I am reading Chapter 12 of Ashcroft and Mermin and I have a great many questions, but one sticks out in particular.
As background, we note that it can be shown quite generally (by applying Born-von Karman boundary conditions to a convenient volume $V$) that the "volume per $k$-state" in $k$-space is (in 3D) $\mathbf{\Delta k} = V/8\pi^3$. Thus, taking the very large $V$ limit and dividing by $V$ we find that the density of $k$-states (per unit real space volume) is $1/8\pi^3$. Now if I were to multiply this by the real and $k$-space volume measures $d\mathbf{k}d\mathbf{r}$ and then integrate over an entire primitive cell (leaving only the $d\mathbf{r}$ measure behind) then I ought to get the total number of $k$-states available (in a given band) at $\mathbf{r}$ in the crystal in a volume $d\mathbf{r}$. This last point (italicized) is what confuses me. When the original derivation for the allowable k-states proceeded (the standard Bloch theorem proof), we showed that some $\mathbf{k}$ (in a given band) indexed a unique electronic state (neglecting spin) throughout the whole crystal, but with the above I seem to have found that all of these $k$-states exist at each $\mathbf{r}$ in the crystal.
I note further the following: the derivation using the BvK boundary conditions proceeded in some finite volume $V$, before letting $V$ go to infinity while dividing by $V$. We then further assert/assume that all quantities (including $k$-space density) per unit volume are the same in the infinite $V$ and any finite $V$ cases. Thus it follows that we can recover the $k$-space density by multiplying the $k$-space density per unit volume (found in the infinite $V$ limit) by whatever the actual crystal $V$ is (actually finite). But here $d\mathbf{r}$ is not the entire crystal volume! It seems as though we are misusing the derivation done earlier in A&M (Ch. 8).
I suppose I am ultimately confused about the nature of density of states., but perhaps someone can explain the above since I am ultimately trying to understand the attached from page 221 of A&M. Equation 12.7 includes the Fermi function, but in a filled band at $T=0$ this evaluates to 1.
Best Answer
When we are computing sums over $k$, we have to include the volume per $k$-state this as part of our integration measure, i.e., $$ \sum_{\vec{k}}\to\frac{V}{(2\pi)^3}\int d^3k\,. $$ This integral on its own (i.e., when the integrand is equal to 1) yields the total number of states, and therefore $$ \frac{V}{(2\pi)^3}d^3k $$ is equal to the number of states in the volume $d^3k$, and hence $$ \frac{d^3k}{(2\pi)^3} $$ is equal to the number of states in the volume $d^3k$ per unit real-space volume. Thus $$ \frac{d^3rd^3k}{(2\pi)^3} $$ is equal to the total number of states in phase-space volume $d^3rd^3k$, or, roughly speaking, the total number of states "at" the phase-space point $(\vec{r},\vec{k})$. (Multiplying by 2 for the two spin states then agrees with that paragraph from A&M.)
Now, if we integrate over just a single primitive unit cell in the reciprocal lattice, we should get $$ \int_k \frac{d^3rd^3k}{(2\pi)^3} = \frac{d^3r}{(2\pi)^3}\frac{(2\pi)^3}{a^3}=\frac{d^3r}{a^3}\,, $$ since$-$assuming a cubic lattice of lattice constant $a$ for convenience$-$${(2\pi)^3}/{a^3}$ is the volume of the primitive unit cell in $k$-space. Since a single band "lives" in this region, what we should be left with is the total number of states in a single band that are "at" the point $\vec{r}$. Note that this quantity is not the total number of states in a single band but rather a fraction of them, namely, the fraction $\frac{d^3r}{a^3}$ of them.
Now, when talking about states, this last paragraph doesn't make a lot of sense, because states don't live in $r$-space, they live in $k$-space, so we can't really talk about states that are "at" $\vec{r}$. However, assuming we have a filled band, we can interpret $\frac{d^3rd^3k}{(2\pi)^3}$ as the total number of electrons in the phase-space volume $d^3rd^3k$ (since all of these states are filled by electrons!), and hence, loosely speaking, $\frac{d^3r}{a^3}$ is on average the number of electrons occupying a particular band and "near" (i.e., within $d^3r$ around) the point $\vec{r}$ in real-space.
Finally, interpreting the paragraph from A&M, we note again that $$ \frac{d^3rd^3k}{(2\pi)^3} $$ as the total number of states in the phase-space volume $d^3rd^3k$, and hence if a band is filled, $$ \frac{1}{(2\pi)^3} $$ is the phase-space density of electrons. On the other hand, if the band is not filled, then, since some of those states are not filled, the electron number-density is therefore smaller than this value.