Schrodinger Equation and Energy Quantization

[malawi_glenn]

the energy of a state:

\Psi _E = a\psi _{E1} + b\psi _{E2} + d\psi _{E4} <br /> <br /> Is EITHER E1, E2 or E4. The energy of the state E is not a*E1+b*E2 + d*E4 or similar. The energy of state E is, as jtbell said, not determined until measurment.<br /> <br /> Now if you get only peaks in the spectroscopy of a gas, measuring thousands of atoms, what does that tell you about the nature of energy in atoms? Well, you'll deduce that the 'energy of an atom' ( = energy of an electron bound to the atomic nucleus) is either E1, E2, .. etc. Spectroscopy is the sum of the energies in the gas, which is the sum of the atomic energies.<br /> <br /> Since you'll don't get any intermediate energy values, the only way your proposal might save you is that photons can loose fractions of its energy equal the eigeneneries of the states in the atom.. which is very unlikley and ad hoc.<br /> <br /> Prepering an experiment with a gas consisting of only one hydrogen atom might be hard to obtain, but I can't see any reason why it is impossible.

 

[Marty]

Perhaps something along these lines?

For the purpose of thinking that the energy levels are discrete, it is not necessary to normalize the wavefunction. So you can excite the modes of the hydrogen atom to any desired value. The unnormalized wavefunction still predicts that anytime you make a measurement of the energy (of a single hydrogen atom), you will get one of several discrete energy levels, and you will never get any values in between.

One way of defining intermediate and indeterminate to be experimentally different might be: if the energy is intermediate, our measurements of the energy will be a Gaussian distribution about one value of energy; if the energy is indeterminate and discrete, the measurements will be Gaussian distributions about multiple discrete energies.

I still don't know what kind of measurement you have in mind. I don't think anyone has proposed a way of measuring the energy of a single hydrogen atom.

 

[atyy]

I still don't know what kind of measurement you have in mind. I don't think anyone has proposed a way of measuring the energy of a single hydrogen atom.

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).

Observing the progressive decoherence of the ''meter'' in a quantum measurement
Brune M, Hagley E, Dreyer J, Maitre X, Maali A, Wunderlich C, Raimond JM, Haroche S
Physical Review Letters 77: 4887-4890, 1996
A mesoscopic superposition of quantum states involving radiation fields with classically distinct phases was created and its progressive decoherence observed. The experiment involved Rydberg atoms interacting one at a time with a few photon coherent field trapped in a high Q microwave cavity. The mesoscopic superposition was the equivalent of an ''atom + measuring apparatus'' system in which the ''meter'' was pointing simultaneously towards two different directions - a ''Schrodinger cat.'' The decoherence phenomenon transforming this superposition into a statistical mixture was observed while it unfolded, providing a direct insight into a process at the heart of quantum measurement.

Trapping and coherent manipulation of a Rydberg atom on a microfabricated device: a proposal
John Mozley, Philippe Hyafil, Gilles Nogues, Michel Brune, Jean-Michel Raimond, Serge Haroche
http://arxiv.org/abs/quant-ph/0506101
We propose to apply atom-chip techniques to the trapping of a single atom in a circular Rydberg state. The small size of microfabricated structures will allow for trap geometries with microwave cut-off frequencies high enough to inhibit the spontaneous emission of the Rydberg atom, paving the way to complete control of both external and internal degrees of freedom over very long times. Trapping is achieved using carefully designed electric fields, created by a simple pattern of electrodes. We show that it is possible to excite, and then trap, one and only one Rydberg atom from a cloud of ground state atoms confined on a magnetic atom chip, itself integrated with the Rydberg trap. Distinct internal states of the atom are simultaneously trapped, providing us with a two-level system extremely attractive for atom-surface and atom-atom interaction studies. We describe a method for reducing by three orders of magnitude dephasing due to Stark shifts, induced by the trapping field, of the internal transition frequency. This allows for, in combination with spin-echo techniques, maintenance of an internal coherence over times in the second range. This method operates via a controlled light shift rendering the two internal states' Stark shifts almost identical. We thoroughly identify and account for sources of imperfection in order to verify at each step the realism of our proposal.

 

[atyy]

Marty, incidentally, and perhaps ironically for me, I believe Haroche and Raimond state that the wavefunction of a single atom has no meaning in their text "Exploring the Quantum: Atoms, Cavities, and Photons" (OUP, 2006). I still think the old fashioned way that a single hydrogen atom has a wavefunction, a view defended (dogmatically) by Gottfried and Tung in "Quantum Mechanics: Fundamentals" (Springer, 2003).

 

[RedX]

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?

 

[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.