How to Use the Method of Iteration to Find μ in Fermi Distribution?

[SadScholar]

Homework Statement

Let $D(\epsilon)$ be an arbitrary but well-behaved (finite at $\epsilon_F$) density of states, and $f(\epsilon,\mu,T)$ be the Fermi distribution. Assume that $\mu\approx\epsilon_F$ at all temperatures of interest. $f(\epsilon,\mu,T=0)-f(\epsilon,\mu,T)$ is an odd function of $\epsilon-\mu$ and it is different from zero only about when $|\epsilon-\mu|\lesssim k_b T$. Use the method of iteration, the Taylor expansion $D(\epsilon)\approx D(\mu)+D'(\mu)(\epsilon-\mu)$ and the equation
$$
\int_{\epsilon_F}^\mu D(\epsilon)\, \mathrm{d}\epsilon = \int_0^\infty D(\epsilon)[f(\epsilon,\mu,T=0)-f(\epsilon,\mu,T)]\, \mathrm{d}\epsilon
$$
to find $\mu$ to the leading order in terms of $D(\epsilon_F)$, $D'(\epsilon_F)$, $k_B$, $T$, and numerical constants.

All relevant equations to my knowledge are included above.

I'm really not very familiar with the method of iteration. I'm just looking for some hints on how to get started on this. I know that $\epsilon_f-\mu$ is going to be a small parameter. I'm not sure if I should use the linear expansion of $D(\epsilon)$ on both sides of the equation or just on one side. I have the feeling that I need to get $\epsilon_F-\mu$ as one side of the equation and do some kind of perturbation. I would love to hear any hints and tips or resources on the method of iteration. Where do I begin?

 

[mark726]

Thank you for your post. The method of iteration is a useful tool for solving equations that cannot be solved analytically. It involves making an initial guess for the solution and then refining it through a series of iterations until the desired accuracy is achieved. In this case, we are trying to find the value of $\mu$ that satisfies the given equation.

To get started, we can use the Taylor expansion of $D(\epsilon)$ to rewrite the integral on the left-hand side of the equation as follows:
$$
\int_{\epsilon_F}^\mu D(\epsilon) \, \mathrm{d}\epsilon = \int_{\epsilon_F}^\mu \left(D(\mu)+D'(\mu)(\epsilon-\mu)+O((\epsilon-\mu)^2)\right) \, \mathrm{d}\epsilon
$$
We can then use the substitution $x = \epsilon - \mu$ to rewrite the integral in terms of $x$, giving us:
$$
\int_0^{\mu-\epsilon_F} D(\mu+x) \, \mathrm{d}x = \int_0^{\mu-\epsilon_F} \left(D(\mu)+D'(\mu)x+O(x^2)\right) \, \mathrm{d}x
$$
Next, we can use the fact that $f(\epsilon,\mu,T=0)-f(\epsilon,\mu,T)$ is an odd function of $\epsilon-\mu$ to rewrite the integral on the right-hand side as:
$$
\int_0^\infty D(\mu+x)\left(f(\mu+x,\mu,T=0)-f(\mu+x,\mu,T)\right) \, \mathrm{d}x = \int_{-\infty}^\infty D(\mu+x)\left(f(\mu+x,\mu,T=0)-f(\mu+x,\mu,T)\right) \, \mathrm{d}x
$$
We can then use the definition of the Fermi distribution to rewrite the integrand as:
$$
D(\mu+x)\left(\frac{1}{1+e^{\frac{\mu+x-\mu}{k_B T}}}-\frac{1}{1+e^{\frac{\mu+x-\mu}{k_B T}}}\right) = D(\mu+x)\left(\frac{1}{