Accelerated electrons emitting radiation

[marlon]

Nice work dextercioby...

marlon

 

[Wes Hughes]

Hi this is Wes Hughes again,

Seems like I am causing a bunch of warnings to be issued. They sent me several letters asking me to preticipate, I thought I was doing the right thing by making an explainatory comment. Sorry about that. I'll stay away from your site in the future.

Wes Hughes

 

[dextercioby]

HYDROGEN ATOM
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

I remember using the term "Laplace equation" for the Kummer equation of the Gauss hypergeometric series,but only in connection with the hypergeometric function,ad not in connection with the confluent/degenerate function whose particular case is Laguerre polynomials,involved in the solution of the H atom radial problem (discrete spectrum).
And that name (Laplace equation) i believe it has only historical reason,since it occurs first time when Pierre Simon (Marquis de) Laplace attempts to solve his equation in spherical coordinates,where he gets the equation for Legendre polynomials/associated Legendre functions.
Anyway,i haven't encountered too often,as the differential equation for the hypergeometrical functions/series is usually related to the name of Kummer.And the functions have been discovered by Carl Friedriech Gauss.

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.

I believe i have done it in my prior posts. And it's pretty good...

Daniel.

 

[seratend]

Hi this is Wes Hughes again,
Seems like I am causing a bunch of warnings to be issued. They sent me several letters asking me to preticipate, I thought I was doing the right thing by making an explainatory comment. Sorry about that. I'll stay away from your site in the future.
Wes Hughes

Please do not hesitate to participate. Personnally I appreciated your post and If you have a link to their demonstration: do not hesitate to post it in this thread!

If you mean that I'm unable to prove that is a Hilbert subspace in the total Hilbert space spanned by the solutions of the spectral equation (for the Hamiltonian of the H atom) ,then u are wrong.
Daniel.

First, it is a real pleasure to exchange something constructive based on some mathematical ground. Thanks a lot, for the correction on the z axis of spherical coordinates rather the r=0 point, however, I am lucky, for the H atom only the r=0 point of the z axis point important ; ).

We need to understand what abstract Hilbert spaces and isomorphism between Hilbert spaces are. First, an Hilbert space is an abstract vector space complete with the associated hermitian product. In addition, we assume the separability property that’s all. See thread Hilbert Space,Dirac Notation,and some other stuff https://www.physicsforums.com/showthread.php?t=44301&page=4 for additionnal information (short description).

The hermitian product does not require the definition of an integral (i.e. we done not define a measure and the associated sigma-algebra, for example the borel sets and the lebesgue measure). It does not exist, a priori, in the abstract Hilbert space (we can induce them by isomorphism thanks to the separable propert, that’s all). That’s the main advantage of the abstract Hilbert spaces: no existence, a priori, of a sigma algebra and a measure.

Now you have an important property: the isomorphism between Hilbert spaces. Because the abstract Hilbert space is separable, you have an isomorphism between this Hilbert space (the one used to define the QM theory, call it H) and the Hilbert space L^2(dx) of “square integrable functions” (where the hermitian product is defined this time by the Lebesgues integral). An isomorphism just states that there is a bijection between the points of the 2 concerned sets that preserves the properties of the Hilbert space and not that the sets are identical, i.e H=L^2 is false.

I will take a simple example: we have |N included in |Q while |N is isomorphic to |Q for the countable property (we have a bijection between the integers and the rational numbers), but we have not |N=|Q !

Your statement about the H_|nlm> is the same thing: you just have an isomorphism between spaces but you have not the equality between the sets. Rather, you may have H_|nlm> included in H_|xyz>.
You have noticed that I am not using the Lebesgues Integral and the L^2 space just because they are peculiar spaces with their own specific properties (we have the danger to attribute a property that depends only on the representation like the H_|nlm> representation).

Your argumentation about the H atom is the same as the following one concerning the moment operator ^p:
In the abstract Hilbert space we have: ^p|p>=p|p> with the spectrum of p is the |R set.
Now, we have H is isomorph to H_|p>. We can define if we want H=H_|p>.
Now choose the integer subset of the ^p eigenvalues, call it |p_n>. We have:
^p|p_n>=p_n|p_n>
Now, we can quickly define the operator P_nat= sum_n p_n |p_n><p_n|
We have ^p =/= P_nat.
However, we can define the Hilbert space defined by the countable eigenvectors of P_nat: H_|p_n>.
H_|p_n> is naturally isomorphic to the l^2 space which is isomorphic to L^2 and therefore to H_|p>. Now if I follow your deduction, I will say that the representation operator p on H_|p_n> has only quantified values however that does not mean that this is the only allowed eigenvalues! you are concluding the same thing when you are using the specific Hilbert space H_|nlm> with a representation of the H atom Hamiltonian on this space (your are retricting the possible eigenvalues of the Hamiltonian).

Please note that I have not used the specific properties of Lebesgues integral to deduce something as they only give properties relatively to the L^2 hilbert space not the abstract Hilbert space (abstract in the sense, without additional properties such as a Lebesgue measure). This is the main power of the Hilbert space tool (you can define by hand a hermitan product and you recover the general theory of integration!).

Question: In GR, we use integrals and the manifold structure is not sufficient to define an integral structure, thus I think GR requires the minimum definition of integrals (a sigma algebra and a measure) to be able to generalise them, but I may be wrong. I am not an expert in GR: it has been a long long time since I have looked at the theory ; ). Any precision about “in GR we don’t have Lebesque integrals” is welcomed for my personnal knowledge (and may be to the benefit of the PF users ; ).

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:
$$u_{l}(r) =C_{1}[\exp(-\frac{r}{2a})](\frac{r}{a})^{l+1} F(l-n+1|2l+2|\frac{r}{a})+C_{2}[\exp(-\frac{r}{2a})](\frac{r}{a})^{-l} F(-l-n|-2l|\frac{r}{a})$$

Thanks for the input on the solutions of eigenvectors H atom SE, I really appreciate (I have not spent enough time with the explicit solutions of the H atom in the q representation). I have failed into transforming the w1-w2 solution into a simpler F(a|b|c) function that are “easier” for analysis ; ). By the way, do not under estimate the messiah subtleties because it is written in an *apparent* simple and accessible mathematical language. (I have made often the mistake ; ).

However, unfortunately for the simplicity of the analysis, I think you have made a mistake with the irregular solution.
I think you have tried to solve the radial equation just picking the regular solution u_l(r) and performing the transformation l into (–l-1) (the radial equation is invariant under such a change) but this does not work. I’ve done an analogue mistake in my quick evaluation in transforming the w1-w2 solution into a F(a|b|c) solution. What you recover, using this method is the same regular solution exp(-r/2a).(r/a)^l.F(l-n+1|2l+2|r/a) not the exp(-r/2a).(r/a)^(1-l).F(-l-n|-2l|r/a):

We have F(a|b|z)= sum_n gamma(a+n)gamma(b)/gamma(a)gamma(b+n).z^n/n!.

Gamma(b) is the well known Eulerian function that is not defined when b is a negative integer => F(-l-n|-2l|r/a) does not exist for positive l,n integers. Now I explain why.

The Laplace equation (or the name you prefer, I do not care, see my previous posts to recover it from the radial equation):
(1) [z(d/dz)^2 + (b-z)d/dz – a]f(z)=0
has *one* regular (at the origin) solution if b is an integer:

* f1(z)=F(a|b|z) if b ≥ 1

* f2(z)= z^(1-b)F(a-b+1|2-b|z) if b ≤ 1 (Messiah QM, 1958)

When you make the transformation of l into “(-l-1)” in the Laplace equation, you obtain a different Laplace equation with the b parameter equal to “-2l” thus you should select the f2(z) as a regular solution that is identical to the f1(z) for “l=-l-1”. And this normal, because the radial equation from, whom the Laplace equation is deduced, remains unchanged by this parameter update.
Thus, if we correct the undefined solution R2(r)= exp(-r/2a).(r/a)^(1-l).F(-l-n|-2l|r/a), we recover R(r)= exp(-r/2a).(r/a)^l.F(l-n+1|2l+2|r/a) the same solution of the bounded states => your solution unfortunatly does not define 2 different solutions but the same solution of the Lapalce equation.

I think we need to use, for the irregular solution as an independent solution, the “awful” form solutions:

(2) R2(r)= exp(-r/a).(r/a)^l.i. [w1(l-n+1|2l+2|2.r/a)- w2(l-n+1|2l+2|2.r/a)]
= exp(-r/a).(r/a)^l.G(l-n+1|2l+2|2.r/a)
Where I have defined G= i(w1-w2)
(I have tried to rewrite the equation with your symbols). However to get the good exponential, I have to define a²=hbar²/2m|E| and not hbar²/8m|E|). Do not hesitate to correct me if someone detects an error.

We can select other mixtures as we have this freedom. We just need to find 2 independent solutions.

However, at the origin we have for this form of solution (singularity for the wave function):

(3a) R2(r)~ (K/r^l)[1+ O(r)] if l ≠ 0
(3b) R2(r)~ (K/r^l)[1+ O(rlnr)] if l = 0

At the infinity, if n is not an integer R2(r) diverges as an exponential as F(l-n+1|2l+2|2.r/a) does. (Messiah QM, 1958, annexe B). However, this statement should be rigorously verified later.

(Note, I am using Messiah book because it is the first available book I found with details on the explicit solutions to the H atom Hamiltonian, both for E>0 and E<0).

I am now studying the case n integer in order to find if there is a bounded solution when r ->+oO . If this is the case, we recover a different quantification rule for the irregular solution However, the “exp(-r/a).(r/a)^l.G(l-n+1|2l+2|2.r/a)” are somewhat more complicated to handle (It has been a long time I have manipulated this functions).

If someone as more date on the properties of [w1(a|b|c)-w2(a|b|c)] function, please do not hesitate to post/mail me.

*********************************************************************
However, I have now a simpler method based on the singularity (1/r^l type) that allows constructing simple “irregular” solutions and leads to a new quantification rule for the singular/irregular states of negative energy (quite similar to the basic methods introduced to solve the H atom in schools).

Let’s assume that we choose now the solutions of the more general form of the Laplace equation (or the name you prefer ; ):

(4) [x(d/dx) ² + (b-x)d/dx – a]f(x)=0

(we have the mapping b= 2l+2 and a= l+1 – (e²/hbar.c)². (m.c²/-2E)½ to recover the usual H atom form and x= k.r where k is the good proportional factor)

(5) y(x)= sum_p≥po c_p x^(-p) with po≥1

We can use general Laurent series, however, we can easily demonstrate that only this form works if we want a convergent series solution of (4) different of the regular solution (we can even use a mix of x^p, x^(-p) and x^pln|x|). Thus, we have enlarged the set of possible solutions to the Laurent series.

If we introduce (5) in (4) we have the following conditions on the coefficients:

(6) x^(-po) coef: (po – a)c_po = 0
(7) x^(-p) coef: (p-1)(p-b).c_p-1 + (p-a).c_p = 0 for p>po

(7) <=> (8) c_p=(p-1)(p-b)/(p-a) .c_p-1

(5) is a non null solution of (4) iff a=po ≥ 1.
(5) is a finite sum iff b ≥ po+1

Now if we use the (a,b) values (integers) issued by the H atom Hamiltonian we have:

(5) is a non null solution <=> l+1 – (e²/hbar.c). (m.c²/-2E)½ = po ≥1

=> (e²/hbar.c)². (m.c²/-2E)½ is a *null* or positive integer that we may call n (we are considering the bounded states E≤ 0)

=> l+1 – po= (e²/hbar.c). (m.c²/-2E)½ ≥ 0 (8)

=> l ≥ po-1 (9) => b=2l+2 ≥ po+1 => (5) is a finite sum !

This result implies that only 1/r singularities are allowed for bounded R(r) orbitals (R(x)=Kx^l.exp(-x/2)y(x)) !

We have the following results for the singular states:
**********************************************************************
(5) y_l_po(x)= sum_p≥po c_p x^(-p) with po≥1 is a singular solution of the eigenvalue problem (E ≤ 0) iff:

(9) l ≥ po-1
(Selection rule for these states)

(10) E_l_po= E_n= - (hbar.c/e²)²/( l+1 – po)²= - (hbar.c/e²)²/n²
(Energy quantification)
***********************************************************************
We have the same energy levels (this is due to this simplified model) with *one main* difference “n=0” is allowed by these new selection rules! <=> E_0= - oO ! (Thus, this model requires the relativistic model in order to be analysed correctly).

We are just recovering the solutions of an infinite quantum well – V(r) potential where E_0= - oO is the reference ground state !
The slopes of the coulombian potential are not sharp enough to get a wave function localized within the well => spatial extension.

This solution and simple model is just amazing because the regular solutions (the well-known bounded states) and irregular solutions are completely decoupled and have the same energy except for the ground state.
To understand it simply, think on a quantum well (with an infinite negative ground state: V=0 for |x|>a and V=-oO for |x|<a), where the bounded wave function is localized within the well. The quantization energy is controlled by the quantum well geometry as in our example. However, we can define wave functions that are outside the quantum well: they are completely decoupled. It is what we encounter with this model: we have the bounded states within the coulombian well (singular solutions) and the bounded states outside the coulombian well (irregular solutions) but coupled “externally” to the quantum well if I am right. (this is new and funny for me ;).

More important, if we want, the states within the well may be interpreted as a new particle with an integer spin that defines its internal energy (l≥ po-1 condition)

If we analyse the E_0= -oO solution, we have (without normalization coeffs):

y_0_0(x)= 1/x => R_0_0(x)= exp(-x/2)/x

=> <psi_0_0|psi_0_0> < +oO

=> <psi_0_0|r^i|psi_0_0> < +oO i, any integer.

if we take pr=(hbar/i)∂r(1/r) => <psi_0_0|pr|psi_0_0> < +oO !


I haven’t verified the other forms in the normal coordinates. However, for a singular state, we already have a wide number of finite hermitian products! Like in a quantum well, we have finite norms.


(In fact, I think we need to enlarge our Hilbert space to a rigged Hilbert space to recover a full mathematic consistency – use of distributions).

Now if we develop the other singular states, we recover orbitals R(x) of the following form:
R_nl(x)= exp(-x/2).(P_nl(x)+1/x) with P_nl(x) a polynomial

=> the higher the energy is the higher the particle in the well is (the spatial extension of the orbital increases following the coulombian potential V(r)).

Now the big question: why can’t we see singular states? Have I done an error in my calculus?

All comments are welcome.

Seratend.

Rediscovering the H atom and the mistery of bounded states .

 

[seratend]

Now if we develop the other singular states, we recover orbitals R(x) of the following form:
R_nl(x)= exp(-x/2).(P_nl(x)+1/x) with P_nl(x) a polynomial

=> the higher the energy is the higher the particle in the well is (the spatial extension of the orbital increases following the coulombian potential V(r)).

Some minor corrections: the orbitals R(x) of singular states are the following:

R_nl(x)= exp(-x/2).P_nl(x) with P_nl(x,1/x) a polynomial of x and 1/x.

(R_nl(x)=exp(-x/2)x^l.y_nl(x), l is the orbital momentum and 0< n <l+2)

Examples:

(l=0,n=1) R_01(x)=exp(-x/2)/x

(l=1,n=1) R_11(x)=exp(-x/2)[1+2/x+2/x^2]
(l=1,n=2) R_12(x)=exp(-x/2)[1/x + 2/x^2]

(l=2,n=1) R_21(x)=exp(-x/2)[x+2+8/x+16/x^2+16/x^3]
(l=2,n=2) R_22(x)=exp(-x/2)[1+6/x+18/x^2+24/x^3]
(l=2,n=3) R_23(x)=exp(-x/2)[1/x+6/x^2+12/x^3]


Seratend

 

[dextercioby]

Some minor corrections: the orbitals R(x) of singular states are the following:

R_nl(x)= exp(-x/2).P_nl(x) with P_nl(x,1/x) a polynomial of x and 1/x.

(R_nl(x)=exp(-x/2)x^l.y_nl(x), l is the orbital momentum and 0< n <l+2)

Examples:

(l=0,n=1) R_01(x)=exp(-x/2)/x

(l=1,n=1) R_11(x)=exp(-x/2)[1+2/x+2/x^2]
(l=1,n=2) R_12(x)=exp(-x/2)[1/x + 2/x^2]

(l=2,n=1) R_21(x)=exp(-x/2)[x+2+8/x+16/x^2+16/x^3]
(l=2,n=2) R_22(x)=exp(-x/2)[1+6/x+18/x^2+24/x^3]
(l=2,n=3) R_23(x)=exp(-x/2)[1/x+6/x^2+12/x^3]


Seratend

Yap,i've read your posts.Quite interesting,i must say. I don't know what's more to be said.If there is anything.My view of the H atom (at least in the Schroedinger equation approach) has been made clear.Apparently you're looking for problems where there aren't any.I was hoping to have been made clear,at least in the radial equation's solution part.I mean,u have a second order differential equation,wouldn't it be natural to have 2 linear independent solutions,incidentally,one wrong and one good??I set it pretty clear why that irregular solution fails.For phyisical reasons and nothing more.And by the way,i did not do that assumption of "l"goes into "-l-1",but a change of variable in the Kummer equation and that was made in good agreement with a book on PDE-s and special functions.I did only the calculus on my own,i followed the steps from a book of which I'm sure it is not wrong.That's why I'M FULLY CONVINCED THAT MY ANALYSIS FOR THE RADIAL EQUATION OF THE H ATOM HAMILTONIAN IS CORRECT.I don't care if u don't believe me.
As for the first part,with the H spaces,i don't know what u found wrong.I assumed a spectral equation and i wanted to find the discrete spectrum.It was natural to assume that the total Hilbert space of eigenstates could be written as a direct sum between a "good" Hilbert space (spanned by the eigenvectors for the discrete spectrum) and a "bad" one (spanned by the eigenvectors for the continuous spectrum).I didn't assume any isomorphism whatsoever.Because they can't be isomorphic one with the other.

Daniel.

PS.Therefore i find no reason to continue this mathematized debate,as apparently both of us think we're right and the other is wrong,and this could honestly go on forever.
Cheers!

PPS.I wonder if u have the same problems with the Dirac equation version of the H atom. I always found that a bit more tricky.

 

[seratend]

PS.Therefore i find no reason to continue this mathematized debate,as apparently both of us think we're right and the other is wrong,and this could honestly go on forever.
Cheers!

PPS.I wonder if u have the same problems with the Dirac equation version of the H atom. I always found that a bit more tricky.

I am not claiming that you are wrong or the books (except for some typos errors ;). I am just saying that the input set hypotheses you use (or the books use) to reduce the possible set of solutions of the hamiltonian are taken a priori (and not that they are wrong).
In my modest opinion, we should consider a larger set of solutions in order to explain logically why the usual bounded ground state is the lowest energy of the H system (at least relatively to this enlarged set). All my remarks concerning your demonstrations are centred on the fact that they depend on the input set.
Considering this input set, the solutions of the equation are included in this set. If you select, a priori, a "small" set, you only have the usual bounded states while if you enlarge this input set you get more solutions (I have demonstrated at least the existence of an additional set of solutions, I have not demonstrated that have no more solutions if we enlarge this input set).

For example, an ODE can be solved on the usual set of derivable functions (your assumption), but it can also be resolved in a larger space: the distribution space for example or may be sobolev spaces (derivation enlarged).

Now, in your argumentation you try to justify this input set to be the set of continuous functions with a finite L2(|R) norm. I have used the argumentation of the operators with a continuous spectrum (in this case the momentum operator - its eigenvector has not a finite norm) to show that this process of selection a priori in a specific representation of the Hilbert space may lead to forget some eigenvalues and eigenvectors.

Now, It could be interesting, if the solutions I have given in the previous post are ok, to show that the usual ground state of the H atom is really the ground state. If we can show that, we have at least a logical demonstration of the ground state of the H atom in a wide set of possible solutions.

In addition, to end, concerning the "physical reasons" concerning bounded vector or continuous vectors: I assume that you accept the infinite quantum well states where the wave function is spatially localized within this well. Therefore, in this case you accept the wave function discontinuities between the interior and outside of the quantum well => <psi|p_allspace|psi> undefined. Now, what are the differences between this usual quantum well and my solutions of a coulombian quantum well? (both models assume a perfect infinite potential).
If your are honest, I think you should accept there is no major difference except for the geometrical form of the potential: i.e. the irregular solutions of the coulombian well are the solutions inside the well, while the regular ones are the ones outside the well.
Now, surely, the answer to the minimum energy is related to the allowed state transitions between the outside states and the interior states. Surely, the relativistic form should had a clearer view on the E=-oO state.

I am sorry if you think, a priori, that the logical explanation of the possibility/impossibility of these states (or the transition to these states) is un-useful , but that's life and we are free to believe what we want .

Seratend.

P.S. concerning the solutions you’ve posted, do you affirm that the solution
$$u_{l}(r) =[\exp(-\frac{r}{2a})](\frac{r}{a})^{l+1} [C_{2}(\frac{r}{a})^{-2l-1} F(-l-n|-2l|\frac{r}{a})]$$
is defined? (I though that gamma functions in F(a|b|c) are not defined for negative integers; b in this case).

 

[godzilla7]

discussion

I have a feeling that experamentalist have a fear of maths and mathemeticians have a fear of experiments; hence string theorists and particle physisists rows; guys evidence is king I can argue that the sky is purple with maths, it doesn't make me right, experimental data can be flawed by unforseen errors I.e the IMAP data, this is all good wrangling stuff; mathemeticians proove it; experimentalists proove it it's all good, thanks guys, my maths is a little weak I'm afraid as I am about to take a maths course but I have read a lot about physics in preperation for my degree; apologize if I come across as a little naive, can't exactly respond to the mathematics, but it seems that perhaps you should look at experimental data and then come up with theories

Have a good Christmas