Let the density of states be given by

$$ g(\epsilon) = \int \frac{\mathop{d^3 q}}{4\pi^3} \delta(\epsilon - \epsilon(\vec{q})), $$

where $\epsilon(\vec{q}) = \frac{\hbar^2}{2m}q_\perp^2 + h_\pm(q_\parallel)$ and where $h_\pm(q_\parallel)$ is a function which depends on $q_\parallel$ alone. $q_\perp$ and $q_\parallel$ denote the components of the vector $\vec{q}$ which are perpendicular/parallel to a vector $\vec{K}$ of the reciprocal lattice. I want to show that $$ g(\epsilon) =\frac{1}{4\pi^2}\left(\frac{2m}{\hbar^2}\right) (q_\parallel^\text{max}-q_\parallel^\text{min})$$ with $q_\parallel^\text{max}$ and $q_\parallel^\text{min}$ the two solutions to the equation $\epsilon = h_\pm(q_\parallel)$.

Unfortunately, I have no idea how to derive the above equation. Hope someone will help me with this.


2 Answers 2


This is what I find. It's not exactly what you were looking for, so maybe there's a mistake, but it looks fine to me. You have $$ g(\epsilon) = \int \frac{d^3\textbf{q}}{4\pi^3} \delta(\epsilon - \epsilon(\textbf{q}) = \frac{1}{4\pi^3} \int d q_{||} \int d^2 \textbf{q}_{\perp} \delta\left(\epsilon - \frac{\hbar^2q^2_{\perp}}{2m} - h_{\pm}(q_{||})\right). $$ You use polar coordinates such as $d^2\textbf{q}_{\perp} = dq_{\perp}d\Omega_q q_{\perp}$, along with the change of variable $\epsilon' = \frac{\hbar^2q^2_{\perp}}{2m}$ implying $q_{\perp} dq_{\perp} = \frac{m}{\hbar^2}$. Thus, $$ g(\epsilon) = \frac{1}{4\pi^3} \int dq_{||} \underbrace{\int d\Omega_q}_{4\pi} \left(\frac{m}{\hbar^2}\right)\underbrace{\int d\epsilon' \delta(\epsilon - \epsilon' - h_{\pm}(q_{||}))}_{1} = \frac{1}{\pi^2}\frac{m}{\hbar^2} \int dq_{||}. $$ Now, you simply evaluate the limits of $q_{||}$, thus $\int dq_{||} = q_{|| \text{max}} - q_{|| \text{min}}$ and you almost have what you were looking for.

Hopefully this is still useful for you, even after a year and a half!


I've done the problem recently. The first answer is quite close but it's missing only one detail; the integral should be done in the cylindrical coordinates.P.S. If you're taking the class PHY730, please try to solve the problem before reading this, as it is a good complement to the class

Which leads to:


Considering the 2 spins : \begin{align} g_\pm (E) &= \frac{2}{V} \int d \mathbf{k} \delta(E - E_\pm ) \nonumber \\ &= \frac{1}{(2 \pi)^3 } \int\limits^{2\pi}_0 d \phi \int\limits^\infty_{-\infty} d k_\parallel \int\limits^{\infty}_{-\infty } d k_\bot k_\bot \delta \left( E- \frac{\hbar^2 k_\bot^2 }{2m} - h_\pm \right) \nonumber \end{align} using $\delta(ac)= \delta(c)/|a|$: \begin{align} g_\pm &= \frac{1}{2 \pi^2 }\frac{2 m}{\hbar^2} \int\limits^\infty_{-\infty} d k_\parallel \int\limits^{\infty}_{-\infty } d k_\bot k_\bot \delta \left( \frac{2m E }{\hbar^2 }- k_\bot^2 - \frac{2m}{\hbar^2 }h_\pm \right) \nonumber \end{align} Knowing that $h_\pm$ does not depend of $k_\bot$ and using \begin{align*} \delta(f(x)) &= \frac{\delta(x-x_i )}{|(\partial f(x) /\partial x)|_{x_i }} \end{align*} with $x_i$ such that $f(x_i ) =0$. One can get: \begin{align} &= \frac{1}{2 \pi^2 } \frac{2m}{\hbar^2 } \int\limits^\infty_{-\infty} d k_\parallel \int\limits^{\infty}_{-\infty }d k_\bot \frac{k_\bot }{2 k_\bot } \delta\left(k_\bot -\left\lbrack \frac{2 m E}{\hbar^2 }- \frac{2 m}{\hbar^2} h_\pm \right\rbrack^{1/2}\right) \end{align} Since both bands present a maxima and a minima, it is possible to write the integral limits of $k_\parallel$ as $k_\parallel^{\text{max}}$ and $k_\parallel^{\text{min}}$. Which gives \begin{align} g_\pm (E) &= \frac{1}{4 \pi^2 } \frac{2m}{\hbar^2 } \int\limits^{k_\parallel^{\text{max}}}_{k_\parallel^{\text{min}}} d k_\parallel \end{align} We finally obtain \begin{align} g_\pm (E) &=\frac{1}{4 \pi^2 } \frac{2m}{\hbar^2 }( k_\parallel^{\text{max}} - k_\parallel^{\text{min}}) \end{align}

Hopefully it will help anyone wondering about the details.

  • $\begingroup$ As Owen Willson said: "Wow" $\endgroup$
    – Laurier
    Commented Feb 23, 2022 at 21:10
  • $\begingroup$ Welcome to Physics! Unfortunately, this looks like a complete answer to a homework-like question, so it's possible that it may be deleted under our policy concerning homework-like questions. That said, given that the question is four years old, perhaps the mods will let it slide. In any event, I hope you'll stick around and answer other questions (just not homework questions!) $\endgroup$ Commented Feb 23, 2022 at 22:08
  • $\begingroup$ I underatand the point, no worries ! It's a question in chapter 9 of the Ashcroft Mermin. I was seeking a hint while I was doing exercices and fell upon the question and thought it would be nice to answer it $\endgroup$
    – Laurier
    Commented Feb 23, 2022 at 22:58

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