Accelerated electrons emitting radiation

[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

 

[niehls]

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

 

[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.

 

[dextercioby]

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

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.

 

[dextercioby]

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

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.

 

[seratend]

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.

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

 

[marlon]

I haven't heard of antenna radiation problem discussed at QED level...

Daniel.

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

 

[dextercioby]

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

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.

 

[seratend]

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.

Daniel.

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.

 

[dextercioby]

Ok, let's reduce our model to the H atom with the coulombian interaction only.

.

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

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).

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

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).

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.

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.

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}.

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.

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.

 

[seratend]

HYDROGEN ATOM

.
Originally Posted by seratend
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).

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

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 : )

.

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).

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).
And, we have the property about this projector P_V --> Identity when V -> +oO.

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.

 

[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

 

[seratend]

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

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.

 

[ZapperZ]

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.

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.

 

[tribdog]

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

 

[ZapperZ]

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

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.

 

[humanino]

it is moving (oscillating) along a line that passes through the nucleus! Neither of these make much sense.

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

 

[humanino]

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

 

[ZapperZ]

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

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.

 

[humanino]

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

 

[seratend]

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).

Are you sure? This is rather a credo than a logical deduction.
I prefer to understand the model and to identify the external hypotheses used in such a model rather than claiming that a solution cannot give a ""physical" answer for anything lower".

Solving a problem with an extrema/varationnal method is one of the possible mathematic tools we have. Extremum problems are always based on an input space. If we restrict the input space in the adequate manner, we can find almost all lower bounds we want. Thus we have to understand/explain the restrictions that are made.

If we look further on the simple classical model of H atom, we may constat that we have a simple toy model to construct what seems a single particle state from 2 particles without using a "blind" scattering method or QFT tools.
I think it should be interesting to explain why this lower energy solution is not taken into account rather than calling for external "physical evidences".

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.

This is also a credo based on the input space you use. In the case of the H atom, that means that you do not accept "singular" states.

Seratend.

 

[ZapperZ]

Are you sure? This is rather a credo than a logical deduction.
I prefer to understand the model and to identify the external hypotheses used in such a model rather than claiming that a solution cannot give a ""physical" answer for anything lower".

This in itself is a "credo". Just because there IS a mathematical solution doesn't mean it represents anything physical. Case in point: solution inside the conductor when one solves the Poisson's equation using the image charge technique. The solution inside the conductor is unphysical. It represents nothing, no matter how much you try to squeeze out the "understanding" out of that model.

Zz.

 

[seratend]

This in itself is a "credo". Just because there IS a mathematical solution doesn't mean it represents anything physical. Case in point: solution inside the conductor when one solves the Poisson's equation using the image charge technique. The solution inside the conductor is unphysical. It represents nothing, no matter how much you try to squeeze out the "understanding" out of that model.

Zz.

I think you should better study the image solution and the boundary limits of poisson's equation ; ). The image model just describes how we can replace the potential created by charge density at the surface of the conductor into a virtual charge. It is the interpretation/explanation of this model that produces the rest.

Seratend.

 

[marlon]

This in itself is a "credo". Just because there IS a mathematical solution doesn't mean it represents anything physical. Case in point: solution inside the conductor when one solves the Poisson's equation using the image charge technique. The solution inside the conductor is unphysical. It represents nothing, no matter how much you try to squeeze out the "understanding" out of that model.

Zz.

Quite right...

As an addendum : look at black holes vs white holes...White holes arise due to time-reversal-symmetry in General Relativity...Just because you replace t by -t or something like that, it may be justified mathematically but that doesn't justify physical relevance...

regards
marlon

 

[seratend]

Quite right...

As an addendum : look at black holes vs white holes...White holes arise due to time-reversal-symmetry in General Relativity...Just because you replace t by -t or something like that, it may be justified mathematically but that doesn't justify physical relevance...

regards
marlon

Except for the big bang model! :tongue2:

 

[ZapperZ]

I think you should better study the image solution and the boundary limits of poisson's equation ; ). The image model just describes how we can replace the potential created by charge density at the surface of the conductor into a virtual charge. It is the interpretation/explanation of this model that produces the rest.

Seratend.

Er.. no. It is the Uniqueness Theorem combined with the Dirichlet/Neumann boundary condition of the Poisson's equation that allows one to use the image method. Without that, there is no guarantee that what you are substituting has any resemblance, or even is the completely identical situation of the original. But this really is besides the point that I was trying to make. You don't pay attention to the solution outside of the actual physical space. The solution of the fields inside the conductor is unphysical. Other than the boundary of the real space with the conductor to be used to find the surface change density, one does nothing else with the fields inside the conductor's surface because these region are meaningless. My point being that just because a solution exists mathematically doesn't mean they actually have any physical relevance.

Zz.

 

[seratend]

Anyway, I want to calm this thread. I am really interested in a good explanation on how we can remove the irregular solutions of the H atom toy model.
All suggestions are welcome :tongue: .
I am trying, now to look deeper (but quiclky as I am almost sure that somewhere there already exists a good explanation) on these solutions in order to propose a reasonnable explanation if I can (I am not sure but I am trying), any help is welcome.

Seratend.

 

[seratend]

Er.. no. It is the Uniqueness Theorem combined with the Dirichlet/Neumann boundary condition of the Poisson's equation that allows one to use the image method. Without that, there is no guarantee that what you are substituting has any resemblance, or even is the completely identical situation of the original. But this really is besides the point that I was trying to make. You don't pay attention to the solution outside of the actual physical space. The solution of the fields inside the conductor is unphysical. Other than the boundary of the real space with the conductor to be used to find the surface change density, one does nothing else with the fields inside the conductor's surface because these region are meaningless. My point being that just because a solution exists mathematically doesn't mean they actually have any physical relevance.

Zz.

It is indeed the field inside the conductor that explains the discontinuities: You can see the discontinuities as an abstract feature or you may try to understand them. In this aspect it is similar to the H atom. You can developp the conductor model in order to understand how this ideal discontinuity is created (you add a thickness, use the charge continuity and ohm's law to define how the field is changed and so on ...).

We must not confuse an ideal mathematical model with a real situation. Understanding the limits of the mathematical model helps in understanding better the real situation (and vice versa). Understanding the limits of the model allow us to say that a solution of this model is not physical. However, for the H atom, I have not found such an understanding. Just external engineering statements to match what is usually seen.

Seratend.

 

[ZapperZ]

We must not confuse an ideal mathematical model with a real situation.

But that is what I've been trying to say here, and by bringing out the image charge situation. Just because there are mathematical solutions in all regions of the problems, doesn't mean they all have any physical attachments to them. Some of them may not make any physical sense.

Understanding the limits of the mathematical model helps in understanding better the real situation (and vice versa). Understanding the limits of the model allow us to say that a solution of this model is not physical. However, for the H atom, I have not found such an understanding. Just external engineering statements to match what is usually seen.

Seratend.

Isn't matching "what is usually seen" typically the final arbiter of what is valid? I don't see this as a problem.

Zz.

 

[seratend]

But that is what I've been trying to say here, and by bringing out the image charge situation. Just because there are mathematical solutions in all regions of the problems, doesn't mean they all have any physical attachments to them. Some of them may not make any physical sense.

It seems this is the main difference between our points of views (even if I think it is just a matter of context ;). If I take what you say (in a very simplified and exagerated version) for true I must say a dipole does not exist. (or the point particle equivalent model that is closer to the charge+conductor image model, if I am wrong in this quick modelling).
The mathematical model says that 2 different source of fields models lead to the same result (the coulombian potential in this case), they do not exclude anything. They rather tell us that we have 2 physical implementations to realize the same result, not that one is virtual and the other is real (i.e. the two models are possible).

In this point of view, it is important, at least for me, to understand the logical assumptions that lead to conclude that the usual ground bounded state of the H atom is really the lowest energy bounded state of the two-particle system.

Claiming that this ground state is the lowest energy of the system because we have not seen a lower state up to now while the theory+model seems to allow it is a strange argumentation.

We can speak a lot on how to use mathematical models and what they say and their interpretation, but this is not the subject.

I think, it is reasonable to question why the irregular states of the H atom are not allowed in our physical environment and to search for a logical explanation (or if they are allowed, how to produce them: the logical selection rules).
If you think the information you have are sufficient to conclude in the impossibility of these states, very good for you! It is not my attention to change your point of view.
I am just looking for a logical argument (for or against it) in the context of QM theory. I have tried to explain in this thread what is missing for a logical argumentation. Up to now, I have not seen an answer that leads to the logical conclusion that such a state is impossible. Therefore, I am looking for it.

However, I am surprised to see such reluctance to question the foundations of a model. I am even more surprised by the implicit rejection of the strange states of a physical model.

Seratend.

 

[ZapperZ]

However, I am surprised to see such reluctance to question the foundations of a model. I am even more surprised by the implicit rejection of the strange states of a physical model.

Seratend.

Who is being reluctant? All I'm saying is that just because you have a mathematical solution, doesn't mean it has any relevance to anything physical. I gave the image charge problem as an example. The defining test is STILL the experimental evidence. I'm sorry if such a criteria is not good enough for you. My training as an experimentalist is rearing its ugly head again.

Zz.

 

[dextercioby]

Reply

HYDROGEN ATOM
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.

My arguments may e sometimes,but I'm trying my best to keep my logics clear,and not circular.If i cannot,then i should do something else.Theoretical phyiscs is all about logics.

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_|nlm> is a Hilbert space.

Yes.I'm saying that $H_{nlm}$ is Hilbert space of states for the Hydrogen atom in total agreement with the first principle QM (Dirac/standard formulation) which says that the states of a given quantum system ar described by a sequence (at most countable):${|\psi_{k}>,p_{k}}$,in which $|\psi_{k}>$ are NORMALIZED VECTORS FROM A SEPARABLE HILBERT SPACE CALLED "THE HILBERT SPACE OF STATES",and $p_{k}$ are real positive numbers satisfying $\sum_{k} p_{k}=1$ which are called "ratios" (I don't know whether this is the most appropriate english translation for the romanian word "pondere") (whose physical significance is given when discussing quantum virtual statistical ensembles).

However,you cannot prove that this subspace is the total H space,

You mean i cannot prove that $H_{nlm}$ is the total space H.Which "total Hilbert space" are you talking about...??

I made use of the first principle to tell me which solutions of the (H atom) Hamiltonian's spectral equation to look for.I'm interested in solutions within the subspace $H_{nlm}$ which is spanned by the basis $|nlm>$ and whose vectors (just like the ones from the base) are NORMALIZABLE.So I impose:
$<n'l'm'|nlm>=\delta_{nn'}\delta_{mm'}\delta_{nn'}$ and ask for a similar relation for every $|\psi>$ from $H_{nlm}$.

If you mean that I'm unable to prove that $H_{nlm}$ is a Hilbert subspace in the total Hilbert space spanned by the solutions of the spectral equation (for the Hamiltonian of the H atom) $\hat{H} |\psi> = E |\psi>$,then u are wrong.H atom's hamiltonian is the closed selfadjoint extension of the hermitean operator defined with agreement to the quantization postulate.The trouble with this (hermitean) operator is that it is unbounded,so it needs a closed selfadjoint extension,else,had it been bounded,it would have required only a selfadjoint extension.It can be proven that $\hat{H}$ is essentially selfadjoint (his adjoint is a selfadjoint operator).So the fact that the hamiltonian has a mixed spectrum (and the general analysis made for arbitrary linear operators),make(s) me write that the total Hilbert space of the solutions to the spectral equation for the H atom's hamiltonian is a direct sum between $H_{nlm}$ and $H_{\alpha s}$,where $\alpha$ is the parameter for the continuous spectrum (for the H atom:the positive real semiaxis) and "s" is the index for the degeneracy of the spectral value $E(\alpha)$.

So i can define veritable (true) orthogonal projectors from the total space to $H_{nlm}$ by chosing that the basis in $H_{nlm}$ to be exactly $|nlm>$,hence the projector is $|nlm><nlm|$,and note this is a real projector (idempotency is easily checked and found to be true).

 

[dextercioby]

Reply2

HYDROGEN ATOM
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.

APPARENTLY,GOOD POINT,MY FRIEND!I have to admit,i have never asked myself this kind of questions (up until now,so i thank you for making me think and allowing me to use all my mathematical knowledge to demonstrate that those books are right,when chosing spherical coordinates ).Whether it is really (mathematically) "fair" to make use of spherical coordinates,knowing (or not) that this fact would alter the legacy of our judgements regarding the behavior of functions in the point "zero".


I can prove to you there is no problem with chosing spherical coordinates. No problems at all,neither with the physics,nor mathematics.

To be strict/correct,the diffeomorphism is defined on $R^{3}-M;M=[A\in R^{3}|x(A)=y(A)=0]$,so the problem should be with the entire Oz axis,not only the origin. So u should not be talking about "the forgotten point '0'",but about the "forgotten axis 0z".

Anyway,apparently this problem occures in states which are not spherically simetric (with $l\neq 0$),where the "orbitals" have such a shape that the probability that the electron is on the Oz axis is zero.
Nonetheless,keep in mind that the probability is a function of the domain of discussion (integration wrt to the volume of R^{3}),and make use of the fact that,since by our choise for $H_{nlm}$ to be the Hilbert space $L^{2}(R^{3}$,there is no problem to restraining the domain to $L^{2}(R^{3}-M)$,if we shall encounter problems with functions defined on all $R^{3}$.I'll explain why.


However,we chose as domain for our functions the entire $R^{3}$,though,as we mentioned,there may be problems with the points on the Oz axis.Why??Because of our particular choise of Hilbert space and scalar product.Keep in mind it's not a Riemann integral involved in the scalar product,but a Lebesgue integral,over a domain in which we have defined the Lebesgue measure.When performing integrals with Lebesgue measure,we can subtract from the domain of integration tricky points like the Oz axis (points with Lebesgue measure null),but only if on the Oz axis our functions (which we integrate) do not explode.So,it makes sense now to exclude Oz axis from the integration and from discussion,while making sure that the integrands do not explode when evaluated on the Oz axis.

I think excluding the Oz axis from our discussion wrt to probabilities implies,as the diffeomorphism is not defined on the Oz axis,that there is no point in calculating probabilities for a point or a line.The probability function is defined on a volume (domain from $R^{3}$) with non-null Lebesgue measure.So it makes no sense to calculate the probability that the electron is found on any point/surface (including the origin and the Oz axis as particular cases) as it is not defined on points/surfaces,which HAVE NULL LEBESGUE MEASURE,when seen as elements of $R^{3}$.

So i believe that u CAN MAKE USE OF SPEHERICAL COORDINATES AS LONG AS U MAKE SURE THAT THE FUNCTIONS WHICH YOU INTEGRATE DO NOT EXPLODE WHEN EVALUATED ON THE NULL LEBESGUE MEASURE DOMAIN ON WHICH THE DIFFEOMORPHISM IS NOT DEFINED,AND THAT,BECAUSE U HAVE CHOSEN THAT THE HILBERT SPACE OF STATES IS $L^{2}(R^{3})$,AND NOT $L^{2} (R^{3}-Oz)$.And even if u had,that would have not changed anything when discussing scalar product (and hence,probabilities),as the Lebesgue measure of integration would have been the same.

I hope to be clear (though,maybe circular (again :tongue2: )),and tell u that in QM there are no black holes,singularities and s***".

Daniel.

PS.I GR,we don't have Lebesgue integrals.

 

[dextercioby]

HYDROGEN ATOM
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.

Yep,i kinda remember this attempt to solving the radial equation from last year.Since I'm not at the library at this very moment and I'm not able to pick Albert Messiah's from the shelf,i cannot follow your equations exactly.

I only want to say,that even though this description from Messiah' book seems to exhaust mathematical analysis into the radial equation (seen as a mathematical entity,without no "a priori" physical relevance),i'm afraid it lacks the absolutely necessary ingredient:the answer to the question:"why on Earth do we refute half of the equation's solution (exactly that irregular part)??"

I remember writing an article (for my use only,not published) last year exactly on this matter:the radial equation of the H atom.I'll extract some results/equations from it:
<<Conclusion:
The radial equation
$$u''_{l}(r)+[\frac{l(l+1)}{r^{2}}-\frac{2\mu\alpha}{\hbar^{2}r}-\frac{2\mu E}{\hbar^{2}}]u_{l}(r) =0$$
,where $u_{l}(r)=r R_{l}(r)$ are the secondary radial functions,admits the solution
$$u_{l}(r) =[\exp(-\frac{r}{2a})](\frac{r}{a})^{l+1} [C_{1} F(l-n+1|2l+2|\frac{r}{a})+C_{2}(\frac{r}{a})^{-2l-1} F(-l-n|-2l|\frac{r}{a})]$$
where $\alpha:=\frac{e^{2}}{4\pi\epsilon_{0}};a^{2}:=\frac{\hbar^{2}}{8\mu E};n:=\frac{2\mu\alpha}{\hbar^{2}} a$
and "l" is the angular momentum quantum number (eigenvalue for $\hat{L}^{2}$),a natural number.

Making the multiplication in the solution,one gets:
$$u_{l}(r) =C_{1}[\exp(-\frac{r}{2a})](\frac{r}{a})^{l+1} F(l-n+1|2l+2|\frac{r}{a})+C_{2}[\exp(-\frac{r}{2a})](\frac{r}{a})^{-l} F(-l-n|-2l|\frac{r}{a})$$

 

[dextercioby]

Now,analysis of the 2 confluent hypergeometrical series involved in the solution reveals 2 aspects:
1)That "n" is found to be the "principal quantum number"and it takes all natural values from "0" till "l+1".Else,"l" takes all possible natural values from "0" till "n-1".That thing is proved by requiring that the hypergeometrical series will not "blow up"when making "r->+oO".
2) The irregular solution $C_{2} [\exp(-\frac{r}{2a})](\frac{r}{a})^{-l} F(-l-n|-2l|\frac{r}{a})$,though correct,mathematically speaking (and linearly independent from the orther one),HAS TO BE REJECTED FROM 2 REASONS:
a) For angular states with $l\neq 0$,it is singular in the origin (the exponential is a constant,and the hypergeometrical function is a constant as well) and therefore would not satisify the boundness and Lebesgue square integrability criterion imposed to the principal radial function $R_{nl}(r)$.This has to do both with mathematics and phyiscs.
b)In the special case in which $l=0$,though the solution is normlizable,the KE integral (expectation value)
[tex] <KE>:=\int_{0}^{+\infty}} R_{nl}(r) (-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}) R_{nl} (r) [/itex]
would diverge.And that's the place where u have to use physics.So far,only mathematics.But the average of the KE operator has to be finite.And that has to be true for every (quantum) observable of the H atom.

Hopefully it's all very clear right now,at least with the H atom analyzed through Shroedinger's methods.If u want to discuss it in relativistic context (Dirac/Schroedinger-Gordon-Klein-Fock),or perturabatively,i'll stay at your disposal.

Daniel.

PS.Messiah's is still a good book,but it lacks many mathematical subtleties.This one with the KE operator average and the one with Lebesgue square integrability criterion imposed to any integrand in a scalar product on $L^{2}(R^{3})$ were just 2 of them.