Schrodinger Equation and Energy Quantization

[atyy]

I think all the books pretty much say you get quantization of energy because of the requirement that the wavefunction be normalizable, or in fancy-speak, the wavefunction must belongs to Hilbert space. So this requirement of normalizeablity amounts to something like a boundary condition.

Anyways, if you take a look at the postulates of quantum mechanics, they all say that the wavefunction must be normalizeable. So it's in the postulates. If there were a deeper reason, then wouldn't that deeper reason replace the postulate?

So I guess this would imply that all the eigenvectors of a Hamiltonian are not used in construction of the Hilbert space: the ones that lead to non-normalizeable wavefunctions are actually thrown out. I haven't verified this yet, so does this sound right, that non-normalizeable eigenvectors are thrown in the garbage?

There is an alternative formulation of the postulates - that a physical state is not a vector in Hilbert space, but a direction in Hilbert space. So I don't think quantization has to do with normalizability. In some sense, quantization is a fundamental postulate when we demand that only the eigenvalues of an operator can be obtained in a measurement (though this does not preclude the existence of some measurements whose operators have continuous eigenvalues).

The main place where people throw out bits of Hilbert space is for systems of identical particles where wavefunctions must be symmetric.

 

[RedX]

I was thinking that you take the time-independent Schrodinger eqn, which is the eigenvalue equation for the Hamiltonian, and in principle you can solve the equation for any numerical E you insert into the equation. However, some values of E cause the solutions of the equation to grow to infinity at +- infinity, making them unnormalizeable and hence not part of Hilbert space. These solutions you disregard.

I've forgotten everything about the theory of differential equations, but I suspect the Schrodinger equation is a simple one in that any value of E you choose to insert there is a mathematical solution to the ODE. For multi-dimensional Schrodinger eqn. is where I've forgotten. I vaguely recall terminology like elliptical and hyperbolic PDEs, but I think a solution exists for the Schrodinger eqn. for any E. But at the very least, I'm pretty sure for the 1-dimensional Schrodinger eqn. any eigenvalue is mathematically possible (the equation is solveable for any E), but the multi-dimensional case I'm forgotten the math.

 

[malawi_glenn]

I've forgotten everything about the theory of differential equations, but I suspect the Schrodinger equation is a simple one in that any value of E you choose to insert there is a mathematical solution to the ODE.

But that are quite opposite to the question asked by the OP. You could always find out what potential and mass etc. which correspond to a given Energy in the schrödinger equation (may not be solvable analytically, but you'll do it numerical). But now the question was why a given set of mass and potential energy only gives you a discrete set of energy eigenvalues.

 

[Marty]

I'm curious what you think of these papers. The first is unfortunately not on arXiv, but the second is. I haven't studied them carefully yet myself (and won't be able to anytime soon).

Without getting into the specifics of the papers you referenced:

I believe you should be able to understand what goes on at the atomic level by taking the wavefunctions seriously. In particular I think you can understand the absorption spectrum of the hydrogen atom by looking at the superposition of different wave functions and analyzing them as little oscillating charge distributions (antennas) interacting with a classical e-m field.

Since you asked, that's what I think.

 

[atyy]

I was thinking that you take the time-independent Schrodinger eqn, which is the eigenvalue equation for the Hamiltonian, and in principle you can solve the equation for any numerical E you insert into the equation. However, some values of E cause the solutions of the equation to grow to infinity at +- infinity, making them unnormalizeable and hence not part of Hilbert space. These solutions you disregard.

I see what you mean. I guess the solutions that grow to infinity are not even part of Hilbert space by definition. Hilbert space is defined to be one where you can take a scalar product of two vectors. For functions, that means that the overlap integral of two functions should make sense.