# Electron waves and the quantum string

#### Prerequisites

Since electrons in quantum mechanics are actually also described by waves, the standing wave on a string turns out to be useful as a toy model system that serves as an introduction for understanding how the discrete energy levels of atoms and molecules arise.

## Modeling a quantum string

Consider a "quantum string" (called, in quantum mechanics courses, "the particle in a box").

When electrons are shared through a chain of covalent bonds in a long molecule or in a metallic crystal (see our discussion of Electric Currents), a reasonable description of the elecitron is that in can move freely in one dimension — along the molecule, say — but can't get out. Since the electron inside a molecule has to be described by waves, we can model it just like a wave on a string with the ends tied down  — a standing wave.

If we consider an electron wave fitting along a molecule of length, $L$, an even number of 1/2 wavelengths have to fit in to give an allowed normal mode as show in the figure below. For the n-th mode the wavelength will have to satisfy

$n \frac{λ_n}{2} = L$    which implies    $λ_n = \frac{2L}{n}$.

The first 4 modes will look like the figure shown at the right — just like the normal modes of an oscillating string tied down at both ends.

If we assume that the PE along the molecule is approximately constant, only the kinetic energy changes for each excitation, so we can take the energy of each mode as just the KE of the electron.

## Connecting the allowed wavelengths to allowed energies

All we need to know about the quantum physics of an electron is that its momentum is what determines its wavelength; and it does it in the same way that the momentum and wavelength are related for a photon:

$$p = h/λ$$

where $h$ is Planck's constant, the parameter that establishes the quantum scale. Since the momentum of an electron is given my $p = mv$, its KE is just

$$E = ½ mv^2 = \frac{(mv)^2}{2m} = \frac{p^2}{2m}$$

So the energy states of the electrons along the molecule will be

$$E = \frac{p^2}{2m} = \frac{1}{2m} \bigg(\frac{h}{\lambda}\bigg)^2 = \frac{h^2}{2m\lambda^2}$$

where $λ$ only takes in the allowed values — the ones that fit in perfectly.

This gives the energy states for the quantum stretched string. Note that it is different from the states of the quantum harmonic oscillator — the energy levels are not at the same separations but get farther apart as you go up.

$$E_n = \frac{h^2}{2m\lambda^2_n} = \frac{h^2}{2m(2L/n)^2} = \bigg(\frac{h^2}{8mL^2}\bigg) n^2.$$

(We have labeled our ground state $n = 1$ so it corresponds to the number of 1/2 wavelengths that fit in. Sometimes the lowest level is labeled $n = 0$.)

## Doing it with numbers

We can calculate our coefficient that goes in front of the $(n/L)^2$. Typically we would do it in Joules, but this involves lots of factors with large powers of 10 and is rather unnatural when working on the atomic scale. A more reasonable way to do it when working with atoms, molecules, and photons, is to introduce some factors of $c$, the speed of light.

The combination, $hc$ has the value of about $hc \approx 1234 \mathrm{eV-nm}$. (It's real value is 1225 but "1234" is a better than 1% mnemonic good for estimations.) So we might want to put a $c^2$ in in the numerator of our expression. We can do it if we also put the same factor in the denominator as well. This gives us the combination $mc^2$. Einstein tells us that this is the energy corresponding to the electron's mass, a useful number to know.

For the electron:  $mc^2 = 511\;\mathrm{keV} \approx 0.5 \times 10^6 \mathrm{eV}$,

or about 1/2 million electron Volts. Putting these numbers together, we get that the energy of the n-th level is given by

$$E_n = \frac{(hc)^2}{2(mc^2)\lambda^2_n} = \frac{(1234\;\mathrm{eV-nm})^2}{2(511\;\mathrm{keV}) \lambda^2_n} \approx \frac{1.5\;\mathrm{eV-nm^2}}{\lambda^2_n} = (0.75\;\mathrm{eV-nm^2})\bigg(\frac{n}{L}\bigg)^2$$

This model gives a reasonable estimate (when combined with some knowledge from chemistry) of the colors of various common chemicals — chromophores.

Joe Redish

4/20/13

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