1. Dec 9, 2004

### niehls

There is this thing i've been wondering about. When deriving Rydberg's formula and the expression for Rydberg's constant (Bohr model of the atom) it mentions the problem of an accelerated electron emitting radiation (Maxwell's theries) and hence would spiral into the nucleus.
It doesn't explain this further. I would need some enlightenment on why this does not occur. Is this yet another "weird" effect of quantum mechanics?

Chris

2. Dec 9, 2004

### niehls

is it that the assumption that accelerated electrons radiates is incorrect and that only oscillating electrons radiate with the frequency of the oscillation?

3. Dec 9, 2004

### imabug

The classical model predicts that orbiting electrons, because they are constantly being accelerated, should eventually radiate away all their energy and spiral into the nucleus.

the fact that they don't do this is what led Bohr to develop his quantum model of the atom, where electrons orbit in quantized discrete energy levels.

4. Dec 9, 2004

### dextercioby

I wouldn't call it "weird".YES,the answer is YES.QM fully explains why the electron moving totally randolmy around the nucleus does not emit radiation,if the atom (actually the electron) is left alone and noninteracting with another atom/radiaton or anything else.
Bohr postulated this thing,he didn't prove it.QM proves it.

To explain it shortly,i'd say that the spectrum of the energies of the electron is (for the bound states) purely discrete and,if the atom (electron) is left alone,it would stay in the fundamental state (which has the smallest possible energy (and therefore is the most stable)) and it would stay there forever,if the H atom would not interract with anything.

Daniel.

5. Dec 9, 2004

### dextercioby

The concept of "accelerated...(anything)" is entirely classical."Acceleration" and all the other words derived from it should not be put in the same text in which QM concepts appear.
Accelerated electric charged particles emit radiation (electromagentic,of course) and should be clearly discussed in a book like Jackson's.(Search for Larmor formula).But the the quantum theory of electrons is called QED (Quantum Electrodynamics) and it successfully describes interactions between electrically charged particles (in general),like the electron and the proton in the H atom.
The theory of oscillating electrons (the theory of radio emisson of an antena for example) is entirely classical (should be described in Jackson as well) and it has nothing to do with the H atom/Bohr model/QM/QED...They radiate energy with the same frequency at which they oscillate...You're right,here...I haven't heard of antenna radiation problem discussed at QED level...

Daniel.

6. Dec 10, 2004

### seratend

Let us try to analyze a little bit more this problem if you want.
Let's take the model of a single hydrogen atom in the universe with the semi classical em field and classical QM.
Can we really say that the fundamental level is the lowest possible Energy level/state?
In other words, what does prevent an "r=0 state" in this semi-classical model?

Seratend

7. Dec 10, 2004

### marlon

That would be one exotic antenna...

marlon....

Just as an addendum : even when non-linear phenomena take place , the classical EM-theory will do a good job at explaining things...QED is not necessary

8. Dec 10, 2004

### dextercioby

Once u have an em field,no matter whether wlassical or quantum described,i think what i said could not apply,and therefore could not be questioned.To get me straight with what u're saying,you're assuming the classical model of a quantum described system (let's call it H atom) which interacts with a classical em field.
Now,the H atom alone,no exterior field /atom to interact with is fully described and solvable using Schroedinger's equation,as the term with the field (the Coulomb term) it time dependent,and moreover ist has spherical symmetry.We all know (or maybe we should) ,how the problem "r=0" is dealt with in the case of a more general central potential.But the this central potential is always assumed to come from a time independent Hamiltonian.So,even for a more complicated (other than Coulomb) potential which preserves spherical symmetry,there is no need for perturbative calculations.Though i'm confident that the radial eq.solution cannot be expressed so easily in terms of confuent hypergeometrical functions.
If the potential is time dependent,we have to use theory of time dependent perturabations.To qoute from my QM course:"The basis problem of time dependent perturbation theory is determining in different orders wrt ot the parameter '\lambda' of the probability with which a certain eigenstate for the unperturbed hamiltonian is achived after time 't' haspassed since the application of the time dependent perturbation".
In this case,we could only calculate the probability of transition between eigenstates of the unperturbed hamiltonian under the absorbtion of a single photon,since absorbtion of more photons requires QED.In the case of spontaneous emission,obviuosly the fundamental level is not the minimum level possible anymore,since a level with E_0-h\nu could possible.

The phenomenology is really complex.But i think that here,QM calculations (even for one photon) would not be correct anymore,since the electron in the H atom would not be isolated from the rest of the universe.Weirdly,u assumed that,but also included a classical em field along with it.If it is a em field to influence eigenvalues and eigenstates of the initial (time independent) Hamiltonian,then the atom is not alone anymore,and what i said in my previuos post would not apply.

Daniel.

9. Dec 10, 2004

### seratend

Ok, let's reduce our model to the H atom with the coulombian interaction only.
Most of the books and courses I know does not answer correclty to the question why the r=0 solution is not possible (why the electron does not fall into the nucleus - the irregular solutions of the H atom SE eigenvalue).

The usual argument concerns the regularity of solutions of the H atom SE (i.e. the wave function must be normalized => selection of the r^(l+1) behavior close to r=0).
However this argument is weak, because we accept the free wavefunctions (with E>0, in the case of the H atom), for example in classical scattering theory.

Now, with this simplified model, what argument can you give to reject the irregular solutions of the H atom (i.e. the Laplace equation)?

Seratend.

10. Dec 10, 2004

### dextercioby

.

Okay,i don't know why you rejected interactions with an em field (classical),but you must have your reasons...

Are u sure...??This is a "heavy" statement,actually an accusation,for which i'm afraid,as to hold,it must be given "evidence".

That's a simplified version which says why the secondary redial functions must have that form close to r=0.
The H atom has an observable (though it cannot be determined experimentally,just theoretically) which is called LOCALIZATION PROBABILITY.It is given by (for a volume of free space V):
$$P(V)=\int\int\int_{V} C_{nlm}R_{nl}^{2} (r) |Y_{lm}|^{2} r^{2} dr d\Omega$$
If V stretches to infinity,$P(R^{3})=1$.
Take a good look at the formula for P(V)...It's a specific example of the more general formula giving the probability in terms of the integration over a certain domain of the probability density.For that integration to be possible,certain conditions are to be imposed over the integrand,right.The fisrt one is continuity over the entire domain of integration.Since that domain can be stretched to infinity and the result of the integration is computed and found to be 1,then the natural condition of continuity over R^{3} is required.Since it can be stretched to infinity and the integration has a bounded value,another constraint upon the probability density emerges:the bounded character over the entire domain and especially in 2 "sensitive" points:0 and infinity.
The conclusions so far transfer really easily to $R_{nl} (r)$,since th spherical harmonics are bounded over the range of the spherical angles and,if $R_{nl}^{2} (r)$ is bounded and continuous,so it will be the case for its square root (it's a real function,no problems extracting square).
So,when solving the H atom,the radial eq. actually,u impose the conditions of continuity and boundness upon the solution.Over the entire (0,+infinity) domain.From here,i'm sure u can discuss,the 0 asimptotic behavior of the solution and find the functions $R_{nl}$
Compute the probability that the electron should be localized in the origin (r=0) in the fundamental state,using the definition of the probability wrt only to the radial function (the spherical harmonics are normed) for the domain (0,+\epsilon).You'llfind the probability of the electron to be found in the sphere determined by the condition (given in spherical coordinates) r=\epsilon.Make epsilon go to zero and U ARE SURE TO GET ZERO.You will find the rigurous QM proof that in its arbitrary movement through space the electron in the H atom WILL NEVER HIT THE PROTON.

In the positive energy spectrum of the H atom,the eigenfunctions of the Hamiltonian will definitely have not the same form as the ones for the bound states.Thy will simply be eigenfunctions of the momentum operator,de Broglie momentum waves.Solve the radial equation for these scatterig states and u will find absolutely no problem wrt to bounding and continuity over (0,+infinity),because the solutions will be in terms of linear combination between sine and cosine.Actually they will be complex exponentials of (r).Compute the probability of a scaterring state electron to be found in origin,by the same method as before.(A sphere whose radius goes to zero).U'll find zero.That's the second part of the proof that ELECTRONS WILL NEVER HIT THE NUCLEUS,EVEN IN SCATTERING STATES.
Actually this second partof the proof i find it to be easier,given the particular form for R_{nl}.

What irregular solutions??U mean discontinuos/unbounded??Yes,then.What Laplace???What does Laplace eq.have to do with H atom???It's generally a Helmholz eq.A stinky one.

Daniel.

11. Dec 13, 2004

### seratend

HYDROGEN ATOM

It is not an accusation (it was not my intention in this post), just a statement of my modest knowledge that is far from being complete. I should have used another word, sorry.
However, I will try to show you the weaknesses I have found in these explanations, as you seem to use an analogue point of view.
I am restricting the H atom model to the coulombian interaction in order to avoid unrelated complications to our local problem. However, we are free to enlarge/complicate the model if required : )

I think you are using the same circular weak argumentation I usually find in the few texts and books I know and I will try to explain why.

Yes, your are defining the probability to detect the H atom in a given volume of the space, i.e. you are defining the observable P_V=|V><V|, where <x,y,z|V> defines the volume: <x,y,z|V>=1 if (x,y,z) belongs to the volume and 0 otherwise. (I have masked the ∫dxdydz of the continuous operator).

Therefore, you are defining P(V)=<psi|P_V|psi> (i.e. the probability to get the atome in the volume V).

Thus <psi|P_V|psi> --> <psi|psi> when V --> +oO.

(I prefer the bra/ket formal view it avoids problems arising with specific basis/representations).

Now, you are defining a general state |psi>=sum_nlm c_nlm|nlm> where |nlm> is the eigen vectors of the Hamiltonian of the H atom, i.e you are selecting the classical bounded eigenvectors of the H atom.
Therefore, what you are saying is that the sub space (call it H_|nlm>) with the basis |nlm> is complete: H_|nlm> is a hilbert space.
However, you cannot prove that this subspace is the total H space if you keep your logical argumentation within this subspace otherwise you are able to demonstrate that a Hilbert subspace strictly included in H is not complete (i.e. it is not an Hilbert space).

You are also using the |nlm> basis that are defined with the spherical coordinates representation (r,θ,φ). The spherical coordinates are very special (we must take a lot of care using them in logical argumentation):
We have a bijection between spherical coordinates and Cartesian coordinates only in |R^3-{0}. This restriction of the domain of spherical coordinates has a direct impact on the spatial translation symmetry generators, the momentum operators, that are by construction ill defined at the point r=0.
Thus you cannot apply reasonable argumentation on the point r=0 using the spherical coordinates as it is not included in these coordinates by *definition*. You always need to use other coordinates well defined on the point r=0 to construct a reasonable argumentation at the point r=0. In other words, what you are constructing implicitly, using the spherical coordinates, is a wave function that is not defined at the point r=0 (and you patch this forgotten point with the external and unjustified continuity argument).
Application: the usual “r.<r,θφ|psi>=0, r-->0” condition of the H atom is completely artificial. The point r=0 does not belong to the spherical coordinates by definition. This condition only means that we restrict our search of solutions to functions that do not care that we remove the point r=0 in the probabilities calculus: i.e. bounded functions at point r=0 (and thus that we can pick up a continuous function that is equal to this function “for almost all x”).

Thus your continuity argumentation in a representation (the spherical coordinates) that does not include the point “r=0” by construction has no meaning. I mean you are free to pick up your solutions for the set of continuous functions but it is an external unjustified restriction. This is also one of the usual problems with the H atom argumentations.

Now, I will try to explain what I think is allowed by the QM model in order to explain why the forgotten point r=0 in the spherical coordinates is important, especially in the H atom.

The non-finite norm vectors in QM are not a problem by themselves, they just tell us, for example:

** Unbounded state:
The probability to detect a particle at the infinity is non-null (eg lim psi(xyz) =/=0 r->+oO; psi(x) continuous => psi(x) does not belong to L^2(|R^3,dxdydz)). This may the case of a particle state crossing the universe (used in scattering theory), e.g. the eigenvector |px> of the momentum px operator.

** Bounded state:
The particle is localized on a single position (e.g. dirac distribution ~ the probability density becomes infinite on “the point of the particle”), e.g. the eigenvector |x> of the x position operator.

We always use them: this is for example the continuous basis decomposition of the vector |psi>.
The “bounded state point” is a singularity (delta function), it can be approximated by several functions (exponentials, etc ...) before its normalisation. Refuting the non-possibility of this state is, for me, like refuting black hole singularities in GR: it is an arbitrary choice rather than a proof. (Note that I am not saying that the “physical world” has such a state, just this is an ideal model as in GR)

What you have demonstrated is what I have found in the few books/papers: a bounded state with a given probability density at r=0+ and r=+o0 defines a Hilbert space. The usual eigenvectors (|nlm>) of the Hatom Hamiltonian are a basis of this Hilbert space. However, this Hilbert space is a subspace of the global Hilbert space, at least this subspace does not include the |x,y,z=0> ray of the Hilbert space (forbidden by the domain of applicability of the spherical coordinates).
With your argumentation, we do not remove the possibility of “r=0 bounded states” (singular states).

Therefore, we need to look at the rest of the “enlarged” eigenvector solutions (i.e. distributions) of the Hamiltonian to find the total Hilbert space (the total Hamiltonian to be more precise).

Now, if we take the simplified H atom equation in spherical coordinates (and not forgetting the singularity of the spherical coordinates), and searching for the eigenvectors of E < 0, we have for the radial part R(r) = y(r)/r (e.g. Messiah, Quantum mechanics, 1958):
(There may be some minor typo errors in the following formulas)

(1) y’’(r)+[ -f² + 2/(d.r) – l(l+1)/r²]y(r)=0

with:

f²= -2mE/hbar² (E <0)
d=(hbar²/ m.e²)

and if we define:

x=2.f.r

and g=1/(f.a)

y(x)=exp(-x/2).x^(l+1).v(x) => R(x)= exp(-x/2).x^l.v(x)

We have the so-called Laplace equation (this is the name I know it, this not the helmotz equation):

(1) <=> (2) [x(d/dx)² + (2l+2-x)d/dx –(l+1-g)]v(x)=0

Where we have the two independent solutions:

(a) v1(x)= w1(l+1-g|2l+2|x)= w1(l+1-g|2l+2|2.f.r)

(b) v2(x)= w2(l+1-g|2l+2|x) = w2(l+1-g|2l+2|2.f.r)

The two solutions are irregular at the point r=0.

However, we can form two other independent solutions with one that is regular at the origine (x^(l+1)) the other irregular (1/x^l):

(c) F(l+1-g|2l+2|x)= w1(l+1-g|2l+2|x)+ w2(l+1-g|2l+2|x): the hyper-geometric series. The one that is regular at the origin. It gives the usual discrete energy solutions if we look for the solutions that have no particle at the infinite.
The remaining solutions between the discrete energy levels may be interpreted as stationary solutions with particles out of the universe (the associated “renormalized probability” density becomes significant only at the infinity): we can keep the coherence if we develop the mathematic consistency.

(d) G(l+1-g|2l+2|x)= w1(l+1-g|2l+2|x)- w2(l+1-g|2l+2|x). This is the irregular solution at the origin.

The rejection of this last solution may lead or not to the possibility of r=0 bounded states.
Currently I do not know any good demonstration that reject or explain that (d) does not correspond to a possible state and you have not demonstrated such impossibility.

Seratend.

12. Dec 13, 2004

### Wes Hughes

Hi, The simplest explanation of why the the electron does not spiral into the proton is this: According to Schroedinger the electron is a wave extending throughout space-time, but most of the wave is very near to the center or proton. The wave is in motion of course so from every point there is generated electromagnetic radiation. However some brave souls have calculated the whole grand total effect. It turns out that as seen from a large distance everything cancels out when you consider phase and strength. So there is no out flowing radiation.This is sort of like in the center of a hollow sphere the gravity field cancels out. So even though the electron is in "motion" it does not radiate as long as it stays in the same state. Hope this helps.

Wesley Hughes

13. Dec 13, 2004

### seratend

Yes, for the stationnary states. However it does not explain the minimum ground state of the H atom, just that stationnary states are stable (or metastable if you want => requires an external interaction to change).

Seratend.

14. Dec 13, 2004

### ZapperZ

Staff Emeritus
But that's a very misleading comment since QM doesn't pretend to explain anything. The fact that these ARE stationary solutions, and that these came out of something resembling Hamilton's formulation of classical mechanics, means that these are "extrema". The lowest energy state simply exists because the solution cannot give a "physical" answer for anything lower (or even negative quantum number).

One can also verify this if one make an educated guess at a possible ground state wavefunction and solve this variationally. Unless we distrust calculus and its ability to find a minimum point, there is always a non-zero energy state corresponding to some non-zero minimum "radius".

Zz.

15. Dec 13, 2004

### tribdog

I think the problem is that so many people still think of the nucleus as the Sun and electrons as planets.

16. Dec 13, 2004

### ZapperZ

Staff Emeritus
Exactly!

It doesn't help that even now, we call the angular part of the solution as "orbitals" due to historical reason. Most people are always amazed when I tell them that the "s" orbital solutions have ZERO angular momentum. If we translate this into classical mechanics, it means this electron either isn't moving, or it is moving (oscillating) along a line that passes through the nucleus! Neither of these make much sense. So the only thing that is at fault here is our classical picture of an atom. Once we get rid of that silly idea, everything else falls into place.

Zz.

17. Dec 13, 2004

### humanino

actually, the elctron in an "s" wave does have a higher probability of being located in the nucleus, which enhances electron capture $$\beta$$-decay

18. Dec 13, 2004

### humanino

...but then of course, gurus are, as usual, right

19. Dec 13, 2004

### ZapperZ

Staff Emeritus
Oh no... let's not start that again.

There is a difference between |<R_nl|R_nl>|^2 and |<R_nl|r|R_nl>|^2. If you look at the rR_nl function plotted out, you'll see that this function is identically zero at r=0. Now whether they have a higher probability to be "closer" to the nucleus, that's a different matter, and no one is disputing this. That's one of the reason why the 4s is filled up first ahead of the 3d.

Zz.

20. Dec 13, 2004

### humanino

Yes you are, again, right
Still, the probability of finding the electron in the nuclei is higher.
s-wave