Wes Hughes said:
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!
dextercioby said:
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 ; ).
dextercioby said:
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})]<br />
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.
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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:
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(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)
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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 .
_|nlm> is a Hilbert space.
No problems at all,neither with the physics,nor mathematics. 
)),and tell u that in QM there are no black holes,singularities and s***".
, but that's life and we are free to believe what we want 
.