Intuitive explanation for the space-dimension dependence of the density of states of a free electron gas If the Schrödinger Equation is solved in different dimensions for an independent electron in an infinitely high potential, different relations are obtained regarding the DOS. These are:
0D: $D(E) \propto \delta E$ (Quantum Point)
1D: $D(E) \propto \frac{1}{\sqrt{E}}$ (Quantum Wire)
2D: $D(E) \propto const$ (Quantum layer/square?)
3D: $D(E) \propto \sqrt{E}$ (Quantum bulk/cube?)
The maths are clear, but I can't get my head around what this "means". Why do energy states come closer together with increasing energy in 3D? Why are the same number of energy states found in every segment dE in 2D? Why are more energy states found in the same segment dE at lower energies than at higher ones at 1D? And weirdest of all, why are their an infinite amount of densities at discrete energy values and every where else there are none in 0D?
This question tried to asked something similar (only 2D), but the answer was again mathematical. Can we imagine why this is so intuitively? Or are they conclusions from a theoretical model that helps us explain what we measure in the lab? 
 A: I think the relationships result from an interplay between the dependence of energy on quantum numbers and the volume element in the space of quantum numbers.
For an $m$ dimensional particle in a box,we can write an eigenvalue for the energy as
$$ E_{n} = k \sum^{i=m}_{i=1}{{n_i}^2 = E n^2}$$ where $k$ is some physical constant and and $n_i$ are the quantum numbers required for the description of the wavefunction.
Now a surface of constant energy corresponds to a $(m-1)$-sphere($S^{m-1}$) in the $m$ dimensional space of quantum numbers $n_i$.
To determine the density of space what we generally do is determinate the number of possible quantum states(possible values of quantum numbers) from energy $E$ to $E+dE$.
This corresponds to the volume element for a "shell" in $m$ dimensions of radius $n = \sqrt{\sum^{i=m}_{i=1}{{n_i}^2}}$  which is given by
$$dV_m = k' n^{m-1} dn$$ where $k'$ is some constant which involves $m$ and $\pi$ and the gamma function.
Using $E = k n^2 \implies n = \sqrt{\frac{E}{k}}$ and $ dn = \frac{dE}{2\sqrt{Ek}}$ in the volume element 
We have the number of states between $E$ and $E+dE$ as dN:
$$ dN = k'' E^{\frac{m-2}{2}} dE $$.This formula explains your four observations mathematically.

Now onto the physics part of the formula.Observe that the energy increases as a consequence of the Schrondiger equation only quadratically as a function of n.
On the other hand,the volume in the space of quantum numbers increases as a function of $n^m$.
So physically we can see that for higher dimensions we can pack more states below a particular energy than we can do in lower dimensions.The system we can say has more "degrees of freedom" to have a particular value of energy and hence the corresponding density is higher for a higher dimension. 
Also in a fixed dimension for higher values of energy/$n$,there is again a tug-of-war in allowed values,$n \propto \sqrt{E}$ and "volume" allowed $\propto E^{\frac{m-1}{2}}$.For $m < 2$ Schrodinger wins and number of states decreases 
with increasing E and for $m>2$ it is the space that wins and we have states for larger $E$.
The exact balance is achieved in two dimensions where both the "volume" of $2$-balls and Energy as a function of $n$ go quadratically.

References:
N-Sphere
