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In the orthodox interpretation, the eigenvalues are the only measurable values of the corresponding operation, i.e., if the total energy of a particle is measured, the only numbers (as outcomes) are the eigenvalues of total energy $H$, $$H\psi = E \psi,$$ where $\psi$ is the eigenfunction and $E$ the eigenvalue.

As a mathematician I'd like to ask why this eigenvalue problem has only quantized (denumerable) solutions (and no uncountable ones), i.e., $$H\psi_n = E_n \psi_n\quad\mbox{for }n=0,1,2,\ldots.$$ Is it a property of the operators used? Hamiltonian, position operator, momentum operator, ... ?

Qmechanic
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Wuestenfux
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2 Answers2

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Your "orthodox" interpretation and your following statement do not imply each other. The counter example are free states. The measurable energy of a free state form a continuum. Systems that create bound states will have quantized energy levels. This isn't always due to boundary conditions. Consider the Hydrogen atom. Or, more generally, an electron in the presence of a proton (or classical potential function caused by the attractive Coulomb law). For Energy levels less than some critical value there will be "quantization", above that the energy is a continuum. The quantization actually comes from the math (not the physics), conditions on the existence of solutions to the PDE that are elements of the set of square integrable functions on R^3. Of course one could argue that our demand that the function be square integrable is driven by the fact that "The probability to find the particle must be normalizable", thus certain math choices were driven by physical reasons. However, to your original question states in QM are not always "denumerable".

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The energies of any particular Hamiltonian are not necessarily discrete. The simplest example is the free Hamiltonian, $H = p^2/2m$, which has the spectrum $\mathbb{R}^{\ge 0}$. (Eigenstates are hard to define in this case since they are not technically states in the Hilbert space as they are not normalizeable, but a generalization is possible)

Many (most?) Hamiltonians have a spectrum with a discrete part and a continuous part. The discrete part is the energies of the bound states, while the continuous part is the energies of the unbound states. In the hydrogen atom, for example, the states with energy less than zero have the electron bound to the proton, but there is also a continuum of states where the electron is not bound to the proton.

As to why discrete spectra occur, it is for the same reason sound waves can form discrete spectra in certain systems (like a guitar string). The energy of a state corresponds to a frequency of oscillation of the wave function. The linearity of the wavefunction and reflection off boundary conditions mean certain discrete frequencies form resonances, and so you may get a discrete spectrum of energies. On the other hand, in the absence of boundaries and reflection sound waves can take on any form, corresponding to continuous spectra.

For more on the mathematics of the difference between continuous and discrete spectra and the generalization of eigenstates you can start with https://en.wikipedia.org/wiki/Spectral_theory

Luke Pritchett
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