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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?

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    $\begingroup$ Could you include the equation and what you've tried? Since we don't all have the book in front of us. This site uses MathJax to render LaTeX equations placed between \$ ... \$ for inline display or \$\$ ... \$\$ for block display. In any case it is probably just the relationship between energy $E(p)$ and momentum $p=\hbar k$ and a regular change of variables. $\endgroup$
    – Michael
    Feb 12, 2013 at 15:27
  • $\begingroup$ I added the equations and the step I can follow. $\endgroup$
    – TMOTTM
    Feb 12, 2013 at 16:10
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    $\begingroup$ So the function you're integrating only depends on $\epsilon(\vec{k})$, which really only depends on the magnitude of $\vec{k}$: $\left|\vec{k}\right|=k$. So the integral is spherically symmetric and you can break it into a radial $k$ part and an angular part over $\theta$ and $\phi$. This is the first step, and is what the book means by "exploiting the fact that...". You've gone from a 3D integral to a 1D integral. The second step is essentially defining the density of states $g(\epsilon)$. You can change the variable of integration from $k$ to $\epsilon$ to find what $g(\epsilon)$ must be. $\endgroup$
    – Michael
    Feb 12, 2013 at 16:20
  • $\begingroup$ "You can change the variable of integration from $k$ to $\epsilon$ to find what $g(ϵ)$ must be" That's exactly my problem: How do I do that? $\endgroup$
    – TMOTTM
    Feb 12, 2013 at 16:36
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    $\begingroup$ @MichaelBrown, you should put your comment in the form of an answer, since it is complete and accurate. $\endgroup$
    – KDN
    Feb 12, 2013 at 20:46

1 Answer 1

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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.

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  • $\begingroup$ Thanks Michael. Yes, it's unusual that the integration extends from minus to plus $\infty$. What is the reason that we do not write $$I = \int_0^\infty d\epsilon \, g(\epsilon)F(\epsilon)$$ and then just let $g(\epsilon) = dk/d\epsilon \, k^2(\epsilon)$? $\endgroup$
    – TMOTTM
    Feb 13, 2013 at 10:05
  • $\begingroup$ Could it be that it's useful to let the integration be from minus to plus $\infty$, because, one might want to carry out integrations as in Eq. 17.44 $$n(\textbf{r}) = \int \frac{d\textbf{k}}{4\pi^3}\frac{1}{\exp[\beta((\hbar^2 k^2/2m) - e\phi(\textbf{r}) - \mu] + 1}$$ $\endgroup$
    – TMOTTM
    Feb 13, 2013 at 10:15
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    $\begingroup$ As long as $\epsilon$ is defined by the relationship $\epsilon = \hbar^2 k^2/2m$ then the range is naturally $\epsilon>0$. In any case the energy must be bounded below. So I really don't know why Ashcroft and Mermin extend the integration - it's probably just to make some contour integration tricks (*cough* I mean "techniques") available later on, but I'm not familiar with the book. $\endgroup$
    – Michael
    Feb 13, 2013 at 11:39
  • $\begingroup$ ok, so is it fair to say that the integration boundaries can be extended to $-\infty$ because $g(\epsilon) = 0 \, \forall \epsilon < 0$? $\endgroup$
    – TMOTTM
    Feb 13, 2013 at 11:52
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    $\begingroup$ $\epsilon(\vec{k})$ only depends on the magnitude of $\vec{k}$. This comes from the form you are given for $\epsilon(\vec{k}) = \hbar^2 k^2 / 2m$, which is the energy for a free particle. It is generally okay to start with this if the kinetic energies of the particles are much greater than the potential energies. You are basically assuming that the potential isn't strong enough to dramatically change the particle wavefunctions, so plane wave states are a good starting point. If you have a system with a strong direction dependent potential (eg. graphene) then you need to change your analysis. $\endgroup$
    – Michael
    Feb 28, 2013 at 8:09

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