Ashcroft Mermin Solid State Physics Eq. 2.60ff

Question

I'm trying to follow the steps in Eq. 2.60 of said book.

What I cant seem to figure out is how to change the integration variables from 'k' to 'E', as they state.

The equation is

$$\int \frac{d\textbf{k}}{4\pi^3} F(\epsilon(\textbf{k})) = \int_0^\infty \frac{k^2 dk}{\pi^2} F(\epsilon(k)) = \int_{-\infty}^\infty d\epsilon \, g(\epsilon) F(\epsilon)$$

I can follow the first transformation (why is $\textbf{k}$ suddendly $k$?),

$$\int\frac{1}{4\pi^3} k^2 F(\epsilon(k)) \, dk \int_0^\pi \sin \theta \, d\theta \int_0^{2\pi} d\phi = \int_0^\infty \frac{k^2 dk}{\pi^2} F(\epsilon(k))$$

But what's happening in the second step is unclear to me.

In the book it says, "one often exploits the fact that the integrand depends on $\textbf{k}$ only through the electronic energy $\epsilon = \hbar^2k^2/2m$,...", but I'm unsure how this is used.

Could anybody point this out to me?

Answer 1

Let's start with your first expression sans some numerical constants:

$$ I \equiv \iiint\mathrm{d}\vec{k}\ F(\epsilon(\vec{k})) $$

where all we care about is that the function $F(\epsilon(\vec{k}))$ is rotationally invariant:

$$ F(\epsilon(\vec{k})) = F(\epsilon(k)) $$

We can seperate the angular integrals, which give a factor of $4\pi$, and we find

$$ I = 4\pi \int_0^\infty \mathrm{d}k\ k^2 F(\epsilon(k)) $$

Now we suppose that $\epsilon(k)$ is an invertable function. In fact, it is a simple quadratic, but we only need invertability and that it be sufficiently smooth. This means we can write

$$ k = k(\epsilon) $$

We also assume that $\epsilon(0)=0$ and $\epsilon(\infty)=\infty$, which is the only reasonable thing and also makes sure that the limits of integration stay trivial. Make a change of variables

$$ \mathrm{d}k = \mathrm{d}\epsilon \frac{\mathrm{d}k}{\mathrm{d}\epsilon} $$

to get

$$ I = 4\pi \int_0^\infty \mathrm{d}\epsilon \frac{\mathrm{d}k}{\mathrm{d}\epsilon} k^2(\epsilon) F(\epsilon) $$

Matching to your next expression

$$ I \propto \int_{-\infty}^\infty \mathrm{d}\epsilon\ g(\epsilon) F(\epsilon) $$

(which we require to hold for all $F$), we obtain

$$ g\left(\epsilon\right)\propto\begin{cases} \frac{\mathrm{d}k}{\mathrm{d}\epsilon}k^{2}(\epsilon) & \epsilon\ge0\\ 0 & \epsilon<0 \end{cases} $$

I leave you to work out the constant of proportionality (the 4s and $\pi$s), but you should get a functional dependence $g(\epsilon)\propto \sqrt{\epsilon}$. The only unusual thing is that they extend the range of integration to $\epsilon<0$ for some reason, but you can do this if you set $g(\epsilon)=0$ for negative $\epsilon$, since there are no states there.