Let us look at a simple model of hydrogen atom: we neglect the spin of the proton and the electron, and relativistic effects. What remains is a point charge in an inverse-square force field-- the classical Kepler problem.
With these simplifications, the states of the hydrogen atom can be specified by giving three integer labels , with:
I should make one refinement to this description of the ``old quantum'' viewpoint. The quantum conditions apply only to so-called stationary orbits. Bohr offered no details for what happened during a transition from one stationary orbit to another (a ``quantum jump''), nor could he explain why these orbits were stationary. Initially, physicists probably regarded these questions as topics for future research.
In the post-1925 reformulation, we have to solve a particular instance of Schrödinger's equation, subject to certain boundary conditions. The space of all solutions is a Hilbert space, with a basis . The vector is simultaneously an eigenvector of three operators:
, the energy
, the magnitude of the orbital angular momentum
, the -component of the orbital angular momentum
So we can say that if the hydrogen atom is in state , then it has a definite energy, and its angular momentum has a definite magnitude and -component. In fact, the eigenvalues for are:
The modern equivalent of the old notion of ``stationary orbit'' is ``eigenstate of the energy operator''. Such eigenstates do not change with time, and they possess a definite value for the energy. (These facts are closely related.) Schrödinger's equation, in fact, amounts to , where is the energy operator.
Figure 1 gives a pictorial representation of the basis in a form known as a term scheme. The horizontal lines stand for basis vectors (or equivalently, stationary quantum states); height gives energy, and transitions are indicated by slanted lines. (Only three transitions are pictured, to avoid clutter.)
From this simple diagram, many treasures flow. The next few paragraphs give a taste.
© 2001 Michael Weiss