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:

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

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

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

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

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it produces the exact same result as the following “dirty trick Hamiltonian” in which the angular derivatives have been replaced by $l(l+1)$:

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

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:

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

(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,

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

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:”

$$4(q+1) \big\langle q\big\rangle - 4 n^2(2q+1) \langle q-1\rangle + n^2 q[(2l+1)^2 - q^2] \langle q-2\rangle = 0$$ (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

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

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

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

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

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

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

$$\big\langle q\big\rangle = \int \rho^q f^2$$

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.

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

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

$$f'' = \left[ \frac{1}{n^2}-\frac{2}{\rho}+\frac{l(l+1)}{\rho^2} \right]f %$$ (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 $\rho$:

$$\int\rho^qf''f = \frac{1}{n^2}\big\langle q\big\rangle - 2 \langle q-1\rangle +l(l+1)\langle q-2\rangle %$$ (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

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

the latter from integr­ation by parts, it follows that

$$\int \rho^q f f' = \frac{1}{2} \rho^q f^2\bigg\vert - \frac{q}{2} \langle q-1\rangle %$$ (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

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

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 $f'f'$:

$$\int\rho^qff'' = \rho^qff'\bigg\vert - \frac{q}{2}\r... ...q+1}{f'}^2\bigg\vert + 2 \int\frac{\rho^{q+1}}{q+1}f'f'' %$$ (D.65)

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

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

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

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

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.