# Intuition for the density of states of the free electron gas model

The density of states as a function of energy for a free electron gas (inside some solid-thing where the electrons are modeled due to the free elecetron gas model) is in:

1D: D(E) ~ $\sqrt[-1/2]{E}$

2D: D(E) ~ $\sqrt[0]{E}$

3D: D(E) ~ $\sqrt[1/2]{E}$

The mathematics is okay but how does this makes sense from the physical point of view? How does the dimension of the system dictate that in 3D we have more states with high energy where as in a 1D the system we get fewer and fewer states for increasing the energy?

The mathematics is okay but how does this makes sense from the physical point of view? How does the dimension of the system dictate that in 3D we have more states with high energy where as in a 1D the system we get fewer and fewer states for increasing the energy?

The density of states just comes from counting the states with energy less that $E$ and identifying the integrand of that count as the density.

In three dimensions the count is given by $$\int d^3r \int d^3p\sim V\int^{\sqrt{2mE}} dp p^2 \sim V\int^E dE' (\sqrt{E'})\;,$$ where the $\sim$ means I've ignored some irrelevant constants. The part in the parenthesis on the far RHS is the density of states.

In two dimensions the count is given by $$\int d^2r \int d^2p\sim V\int^{\sqrt{2mE}} dp p\sim \int ^E dE\;,$$ where you may notice that there is only 1 in the integrand.

In d-dimensions the count is given by $$\int d^dr \int d^dp\sim V\int^{\sqrt{2mE}} dp p^{d-1}\sim \int^E dE (E^{(d-2)/2})\;.$$

So, the density of states for the free particle is always proportional to $$E^{(d-2)/2}\;,$$ just due to counting states and the fact that E is a quadratic function of p.

• In the book I'm reading there is only an approximate explanation/calculation - your stuff seems more enlightening to me. Thank you. – Thomas Elliot May 11 '15 at 7:58
• @ThomasElliot, you are very welcome. Good luck. – hft May 11 '15 at 8:32