D.84 Expectation powers of r for hydrogen

This note derives the expectation values of the powers of $r$ for the hydrogen energy eigenfunctions $\psi_{nlm}$. The various values to be be derived are:

\begin{displaymath}
\begin{array}{l}
\ldots \\
\displaystyle\frac{\strut}...
...
=\frac{n^2(5n^2-3l(l+1)+1)}{2} \\
\ldots
\end{array} %
\end{displaymath} (D.60)

where $a_0$ is the Bohr radius, about 0.53 Å. Note that you can get the expectation value of a more general function of $r$ by summing terms, provided that the function can be expanded into a Laurent series. Also note that the value of $m$ does not make a difference: you can combine $\psi_{nlm}$ of different $m$ values together and it does not change the above expectation values. And watch it, when the power of $r$ becomes too negative, the expectation value will cease to exist. For example, for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 the expectation values of $(a_0/r)^3$ and higher powers are infinite.

The trickiest to derive is the expectation value of $(a_0/r)^2$, and that one will be done first. First recall the hydrogen Hamiltonian from chapter 4.3,

\begin{displaymath}
H =
- \frac{\hbar^2}{2m_{\rm e}r^2}
\left\{
\frac{\p...
...tial \phi^2}
\right\}
- \frac{e^2}{4\pi\epsilon_0}\frac1r
\end{displaymath}

Its energy eigenfunctions of given square and $z$ angular momentum and their energy are

\begin{displaymath}
\psi_{nlm} = R_{nl}(r)Y^m_l(\theta,\phi)
\qquad
E_n = ...
...^2}
\qquad
a_0=\frac{4\pi\epsilon_0\hbar^2}{m_{\rm e}e^2}
\end{displaymath}

where the $Y_l^m$ are called the spherical harmonics.

When this Hamiltonian is applied to an eigenfunction $\psi_{nlm}$, it produces the exact same result as the following dirty trick Hamiltonian in which the angular derivatives have been replaced by $l(l+1)$:

\begin{displaymath}
H_{\rm DT} =
- \frac{\hbar^2}{2m_{\rm e}r^2}
\left\{
...
...)
- l(l+1)
\right\}
- \frac{e^2}{4\pi\epsilon_0}\frac1r
\end{displaymath}

The reason is that the angular derivatives are essentially the square angular momentum operator of chapter 4.2.3. Now, while in the hydrogen Hamiltonian the quantum number $l$ has to be an integer because of its origin, in the dirty trick one $l$ can be allowed to assume any value. That means that you can differentiate the Hamiltonian and its eigenvalues $E_n$ with respect to $l$. And that allows you to apply the Hellmann-Feynman theorem of section A.37.1:

\begin{displaymath}
\frac{\partial E_{n,{\rm DT}}}{\partial l} =
\bigg\lang...
... \frac{\partial H_{\rm DT}}{\partial l}\psi_{nlm}\bigg\rangle
\end{displaymath}

(Yes, the eigenfunctions $\psi_{nlm}$ are good, because the purely radial $H_{\rm {DT}}$ commutes with both $\L _z$ and $\L ^2$, which are angular derivatives.) Substituting in the dirty trick Hamiltonian,

\begin{displaymath}
\frac{\partial E_{n,{\rm DT}}}{\partial l} =
\frac{\hba...
...gg\vert \left(\frac{a_0}{r}\right)^2
\psi_{nlm}\bigg\rangle
\end{displaymath}

So, if you can figure out how the dirty trick energy changes with $l$ near some desired integer value $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l_0$, the desired expectation value of $(a_0/r)^2$ at that integer value of $l$ follows. Note that the eigenfunctions of $H_{\rm {DT}}$ can still be taken to be of the form $R_{nl}(r)Y_{l_0}^m(\theta,\phi)$, where $Y_{l_0}^m$ can be divided out of the eigenvalue problem to give $H_{\rm {DT}}R_{nl}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{DT}R_{nl}$. If you skim back through chapter 4.3 and its note, you see that that eigenvalue problem was solved in derivation {D.15}. Now, of course, $l$ is no longer an integer, but if you skim through the note, it really makes almost no difference. The energy eigenvalues are still $E_{n,\rm {DT}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$\hbar^2$$\raisebox{.5pt}{$/$}$$2n^2{m_{\rm e}}a_0^2$. If you look near the end of the note, you see that the requirement on $n$ is that $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $q+l+1$ where $q$ must remain an integer for valid solutions, hence must stay constant under small changes. So ${\rm d}{n}$$\raisebox{.5pt}{$/$}$${\rm d}{l}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, and then according to the chain rule the derivative of $E_{\rm {DT}}$ is $\hbar^2$$\raisebox{.5pt}{$/$}$$n^3{m_{\rm e}}a_0^2$. Substitute it in and there you have that nasty expectation value as given in (D.60).

All other expectation values of $(r/a_0)^q$ for integer values of $q$ may be found from the “Kramers relation,” or “(second) Pasternack relation:”

\begin{displaymath}
4(q+1) \big\langle q\big\rangle
- 4 n^2(2q+1) \langle q-1\rangle
+ n^2 q[(2l+1)^2 - q^2] \langle q-2\rangle
= 0
\end{displaymath} (D.61)

where $\langle{q}\rangle$ is shorthand for the expectation value $\langle\psi_{nlm}\vert(r/a_0)^q\psi_{nlm}\rangle$.

Substituting $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 into the Kramers-Pasternack relation produces the expectation value of $a_0$$\raisebox{.5pt}{$/$}$$r$ as in (D.60). It may be noted that this can instead be derived from the virial theorem of chapter 7.2, or from the Hellmann-Feynman theorem by differentiating the hydrogen Hamiltonian with respect to the charge $e$. Substituting in $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, ...produces the expectation values for $r/a_0$, $(r/a_0)^2$, .... Substituting in $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1 and the expectation value for $(a_0/r)^2$ from the Hellmann-Feynman theorem gives the expectation value for $(a_0/r)^3$. The remaining negative integer values $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$2, $\vphantom0\raisebox{1.5pt}{$-$}$3, ...produce the remaining expectation values for the negative integer powers of $r$$\raisebox{.5pt}{$/$}$$a_0$ as the $\langle{q}-2\rangle$ term in the equation.

Note that for a sufficiently negative powers of $r$, the expectation value becomes infinite. Specifically, since $\psi_{nlm}$ is proportional to $r^l$, {D.15}, it can be seen that $\langle{q-2}\rangle$ becomes infinite when $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-2l-1$. When that happens, the coefficient of the expectation value in the Kramers-Pasternack relation becomes zero, making it impossible to compute the expectation value. The relationship can be used until it crashes and then the remaining expectation values are all infinite.

The remainder of this note derives the Kramers-Pasternack relation. First note that the expectation values are defined as

\begin{displaymath}
\big\langle q\big\rangle \equiv
\langle\psi_{nlm}\vert(r...
...}}(r/a_0)^q \vert R_{nl}Y_l^m\vert^2{\,\rm d}^3{\skew0\vec r}
\end{displaymath}

When this integral is written in spherical coordinates, the integration of the square spherical harmonic over the angular coordinates produces one. So, the expectation value simplifies to

\begin{displaymath}
\big\langle q\big\rangle = \int_{r=0}^\infty (r/a_0)^q R_{nl}^2 r^2 {\,\rm d}r
\end{displaymath}

To simplify the notations, a nondi­men­sion­al radial coordinate $\rho$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r$$\raisebox{.5pt}{$/$}$$a_0$ will be used. Also, a new radial function $f$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ $\sqrt{a_0^3}{\rho}R_{nl}$ will be defined. In those terms, the expression above for the expectation value shortens to

\begin{displaymath}
\langle q \rangle = \int_0^\infty \rho^q f^2 {\,\rm d}\rho
\end{displaymath}

To further shorten the notations, from now on the limits of integration and ${\rm d}\rho$ will be omitted throughout. In those notations, the expectation value of $(r/a_0)^q$ is

\begin{displaymath}
\big\langle q\big\rangle = \int \rho^q f^2
\end{displaymath}

Also note that the integrals are improper. It is to be assumed that the integrations are from a very small value of $r$ to a very large one, and that only at the end of the derivation, the limit is taken that the integration limits become zero and infinity.

According to derivation {D.15}, the function $R_{nl}$ satisfies in terms of $\rho$ the ordinary differential equation.

\begin{displaymath}
- \rho^2 R_{nl}'' - 2\rho R_{nl}'
+ \left[l(l+1)-2\rho+\frac{1}{n^2}\rho^2\right]R_{nl} = 0
\end{displaymath}

where primes indicate derivatives with respect to $\rho$. Substituting in $R_{nl}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f$$\raisebox{.5pt}{$/$}$$\sqrt{a_0^3}{\rho}$, you get in terms of the new unknown function $f$ that
\begin{displaymath}
f'' =
\left[
\frac{1}{n^2}-\frac{2}{\rho}+\frac{l(l+1)}{\rho^2}
\right]f %
\end{displaymath} (D.62)

Since this makes $f''$ proportional to $f$, forming the integral $\int\rho^qf''f$ produces a combination of terms of the form $\int\rho^{\rm {power}}f^2$, hence of expectation values of powers of $\rho$:
\begin{displaymath}
\int\rho^qf''f = \frac{1}{n^2}\big\langle q\big\rangle - 2 \langle q-1\rangle
+l(l+1)\langle q-2\rangle %
\end{displaymath} (D.63)

The idea is now to apply integration by parts on $\int\rho^qf''f$ to produce a different combination of expectation values. The fact that the two combinations must be equal will then give the Kramers-Pasternack relation.

Before embarking on this, first note that since

\begin{displaymath}
\int \rho^q f f' = \int \rho^q \left({\textstyle\frac{1}{2...
...^2\bigg\vert - \int q \rho^{q-1} {\textstyle\frac{1}{2}} f^2,
\end{displaymath}

the latter from integration by parts, it follows that
\begin{displaymath}
\int \rho^q f f' =
\frac{1}{2} \rho^q f^2\bigg\vert
- \frac{q}{2} \langle q-1\rangle %
\end{displaymath} (D.64)

This result will be used routinely in the manipulations below to reduce integrals of that form.

Now an obvious first integration by parts on $\int\rho^qf''f$ produces

\begin{displaymath}
\int\rho^qf\,f'' = \rho^qff'\bigg\vert - \int\left(\rho^qf...
... = \rho^qff'\bigg\vert - \int q\rho^{q-1}ff' - \int\rho^qf'f'
\end{displaymath}

The first of the two integrals reduces to an expectation value of $\rho^{q-2}$ using (D.64). For the final integral, use another integration by parts, but make sure you do not run around in a circle because if you do you will get a trivial expression. What works is integrating $\rho^q$ and differentiating $f'f'$:
\begin{displaymath}
\int\rho^qff''
=
\rho^qff'\bigg\vert
- \frac{q}{2}\r...
...{q+1}{f'}^2\bigg\vert
+ 2 \int\frac{\rho^{q+1}}{q+1}f'f'' %
\end{displaymath} (D.65)

In the final integral, according to the differential equation (D.62), the factor $f''$ can be replaced by powers of $\rho$ times $f$:

\begin{displaymath}
2 \int\frac{\rho^{q+1}}{q+1}f'f'' =
2 \int\frac{\rho^{q+...
...[\frac{1}{n^2}-\frac{2}{\rho}+\frac{l(l+1)}{\rho^2}\right]ff'
\end{displaymath}

and each of the terms is of the form (D.64), so you get

\begin{eqnarray*}
\lefteqn{2\int\frac{\rho^{q+1}}{q+1}f'f'' =
\frac{1}{(q+1)...
...angle q-1\rangle
- \frac{l(l+1)(q-1)}{q+1} \langle q-2\rangle
\end{eqnarray*}

Plugging this into (D.65) and then equating that to (D.63) produces the Kramers-Pasternack relation. It also gives an additional right hand side

\begin{displaymath}
\rho^qff'\bigg\vert
- \frac{q\rho^{q-1}}{2}f^2\bigg\vert...
...1}f^2\bigg\vert
+ \frac{l(l+1)\rho^{q-1}}{q+1}f^2\bigg\vert
\end{displaymath}

but that term becomes zero when the integration limits take their final values zero and infinity. In particular, the upper limit values always become zero in the limit of the upper bound going to infinity; $f$ and its derivative go to zero exponentially then, beating out any power of $\rho$. The lower limit values also become zero in the region of applicability that $\langle{q}-2\rangle$ exists, because that requires that $\rho^{q-1}f^2$ is for small $\rho$ proportional to a power of $\rho$ greater than zero.

The above analysis is not valid when $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1, since then the final integration by parts would produce a logarithm, but since the expression is valid for any other $q$, not just integer ones you can just take a limit $q\to-1$ to cover that case.