Sub­sec­tions


A.39 The rel­a­tivis­tic hy­dro­gen atom

The de­scrip­tion of the hy­dro­gen atom given ear­lier in chap­ter 4.3 is very ac­cu­rate by en­gi­neer­ing stan­dards. How­ever, it is not ex­act. This ad­den­dum ex­am­ines var­i­ous rel­a­tivis­tic ef­fects that were ig­nored in the analy­sis.

The ap­proach will be to take the re­sults of chap­ter 4.3 as the start­ing point. Then cor­rec­tions are ap­plied to them us­ing per­tur­ba­tion the­ory as de­scribed in ad­den­dum {A.38}.


A.39.1 In­tro­duc­tion

Ac­cord­ing to the de­scrip­tion of the hy­dro­gen atom given in chap­ter 4.3, all en­ergy eigen­func­tions $\psi_{nlm}{\updownarrow}$ with the same value of $n$ have the same en­ergy $E_n$. There­fore they should show up as a sin­gle line in an ex­per­i­men­tal line spec­trum. But ac­tu­ally, when these spec­tra are ex­am­ined very pre­cisely, the $E_n$ en­ergy lev­els for a given value of $n$ are found to con­sist of sev­eral closely spaced lines, rather than a sin­gle one. That is called the hy­dro­gen atom fine struc­ture. It means that eigen­func­tions that all should have ex­actly the same en­ergy, don’t.

To ex­plain why, the so­lu­tion of chap­ter 4.3 must be cor­rected for a va­ri­ety of rel­a­tivis­tic ef­fects. Be­fore do­ing so, it is help­ful to ex­press the non­rel­a­tivis­tic en­ergy lev­els of that chap­ter in terms of the rest mass en­ergy ${m_{\rm e}}c^2$ of the elec­tron, as fol­lows:

\begin{displaymath}
\fbox{$\displaystyle
E_n = - \frac{\alpha^2}{2 n^2} m_{\rm...
...= \frac{e^2}{4\pi\epsilon_0\hbar c}
\approx \frac1{137}
$} %
\end{displaymath} (A.247)

The con­stant $\alpha$ is called the “fine struc­ture con­stant.” It com­bines the con­stants $e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0$ from elec­tro­mag­net­ism, $\hbar$ from quan­tum me­chan­ics, and the speed of light $c$ from rel­a­tiv­ity into one nondi­men­sion­al num­ber. It is with­out doubt the sin­gle most im­por­tant num­ber in all of physics, [19].

No­body knows why it has the value that it has. Still, ob­vi­ously it is a mea­sur­able value, so, fol­low­ing the stated ideas of quan­tum me­chan­ics, maybe the uni­verse mea­sured this value dur­ing its early for­ma­tion by a process that we may never un­der­stand, (since we do not have other mea­sured val­ues for $\alpha$ to de­duce any prop­er­ties of that process from.) If you have a demon­stra­bly bet­ter ex­pla­na­tion, Swe­den awaits you.

In any case, for en­gi­neer­ing pur­poses it is a small num­ber, less than 1%. That makes the hy­dro­gen en­ergy lev­els re­ally small com­pared to the rest mass en­ergy of the elec­tron, be­cause they are pro­por­tional to the square of $\alpha$, which is as small as 0.005%. In sim­ple terms, the elec­tron in hy­dro­gen stays well clear of the speed of light.

And that in turn means that the rel­a­tivis­tic er­rors in the hy­dro­gen en­ergy lev­els are small. Still, even small er­rors can some­times be very im­por­tant. The re­quired cor­rec­tions are listed be­low in or­der of de­creas­ing mag­ni­tude.

The fol­low­ing sub­sec­tions dis­cuss each cor­rec­tion in more de­tail.


A.39.2 Fine struc­ture

From the Dirac equa­tion, it can be seen that three terms need to be added to the non­rel­a­tivis­tic Hamil­ton­ian of chap­ter 4.3 to cor­rect the en­ergy lev­els for rel­a­tivis­tic ef­fects. The three terms are worked out in de­riva­tion {D.81}. But that math­e­mat­ics re­ally pro­vides very lit­tle in­sight. It is much more in­struc­tive to try to un­der­stand the cor­rec­tions from a more phys­i­cal point of view.

The first term is rel­a­tively easy to un­der­stand. Con­sider Ein­stein’s fa­mous re­la­tion $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$, where $E$ is en­ergy, $m$ mass, and $c$ the speed of light. Ac­cord­ing to this re­la­tion, the ki­netic en­ergy of the elec­tron is not ${\textstyle\frac{1}{2}}{m_{\rm e}}v^2$, with $v$ the ve­loc­ity, as New­ton­ian physics says. In­stead it is the dif­fer­ence be­tween the en­ergy $m_{{\rm {e}},v}c^2$ based on the mass $m_{{\rm {e}},v}$ of the elec­tron in mo­tion and the en­ergy ${m_{\rm e}}c^2$ based on the mass $m_{\rm e}$ of the elec­tron at rest. In terms of mo­men­tum $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{{\rm {e}},v}v$, chap­ter 1.1.2,

\begin{displaymath}
T = m_{\rm e}c^2 \sqrt{1 + \frac{p^2}{m_{\rm e}^2c^2}} - m_{\rm e}c^2
\end{displaymath} (A.248)

Since the speed of light is large com­pared to the typ­i­cal speed of the elec­tron, the square root can be ex­panded in a Tay­lor se­ries, [40, 22.12], to give:

\begin{displaymath}
T \approx \frac{p^2}{2m_{\rm e}} - \frac{p^4}{8m_{\rm e}^3c^2} + \ldots
\end{displaymath}

The first term cor­re­sponds to the ki­netic en­ergy op­er­a­tor used in the non­rel­a­tivis­tic quan­tum so­lu­tion of chap­ter 4.3. (It may be noted that the rel­a­tivis­tic mo­men­tum ${\skew0\vec p}$ is based on the mov­ing mass of the elec­tron, not its rest mass. It is this rel­a­tivis­tic mo­men­tum that cor­re­sponds to the op­er­a­tor ${\skew 4\widehat{\skew{-.5}\vec p}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$. So the Hamil­ton­ian used in chap­ter 4.3 was a bit rel­a­tivis­tic al­ready, be­cause in re­plac­ing ${\skew0\vec p}$ by $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$, it used the rel­a­tivis­tic ex­pres­sion.) The sec­ond term in the Tay­lor se­ries ex­pan­sion above is the first of the cor­rec­tions needed to fix up the hy­dro­gen en­ergy lev­els for rel­a­tiv­ity. Rewrit­ten in terms of the square of the clas­si­cal ki­netic en­ergy op­er­a­tor, the Bohr ground state en­ergy $E_1$ and the fine struc­ture con­stant $\alpha$, it is
\begin{displaymath}
H_{1,{\rm Einstein}}= - \frac{\alpha^{\,2}}{4\vert E_1\vert}
\left(\frac{{\widehat p}^2}{2m_{\rm e}}\right)^2 %
\end{displaymath} (A.249)

The sec­ond cor­rec­tion that must be added to the non­rel­a­tivis­tic Hamil­ton­ian is the so-called spin-or­bit in­ter­ac­tion. In clas­si­cal terms, it is due to the spin of the elec­tron, which makes it into a mag­netic di­pole. Think of it as a mag­net of in­fin­i­tes­i­mally small size, but with in­fi­nitely strong north and south poles to make up for it. The prod­uct of the in­fin­i­tes­i­mal vec­tor from south to north pole times the in­fi­nite strength of the poles is fi­nite, and de­fines the mag­netic di­pole mo­ment $\vec\mu$. By it­self, it is quite in­con­se­quen­tial since the mag­netic di­pole does not in­ter­act di­rectly with the elec­tric field of the nu­cleus. How­ever, mov­ing mag­netic poles cre­ate an elec­tric field just like the mov­ing elec­tric charges in an elec­tro­mag­net cre­ate a mag­netic field. The elec­tric fields gen­er­ated by the mov­ing mag­netic poles of the elec­tron are op­po­site in strength, but not quite cen­tered at the same po­si­tion. There­fore they cor­re­spond to a mo­tion-in­duced elec­tric di­pole. And an elec­tric di­pole does in­ter­act with the elec­tric field of the nu­cleus; it wants to align it­self with it. That is just like the mag­netic di­pole wanted to align it­self with the ex­ter­nal mag­netic field in the Zee­man ef­fect.

So how big is this ef­fect? Well, the en­ergy of an elec­tric di­pole $\vec\wp$ in an elec­tric field $\skew3\vec{\cal E}$ is

\begin{displaymath}
E_{1,\mbox{\scriptsize spin-orbit}} = -\vec\wp \cdot \skew3\vec{\cal E}
\end{displaymath}

As you might guess, the elec­tric di­pole gen­er­ated by the mag­netic poles of the mov­ing elec­tron is pro­por­tional to the speed of the elec­tron $\vec{v}$ and its mag­netic di­pole mo­ment $\vec\mu$. More pre­cisely, the elec­tric di­pole mo­ment $\vec\wp$ will be pro­por­tional to $\vec{v}$ $\times$ $\vec\mu$ be­cause if the vec­tor con­nect­ing the south and north poles is par­al­lel to the mo­tion, you do not have two neigh­bor­ing cur­rents of mag­netic poles, but a sin­gle cur­rent of both neg­a­tive and pos­i­tive poles that com­pletely can­cel each other out. Also, the elec­tric field $\skew3\vec{\cal E}$ of the nu­cleus is mi­nus the gra­di­ent of its po­ten­tial $e$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0r$, so

\begin{displaymath}
E_{1,\mbox{\scriptsize spin-orbit}} \propto
(\vec v\times\vec\mu) \cdot \frac{e}{4\pi\epsilon_0 r^3}{\skew0\vec r}
\end{displaymath}

Now the or­der of the vec­tors in this triple prod­uct can be changed, and the di­pole strength $\vec\mu$ of the elec­tron equals its spin $\vec{S}$ times the charge per unit mass $\vphantom{0}\raisebox{1.5pt}{$-$}$$e$$\raisebox{.5pt}{$/$}$$m_{\rm e}$, so

\begin{displaymath}
E_{1,\mbox{\scriptsize spin-orbit}} \propto
\frac{e^2}{m_{...
...0 r^3}
\left({\skew0\vec r}\times\vec{v}\right) \cdot \vec{S}
\end{displaymath}

The ex­pres­sion be­tween the paren­the­ses is the an­gu­lar mo­men­tum $\vec{L}$ save for the elec­tron mass. The con­stant of pro­por­tion­al­ity is worked out in de­riva­tion {D.82}, giv­ing the spin-or­bit Hamil­ton­ian as
\begin{displaymath}
H_{1,\mbox{\scriptsize spin-orbit}} =
\alpha^2 \vert E_1\v...
...r^2} {\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}} %
\end{displaymath} (A.250)

The fi­nal cor­rec­tion that must be added to the non­rel­a­tivis­tic Hamil­ton­ian is the so-called “Dar­win term:”

\begin{displaymath}
H_{1,{\rm Darwin}} = \alpha^2 \vert E_1\vert\; \pi a_0^3\delta^3({\skew0\vec r}) %
\end{displaymath} (A.251)

Ac­cord­ing to its de­riva­tion in {D.81}, it is a crude fix-up for an in­ter­ac­tion with a vir­tual positron that sim­ply can­not be in­cluded cor­rectly in a non­rel­a­tivis­tic analy­sis.

If that is not very sat­is­fac­tory, the fol­low­ing much more de­tailed de­riva­tion can be found on the web. It does suc­ceed in ex­plain­ing the Dar­win term fully within the non­rel­a­tivis­tic pic­ture alone. First as­sume that the elec­tric po­ten­tial of the nu­cleus does not re­ally be­come in­fi­nite as 1$\raisebox{.5pt}{$/$}$$r$ at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, but is smoothed out over some fi­nite nu­clear size. Also as­sume that the elec­tron does not see this po­ten­tial sharply, but per­ceives of its fea­tures a bit vaguely, as dif­fused out sym­met­ri­cally over a typ­i­cal dis­tance equal to the so-called Comp­ton wave length $\hbar$$\raisebox{.5pt}{$/$}$${m_{\rm e}}c$. There are sev­eral plau­si­ble rea­sons why it might: (1) the elec­tron has il­le­gally picked up a chunk of a neg­a­tive rest mass state, and it is trem­bling with fear that the un­cer­tainty in en­ergy will be noted, mov­ing rapidly back and for­wards over a Comp­ton wave length in a so-called Zit­ter­be­we­gung; (2) the elec­tron has de­cided to move at the speed of light, which is quite pos­si­ble non­rel­a­tivis­ti­cally, so its un­cer­tainty in po­si­tion is of the or­der of the Comp­ton wave length, and it just can­not fig­ure out where the right po­ten­tial is with all that un­cer­tainty in po­si­tion and light that fails to reach it; (3) the elec­tron needs glasses. Fur­ther as­sume that the Comp­ton wave length is much smaller than the size over which the nu­clear po­ten­tial is smoothed out. In that case, the po­ten­tial within a Comp­ton wave length can be ap­prox­i­mated by a sec­ond or­der Tay­lor se­ries, and the dif­fu­sion of it over the Comp­ton wave length will pro­duce an er­ror pro­por­tional to the Lapla­cian of the po­ten­tial (the only fully sym­met­ric com­bi­na­tion of de­riv­a­tives in the sec­ond or­der Tay­lor se­ries.). Now if the po­ten­tial is smoothed over the nu­clear re­gion, its Lapla­cian, giv­ing the charge den­sity, is known to pro­duce a nonzero spike only within that smoothed nu­clear re­gion, fig­ure 13.7 or (13.30). Since the nu­clear size is small com­pared to the elec­tron wave func­tions, that spike can then be ap­prox­i­mated as a delta func­tion. Tell all your friends you heard it here first.

The key ques­tion is now what are the changes in the hy­dro­gen en­ergy lev­els due to the three per­tur­ba­tions dis­cussed above. That can be an­swered by per­tur­ba­tion the­ory as soon as the good eigen­func­tions have been iden­ti­fied. Re­call that the usual hy­dro­gen en­ergy eigen­func­tions $\psi_{nlm}{\updownarrow}$ are made unique by the square an­gu­lar mo­men­tum op­er­a­tor $\L ^2$, giv­ing $l$, the $z$ an­gu­lar mo­men­tum op­er­a­tor $\L _z$, giv­ing $m$, and the spin an­gu­lar mo­men­tum op­er­a­tor ${\widehat S}_z$ giv­ing the spin quan­tum num­ber $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm\frac12$ for spin up, re­spec­tively down. The de­ci­sive term whether these are good or not is the spin-or­bit in­ter­ac­tion. If the in­ner prod­uct in it is writ­ten out, it is

\begin{displaymath}
H_{1,\mbox{\scriptsize spin-orbit}} =
\alpha^2 \vert E_1\v...
...x{\widehat S}_x+\L _y{\widehat S}_y+\L _z{\widehat S}_z\right)
\end{displaymath}

The ra­dial fac­tor is no prob­lem; it com­mutes with every or­bital an­gu­lar mo­men­tum com­po­nent, since these are purely an­gu­lar de­riv­a­tives, chap­ter 4.2.2. It also com­mutes with every com­po­nent of spin be­cause all spa­tial func­tions and op­er­a­tors do, chap­ter 5.5.3. As far as the dot prod­uct is con­cerned, it com­mutes with $\L ^2$ since all the com­po­nents of ${\skew 4\widehat{\vec L}}$ do, chap­ter 4.5.4, and since all the com­po­nents of ${\skew 6\widehat{\vec S}}$ com­mute with any spa­tial op­er­a­tor. But un­for­tu­nately, $\L _x$ and $\L _y$ do not com­mute with $\L _z$, and ${\widehat S}_x$ and ${\widehat S}_y$ do not com­mute with ${\widehat S}_z$ (chap­ters 4.5.4 and 5.5.3):

\begin{displaymath}[\L _x,\L _z]= - {\rm i}\hbar\L _y \quad
[\L _y,\L _z] = {\r...
...
[{\widehat S}_y,{\widehat S}_z] = {\rm i}\hbar{\widehat S}_x
\end{displaymath}

The quan­tum num­bers $m$ and $m_s$ are bad.

For­tu­nately, ${\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}$ does com­mute with the net $z$ an­gu­lar mo­men­tum ${\widehat J}_z$, de­fined as $\L _z+{\widehat S}_z$. In­deed, us­ing the com­mu­ta­tors above and the rules of chap­ter 4.5.4 to take apart com­mu­ta­tors:

\begin{eqnarray*}
& & [\L _x{\widehat S}_x,\L _z+{\widehat S}_z] = [\L _x,\L _z...
...,\L _z]{\widehat S}_z + \L _z[{\widehat S}_z,{\widehat S}_z] = 0
\end{eqnarray*}

and adding it all up, you get $[{\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}},{\widehat J}_z]$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The same way of course ${\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}$ com­mutes with the other com­po­nents of net an­gu­lar mo­men­tum ${\skew 6\widehat{\vec J}}$, since the $z$-​axis is ar­bi­trary. And if ${\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}$ com­mutes with every com­po­nent of ${\skew 6\widehat{\vec J}}$, then it com­mutes with their sum of squares ${\widehat J}^2$. So, eigen­func­tions of $\L ^2$, ${\widehat J}^2$, and ${\widehat J}_z$ are good eigen­func­tions.

Such good eigen­func­tions can be con­structed from the $\psi_{nlm}{\updownarrow}$ by form­ing lin­ear com­bi­na­tions of them that com­bine dif­fer­ent $m$ and $m_s$ val­ues. The co­ef­fi­cients of these good com­bi­na­tions are called Cleb­sch-Gor­dan co­ef­fi­cients and are shown for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 in fig­ure 12.5. Note from this fig­ure that the quan­tum num­ber $j$ of net square mo­men­tum can only equal $l+\frac12$ or $l-\frac12$. The half unit of elec­tron spin is not big enough to change the quan­tum num­ber of square or­bital mo­men­tum by more than half a unit. For the rest, how­ever, the de­tailed form of the good eigen­func­tions is of no in­ter­est here. They will just be in­di­cated in ket no­ta­tion as ${\left\vert nljm_j\right\rangle}$, in­di­cat­ing that they have un­per­turbed en­ergy $E_n$, square or­bital an­gu­lar mo­men­tum $l(l+1)\hbar^2$, square net (or­bital plus spin) an­gu­lar mo­men­tum $j(j+1)\hbar^2$, and net $z$ an­gu­lar mo­men­tum $m_j\hbar$.

As far as the other two con­tri­bu­tions to the fine struc­ture are con­cerned, ac­cord­ing to chap­ter 4.3.1 ${\skew 4\widehat{\skew{-.5}\vec p}}^{\,2}$ in the Ein­stein term con­sists of ra­dial func­tions and ra­dial de­riv­a­tives plus $\L ^2$. These com­mute with the an­gu­lar de­riv­a­tives that make up the com­po­nents of ${\skew 4\widehat{\vec L}}$, and as spa­tial func­tions and op­er­a­tors, they com­mute with the com­po­nents of spin. So the Ein­stein Hamil­ton­ian com­mutes with all com­po­nents of ${\skew 4\widehat{\vec L}}$ and ${\skew 6\widehat{\vec J}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew 4\widehat{\vec L}}+{\skew 6\widehat{\vec S}}$, hence with $\L ^2$, ${\widehat J}^2$, and ${\widehat J}_z$. And the delta func­tion in the Dar­win term can be as­sumed to be the limit of a purely ra­dial func­tion and com­mutes in the same way. The eigen­func­tions ${\left\vert nljm_j\right\rangle}$ with given val­ues of $l$, $j$, and $m_j$ are good ones for the en­tire fine struc­ture Hamil­ton­ian.

To get the en­ergy changes, the Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients

\begin{displaymath}
\langle m_jjln \vert
H_{1,{\rm Einstein}}+H_{1,\mbox{\scriptsize spin-orbit}}+H_{1,{\rm Darwin}}
\vert nljm_j \rangle
\end{displaymath}

must be found. Start­ing with the Ein­stein term, it is

\begin{displaymath}
\langle m_jjln\vert H_{1,{\rm Einstein}}\vert nljm_j\rangle...
...vert\frac{{\widehat p}^{\,4}}{4m_{\rm e}^2}\vert nljm_j\rangle
\end{displaymath}

Un­like what you may have read else­where, ${\widehat p}^{\,4}$ is in­deed a Her­mit­ian op­er­a­tor, but ${\widehat p}^{\,4}\vert nljm_j\rangle$ may have a delta func­tion at the ori­gin, (13.30), so watch it with blindly ap­ply­ing math­e­mat­i­cal ma­nip­u­la­tions to it. The trick is to take half of it to the other side of the in­ner prod­uct, and then use the fact that the eigen­func­tions sat­isfy the non­rel­a­tivis­tic en­ergy eigen­value prob­lem:

\begin{eqnarray*}
\langle m_jjln \vert \frac{{\widehat p}^2}{2m_{\rm e}} \bigg\...
...
& = & \langle m_jjln\vert E_n^2-2VE_n +V^2\vert nljm_j\rangle
\end{eqnarray*}

Not­ing from chap­ter 4.3 that $E_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_1$$\raisebox{.5pt}{$/$}$$n^2$, $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2E_1a_0$$\raisebox{.5pt}{$/$}$$r$ and that the ex­pec­ta­tion val­ues of $a_0$$\raisebox{.5pt}{$/$}$$r$ and $(a_0/r)^2$ are given in de­riva­tion {D.83}, you find that

\begin{displaymath}
\langle m_jjln\vert H_{1,{\rm Einstein}}\vert nljm_j\rangle...
...lpha^2}{4n^2}\left(\frac{4n}{l+\frac12}-3\right)\vert E_n\vert
\end{displaymath}

The spin-or­bit en­ergy cor­rec­tion is

\begin{displaymath}
\langle m_jjln\vert H_{1,\mbox{\scriptsize spin-orbit}}\ver...
...ehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}\vert nljm_j\rangle
\end{displaymath}

For states with no or­bital an­gu­lar mo­men­tum, all com­po­nents of ${\skew 4\widehat{\vec L}}$ pro­duce zero, so there is no con­tri­bu­tion. Oth­er­wise, the dot prod­uct ${\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}$ can be rewrit­ten by ex­pand­ing

\begin{displaymath}
{\widehat J}^2 = ({\skew 4\widehat{\vec L}}+{\skew 6\wideha...
... S}^2+2{\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}
\end{displaymath}

to give

\begin{eqnarray*}
{\skew 4\widehat{\vec L}}\cdot{\skew 6\widehat{\vec S}}\vert ...
...rac{1}{2}} (1+{\textstyle\frac{1}{2}}) \Big) \vert nljm_j\rangle
\end{eqnarray*}

That leaves only the ex­pec­ta­tion value of $(a_0/r)^3$ to be de­ter­mined, and that can be found in de­riva­tion {D.83}. The net re­sult is

\begin{displaymath}
\langle m_jjln\vert H_{1,\mbox{\scriptsize spin-orbit}}\ver...
...2}})}{l(l+\frac12)(l+1)}\vert E_n\vert
\quad\mbox{if }l \ne 0
\end{displaymath}

or zero if $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

Fi­nally the Dar­win term,

\begin{displaymath}
\langle m_jjln\vert H_{1,{\rm Darwin}}\vert nljm_j\rangle =...
...\langle m_jjln\vert\delta^3({\skew0\vec r})\vert nljm_j\rangle
\end{displaymath}

Now a delta func­tion at the ori­gin has the prop­erty to pick out the value at the ori­gin of what­ever func­tion it is in an in­te­gral with, com­pare chap­ter 7.9.1. De­riva­tion {D.15}, (D.9), im­plies that the value of the wave func­tions at the ori­gin is zero un­less $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, and then the value is given in (D.10). So the Dar­win con­tri­bu­tion be­comes

\begin{displaymath}
\langle m_jjln\vert H_{1,{\rm Darwin}}\vert nljm_j\rangle =
\frac{\alpha^2}{4n^2} 4n \vert E_n\vert \qquad \mbox{if $l=0$}
\end{displaymath}

To get the to­tal en­ergy change due to fine struc­ture, the three con­tri­bu­tions must be added to­gether. For $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, add the Ein­stein and Dar­win terms. For $l$ $\raisebox{.2pt}{$\ne$}$ 0, add the Ein­stein and spin-or­bit terms; you will need to do the two pos­si­bil­i­ties that $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l+{\textstyle\frac{1}{2}}$ and $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l-{\textstyle\frac{1}{2}}$ sep­a­rately. All three pro­duce the same fi­nal re­sult, any­way:

\begin{displaymath}
\fbox{$\displaystyle
E_{nljm_j,1}= - \left(\frac{1}{n(j+\f...
...\frac{3}{4}\frac{1}{n^2}\right)
\alpha^2 \vert E_n\vert
$} %
\end{displaymath} (A.252)

Since $j+{\textstyle\frac{1}{2}}$ is at most $n$, the en­ergy change due to fine struc­ture is al­ways neg­a­tive. And it is the biggest frac­tion of $E_n$ for $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}$ and $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, where it is $-\frac{5}{16}\alpha^2\vert E_n\vert$, still no more than a sixth of a per­cent of a per­cent change in en­ergy.

In the ground state $j$ can only be one half, (the elec­tron spin), so the ground state en­ergy does not split into two due to fine struc­ture. You would of course not ex­pect so, be­cause in empty space, both spin di­rec­tions are equiv­a­lent. The ground state does show the largest ab­solute change in en­ergy.

Woof.


A.39.3 Weak and in­ter­me­di­ate Zee­man ef­fect

The weak Zee­man ef­fect is the ef­fect of a mag­netic field that is suf­fi­ciently weak that it leaves the fine struc­ture en­ergy eigen­func­tions al­most un­changed. The Zee­man ef­fect is then a small per­tur­ba­tion on a prob­lem in which the un­per­turbed (by the Zee­man ef­fect) eigen­func­tions ${\left\vert nljm_j\right\rangle}$ de­rived in the pre­vi­ous sub­sec­tion are de­gen­er­ate with re­spect to $l$ and $m_j$.

The Zee­man Hamil­ton­ian

\begin{displaymath}
H_1 = \frac{e}{2m_{\rm e}}{\cal B}_{\rm ext}\left(\L _z + 2 {\widehat S}_z\right)
\end{displaymath}

com­mutes with both $\L ^2$ and ${\widehat J}_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\widehat S}_z+\L _z$, so the eigen­func­tions ${\left\vert nljm_j\right\rangle}$ are good. There­fore, the en­ergy per­tur­ba­tions can be found as

\begin{displaymath}
\frac{e}{2m_{\rm e}} {\cal B}_{\rm ext} \langle m_jjln\vert \L _z+2{\widehat S}_z \vert nljm_j\rangle
\end{displaymath}

To eval­u­ate this rig­or­ously would re­quire that the ${\left\vert nljm_j\right\rangle}$ state be con­verted into the one or two $\psi_{nlm}{\updownarrow}$ states with $\vphantom{0}\raisebox{1.5pt}{$-$}$$l$ $\raisebox{-.3pt}{$\leqslant$}$ $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_j\pm{\textstyle\frac{1}{2}}$ $\raisebox{-.3pt}{$\leqslant$}$ $l$ and $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\mp\frac12$ us­ing the ap­pro­pri­ate Cleb­sch-Gor­dan co­ef­fi­cients from fig­ure 12.5.

How­ever, the fol­low­ing sim­plis­tic de­riva­tion is usu­ally given in­stead, in­clud­ing in this book. First get rid of $L_z$ by re­plac­ing it by ${\widehat J}_z-{\widehat S}_z$. The in­ner prod­uct with ${\widehat J}_z$ can then be eval­u­ated as be­ing $m_j\hbar$, giv­ing the en­ergy change as

\begin{displaymath}
\frac{e}{2m_{\rm e}} {\cal B}_{\rm ext}
\left[m_j\hbar + \langle m_jjln\vert{\widehat S}_z \vert nljm_j\rangle\right]
\end{displaymath}

For the fi­nal in­ner prod­uct, make a semi-clas­si­cal ar­gu­ment that only the com­po­nent of ${\skew 6\widehat{\vec S}}$ in the di­rec­tion of $\vec{J}$ gives a con­tri­bu­tion. Don’t worry that $\vec{J}$ does not ex­ist. Just note that the com­po­nent in the di­rec­tion of $\vec{J}$ is con­strained by the re­quire­ment that ${\skew 4\widehat{\vec L}}$ and ${\skew 6\widehat{\vec S}}$ must add up to ${\skew 6\widehat{\vec J}}$, but the com­po­nent nor­mal to $\vec{J}$ can be in any di­rec­tion and pre­sum­ably av­er­ages out to zero. Dis­miss­ing this com­po­nent, the com­po­nent in the di­rec­tion of $\vec{J}$ is

\begin{displaymath}
{\skew 6\widehat{\vec S}}_J = \frac{1}{J^2}({\skew 6\wideha...
...ec S}}\cdot{\skew 6\widehat{\vec J}}){\skew 6\widehat{\vec J}}
\end{displaymath}

and the dot prod­uct in it can be found from ex­pand­ing

\begin{displaymath}
\L ^2 = {\skew 4\widehat{\vec L}}\cdot{\skew 4\widehat{\vec...
...2 {\skew 6\widehat{\vec J}}\cdot{\skew 6\widehat{\vec S}}+ S^2
\end{displaymath}

to give

\begin{displaymath}
{\skew 6\widehat{\vec S}}_J = \frac{J^2 - \L ^2 + S^2}{2J^2}{\skew 6\widehat{\vec J}}
\end{displaymath}

For a given eigen­func­tion ${\left\vert nljm_j\right\rangle}$, $J^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2j(j+1)$, $\L ^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2l(l+1)$, and $S^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2s(s+1)$ with $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$.

If the $z$-​com­po­nent of ${\skew 6\widehat{\vec S}}_J$ is sub­sti­tuted for ${\widehat S}_z$ in the ex­pres­sion for the Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients, the en­ergy changes are

\begin{displaymath}
\left[1+\frac{j(j+1)-l(l+1)+s(s+1)}{2j(j+1)}\right]
\frac{e\hbar}{2m_{\rm e}} {\cal B}_{\rm ext} m_j %
\end{displaymath} (A.253)

(Rig­or­ous analy­sis us­ing fig­ure 12.5, or more gen­er­ally item 2 in chap­ter 12.8, pro­duces the same re­sults.) The fac­tor within the brack­ets is called the “Landé $g$-​fac­tor.” It is the fac­tor by which the mag­netic mo­ment of the elec­tron in the atom is larger than for a clas­si­cal par­ti­cle with the same charge and to­tal an­gu­lar mo­men­tum. It gen­er­al­izes the $g$-​fac­tor of the elec­tron in iso­la­tion to in­clude the ef­fect of or­bital an­gu­lar mo­men­tum. Note that it equals 2, the Dirac $g$-​fac­tor, if there is no or­bital mo­men­tum, and 1, the clas­si­cal value, if the or­bital mo­men­tum is so large that the half unit of spin can be ig­nored.

In the in­ter­me­di­ate Zee­man ef­fect, the fine struc­ture and Zee­man ef­fects are com­pa­ra­ble in size. The dom­i­nant per­tur­ba­tion Hamil­ton­ian is now the com­bi­na­tion of the fine struc­ture and Zee­man ones. Since the Zee­man part does not com­mute with ${\widehat J}^2$, the eigen­func­tions ${\left\vert nljm_j\right\rangle}$ are no longer good. Eigen­func­tions with the same val­ues of $l$ and $m_j$, but dif­fer­ent val­ues of $j$ must be com­bined into good com­bi­na­tions. For ex­am­ple, if you look at $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, the eigen­func­tions ${\left\vert 21{\textstyle\frac{3}{2}}{\textstyle\frac{1}{2}}\right\rangle}$ and ${\left\vert 21{\textstyle\frac{1}{2}}{\textstyle\frac{1}{2}}\right\rangle}$ have the same un­per­turbed en­ergy and good quan­tum num­bers $l$ and $m_j$. You will have to write a two by two ma­trix of Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients for them, as in ad­den­dum {A.38.3}, to find the good com­bi­na­tions and their en­ergy changes. And the same for the ${\left\vert 21{\textstyle\frac{3}{2}}\rule[2.5pt]{5pt}{.5pt}{\textstyle\frac{1}{2}}\right\rangle}$ and ${\left\vert 21{\textstyle\frac{1}{2}}\rule[2.5pt]{5pt}{.5pt}{\textstyle\frac{1}{2}}\right\rangle}$ eigen­func­tions. To ob­tain the ma­trix co­ef­fi­cients, use the Cleb­sch-Gor­dan co­ef­fi­cients from fig­ure 12.5 to eval­u­ate the ef­fect of the Zee­man part. The fine struc­ture con­tri­bu­tions to the ma­tri­ces are given by (A.252) when the $j$ val­ues are equal, and zero oth­er­wise. This can be seen from the fact that the en­ergy changes must be the fine struc­ture ones when there is no mag­netic field; note that $j$ is a good quan­tum num­ber for the fine struc­ture part, so its per­tur­ba­tion co­ef­fi­cients in­volv­ing dif­fer­ent $j$ val­ues are zero.


A.39.4 Lamb shift

A fa­mous ex­per­i­ment by Lamb & Rether­ford in 1947 showed that the hy­dro­gen atom state $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, also called the 2S$_{1/2}$ state, has a some­what dif­fer­ent en­ergy than the state $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$, also called the 2P$_{1/2}$ state. That was un­ex­pected, be­cause even al­low­ing for the rel­a­tivis­tic fine struc­ture cor­rec­tion, states with the same prin­ci­pal quan­tum num­ber $n$ and same to­tal an­gu­lar mo­men­tum quan­tum num­ber $j$ should have the same en­ergy. The dif­fer­ence in or­bital an­gu­lar mo­men­tum quan­tum num­ber $l$ should not af­fect the en­ergy.

The cause of the un­ex­pected en­ergy dif­fer­ence is called Lamb shift. To ex­plain why it oc­curs would re­quire quan­tum elec­tro­dy­nam­ics, and that is well be­yond the scope of this book. Roughly speak­ing, the ef­fect is due to a va­ri­ety of in­ter­ac­tions with vir­tual pho­tons and elec­tron/positron pairs. A good qual­i­ta­tive dis­cus­sion on a non­tech­ni­cal level is given by Feyn­man [19].

Here it must suf­fice to list the ap­prox­i­mate en­ergy cor­rec­tions in­volved. For states with zero or­bital an­gu­lar mo­men­tum, the en­ergy change due to Lamb shift is

\begin{displaymath}
E_{{\vec n},1,{\rm Lamb}} = -\frac{\alpha^3}{2n} k(n,0) E_n
\qquad \mbox{if } l = 0 %
\end{displaymath} (A.254)

where $k(n,0)$ is a nu­mer­i­cal fac­tor that varies a bit with $n$ from about 12.7 to 13.2. For states with nonzero or­bital an­gu­lar mo­men­tum,
\begin{displaymath}
E_{{\vec n},1,{\rm Lamb}} = -\frac{\alpha^3}{2n}
\left[k(n...
...f } l \ne 0
\mbox { and } j = l \pm {\textstyle\frac{1}{2}} %
\end{displaymath} (A.255)

where $k(n,l)$ is less than 0.05 and varies some­what with $n$ and $l$.

It fol­lows that the en­ergy change is re­ally small for states with nonzero or­bital an­gu­lar mo­men­tum, which in­cludes the 2P$_{1/2}$ state. The change is biggest for the 2S$_{1/2}$ state, the other state in the Lamb & Rether­ford ex­per­i­ment. (True, the cor­rec­tion would be big­ger still for the ground state $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, but since there are no states with nonzero an­gu­lar mo­men­tum in the ground state, there is no split­ting of spec­tral lines in­volved there.)

Qual­i­ta­tively, the rea­son that the Lamb shift is small for states with nonzero an­gu­lar mo­men­tum has to do with dis­tance from the nu­cleus. The non­triv­ial ef­fects of the cloud of vir­tual par­ti­cles around the elec­tron are most pro­nounced in the strong elec­tric field very close to the nu­cleus. In states of nonzero an­gu­lar mo­men­tum, the wave func­tion is zero at the nu­cleus, (D.9). So in those states the elec­tron is un­likely to be found very close to the nu­cleus. In states of zero an­gu­lar mo­men­tum, the square mag­ni­tude of the wave func­tion is 1$\raisebox{.5pt}{$/$}$$n^3{\pi}a_0^3$ at the nu­cleus, re­flected in both the much larger Lamb shift as well as its ap­prox­i­mate 1$\raisebox{.5pt}{$/$}$$n^3$ de­pen­dence on the prin­ci­pal quan­tum num­ber $n$.


A.39.5 Hy­per­fine split­ting

Hy­per­fine split­ting of the hy­dro­gen atom en­ergy lev­els is due to the fact that the nu­cleus acts as a lit­tle mag­net just like the elec­tron. The sin­gle-pro­ton nu­cleus and elec­tron have mag­netic di­pole mo­ments due to their spin equal to

\begin{displaymath}
\vec \mu_{\rm {p}} = \frac{g_pe}{2m_{\rm p}} {\skew 6\wideh...
... - \frac{g_ee}{2m_{\rm e}} {\skew 6\widehat{\vec S}}_{\rm {e}}
\end{displaymath}

in which the $g$-​fac­tor of the pro­ton is about 5.59 and that of the elec­tron 2. The mag­netic mo­ment of the nu­cleus is much less than the one of the elec­tron, since the much greater pro­ton mass ap­pears in the de­nom­i­na­tor. That makes the en­ergy changes as­so­ci­ated with hy­per­fine split­ting re­ally small com­pared to other ef­fects such as fine struc­ture.

This dis­cus­sion will re­strict it­self to the ground state, which is by far the most im­por­tant case. For the ground state, there is no or­bital con­tri­bu­tion to the mag­netic field of the elec­tron. There is only a spin-spin cou­pling be­tween the mag­netic mo­ments of the elec­tron and pro­ton, The en­ergy in­volved can be thought of most sim­ply as the en­ergy $-\vec\mu_{\rm {e}}\cdot\skew2\vec{\cal B}_p$ of the elec­tron in the mag­netic field $\skew2\vec{\cal B}_{\rm {p}}$ of the nu­cleus. If the nu­cleus is mod­elled as an in­fin­i­tes­i­mally small elec­tro­mag­net, its mag­netic field is that of an ideal cur­rent di­pole as given in ta­ble 13.2. The per­tur­ba­tion Hamil­ton­ian then be­comes

\begin{displaymath}
H_{1,\mbox{\scriptsize spin-spin}} =
\frac{g_{\rm {p}}g_{\...
...dehat{\vec S}}_{\rm {e}})}{3}\delta^3({\skew0\vec r})
\right]
\end{displaymath}

The good states are not im­me­di­ately self-ev­i­dent, so the four un­per­turbed ground states will just be taken to be the ones which the elec­tron and pro­ton spins com­bine into the triplet or sin­glet states of chap­ter 5.5.6:

\begin{displaymath}
\mbox{triplet:}\quad\psi_{100}{\left\vert 1\:1\right\rangle...
...quad\mbox{singlet:}\quad\psi_{100}{\left\vert\:0\right\rangle}
\end{displaymath}

or $\psi_{100}{\left\vert s_{\rm {net}}m_{\rm {net}}\right\rangle}$ for short, where $s_{\rm {net}}$ and $m_{\rm {net}}$ are the quan­tum num­bers of net spin and its $z$-​com­po­nent. The next step is to eval­u­ate the four by four ma­trix of Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients

\begin{displaymath}
\langle {\underline m}_{\rm net}{\underline s}_{\rm net}\ve...
...ze spin-spin}}
\psi_{100} \vert s_{\rm net}m_{\rm net}\rangle
\end{displaymath}

us­ing these states.

Now the first term in the spin-spin Hamil­ton­ian does not pro­duce a con­tri­bu­tion to the per­tur­ba­tion co­ef­fi­cients. The rea­son is that the in­ner prod­uct of the per­tur­ba­tion co­ef­fi­cients writ­ten in spher­i­cal co­or­di­nates in­volves an in­te­gra­tion over the sur­faces of con­stant $r$. The ground state eigen­func­tion $\psi_{100}$ is con­stant on these sur­faces. So there will be terms like $3{\widehat S}_{{\rm {p}},x}{\widehat S}_{{\rm {e}},y}xy$ in the in­te­gra­tion, and those are zero be­cause $x$ is just as much neg­a­tive as pos­i­tive on these spher­i­cal sur­faces, (as is $y$). There will also be terms like $3{\widehat S}_{{\rm {p}},x}{\widehat S}_{{\rm {e}},x}x^2-{\widehat S}_{{\rm {p}},x}{\widehat S}_{{\rm {e}},x}r^2$ in the in­te­gra­tion. These will be zero too be­cause by sym­me­try the av­er­ages of $x^2$, $y^2$, and $z^2$ are equal on the spher­i­cal sur­faces, each equal to one third the av­er­age of $r^2$.

So only the sec­ond term in the Hamil­ton­ian sur­vives, and the Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients be­come

\begin{displaymath}
\frac{g_{\rm {p}}g_{\rm {e}}e^2}{6m_{\rm e}m_{\rm p}\epsilo...
...\skew0\vec r})
\psi_{100} \vert s_{\rm net}m_{\rm net}\rangle
\end{displaymath}

The spa­tial in­te­gra­tion in this in­ner prod­uct merely picks out the value $\psi_{100}^2(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1$\raisebox{.5pt}{$/$}$${\pi}a_0^3$ at the ori­gin, as delta func­tions do. That leaves the sum over the spin states. Ac­cord­ing to ad­den­dum {A.10},

\begin{displaymath}
\mbox{triplet:}\quad {\skew 6\widehat{\vec S}}_{\rm {p}}\cd...
... -{\textstyle\frac{3}{4}} \hbar^2 {\left\vert\,0\right\rangle}
\end{displaymath}

Since the triplet and sin­glet spin states are or­tho­nor­mal, only the Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients for which ${\underline s}_{\rm {net}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $s_{\rm {net}}$ and ${\underline m}_{\rm {net}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {net}}$ sur­vive, and these then give the lead­ing or­der changes in the en­ergy.

Plug­ging it all in and rewrit­ing in terms of the Bohr en­ergy and fine struc­ture con­stant, the en­ergy changes are:

\begin{displaymath}
\mbox{triplet: } E_{1,\mbox{\scriptsize spin-spin}} =
{\te...
... g_{\rm {e}} \frac{m_{\rm e}}{m_{\rm p}}\alpha^2\vert E_1\vert
\end{displaymath} (A.256)

The en­ergy of the triplet states is raised and that of the sin­glet state is low­ered. There­fore, in the true ground state, the elec­tron and pro­ton spins com­bine into the sin­glet state. If they some­how get kicked into a triplet state, they will even­tu­ally tran­si­tion back to the ground state, say af­ter 10 mil­lion years or so, and re­lease a pho­ton. Since the dif­fer­ence be­tween the two en­er­gies is so tiny on ac­count of the very small val­ues of both $\alpha^2$ and $m_{\rm e}$$\raisebox{.5pt}{$/$}$$m_{\rm p}$, this will be a very low en­ergy pho­ton. Its wave length is as long as 0.21 m, pro­duc­ing the 21 cm hy­dro­gen line.