In his 1926 paper Fermi states without further explanation that it follows from the Thomas-Fermi equation $$\frac{d^2\psi(x)}{dx^2}=\frac{\psi(x)^{3/2}}{\sqrt{x}},\label{1} \tag{1}$$ and boundary conditions $$\psi(0)=1,\quad\psi(\infty)=0\label{2}\tag{2},$$ that $$\int\limits_0^\infty{\frac{\psi^{5/2}(x)}{\sqrt{x}}}dx=-\frac{5}{7}\psi^\prime(0),\label{3} \tag{3}$$ where $\psi^\prime(x)=\frac{d\psi(x)}{dx}$. It seems Fermi considered this inference trivial. However, the proof given below and inspired by Kleinert, p. 429, although simple, is not at all trivial.

The Thomas-Fermi equation \eqref{1} can be considered as a Euler-Lagrange equation coresponding to the action principle $$ \delta S=0,\;\; S=\int\limits_0^\infty\left [\frac{1}{2}\left (\frac{d\psi}{dx}\right)^2+\frac{2}{5}\frac{\psi^{5/2}(x)}{\sqrt{x}}\right ]dx.\label{4}\tag{4} $$ The following infinitesimal deformation of the "coordinate" function $$\psi(x)\to \bar\psi(x)=\psi(\lambda x),\;\lambda=1+\epsilon,\;\epsilon\ll 1,$$ respects the boundary conditions \eqref{2} and changes the action functional \eqref{4} to $$\bar S=\int\limits_0^\infty\left [\frac{1}{2}\left (\frac{d\bar\psi(x)}{dx}\right)^2+\frac{2}{5}\frac{\bar\psi^{5/2}(x)}{\sqrt{x}}\right ]dx=\int\limits_0^\infty\left [\frac{\lambda}{2}\left (\frac{d\psi(y)}{dy}\right)^2+\frac{2}{5}\frac{\psi^{5/2}(y)}{\sqrt{\lambda y}}\right ]dy,$$ where $y=\lambda x$. Therefore, using $\lambda =1+\epsilon,\;\lambda^{-1/2}\approx 1-\epsilon/2$, we get $$\bar S =S+\epsilon \int\limits_0^\infty\left [\frac{1}{2}\left (\frac{d\psi(y)}{dy}\right)^2-\frac{1}{5}\frac{\psi^{5/2}(y)}{\sqrt{y}}\right ]dy,$$ and since we must have $\delta S=\bar S-S=0$ for any $\epsilon\ll 1$, we conclude $$\int\limits_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx=\frac{2}{5}\int\limits_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}} dx.\label{5} \tag{5}$$ On the other hand, using \eqref{1}, \eqref{2} and integration by parts, we have $$ \begin{split} \int\limits_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}} dx &=\int\limits_0^\infty \psi(x)\frac{d^2\psi(x)}{dx^2}dx\\ &=\left .\psi(x)\frac{d\psi(x)}{dx}\right |_0^\infty-\int\limits_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx\\ & =-\left .\frac{d\psi(x)}{dx}\right |_{x=0}-\int\limits_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx, \end{split}$$ and the Fermi result \eqref{3} immediately follows from \eqref{5}.

Is there a simpler proof of \eqref{3} that Fermi might have had in mind?


1 Answer 1


The way this is taught in text books,$^\ast$ which is likely the way Fermi reasoned, is to compare two alternative integrations by parts. One the one hand, $$\int_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}}\,dx=-5\int_0^\infty \psi'(x)\psi^{3/2}(x)\sqrt{x}\,dx$$ $$\qquad\qquad=-\frac{5}{2}\int_0^\infty x\frac{d}{dx}[\psi'(x)]^2\,dx=\frac{5}{2}\int_0^\infty [\psi'(x)]^2\,dx.$$ On the other hand, $$\int_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}}\,dx=\int_0^\infty \psi(x)\psi''(x)\,dx=-\psi'(0)-\int_0^\infty[\psi'(x)]^2\,dx.$$ Hence $$\int_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}}\,dx=-\frac{5}{7}\psi'(0).$$

$^\ast$ See, for, example, page 17 of Roi Baer's notes

  • $\begingroup$ In the final step for the first integration by parts, doesn't the identity $-\frac{5}{2} \int_0^{\infty}x\frac{d}{dx}[\psi'(x)]^2 dx = \frac{5}{2}\int_0^\infty [\psi'(x)]^2dx$ assume that $\lim_{x\to \infty}x \psi'(x)^2 = 0$? $\endgroup$ Mar 18 at 18:59
  • 9
    $\begingroup$ @ViníciusNovelli: the question asks "How did Fermi calculate this integral", and not whether the calculation that Fermi did was fully justified. :-) Even in the first step there is the assumption that the limit $\lim \psi^{5/2}(x) \sqrt{x} = 0$ which is more than (2) assumes. $\endgroup$ Mar 18 at 22:49
  • 2
    $\begingroup$ @ViníciusNovelli -- yes indeed, this assumption is not stated explicitly by Fermi, who merely writes $\psi(x)\rightarrow 0$ for $x\rightarrow\infty$; one can infer the more precise limit $\lim_{x\rightarrow\infty} x^3\psi(x)=144$ from the differential equation for $\psi$, which is sufficient to cancel the boundary terms at infinity in the partial integration. (The more explicit treatment in Roi Baer's notes states this condition in eq. 2.6.12.) $\endgroup$ Mar 19 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.