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We know that $$ u=\int_{-\infty}^{+\infty} D(\mathcal{E})\mathcal{E}f(\mathcal{E})d\mathcal{E} $$ where $D(\mathcal{E})$ is called density of levels per unit volume. My textbook (Kittel) says that:

  • $D(\mathcal{E})$ is the density of single-particle states as function of energy
  • $f(\mathcal{E})D(\mathcal{E})$ is the density of filled orbitals

I don't understand the difference. And orbital and single.particle state are the same thing? Each orbital is defined by all the quantum numbers and only one electron can occupy it. So the density of energy state D(E) is the number of suitable orbitals in that range of energy?

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  • $\begingroup$ $D$ is the density of levels, whether or not they are occupied. $f$ is the probability that a level is occupied. Does that help? $\endgroup$ – garyp Jul 10 at 20:16
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orbital and single.particle state are the same thing?

A part for a factor $2$, due to the spin degeneracy, yes they are the same thing.

The difference between $D(\mathcal{E})$ and $f(\mathcal{E})D(\mathcal{E})$ is entirely in the adjective filled.

The density of states is a purely mechanic concept depending only on the spectrum of the eigenvalues. Instead, physical properties depend heavily on how the available states are occupied. Which is, in turn, a function of the temperature of the system.

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  • $\begingroup$ Ok. I totally missed the adjective filled. $\endgroup$ – JBach Jul 10 at 20:20
  • $\begingroup$ but, searching better in the textbook, I found this: "Strictly, D(E) is the density of one-particle states, or density of orbitals". So, at least in the textbook, orbital and single-particle state are the same $\endgroup$ – JBach Jul 10 at 20:22
  • $\begingroup$ What do you think? Do you agree? $\endgroup$ – JBach Jul 10 at 20:47
  • $\begingroup$ @JBach I agree. Orbital = wavefunction solution of Schroedinger equation = one particle state (actually two-particle, taking into account the spin). Of course, the filling of such orbitals/states is a different story. The state may be filled or empty, depending on how many electrons are thee in total and on the precise external conditions. $\endgroup$ – GiorgioP Jul 11 at 11:28

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