I'm confused on the difference in results I'm seeing for the density of states for a free electron (for example, a conduction electron in a metal). For one textbook (Kittel), I'm seeing that the density of states depends on the volume of the system, whereas in another, the density of states does not depend on the volume. I think this may be due to a differing definition of density of states but I can't quite figure this out. My question is two-fold:

  1. Where $N_{\text{tot}}$ is defined as the total number of states between $E$ and $E+dE$, is the density of states defined to be $D(E) = \dfrac{dN_{\text{tot}}}{dE}$ or $D(E) = \dfrac{1}{V}\dfrac{dN_{\text{tot}}}{dE}$ where $V$ is the volume of the system? It seems like the derivations below present inconsistent definitions.
  2. Are these two results consistent, but define the density of states differently, or are they inconsistent?

Below are derivations in two textbooks showing the derivation for the density states of a free electron confined to some volume V.

From Kittel - Introduction to Solid State Physics ![enter image description here enter image description here

From Chenming- Modern Semiconductor Devices enter image description here


1 Answer 1


I think the difference here is that Kittel gives the number of states per unit energy, while Chenming is giving the number of states per unit energy per unit volume. Take Kittel's expression and divide through by the volume to get the same units as Chenming and one gets the following: $$\frac{\frac{dN}{d\epsilon}}{V}=\frac{(2m)^{1.5}\epsilon^{0.5}}{2\pi^2\hbar^3}=\frac{(8\pi^3)(2m)^{1.5}\epsilon^{0.5}}{2\pi^2 h^3}=\frac{4\pi(2m)^{1.5}\epsilon^{0.5}}{h^3} = \frac{8\pi m(2m\epsilon)^{0.5}}{h^3}$$ which is the same as Chenming's expression.

  • $\begingroup$ Hmm yeah this seems right. However, I'm not seeing how you go from the second to third expression here, did you forget a term? What is the unit volume defined as? Kittel defines V as $V = L^3$, the volume of the system. $\endgroup$ Commented Jul 25, 2020 at 18:02
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    $\begingroup$ OK no problem. $\hbar$ is h divided by 2$\pi$. So that's where the factor of 8 $\pi^3$ comes from in the numerator. (Note that I don't have $\hbar$ in the denominator anymore in the third expression, but just h.) Yes, V=$L^3$ so when you divide Kittel's expression by V you get an expression "per unit volume". $\endgroup$
    – CGS
    Commented Jul 25, 2020 at 18:22
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    $\begingroup$ Ah thank you, totally missed the difference of $h$ and $\hbar$ in the two expressions =) $\endgroup$ Commented Jul 25, 2020 at 18:31

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