Density of states from $k$ to $E$ Speaking about Quantum mechanics, considering the "particle in a box" condition as an approximation of the electrons condition in a semiconductor, let the material be represented by a volume $V$ with sides $L_x, L_y, L_z$.
In the $k$ space, each allowed state "occupies" a volume $V = \pi^3/(L_x L_y L_z)$. The density of states in the $k$ space is therefore constant and always and everywhere equal to
$$\delta = \frac{L_x L_y L_z}{\pi^3}$$
But in order to obtain the density of space with respect to energy instead of $k$, we must before evaluate the number of states in a shell between the sphere of radius $k$ and the sphere of radius $k + dk$. The volume of this shell (just considering one octant) is
$$\frac{1}{2} \pi k^2 dk$$
and so the number of states there contained is
$$2 \delta \frac{1}{2} \pi k^2 dk = \frac{L_x L_y L_z}{\pi^2} k^2 dk = \rho(k) dk$$
(taking into account the fact that two electrons can occupy the same state with opposite spin).
So, when evaluated against $k$, the density is no more constant: it is $\rho(k) = (L_xL_yL_z k^2) / \pi^2$. My question is: why?
Consider a sphere shell of thickness $k_1 + \Delta k$ and volume $V_1$: it will contain $2 \delta V_1$ states. Stating that $\rho(k)$ is not constant is equivalent to state that a sphere shell of thickness $k_2 + \Delta k$, volume $V_2$, with $k_2 > k_1$ will contain a number of states which is different from $2 \delta V_2$ states. But I can't understand why, because I would still compute the number of states in the second sphere shell as $2 \delta V_2$ and this would mean that the density is constant and not dependent on $k$.
 A: Your confusion arises from the fact that you are confusing scalars and vectors. Scalars, are like numbers, and they have only magnitude. Vectors on the other hand have direction in addition to magnitude.
In your question, you mention the wave vector, which, as its name suggests, is a vector. Typically vectors are written in bold or with an arrow over them; that is, the wave vector would be written as $\mathbf{k}$ or $\vec{k}$. I will write it as $\mathbf{k}$.
Now we may take this wave vector and obtain a scalar by taking its magnitude. I will denote the magnitude by $k$ so that $k=|\mathbf{k}|$.
Now we may define two different densities. The first is a density in the three dimensional space of $\mathbf{k}$'s. To be clear, I will call this density $\rho_3(\mathbf{k})$. $\rho_3(\mathbf{k})$ is defined by the condition that $\rho_3(\mathbf{k})d\mathbf{k}$ give the number of states contained in a small volume $d\mathbf{k}$ centered at $\mathbf{k}$. As you say $\rho_3(\mathbf{k})$ is uniform, or in other words, it does not depend on $\mathbf{k}$.
The second density we may define is a density in the space $k$'s, that is, in the space of wave vector magnitudes. I will denote this function as $\rho_1(k)$. $\rho_1(k)$ is defined by the condition that $\rho_1(k)dk$ give the number of states contained in a small spherical shell having inner radius $k$ and outer radius $k+dk$. $\rho_1(k)$ can be related to $\rho_3(\mathbf{k})$ by the formula $$\rho_1(k)=\frac{d}{dk}\int_0^k \int_0^\pi \int_0^{2\pi} \rho_3(\mathbf{k}) k^2 \sin \theta d \phi d\theta dk.$$ Since $\rho_3(\mathbf{k})$ is a uniform, $\rho_1(k)$ goes like $k^2$ as you say.
Now one thing I should say is that people are sloppy with notation and they write $\rho_1(k)$ ad $\rho(k)$ and they write $\rho_3(\mathbf{k})$ as $\rho(\mathbf{k})$. You are supposed to know that the $\rho$ in $\rho(k)$ refers to $\rho_1$ because $\rho(k)$ has a scalar argument, while the $\rho$ in $\rho(\mathbf{k})$ refers to $\rho_3$ because $\rho(\mathbf{k})$ has a vector argument
