglect to include the mechanism (field equations) whereby radiation is emitted (or just skip to the result, i.e. the Larmor formula), then it won't radiate away any energy. But the model would be at odds with known physics.

If you do it anyway, then you get a continuum of stable orbits and energy states, which isn't true for the atom. Restrict the system to only allow certain values for angular momentum and you have the Bohr model. In both cases you'll be missing the fact that the hydrogen atom's electron has zero angular momentum in its ground state. It's actually not 'orbiting' at all! (nor accelerating)

 

[JK423]

If you do it anyway, then you get a continuum of stable orbits and energy states, which isn't true for the atom. Restrict the system to only allow certain values for angular momentum and you have the Bohr model. In both cases you'll be missing the fact that the hydrogen atom's electron has zero angular momentum in its ground state. It's actually not 'orbiting' at all! (nor accelerating)

So, what are you saying? If we include the field equation (for radiation) we get the wrong results?? How can that be?
It doesn't seem 'more correct' to just neglect the radiation problem and just put a coulomb potential in the Schrodinger equation..!

Do you have a source on this? It's really interesting

 

[eoghan]

Well, sure - if you neglect to include the mechanism (field equations) whereby radiation is emitted (or just skip to the result, i.e. the Larmor formula), then it won't radiate away any energy. But the model would be at odds with known physics.

If you do it anyway, then you get a continuum of stable orbits and energy states, which isn't true for the atom. Restrict the system to only allow certain values for angular momentum and you have the Bohr model. In both cases you'll be missing the fact that the hydrogen atom's electron has zero angular momentum in its ground state. It's actually not 'orbiting' at all! (nor accelerating)

But in an excited state the electron has a non-zero angular momentum so it is accelerated (whatever it means to be accelerated in QM), but that is still a steady state! Why the electron in a p orbital doesn't radiate away its energy and falls in an s orbital?

 

[alxm]

So, what are you saying? If we include the field equation (for radiation) we get the wrong results?? How can that be?

I was talking about the classical case.

It doesn't seem 'more correct' to just neglect the radiation problem and just put a coulomb potential in the Schrodinger equation..!

If you put a classic EM field (Coulomb gauge) into the S.E. you end up with the Coulomb potential for stationary states.

 

[JK423]

Any references where they deal with the *real* hydrogen atom and not with the *radiation free by hand* hydrogen atom?

 

[tom.stoer]

It is true that one neglects radiative corrections when treating the hydrogen atom in quantum mechanics. In principle one has to quantize the electromagnetic field as well. This is done in quantum electrodynamics which treats both the matter fields (electron, positron, ...) and the electromagnic field as quantum fields (whereas in standard QM the el.-mag. field = the Coulomb potential in the hydrogen atom stays classical).

The result is that quantum electrodynamics generates radiative corrections which change the energy levels slightly. This effect is known as Lamb shift http://en.wikipedia.org/wiki/Lamb_shift

 

[jfy4]

So, what are you saying? If we include the field equation (for radiation) we get the wrong results?? How can that be?
It doesn't seem 'more correct' to just neglect the radiation problem and just put a coulomb potential in the Schrodinger equation..!

Do you have a source on this? It's really interesting

But in an excited state the electron has a non-zero angular momentum so it is accelerated (whatever it means to be accelerated in QM), but that is still a steady state! Why the electron in a p orbital doesn't radiate away its energy and falls in an s orbital?

Why do you think including the radiation emitted from a orbiting electron is correct? what gives you the idea that we are ignoring anything in the first place? These are questions you should answer first.

The radiation emitted from excited atoms is extremely well studied experimentally and theoretically, however has nothing to do with acceleration... The topic in its full detail is described by the field of Quantum Optics. A fantastic text is M. Fox Introduction to Quantum Optics, and I highly recommend reading it.

 

[JK423]

Why do you think including the radiation emitted from a orbiting electron is correct? what gives you the idea that we are ignoring anything in the first place? These are questions you should answer first.

The radiation emitted from excited atoms is extremely well studied experimentally and theoretically, however has nothing to do with acceleration... The topic in its full detail is described by the field of Quantum Optics. A fantastic text is M. Fox Introduction to Quantum Optics, and I highly recommend reading it.

Your first 2 questions has been answered by the original post of eoghan and also by tom.stoer's post. If you don't consider the electromagnetic field and the momentum it can carry, then its logical that you'll get no radiation! And remember: The QM theory was invented to solve the radiation problem. And how does it solve it? By ignoring the electromagnetic field?
An answer has been given by tom.stoer, and i will research on this a bit more..
I`ll also check out M. Fox's text, thanks for the source

 

[jfy4]

Your first 2 questions has been answered by the original post of eoghan and also by tom.stoer's post. If you don't consider the electromagnetic field and the momentum it can carry, then its logical that you'll get no radiation! And remember: The QM theory was invented to solve the radiation problem. And how does it solve it? By ignoring the electromagnetic field?
An answer has been given by tom.stoer, and i will research on this a bit more..
I`ll also check out M. Fox's text, thanks for the source

In my last post, the point I was trying to get at was that there was no radiation problem. Atoms do not radiate away themselves. The idea that atoms radiate away themselves and die is wrong, that is not happening. So in formulating the correct theory of atoms, we are not looking to explain how atoms radiate away themselves due to acceleration. I was asking you to ask yourself, why you would want to include a phenomenon into quantum mechanics that isn't physically happening, (or why you want to include the energy radiated away when no such thing is happening)..? Am I being clear here?

 

[JK423]

Yes i understand what you are trying to say. But my problem is a little different. In the theory, i want the electromagnetic field to be included. I want everything to be included. Then, if the theory accounts for all that and still gives us NO radiation then so be it! It's correct!
Imagine a classical electron going around a proton, and in your classical theory you account only the coulomb potential. If you DONT account the electromagnetic field and the momentum it can carry, then NO radiation problem!
Do you understand what my problem is?

 

[jfy4]

Im sorry, I don't understand. Could you explain a little more in detail? Why do you want to talk about energy (potential) and field quantities together. Wouldn't working in the energy (Hamiltonian) formalism be just as acceptable?

 

[JK423]

When you have a charge moving in an electromagnetic field, as you probably know, its momentum is not conserved. You must also account for the EM field, the field's momentum. So, in order for your Hamiltonian to be complete, it shouldn't account only for the charge but also for the field. If you don't do this, then in principle your charge won't be able to give momentum to the field --> no radiation. The question is, what happens if we include the electromagnetic field?
And here is where tom.stoer's post comes in.

 

[alxm]

Yes i understand what you are trying to say. But my problem is a little different. In the theory, i want the electromagnetic field to be included. I want everything to be included. Then, if the theory accounts for all that and still gives us NO radiation then so be it! It's correct!

Well jfy4 seems to be trying to justify empirically why you wouldn't include the field in the QM solution. As I said in my post, there is no problem doing so. But at the non-relativistic limit in the Coulomb gauge, you end up with the coulomb potential. Any rigorous derivation of the Coulomb potential starts from the (relativistic) EM field equations, not from Coulomb's law.

If you solve the Schrödinger equation with coupling to an EM field (classical or not, relativistic or not), the ground state will not radiate. Moreover, if you solve it for an excited state non-relativistically, even excited states won't radiate. (It requires a perturbation of the EM field to occur) If you solve it classically and include coupling to the field, it will radiate. But perhaps more important is the simple fact that classically, the electron can have zero momentum, so regardless of the mechanism by which it gets there, it should end up at r=0 anyway.

 

[JK423]

Well jfy4 seems to be trying to justify empirically why you wouldn't include the field in the QM solution. As I said in my post, there is no problem doing so. But at the non-relativistic limit in the Coulomb gauge, you end up with the coulomb potential. Any rigorous derivation of the Coulomb potential starts from the (relativistic) EM field equations, not from Coulomb's law.

If you solve the Schrödinger equation with coupling to an EM field (classical or not, relativistic or not), the ground state will not radiate. Moreover, if you solve it for an excited state non-relativistically, even excited states won't radiate. (It requires a perturbation of the EM field to occur)

Thanks alxm, that's what i wanted to hear..
Do you have a source where the full procedure is followed? Without any simplifications?
All the texts that I've looked at, do not treat the problem like this

 

[alxm]

Thanks alxm, that's what i wanted to hear..
Do you have a source where the full procedure is followed? Without any simplifications?
All the texts that I've looked at, do not treat the problem like this

Hmm, "Atom-Photon Interactions" by Cohen-Tannoudji, for instance, is wholly dedicated to the issue.