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Quantum string (2013)

Page history last edited by Mark Eichenlaub 8 years, 2 months ago

Working Content > Oscillations and waves > Harmonic oscillation > Mass on a spring

 

Prerequisites:

 

Another simple quantum system that serves as a useful simple model for understanding the energy levels of atoms and molecules is the "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 elecittron 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. For the n-th mode the wavelength will have to satisfy

 

n n/2) = L    which implies    λn = 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.

 

All we need to know about the quantum physics of an electron is that its momentum is related to its wavelength in the same way that a photon is

 

p = h/λ

 

and, since momentum is p = mv, that its KE is just

 

 

 

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

 

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 distance but get farther apart as you go up.

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

 

We can calculate our coefficient the goes in front of the (n/L)2. Typically we would do it in Joules, but this involves lots of large powers of 10 terms and is rather unnatural. 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.

 

We know the combination, hc ~ 1234 eV-nm. So we might want to put a c2 in in the numerator of our expression. We could do it if we also put the same factor in the denominator as well. This gives us the combination mc2. Now Einstein tells us that this is the energy corresponding to the electron's mass. This is a useful number to know:

 

For the electron:  mc2 = 511 keV ~ 0.5 x 106 eV,

 

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

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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