Question about equation 2.73 in Ashcroft and Mermin

Question

I don't understand the integral expansion done for equation 2.73 in Solid State Physics by Ashcroft and Mermin.

Specifically the equation deals with the Sommerfeld Expansion and looks like this: $$\int_0^\mu H(\epsilon) d\epsilon = \int_0^{\epsilon_f} H(\epsilon) d\epsilon + (\mu-\epsilon_f)H(\epsilon_f) + ...$$

I am trying to extend this to:

$$\int_0^\mu H(\epsilon) d\epsilon = \int_0^{\epsilon_f} H(\epsilon) d\epsilon + (\mu-\epsilon_f)H(\epsilon_f) + \frac12(\mu-\epsilon_f)^2H'(\epsilon_f) + ...$$

But need to understand the source of the first equation before I can do so. It feels like I'm overthinking this or missing something obvious! Am I just being dumb?!

Answer 1

Define the function $$\boxed{f(x) := \int_0^x H(\epsilon) d\epsilon}$$

Suppose we want to expand this function around the value $\epsilon_f$, i.e. $x = \epsilon_f + \delta x$ with $\delta x$ small. Then of course $$ f(x) = f(\epsilon_f) + \delta x \; f'(\epsilon_f) + \frac{\left(\delta x\right)^2}{2} \; f''(\epsilon_f) + \mathcal O \left( (\delta x)^3 \right)$$

Note that $f'(x) = H(x)$ (which also directly gives $f''(x) = H'(x)$). In case this is not clear, it is just a one-liner using the definition $f'(x) = \lim_{\alpha \to 0} \frac{f(x+\alpha) - f(x)}{\alpha} $.

Plugging this in, we get $$\boxed{f(x) = \int_0^{\epsilon_f} H(\epsilon) d\epsilon + \delta x \; H(\epsilon_f) + \frac{\left(\delta x\right)^2}{2} \; H'(\epsilon_f) + \mathcal O \left( (\delta x)^3 \right) }$$

Simply setting $x = \mu$ (and hence $\delta x = \mu -\epsilon_f$), we're done.

Answer 2

I think that the idea is: $$\int_0^{\mu}{H(\epsilon)d\epsilon}=\int_0^{\epsilon_f}{H(\epsilon)d\epsilon}+\int_{\epsilon_f}^{\mu}{H(\epsilon)d\epsilon}\approx\int_0^{\epsilon_f}{H(\epsilon)d\epsilon}+(\mu-\epsilon_f)H(\epsilon_f)$$ where the last step is motivated by the fact that the interval $\mu-\epsilon_f$ is very narrow and so the function $H$ can be assumed to be constant and equal to its value at $\epsilon_f$ in that region. I imagine that the book is showing how to do a low $T$ approximation, and so the chemical potential is only slightly different from the Fermi energy.