HYDROGEN ATOM
dextercioby said:
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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 : )
dextercioby said:
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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):
[tex]P(V)=\int\int\int_{V} C_{nlm}R_{nl}^{2} (r) |Y_{lm}|^{2} r^{2} dr d\Omega[/tex]
If V stretches to infinity,[itex]P(R^{3})=1[/itex].
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.
