I'm trying to find something called the density of states and the model that I am using specifies

$$E = \frac{h^2}{2 m} k^2$$

where $k = |\bf{k}|$.

The quantity I am trying to calculate is

$$D(E) = \int{(\nabla_k E)^{-1} \mathrm{d}E}$$.

I think how to simplify this is to substitute

$$\mathrm{d}E = \frac{h^2}{m} \mathrm{d}k$$


$$\nabla_k E = \nabla_k [\frac{h^2}{2 m} (k_x^2 + k_y^2 + k_z^2)].$$


$$\nabla_k E = \frac{h^2}{m}(k_x, k_y, k_z).$$

However this leaves me with

$$D(E) = \int{\frac{1}{(k_x, k_y, k_z)} \mathrm{d}k},$$

which I am sure how to solve. Any idea how to solve this with a vector in the denominator?


I realized that $D(E) = \int{(\nabla_k E)^{-1} \mathrm{d}E}$ is actually

$$D(E) = \int{(\nabla_k E)^{-1} \mathrm{d}\bf{S}}.$$

  • $\begingroup$ Shouldn't $ \ E \ $ be a scalar quantity, since $ \ k^2 = \overrightarrow{k} \cdot \overrightarrow{k} \ $ ? Also, the divergence should also be a scalar, so you aren't going to have a vector in the denominator. $\endgroup$ Jun 6, 2013 at 20:57
  • $\begingroup$ $E$ is a scalar, but then I'm taking the gradient of $E$ so end up with a vector. @RecklessReckoner $\endgroup$ Jun 6, 2013 at 21:04
  • $\begingroup$ Who gave you a formula with an inverse of a vector in it? $\endgroup$
    – Muphrid
    Jun 6, 2013 at 21:05
  • $\begingroup$ Sorry, for some reason, I locked on "divergence" and read the operation that way. But the reciprocal of a gradient vector can't be the intent there, because it isn't physically meaningful. Hunting down some notes on density of states, it looks like you're supposed to have something more like $ \ \frac{dk}{dE} \ $ in the integral. $\endgroup$ Jun 6, 2013 at 21:09
  • $\begingroup$ The formula with the inverse is in "Band-Structure Effects in the Field-Induced Tunneling of Electrons from Metals" Gadzuk 1969 @Murphrid $\endgroup$ Jun 6, 2013 at 22:14

1 Answer 1


This is truly warped notation!

Let me explain this more straightforwardly:

  • The wavevectors $\mathbf k \in \mathbb R^3$ are assumed to be spaced uniformly, like $(0,0,0),(0,0,\Delta),(0,\Delta,0),(\Delta,0,0),\cdots$ for instance. (This is usually justified by a particle in a box argument.)
  • We want to calculate sums of functions $f(\mathbf k)$ over all states (with some uninteresting overall scaling by a partition function). We approximate the sum by an integral because it's easier. The sum should be approximated via $$\sum_{\text{integer vectors }\mathbf n} f(\Delta \mathbf n) \propto \int \mathrm d^3 \mathbf n \quad f(\Delta \mathbf n)$$for small $\Delta$.
  • Then changing variables we get $$\propto\int \mathrm d^3 \mathbf k \quad f(\mathbf k)$$ Assuming the functions we calculate depend only on $k=|\mathbf k|$, notice that spherical coordinates give $$I = \int \mathrm d^3 \mathbf k \; f(k) = 4\pi \int_0^\infty \mathrm d k \; k^2 f(k)$$ Now we have a one-dimensional integral and everything is nice.
  • We can change variables to $E(k)=\frac{\hbar^2}{2m}k^2$ in the usual way: $$k=\sqrt{2mE/\hbar^2} \implies \mathrm dk=\sqrt{m/2\hbar^2E}\, \mathrm dE \implies I = 4\pi \int_0^\infty \mathrm d E \; \sqrt{\frac{m}{2\hbar^2E}} \frac{2mE}{\hbar^2} f(k(E))$$ or simplifying $$I = \int_0^\infty \mathrm d E \; \hat f(E) \times \sqrt E\times 2\pi \left(\frac{2m}{\hbar^2}\right)^{3/2}$$
  • The thing you are supposed to work out is what you get from setting $$\hat f(E)=\cases{1 & $E<E_0$ \\ 0 & $E\ge E_0$}$$ i.e. integrate from $E=0$ up to $E_0$.
  • What the strange expression you had meant was "First, switch to $k=|\mathbf k|$, then use $\mathrm d k = k'(E) \mathrm d E = 1/E'(k) \mathrm d E$."
  • $\begingroup$ Nice answer. I was wondering why there is an inverse there earlier... +1 $\endgroup$
    – Shuhao Cao
    Jun 7, 2013 at 2:20

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