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 $k$.

  • $\begingroup$ I don't understand what you are asking. You've shown very nicely why $\rho(k)$ depends on $k$, i.e. because the volume of the spherical shell from $k$ to $k+dk$ depends on $k$. Your last paragraph seems to just restate this. Can you clarify what exactly you are asking? $\endgroup$ Jun 21, 2016 at 16:55
  • $\begingroup$ @JohnRennie Do you really think I showed why $\rho(k)$ depends on $k$? It was exactly my question. I edited the last two paragraphs trying to better reformulate it. $\endgroup$
    – BowPark
    Jun 21, 2016 at 17:05
  • $\begingroup$ The density per unit volume (in $k$-space) is constant, and the volume (in $k$-space) of the shell $V_2$ is greater than volume of the shell $V_1$. Therefore the number of states inside the shell $2$ is greater than the number of states inside the shell $1$. $\endgroup$ Jun 21, 2016 at 17:19
  • $\begingroup$ @JohnRennie Yes, exactly. But the number of states inside the shell 2 is greater than that in shell 1, not the density. This is the fact which is confusing me. How is the density $\rho(k)$ related to the number of states inside a shell? $\endgroup$
    – BowPark
    Jun 21, 2016 at 17:27
  • 1
    $\begingroup$ Ah I think I see what you mean. The $k$ in the equation is the modulus of the vector $\mathbf k$ i.e. $|\mathbf{k}|$. So there are many different vectors $\mathbf k$ that have the same modulus. The density is the density as a function of $|\mathbf{k}|$. $\endgroup$ Jun 21, 2016 at 18:30

1 Answer 1


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

  • $\begingroup$ Thanks so much for your answer. I can follow what you are saying, but I can't understand the expression you gave for $\rho_1(k)$. You used spherical coordinates, as the spherical geometry requires. But why you used $\rho_3(\mathbf{k})$ and - more important - why you used the derivative? If you find it useful, maybe an example in 1D can help. $\endgroup$
    – BowPark
    Jun 22, 2016 at 8:50
  • $\begingroup$ The integral without the derivative is the number of states that are available if you have an energy of $c \hbar k$, expressed in spherical coordinates. You can interpret this as a function that gives you the number of states as a function of the magnitude $k$. Taking the derivative gives you the change of the number of states with the magnitude of $k$, which is the desired density of states. The expression is also a bit sloppy since the integration variable should be called $k'$, and shoudn't be the same symbol as the integration limit. You must take the derivative with regard to the limit. $\endgroup$
    – Hans Wurst
    Aug 20, 2020 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.