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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 $2dk$?

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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.

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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

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

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