The results for deriving the density of states in different dimensions is as follows:

  • 3D: $g(k)dk = 1/(2\pi)^3 4 \pi k^2 dk$
  • 2D: $g(k)dk = 1/(2\pi)^2 2 \pi k dk$
  • 1D: $g(k)dk = 1/(2\pi) 2 dk$

I get for the 3d one the $4\pi k^2 dk$ is the volume of a sphere between $k$ and $k + dk$. Similarly for 2D we have $2\pi kdk$ for the area of a sphere between $k$ and $k + dk$. However I am unsure why for 1D it is $2dk$ as opposed to $2 \pi dk$. So could someone explain to me why the factor is $2dk$?


The factor of pi comes in because in 2 and 3 dim you are looking at a thin circular or spherical shell in that dimension, and counting states in that shell. In 1-dim there is no real "hyper-sphere" or to be more precise the logical extension to 1-dim is the set of disjoint intervals, {-dk, dk}. Pardon my notation, this represents an interval dk symmetrically placed on each side of k = 0 in k-space. Hence the differential hyper-volume in 1-dim is 2*dk. The factor of 2 because you must count all states with same energy (or magnitude of k). In 2-dim the shell of constant E is 2*pikdk, and so on.


The volume of an $n$-dimensional sphere of radius $k$, also called an "n-ball", is

$$ V_n(k) = \frac{\pi^{n/2} k^n}{\Gamma(n/2+1)} $$

The volume of an infinitesimal spherical shell of thickness $dk$ is

$$ S_n(k) dk = \frac{d V_{n} (k)}{dk} dk = \frac{n \ \pi^{n/2} k^{n-1}}{\Gamma(n/2+1)} dk $$

where $S_n$ is the surface area.

For example, for $n=3$ we have the usual 3D sphere. Its volume is

$$ V_3(k) = \frac{\pi^{3/2} k^3}{\Gamma(3/2+1)} = \frac{\pi \sqrt \pi}{\frac{3 \sqrt \pi}{4}} k^3 = \frac 4 3 \pi k^3 $$

The surface area is

$$ S_3(k) = \frac {d}{dk} \left( \frac 4 3 \pi k^3 \right) = 4 \pi k^2 $$

and the thickness of the infinitesimal shell is

$$ S_3(k) dk = 4 \pi k^2 dk $$

In 1D, the "sphere" of radius $k$ is a segment of length $2k$ (why? think about the general definition of a sphere, or more precisely a ball...). The above equations give you

$$ V_1(k) = 2k\\ S_1(k) = 2\\ S_1(k) dk = 2dk\\ $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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