Ashcroft Mermin Eq. 17.47ff

Question

In "Solid State Physics" by said authors, Eq. 17.46 is $$ \rho^{ind}(\textbf{r}) = - e[n_0(\mu + e\phi(\textbf{r})) - n_0(\mu)]$$ and then the authors write

In the present case we assume that $\phi$ is small enough for Eq. 17.46 to be expanded to give in leading order

$$\rho^{ind}(\textbf{r}) = -e^2 \frac{\partial n_0}{\partial \mu} \phi(\textbf{r}) \hspace{1cm} \mbox{[Eq. 17.47]}$$

I wonder how to formally exact write this expansion. To understand this, I would like to know

1. What is the independent variable in Eq. 17.46? Is it $\phi$, $\mu$ or $(e\phi + \mu)$?

2. Does it make sense to write it as $$\rho^{ind}(\textbf{r}) = -e \left[ n_0(\mu) + \left . e\phi(\textbf{r})\frac{\partial n_0(e\phi+\mu)}{\partial \mu}\right|_{e\phi=0} - n_0(\mu) \right]$$ But then, if I set $e\phi = 0$, would the middle term not collapse to zero?

Answer 1

Your quotation of the equation is actually wrong. The correct version is given by

$\rho^{ind}(r)=-e[n_0(\mu+e\phi(r))-n_0(\mu)].$

Now you have to expand this expression around $\phi=0$

$\rho^{ind}(r)=\rho^{ind}(r)\mid_{\phi=0}+\frac{\partial}{\partial \phi}\rho^{ind}\mid_{\phi=0}\phi+\ldots$

Now we have to evaluate the zero and first order terms:

$\rho^{ind}(r)\mid_{\phi=0}=e[n_0(\mu)-n_0(\mu)]=0,$

$\frac{\partial}{\partial \phi}\rho^{ind}\mid_{\phi=0}\phi(r)=-e\frac{\partial n_0}{\partial(\mu+e\phi(r))}\frac{\partial}{\partial\phi}(\mu+e\phi(r))\mid_{\phi=0}\phi=-e^2\frac{\partial n_0}{\partial\mu}\phi(r).$

As you can see, the last line yields the desired expression by application of the chain rule.

Answer 2

Let's answer both questions at once. You're interested in the behaviour of the function $n_0(\mu + e\phi)$ as you vary $\phi$, especially in the limit $e \phi \rightarrow 0.$ This suggests that you expand $n_0(\mu + e\phi)$ around $e\phi = 0$:

$$n_0(\mu + e\phi) = n_0(\mu) + e\phi \, \frac{\partial n_0(\mu)}{\partial \mu} + \mathcal{O}\left[(e\phi)^2\right].$$ (This is precisely your second equation, which is perfectly fine.)

Indeed if you brutally set $e\phi \rightarrow 0$, you'll end up with zero; that's why you keep the leading term in $e\phi.$

If it's still confusing, go back to the usual Taylor expansion of $f(x+a)$ at small $a$: $f(x+a) \simeq f(x) + a f'(x) + \ldots,$ so $$f(x+a) - f(x) \simeq a f'(x) + \mathcal{O}(a^2).$$