sophiecentaur said:
There is no confusion. I should be grateful if you would point out why, for simple cases of just a few charges interacting, you get integer values of quantum numbers and, hence discrete 'levels'. And yet, increasing the number involved, changes that 'in principle'. Of course, you can't do the same sums for large quantities and naturally go to statistics and calculus to get an answer.
Tell me where the 'gear change' happens. Is it for 10 atoms, 1000, 100,000?
We are surely just looking at two ends of a range of situations. It would be daft to consider the actual value of energy gaps when a band model is more suitable just as it would be daft to assume a continuum for just a few atoms. we are not dealing with reality. We are applying the most convenient model to describe things. How is it more than that?
Now I see that you had made a conceptual mistake. You claim that even a macroscopic body has a discrete spectrum, but the levels are very finely spaced.
While technically this is true, it is as useless as the idealization of an infinite crystal, which gives continuous bands
exactly.
The problem comes when you try to define your system! Namely, you need to make the system isolated, so that it does not exchange particles nor energy with the environment, to really calculate energy eigenvalues. But, every system is essentially open. Because of the fine spacing between the levels, you can never claim that you isolated your system so well to be able to distinguish individual discrete eigenvalues.
Let us estimate the difference between subsequent energy levels. For this, assume a crysta lsystem with a cubic unit cell with side
a. The total number of atoms in the system is
N, so that N = (L/a)^3. The discrete values for the wave vector are:
<br />
k_{i} = \frac{2 \pi \, n_i}{L}, \ (i = x, y, z)<br />
Assume the free electron model so that the energy eigenvalues are:
<br />
E = \frac{\hbar^2 \, k^2}{2 m}<br />
Then, the uncertainty in energy due to a neighboring level is:
<br />
\Delta E = \frac{\hbar^2 \, \vert k_i \vert}{2 m} \, \Delta k_i<br />
The order of magnitude for the wave vector corresponds to the Fermi wavevector, found from the total electron density
n:
<br />
\frac{N_e}{a^3} = 2 \frac{1}{(2\pi)^3} \, \frac{4 \pi k^{3}_{F}}{3} = \frac{k^{3}_{F}}{3 \pi^2} \Rightarrow k_F = \frac{1}{a} \, \left( 3 \pi^2 \, N_e \right)^{\frac{1}{3}}<br />
where N_e is the number of electrons per atom.
The uncertainty in the wave vector is given by:
<br />
\Delta k_i = \frac{2 \pi}{L} = \frac{2 \pi}{a} \, N^{-\frac{1}{3}}<br />
Then, we have:
<br />
\Delta E = \frac{\hbar^2}{2 m} \, \frac{1}{a} \, \left( 3 \pi^2 \, N_e \right)^{\frac{1}{3}} \, \frac{2 \pi}{a} \, N^{-\frac{1}{3}}<br />
<br />
\Delta E = \frac{1}{4} \, (\frac{3}{\pi})^{1/3} \, \frac{h^2}{m \, a^2} \, \left( \frac{N_e}{N} \right)^{\frac{1}{3}}<br />
For typical crystals a \sim 5 \stackrel{o}{A}. Take an Avogadro number of atoms, and one electron per atom (N_e = 1). We have:
<br />
\Delta E = 0.246 \times \frac{(6.626 \times 10^{-34})^{2}}{9.11 \times 10^{-31} \times (5 \times 10^{-10})^2} \times (6.022 \times 10^{23})^{\frac{-1}{3}} \, \mathrm{J} \times \frac{1 \, \mathrm{eV}}{1.602 \times 10^{-19} \, \mathrm{J}} = 2.9 \times 10^{-5} \, \mathrm{eV}<br />
