D.83 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:

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

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

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

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

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 $l(l+1)$:

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

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.38.1:

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

(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,

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

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 $E_{n,\rm {DT}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar^2$$\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:”

$$4(q+1) \left\langle{q}\right\rangle - 4 n^2(2q+1) \langle q-1\rangle + n^2 q[(2l+1)^2 - q^2] \langle q-2\rangle = 0$$ (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$$\raisebox{.5pt}{$/$}$​$r$ as in (D.60). It may be noted that this can in­stead be de­rived from the vir­ial the­o­rem of chap­ter 7.2, or from the Hell­mann-Feyn­man the­o­rem by dif­fer­en­ti­at­ing the hy­dro­gen Hamil­ton­ian with re­spect to the charge $e$. Sub­sti­tut­ing in $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, ...pro­duces the ex­pec­ta­tion val­ues for $r/a_0$, $(r/a_0)^2$, .... Sub­sti­tut­ing in $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$1 and the ex­pec­ta­tion value for $(a_0/r)^2$ from the Hell­mann-Feyn­man the­o­rem gives the ex­pec­ta­tion value for $(a_0/r)^3$. The re­main­ing neg­a­tive in­te­ger val­ues $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$2, $\vphantom{0}\raisebox{1.5pt}{$-$}$3, ...pro­duce the re­main­ing ex­pec­ta­tion val­ues for the neg­a­tive in­te­ger pow­ers of $r$$\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

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

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

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

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

$$\left\langle{q}\right\rangle = \int \rho^q f^2$$

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 $\rho$:

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

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

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 $f'f'$:

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

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

$$2 \int\frac{\rho^{q+1}}{q+1}f'f'' = 2 \int\frac{\rho^{q+1}... ...t[\frac{1}{n^2}-\frac{2}{\rho}+\frac{l(l+1)}{\rho^2}\right]ff'$$

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

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

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

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

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 $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$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.