The Feynman-Hellmann theorem (Problem 6.32) can be used to determine the expectation values of$1/r$and$1/r^2$for hydrogen.${}^{23}$The effective Hamiltonian for the radial wave functions is (Equation4.53)

$\frac{{\hslash }^2}{2m}\frac{d^2}{d{r}^2}+\frac{{\hslash }^2}{2m}\frac{l(l+1)}{r^2}-\frac{{\displaystyle e^2}}{4\mathrm{\pi }{\in }_0}\frac{1}{r}$

And the eigenvalues (expressed in terms of$l{)}^{24}$are (Equation 4.70)

$E_n=-\frac{{\displaystyle m{e}^4}}{{\displaystyle 32{\mathrm{\pi }}^2{\in }_0^2{\mathrm{h}}^2{\left({\mathrm{j}}_{\max }+\mathrm{l}+1\right)}^2}}$

(a) Use $\lambda =e$in the Feynman-Hellmann theorem to obtain $⟨1/r⟩$. Check your result against Equation 6.55.

(b) Use $\lambda =l$to obtain . Check your answer with Equation6.56.

The Hellmann–Feynman theorem connects the derivative of total energy with respect to a parameter with the expectation value of the Hamiltonian's derivative with respect to the same parameter. All the forces in the system can be estimated using classical electrostatics once the spatial distribution of the electrons has been known by solving the Schrödinger equation, according to the theorem.

Hamiltonian and associated energies are accessible:

$$\begin{gathered} H=-\frac{\hslash }{2m}\frac{d^2}{d{r}^2}+\frac{\hslash }{2m}\frac{l(l+1)}{r^2}-\frac{e^2}{4\pi {\epsilon }_0}\frac{1}{r} \\ {E}_n=-\frac{m{e}^4}{32{\pi }^2{\epsilon }_0^2{\hslash }^2{\left(j_{\max }+l+1\right)}^2} \end{gathered}$$

a) $\lambda =e$

$$\begin{gathered} \frac{\partial E_n}{\partial e}=\psi \left|\frac{\partial H}{\partial e}\right|\psi \\ -\frac{m{e}^3}{8{\pi }^2{\epsilon }_0^2{\hslash }^2{\left(j_{\max }+l+1\right)}^2}=\psi \left|-\frac{e}{2\pi {\epsilon }_0}\frac{1}{r}\right|\psi \end{gathered}$$

$$\begin{gathered} -\frac{m{e}^3}{8{\pi }^2{\epsilon }_0^2{\hslash }^2{\left(j_{\max }+l+1\right)}^2}=-\frac{e}{2\pi {\epsilon }_0}\frac{1}{r} \\ \frac{1}{r}=\frac{m{e}^2}{4\pi {\epsilon }_0{\hslash }^2{\left(j_{\max }+l+1\right)}^2}a=\frac{4\pi \epsilon {\hslash }^2}{m{e}^2} \\ \Rightarrow \frac{1}{r}=\frac{1}{a{n}^2} \end{gathered}$$

The Hamiltonian$\lambda =e$value of$\frac{1}{r}=\frac{1}{a{n}^2}$

b) $\lambda =l$

$$\begin{gathered} \frac{\partial E_n}{\partial l}=\psi \left|\frac{\partial H}{\partial l}\right|\psi \\ \frac{m{e}^4}{16{\pi }^2{\epsilon }_0^2{\hslash }^2{\left(j_{\max }+l+1\right)}^3}=\psi \left|\frac{{\hslash }^2}{2m{r}^2}(2l+1)\right|\psi \\ \frac{m{e}^4}{16{\pi }^2{\epsilon }_0^2{\hslash }^2{\left(j_{\max }+l+1\right)}^3}=\frac{{\hslash }^2(2l+1)}{2m}\frac{1}{r^2} \end{gathered}$$

The Hamiltonian$\lambda =l$value of$\frac{1}{r^2}=\frac{1}{a^2{n}^{3}l+\frac{1}{2}}$