The usual place to start is with the simplest form of matter possible: a dilute gas containing a single species of atom. Each atom consists of a positively charged nucleus orbited by electrons. Classical electromagnetic theory tells us that accelerated charge produces electromagnetic radiation, so we might expect that the orbiting electrons would continuously radiate, as they’re constantly being accelerated towards the nucleus. However, due to energy conservation, if the electrons were constantly radiating electromagnetic energy they would lose their own kinetic energy and fall into the nucleus. If this happened to all atoms in the universe, it would be the end of everything. Thank goodness that hasn’t happened!

Quantum mechanics provides us with a different picture of the atom, which we accept because it accurately reflects the behavior we observe, even though it’s counter-intuitive. Quantum theory says that the behavior of the electron is described by a probability wave (wavefunction), which is a solution of the Schrodinger wave equation:

$latex \displaystyle i \hbar \frac{\partial \psi}{\partial t} = \left( \frac{-\hbar^2}{2 m} \nabla^2 + V \right) \psi $

where $latex \hbar = h / 2 \pi$ is Planck’s constant divided by 2 pi, $latex \psi$ is the wavefunction of the electron, and $latex V$ is the potential energy as a function of position and time (in the case of our atom, the electrostatic potential created by the nucleus and other electrons). Unlike an electromagnetic wave, this wavefunction is scalar and complex. The physical interpretation of the wavefunction is that its magnitude squared $latex |\psi|^2$ gives the probability of finding the electron at some point of time and space, which is something we can measure (and therefore must be real). This is the major distinction between quantum mechanics and classical mechanics: quantum mechanics is probabilistic. In classical mechanics, if you have complete knowledge of all of the forces acting on an object and where that object started, you can predict exactly where that object will be found at any later time. In quantum mechanics, even with complete knowledge of the system, you will only be able to calculate relative probabilities of finding an electron at various positions. There has been a lot of thinking and discussion about this issue by very smart people, and I encourage you to read further on it if you’re curious.

Some of the work we did with electromagnetic waves can carry over to Schrodinger waves: we can seek time harmonic solutions, find their velocity, etc. One difference is that the Schrodinger equation is first order in time, unlike the Helmholtz equation which was second order in time; also, the solutions are complex. If the potential energy $latex V$ is a constant independent of time and space (which is not true in general), the harmonic solutions to the Schrodinger equation resemble the phasor solutions to the Helmholtz equation:

$latex \displaystyle \psi = A e^{i(k z – \omega t)} $

which is a wave propagating in the $latex z$ direction, and $latex k$ and $latex \omega$ are the wavenumber and angular frequency, as before. Plugging this back into the Schrodinger equation, we find that

$latex \displaystyle \hbar \omega = \frac{\hbar^2 k^2}{2 m} + V$

The quantity $latex \hbar k$ is associated with the electron’s momentum in quantum mechanics, and $latex \hbar^2 k^2 / 2 m$ is therefore the kinetic energy. The above equation tells us that $latex \hbar \omega$ equals the kinetic energy plus the potential energy; therefore we identify the total energy of the electron $latex E = \hbar \omega$.

Similar to the plane wave in electromagnetics, we can generalize this wave to propagate in any direction:

$latex \displaystyle \psi = A e^{i ( \vec{k} \cdot \vec{r} – \omega t)} $

in which case the wavevector $latex \vec{k} = k \hat{a}_n$ equals the wavenumber times the direction of propagation.