Sub­sec­tions


7.4 Con­ser­va­tion Laws in Emis­sion

Con­ser­va­tion laws are very use­ful for un­der­stand­ing emis­sion or ab­sorp­tion of ra­di­a­tion of var­i­ous kinds, as well as nu­clear re­ac­tions, col­li­sion processes, etcetera. As an ex­am­ple, this sec­tion will ex­am­ine what con­ser­va­tion laws say about the spon­ta­neous emis­sion of a pho­ton of light by an ex­cited atom. While this ex­am­ple is rel­a­tively sim­ple, the con­cepts dis­cussed here ap­ply in es­sen­tially the same way to more com­plex sys­tems.

Fig­ure 7.3: Crude con­cept sketch of the emis­sion of an elec­tro­mag­netic pho­ton by an atom. The ini­tial state is left and the fi­nal state is right.
\begin{figure}\centering
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\begin{picture}(400,67...
...put(171,0){\makebox(0,0)[b]{$E_\gamma=\hbar\omega$}}
\end{picture}
\end{figure}

Fig­ure 7.3 gives a sketch of the emis­sion process. The atom is ini­tially in an high en­ergy, or ex­cited, state that will be called $\psi_{\rm {H}}$. Af­ter some time, the atom re­leases a pho­ton and re­turns a lower en­ergy state that will be called $\psi_{\rm {L}}$. As a sim­ple ex­am­ple, take an hy­dro­gen atom. Then the ex­cited atomic state could be the 2p$_z$ $\psi_{210}$ state, chap­ter 4.3. The fi­nal atomic state will then be the 1s ground state $\psi_{100}$.

The emit­ted pho­ton has an en­ergy given by the Planck-Ein­stein re­la­tion

\begin{displaymath}
E_\gamma = \hbar\omega
\end{displaymath}

where $\omega$ is the fre­quency of the elec­tro­mag­netic ra­di­a­tion cor­re­spond­ing to the pho­ton. Note that $\gamma$ (gamma, think gamma de­cay) is the stan­dard sym­bol used to in­di­cate a pho­ton.


Key Points
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Atoms can tran­si­tion to a lower elec­tronic en­ergy level while emit­ting a pho­ton of elec­tro­mag­netic ra­di­a­tion.

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The Planck-Ein­stein re­la­tion gives the en­ergy of a pho­ton in terms of its fre­quency.


7.4.1 Con­ser­va­tion of en­ergy

The first con­ser­va­tion law that is very use­ful for un­der­stand­ing the emis­sion process is con­ser­va­tion of en­ergy. The fi­nal atom and pho­ton should have the ex­act same en­ergy as the ini­tial ex­cited atom. So the dif­fer­ence be­tween the atomic en­er­gies $E_{\rm {H}}$ and $E_{\rm {L}}$ must be the en­ergy $\hbar\omega$ of the pho­ton. There­fore, the emit­ted pho­ton must have a very pre­cise fre­quency $\omega$. That means that it has a very pre­cise color. For ex­am­ple, for the 2p$_z$ to 1s tran­si­tion of a hy­dro­gen atom, the emit­ted pho­ton is a very spe­cific ul­tra­vi­o­let color.

It should be pointed out that the fre­quency of the emit­ted pho­ton does have a very slight vari­a­tion. The rea­son can be un­der­stood from the fact that the ex­cited state de­cays at all. En­ergy eigen­states should be sta­tion­ary, sec­tion 7.1.4.

The very fact that a state de­cays shows that it is not truly an en­ergy eigen­state.

The big prob­lem with the analy­sis of the hy­dro­gen atom in chap­ter 4.3 was that it ig­nored any am­bi­ent ra­di­a­tion that the elec­tron might be ex­posed to. It turns out that there is al­ways some per­turb­ing am­bi­ent ra­di­a­tion, even if the atom is in­side a black box at ab­solute zero tem­per­a­ture. This is re­lated to the fact that the elec­tro­mag­netic field has quan­tum un­cer­tainty. Ad­vanced quan­tum analy­sis is needed to take that into ac­count, {A.23}. For­tu­nately, the un­cer­tainty in en­ergy is ex­tremely small for the typ­i­cal ap­pli­ca­tions con­sid­ered here.

As a mea­sure of the un­cer­tainty in en­ergy of a state, physi­cists of­ten use the so-called “nat­ural width”

\begin{displaymath}
\fbox{$\displaystyle
\Gamma=\frac{\hbar}{\tau}
$} %
\end{displaymath} (7.10)

Here $\tau$ is the mean life­time of the state, the av­er­age time it takes for the pho­ton to be emit­ted.

The claim that this width gives the un­cer­tainty in en­ergy of the state is usu­ally jus­ti­fied us­ing the all-pow­er­ful en­ergy-time un­cer­tainty equal­ity (7.9). A dif­fer­ent ar­gu­ment will be given at the end of sec­tion 7.6.1. In any case, the bot­tom line is that $\Gamma$ does in­deed give the ob­served un­cer­tainty in en­ergy for iso­lated atoms, [51, p. 139], and for nu­clei, [30, p. 40, 167].

As an ex­am­ple, the hy­dro­gen atom 2p$_z$ state has a life­time of 1.6 nanosec­onds. (The life­time can be com­puted us­ing the data in ad­den­dum {A.25.8}.) That makes its width about 4 10$\POW9,{-7}$ eV. Com­pared to the 10 eV en­ergy of the emit­ted pho­ton, that is ob­vi­ously ex­tremely small. En­ergy con­ser­va­tion in atomic tran­si­tions may not be truly ex­act, but it is def­i­nitely an ex­cel­lent ap­prox­i­ma­tion.

Still, since a small range of fre­quen­cies can be emit­ted, the ob­served line in the emis­sion spec­trum is not go­ing to be a math­e­mat­i­cally ex­act line, but will have a small width. Such an ef­fect is known as “spec­tral line broad­en­ing.”

The nat­ural width of a state is usu­ally only a small part of the ac­tual line broad­en­ing. If the atom is ex­posed to an in­co­her­ent am­bi­ent elec­tro­mag­netic field, it will in­crease the un­cer­tainty in en­ergy. (The evo­lu­tion of atoms in an in­co­her­ent elec­tro­mag­netic field will be an­a­lyzed in {D.41}.) Fre­quent in­ter­ac­tions with sur­round­ing atoms or other per­tur­ba­tions will also in­crease the un­cer­tainty in en­ergy, in part for rea­sons dis­cussed at the end of sec­tion 7.6.1. And any­thing else that changes the atomic en­ergy lev­els will of course also change the emit­ted fre­quen­cies.

An im­por­tant fur­ther ef­fect that causes spec­tral line de­vi­a­tions is atom mo­tion, ei­ther ther­mal mo­tion or global gas mo­tion. It pro­duces a Doppler shift in the ra­di­a­tion. This is not nec­es­sar­ily bad news in as­tron­omy; line broad­en­ing can pro­vide an hint about the tem­per­a­ture of the gas you are look­ing at, while line dis­place­ment can pro­vide a hint of its over­all mo­tion away from you.

It may also be men­tioned that the nat­ural width is not al­ways small. If you start look­ing at ex­cited nu­clear par­ti­cles, the un­cer­tainty in en­ergy can be enor­mous. Such par­ti­cles may have an un­cer­tainty in en­ergy that is of the or­der of 10% of their rel­a­tivis­tic rest mass en­ergy. And as you might there­fore guess, they are hardly sta­tion­ary states. Typ­i­cally, they sur­vive for only about 10$\POW9,{-23}$ sec­onds af­ter they are cre­ated. Even mov­ing at a speed com­pa­ra­ble to the speed of light, such par­ti­cles will travel only a dis­tance com­pa­ra­ble to the di­am­e­ter of a pro­ton be­fore dis­in­te­grat­ing.

Gen­er­ally speak­ing, the shorter the life­time of a state, the larger its un­cer­tainty in en­ergy, and vice-versa.
(To be fair, physi­cists do not ac­tu­ally man­age to see these par­ti­cles dur­ing their in­fin­i­tes­i­mal life­time. In­stead they in­fer the life­time from the vari­a­tion in en­ergy of the re­sult­ing state.)


Key Points
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In a tran­si­tion, the dif­fer­ence in atomic en­ergy lev­els gives the en­ergy, and so the fre­quency, of the emit­ted pho­ton.

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Un­sta­ble states have some un­cer­tainty in en­ergy, but it is usu­ally very small. For ex­tremely un­sta­ble par­ti­cles, the un­cer­tainty can be a lot.

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The width of a state is $\Gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar$$\raisebox{.5pt}{$/$}$$\tau$ with $\tau$ the mean life­time. It is a mea­sure for the min­i­mum ob­served vari­a­tion in en­ergy of the fi­nal state.


7.4.2 Com­bin­ing an­gu­lar mo­menta and par­i­ties

Con­ser­va­tion of an­gu­lar mo­men­tum and par­ity is eas­ily stated:

The an­gu­lar mo­men­tum and par­ity of the ini­tial atomic state must be the same as the com­bined an­gu­lar mo­men­tum and par­ity of the fi­nal atomic state and pho­ton.
The ques­tion is how­ever, how do you com­bine an­gu­lar mo­menta and par­ity val­ues? Even com­bin­ing an­gu­lar mo­menta is not triv­ial, be­cause an­gu­lar mo­menta are quan­tized.

Fig­ure 7.4: Ad­di­tion of an­gu­lar mo­menta in clas­si­cal physics.
\begin{figure}\centering
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\begin{picture}(155,84...
...}
\put(116.5,-1){\makebox(0,0)[l]{$\vec J_\gamma$}}
\end{picture}
\end{figure}

To get an idea of how an­gu­lar mo­menta com­bine, first con­sider what would hap­pen in clas­si­cal physics. Con­ser­va­tion of an­gu­lar mo­men­tum would say that

\begin{displaymath}
\vec J_{\rm {H}} = \vec J_{\rm {L}} + \vec J_\gamma
\end{displaymath}

Here $\vec{J}_{\rm {H}}$ is the an­gu­lar mo­men­tum vec­tor of the ini­tial high en­ergy atomic state, and $\vec{J}_{\rm {L}}$ and $\vec{J}_\gamma$ are those of the fi­nal low en­ergy atomic state and the emit­ted pho­ton. The con­ser­va­tion law is shown graph­i­cally in fig­ure 7.4.

Fig­ure 7.5: Longest and short­est pos­si­ble fi­nal atomic an­gu­lar mo­menta in clas­si­cal physics.
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\begin{picture}(404,35...
...$}}
\put(367,-2){\makebox(0,0)[t]{$\vec J_\gamma$}}
\end{picture}
\end{figure}

Now con­sider what pos­si­ble lengths the vec­tor $\vec{J}_{\rm {L}}$ of the fi­nal atomic state can have. As fig­ure 7.4 shows, the length of $\vec{J}_{\rm {L}}$ is the dis­tance be­tween the start­ing points of the other two vec­tors. So the max­i­mum length oc­curs when the two vec­tors point in op­po­site di­rec­tion, with their noses touch­ing, like to the left in fig­ure 7.5. In that case, the length of $\vec{J}_{\rm {L}}$ is the sum of the lengths of the other two vec­tors. The min­i­mum length for $\vec{J}_{\rm {L}}$ oc­curs when the other two vec­tors are in the same di­rec­tion, still point­ing at the same point, like to the right in fig­ure 7.5. In that case the length of $\vec{J}_{\rm {L}}$ is the dif­fer­ence in length be­tween the other two vec­tors.

All to­gether:

\begin{displaymath}
\mbox{classical physics:} \qquad
\vert J_{\rm {H}}-J_\gamm...
...}} \mathrel{\raisebox{-.7pt}{$\leqslant$}}J_{\rm {H}}+J_\gamma
\end{displaymath}

Note that the omis­sion of a vec­tor sym­bol in­di­cates that the length of the vec­tor is meant, rather than the vec­tor it­self. The sec­ond in­equal­ity is the fa­mous “tri­an­gle in­equal­ity.” (The first in­equal­ity is a rewrit­ten tri­an­gle in­equal­ity for the longer of the two vec­tors in the ab­solute value.) The bot­tom line is that ac­cord­ing to clas­si­cal physics, the length of the fi­nal atomic an­gu­lar mo­men­tum can take any value in the range given above.

How­ever, in quan­tum me­chan­ics an­gu­lar mo­men­tum is quan­tized. The length of an an­gu­lar mo­men­tum vec­tor $\vec{J}$ must be $\sqrt{j(j+1)}\hbar$. Here the “az­imuthal quan­tum num­ber” $j$ must be a non­neg­a­tive in­te­ger or half of one. For­tu­nately, the tri­an­gle in­equal­ity above still works if you re­place lengths by az­imuthal quan­tum num­bers. To be pre­cise, the pos­si­ble val­ues of the fi­nal atomic an­gu­lar mo­men­tum quan­tum num­ber are:

\begin{displaymath}
\fbox{$\displaystyle
j_{\rm{L}} =
\vert j_{\rm{H}}{-}j_\g...
...rm{H}}{+}j_\gamma{-}1\mbox{, or }
j_{\rm{H}}{+}j_\gamma
$} %
\end{displaymath} (7.11)

In other words, the pos­si­ble val­ues of $j_{\rm {L}}$ in­crease from $\vert j_{\rm {H}}{-}j_\gamma\vert$ to $j_{\rm {H}}{+}j_\gamma$ in steps of 1. To show that an­gu­lar mo­men­tum quan­tum num­bers sat­isfy the tri­an­gle in­equal­ity in this way is not triv­ial; that is a ma­jor topic of chap­ter 12.

Clas­si­cal physics also says that com­po­nents of vec­tors can be added and sub­tracted as or­di­nary num­bers. Quan­tum physics agrees, but adds that for nonzero an­gu­lar mo­men­tum only one com­po­nent can be cer­tain at a time. That is usu­ally taken to be the $z$-​com­po­nent. Also, the com­po­nent can­not have any ar­bi­trary value; it must have a value of the form $m\hbar$. Here the “mag­netic quan­tum num­ber” $m$ can only have val­ues that range from $\vphantom{0}\raisebox{1.5pt}{$-$}$$j$ to $j$ in in­cre­ments of 1.

If you can add and sub­tract com­po­nents of an­gu­lar mo­men­tum, then you can also add and sub­tract mag­netic quan­tum num­bers. Af­ter all, they are only dif­fer­ent from com­po­nents by a fac­tor $\hbar$. There­fore, the con­ser­va­tion of an­gu­lar mo­men­tum in the $z$-​di­rec­tion be­comes

\begin{displaymath}
m_{\rm {L}} = m_{\rm {H}} - m_\gamma
\end{displaymath}

Putting in the pos­si­ble val­ues of the mag­netic quan­tum num­ber of the pho­ton gives for the fi­nal atomic mag­netic quan­tum num­ber:
\begin{displaymath}
\fbox{$\displaystyle
m_{\rm{L}} =
m_{\rm{H}}{-}j_\gamma\m...
...rm{H}}{+}j_\gamma{-}1\mbox{, or }
m_{\rm{H}}{+}j_\gamma
$} %
\end{displaymath} (7.12)

To be sure, $m_{\rm {L}}$ is also con­strained by the fact that its mag­ni­tude can­not ex­ceed $j_{\rm {L}}$.

Next con­sider con­ser­va­tion of par­ity. Re­call from sec­tion 7.3 that par­ity is the fac­tor by which the wave func­tion changes when the pos­i­tive di­rec­tion of all three co­or­di­nate axes is in­verted. That re­places every po­si­tion vec­tor ${\skew0\vec r}$ by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. Par­ity can have only two val­ues, 1 or $\vphantom{0}\raisebox{1.5pt}{$-$}$1. Par­ity is com­monly in­di­cated by $\pi$, which is the Greek let­ter for p. Par­ity starts with a p and may well be Greek. Also, the sym­bol avoids con­fu­sion, as­sum­ing that $\pi$ is not yet used for any­thing else in sci­ence.

Con­ser­va­tion of par­ity means that the ini­tial and fi­nal par­i­ties must be equal. The par­ity of the ini­tial high en­ergy atom must be the same as the com­bined par­ity of the fi­nal low en­ergy atom and pho­ton:

\begin{displaymath}
\pi_{\rm {H}} = \pi_{\rm {L}} \pi_\gamma
\end{displaymath}

Note that par­ity is a mul­ti­plica­tive quan­tity. You get the com­bined par­ity of the fi­nal state by mul­ti­ply­ing the par­i­ties of atom and pho­ton; you do not add them.

(Just think of the sim­plest pos­si­ble wave func­tion of two par­ti­cles, $\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_1({\skew0\vec r}_1)\psi_2({\skew0\vec r}_2)$. If $\psi_1$ changes by a fac­tor $\pi_1$ when ${\skew0\vec r}_1\to-{\skew0\vec r}_1$ and $\psi_2$ changes by a fac­tor $\pi_2$ when ${\skew0\vec r}_2\to-{\skew0\vec r}_2$, then the to­tal wave func­tion $\Psi$ changes by a fac­tor $\pi_1\pi_2$. Ac­tu­ally, it is an­gu­lar mo­men­tum, not par­ity, that is the weird case. The rea­son that an­gu­lar mo­menta must be added to­gether in­stead of mul­ti­plied to­gether is be­cause an­gu­lar mo­men­tum is de­fined by tak­ing a log­a­rithm of the nat­ural con­served quan­tity. For de­tails, see ad­den­dum {A.19}.)

The par­ity of the atom is re­lated to the or­bital an­gu­lar mo­men­tum of the elec­tron, and in par­tic­u­lar to its az­imuthal quan­tum num­ber $l$. If you check out the ex­am­ple spher­i­cal har­mon­ics in ta­ble 4.3, you see that those with even val­ues of $l$ only con­tain terms that are square in the po­si­tion co­or­di­nates. So these states do not change when ${\skew0\vec r}$ is re­placed by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. In other words, they change by a triv­ial fac­tor 1. That makes the par­ity 1, or even, or pos­i­tive. The spher­i­cal har­mon­ics for odd $l$ change sign when ${\skew0\vec r}$ is re­placed by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. In other words, they get mul­ti­plied by a fac­tor $\vphantom{0}\raisebox{1.5pt}{$-$}$1. That makes the par­ity $\vphantom{0}\raisebox{1.5pt}{$-$}$1, or odd, or neg­a­tive. These ob­ser­va­tions ap­ply for all val­ues of $l$, {D.14}.

The par­ity can there­fore be writ­ten for any value of $l$ as

\begin{displaymath}
\fbox{$\displaystyle
\pi = (-1)^l
$} %
\end{displaymath} (7.13)

This is just the par­ity due to or­bital an­gu­lar mo­men­tum. If the par­ti­cle has neg­a­tive in­trin­sic par­ity, you need to mul­ti­ply by an­other fac­tor $\vphantom{0}\raisebox{1.5pt}{$-$}$1. How­ever, an elec­tron has pos­i­tive par­ity, as does a pro­ton. (Positrons and an­tipro­tons have neg­a­tive par­ity. That is partly a mat­ter of con­ven­tion. Con­ser­va­tion of par­ity would still work if it was the other way around.)

It fol­lows that par­ity con­ser­va­tion in the emis­sion process can be writ­ten as

\begin{displaymath}
\fbox{$\displaystyle
(-1)^{l_{\rm{H}}} = (-1)^{l_{\rm{L}}} \pi_\gamma
$} %
\end{displaymath} (7.14)

There­fore, if the par­ity of the pho­ton is even, (i.e. 1), then $l_{\rm {H}}$ and $l_{\rm {L}}$ are both even or both odd. In other words, the atomic par­ity stays un­changed. If the par­ity of the pho­ton is odd, (i.e. $\vphantom{0}\raisebox{1.5pt}{$-$}$1), then one of $l_{\rm {H}}$ and $l_{\rm {L}}$ is even and the other odd. The atomic par­ity flips over.

To ap­ply the ob­tained con­ser­va­tion laws, the next step must be to fig­ure out the an­gu­lar mo­men­tum and par­ity of the pho­ton.


Key Points
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The rules for com­bin­ing an­gu­lar mo­menta and par­i­ties were dis­cussed.

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An­gu­lar mo­men­tum and par­ity con­ser­va­tion lead to con­straints on the atomic emis­sion process given by (7.11), (7.12), and (7.14).


7.4.3 Tran­si­tion types and their pho­tons

The con­ser­va­tion laws of an­gu­lar mo­men­tum and par­ity re­strict the emis­sion of a pho­ton by an ex­cited atom. But for these laws to be use­ful, there must be in­for­ma­tion about the spin and par­ity of the pho­ton.

This sec­tion will just state var­i­ous needed pho­ton prop­er­ties. De­riva­tions are given in {A.21.7} for the brave. In any case, the main con­clu­sions reached about the pho­tons as­so­ci­ated with atomic tran­si­tions will be ver­i­fied by more de­tailed analy­sis of tran­si­tions in later sec­tions.

There are two types of tran­si­tions, elec­tric ones and mag­netic ones. In elec­tric tran­si­tions, the elec­tro­mag­netic field pro­duced by the pho­ton at the atom is pri­mar­ily elec­tric, {A.21.7}. In mag­netic tran­si­tions, it is pri­mar­ily mag­netic. Elec­tric tran­si­tions are eas­i­est to un­der­stand phys­i­cally, and will be dis­cussed first.

A pho­ton is a par­ti­cle with spin $s_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and in­trin­sic par­ity $\vphantom{0}\raisebox{1.5pt}{$-$}$1. Also, as­sum­ing that the size of the atom is neg­li­gi­ble, in the sim­plest model the pho­ton will have zero or­bital an­gu­lar mo­men­tum around the cen­ter of the atom. That is most eas­ily un­der­stood us­ing clas­si­cal physics: a par­ti­cle that trav­els along a line that comes out of a point has zero an­gu­lar mo­men­tum around that point. For equiv­a­lent quan­tum ar­gu­ments, see {N.10} or {A.21.7}. It means in terms of quan­tum me­chan­ics that the pho­ton has a quan­tum num­ber $l_\gamma$ of or­bital an­gu­lar mo­men­tum that is zero. That makes the to­tal an­gu­lar mo­men­tum quan­tum num­ber $j_\gamma$ of the pho­ton equal to the spin $s_\gamma$, 1.

The nor­mal, ef­fi­cient kind of atomic tran­si­tion does in fact pro­duce a pho­ton like that. Since the term nor­mal is too nor­mal, such a tran­si­tion is called “al­lowed.” For rea­sons that will even­tu­ally be ex­cused for in sec­tion 7.7.2, al­lowed tran­si­tions are also more tech­ni­cally called “elec­tric di­pole” tran­si­tions. Ac­cord­ing to the above, then, the pho­ton net an­gu­lar mo­men­tum and par­ity are:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{for electric dipole transitions:}\qquad
j_\gamma = 1 \qquad \pi_\gamma = -1
$} %
\end{displaymath} (7.15)

Tran­si­tions that can­not hap­pen ac­cord­ing to the elec­tric di­pole mech­a­nism are called “for­bid­den.” That does not mean that these tran­si­tions can­not oc­cur at all; just for­bid your kids some­thing. But they are much more awk­ward, and there­fore nor­mally very much slower, than al­lowed tran­si­tions.

One im­por­tant case of a for­bid­den tran­si­tion is one in which the atomic an­gu­lar mo­men­tum changes by 2 or more units. Since the pho­ton has only 1 unit of spin, in such a tran­si­tion the pho­ton must have nonzero or­bital an­gu­lar mo­men­tum. Tran­si­tions in which the pho­ton has more than 1 unit of net an­gu­lar mo­men­tum are called “mul­ti­pole tran­si­tions.” For ex­am­ple, in a “quadru­pole” tran­si­tion, the net an­gu­lar mo­men­tum of the pho­ton $j_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2. In an oc­tu­pole tran­si­tion, $j_\gamma$ = 3 etcetera. In all these tran­si­tions, the pho­ton has at least $j_\gamma-1$ units of or­bital an­gu­lar mo­men­tum.

To roughly un­der­stand how or­bital an­gu­lar mo­men­tum arises, re­con­sider the sketch of the emis­sion process in fig­ure 7.3. As shown, the pho­ton has no or­bital an­gu­lar mo­men­tum around the cen­ter of the atom, clas­si­cally speak­ing. But the pho­ton does not have to come from ex­actly the cen­ter of the atom. If the atom has a typ­i­cal ra­dius $R$, then the pho­ton could come from a point at a dis­tance com­pa­ra­ble to $R$ away from the cen­ter. That will give it an or­bital an­gu­lar mo­men­tum of or­der $Rp$ around the cen­ter, where $p$ is the lin­ear mo­men­tum of the pho­ton. And ac­cord­ing to rel­a­tiv­ity, (1.2), the pho­ton’s mo­men­tum is re­lated to its en­ergy, which is in turn re­lated to its fre­quency by the Planck-Ein­stein re­la­tion. That makes the clas­si­cal or­bital an­gu­lar mo­men­tum of the pho­ton of or­der $R\hbar\omega$$\raisebox{.5pt}{$/$}$$c$. But $c$$\raisebox{.5pt}{$/$}$$\omega$ is the wave length $\lambda$ of the pho­ton, within a fac­tor $2\pi$. That fac­tor is not im­por­tant for a rough es­ti­mate. So the typ­i­cal clas­si­cal or­bital an­gu­lar mo­men­tum of the pho­ton is

\begin{displaymath}
L \sim \frac{R}{\lambda} \hbar
\end{displaymath}

The frac­tion is typ­i­cally small. For ex­am­ple, the wave length $\lambda$ of vis­i­ble light is about 5 000 Å and the size $R$ of an atom is about an Å. So the or­bital an­gu­lar mo­men­tum above is a very small frac­tion of $\hbar$.

But ac­cord­ing to quan­tum me­chan­ics, the or­bital an­gu­lar mo­men­tum can­not be a small frac­tion of $\hbar$. If the quan­tum num­ber $l_\gamma$ is zero, then so is the or­bital an­gu­lar mo­men­tum. And if $l_\gamma$ is 1, then the or­bital an­gu­lar mo­men­tum is $\sqrt{2}\hbar$. There is noth­ing in be­tween. The above clas­si­cal or­bital an­gu­lar mo­men­tum should be un­der­stood to mean that there is quan­tum un­cer­tainty in or­bital an­gu­lar mo­men­tum. That the pho­ton has al­most cer­tainly zero or­bital an­gu­lar mo­men­tum, but that there re­mains a small prob­a­bil­ity of $l_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. In par­tic­u­lar, if you take the ra­tio $R$$\raisebox{.5pt}{$/$}$$\lambda$ to be the co­ef­fi­cient of the $l_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 state, then the prob­a­bil­ity of the pho­ton com­ing out with $l_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is the square of that, $(R/\lambda)^2$. That will be a very small prob­a­bil­ity. But still, there is a slight prob­a­bil­ity that the net pho­ton an­gu­lar mo­men­tum $j_\gamma$ will be in­creased from 1 to 2 by a unit’s worth of or­bital an­gu­lar mo­men­tum. That will then pro­duce a quadru­pole tran­si­tion. And of course, two units of or­bital an­gu­lar mo­men­tum can in­crease the net pho­ton an­gu­lar mo­men­tum to $j_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, the oc­tu­pole level. But that re­duces the prob­a­bil­ity by an­other fac­tor $(R\lambda)^2$, so don’t hold you breath for these higher or­der mul­ti­pole tran­si­tions to oc­cur.

(If the above ran­dom mix­ture of un­jus­ti­fied clas­si­cal and quan­tum ar­gu­ments is too un­con­vinc­ing, there is a quan­tum ar­gu­ment in {N.10} that may be more be­liev­able. If you are brave, see {A.21.7} for a pre­cise analy­sis of the rel­e­vant pho­ton mo­menta and their prob­a­bil­i­ties in an in­ter­ac­tion with an atom or nu­cleus. But the bot­tom line is that the above ideas do de­scribe what hap­pens in tran­si­tion processes. That fol­lows from a com­plete analy­sis of the tran­si­tion process, as dis­cussed in later sec­tions and notes like {A.25} and {D.39}.)

So far, only elec­tric mul­ti­pole tran­si­tions have been dis­cussed, in which the elec­tro­mag­netic field at the atom is pri­mar­ily elec­tric. In mag­netic mul­ti­pole tran­si­tions how­ever, it is pri­mar­ily mag­netic. In a “mag­netic di­pole” tran­si­tion, the pho­ton comes out with one unit of net an­gu­lar mo­men­tum just like in an elec­tric di­pole one. How­ever, the par­ity of the pho­ton is now even:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{for magnetic dipole transitions:}\qquad
j_\gamma = 1 \qquad \pi_\gamma = 1
$} %
\end{displaymath} (7.16)

You might won­der how the pos­i­tive par­ity is pos­si­ble if the pho­ton has neg­a­tive in­trin­sic par­ity and no or­bital an­gu­lar mo­men­tum. The rea­son is that in a mag­netic di­pole tran­si­tion, the pho­ton does have a unit of or­bital an­gu­lar mo­men­tum. Re­call from the pre­vi­ous sub­sec­tion that it is quite pos­si­ble for one unit of spin and one unit of or­bital an­gu­lar mo­men­tum to com­bine into still only one unit of net an­gu­lar mo­men­tum.

In view of the crude dis­cus­sion of or­bital an­gu­lar mo­men­tum given above, this may still seem weird. How come that an atom of van­ish­ing size does sud­denly man­age to read­ily pro­duce a unit of or­bital an­gu­lar mo­men­tum in a mag­netic di­pole tran­si­tion? The ba­sic rea­son is that the mag­netic field acts in some way as if it has one unit of or­bital an­gu­lar mo­men­tum less than the pho­ton, {A.21.7}. It is un­pex­pect­edly strong at the atom. This al­lows a mag­netic atom state to get a solid grip on a pho­ton state of unit or­bital an­gu­lar mo­men­tum. It is some­what like hit­ting a rapidly spin­ning ball with a bat in base­ball; the re­sult­ing mo­tion of the ball can be weird. And in a sense the or­bital an­gu­lar mo­men­tum comes at the ex­pense of the spin; the net an­gu­lar mo­men­tum $j_\gamma$ of a pho­ton in a mag­netic di­pole tran­si­tion will not be 2 de­spite the or­bital an­gu­lar mo­men­tum.

Cer­tainly this sort of com­pli­ca­tions would not arise if the pho­ton had no spin. With­out dis­cus­sion, the pho­ton is one of the most ba­sic par­ti­cles in physics. But it is sur­pris­ingly com­plex for such an el­e­men­tary par­ti­cle. This also seems the right place to con­fess to the fact that elec­tric mul­ti­pole pho­tons have un­cer­tainty in or­bital an­gu­lar mo­men­tum. For ex­am­ple, an elec­tric di­pole pho­ton has a prob­a­bil­ity for $l_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 in ad­di­tion to $l_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. How­ever, this ad­di­tional or­bital an­gu­lar mo­men­tum comes cour­tesy of the in­ter­nal me­chan­ics, and es­pe­cially the spin, of the pho­ton. It does not give the pho­ton a prob­a­bil­ity for net an­gu­lar mo­men­tum $j_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2. So it does not re­ally change the given dis­cus­sion.

All else be­ing the same, the prob­a­bil­ity of a mag­netic di­pole tran­si­tion is nor­mally much smaller than an elec­tric di­pole one. The prin­ci­pal rea­son is that the mag­netic field is re­ally a rel­a­tivis­tic ef­fect. That can be un­der­stood, for ex­am­ple, from how the mag­netic field popped up in the de­scrip­tion of the rel­a­tivis­tic mo­tion of charged par­ti­cles, chap­ter 1.3.2. So you would ex­pect the ef­fect of the mag­netic field to be mi­nor un­less the atomic elec­tron or nu­cleon in­volved in the tran­si­tion has a ki­netic en­ergy com­pa­ra­ble to its rest mass en­ergy. In­deed, it turns out that the prob­a­bil­ity of a mag­netic tran­si­tion is smaller than an elec­tric one by a fac­tor of or­der $T$$\raisebox{.5pt}{$/$}$$mc^2$, where $T$ is the ki­netic en­ergy of the par­ti­cle and $mc^2$ its rest mass en­ergy, {A.25.4}. For the elec­tron in a hy­dro­gen atom, and for the outer elec­trons in atoms in gen­eral, this ra­tio is very much less than one. The same holds for the nu­cle­ons in nu­clei. It fol­lows that mag­netic di­pole tran­si­tions will nor­mally take place much slower than elec­tric di­pole ones.

In mag­netic mul­ti­pole tran­si­tions, the pho­ton re­ceives ad­di­tional an­gu­lar mo­men­tum. Like for elec­tric mul­ti­pole tran­si­tions, there is one ad­di­tional unit of an­gu­lar mo­men­tum for each ad­di­tional mul­ti­pole or­der. And there is a cor­re­spond­ing slow down of the tran­si­tions.


Ta­ble 7.1: Prop­er­ties of pho­tons emit­ted in elec­tric and mag­netic mul­ti­pole tran­si­tions.
\begin{table}\begin{displaymath}
\renewedcommand{arraystretch}{1.5}
\setlength{...
...}{$\geqslant$}}1$}}
\\ \hline\hline
\end{array} \end{displaymath}
\end{table}


Ta­ble 7.1 gives a sum­mary of the pho­ton prop­er­ties in mul­ti­pole tran­si­tions. It is con­ven­tional to write elec­tric mul­ti­pole tran­si­tions as ${\rm {E}}\ell$ and mag­netic ones as ${\rm {M}}\ell$ where $\ell$, (or L, but never $j$), is the net pho­ton an­gu­lar mo­men­tum $j_\gamma$. So an elec­tric di­pole tran­si­tion is $\rm {E}1$ and a mag­netic di­pole one $\rm {M}1$. In agree­ment with the pre­vi­ous sec­tion, each unit in­crease in the or­bital an­gu­lar mo­men­tum pro­duces an ad­di­tional fac­tor $\vphantom{0}\raisebox{1.5pt}{$-$}$1 in par­ity.

The col­umn slow down gives an or­der of mag­ni­tude es­ti­mate by what fac­tor a tran­si­tion is slower than an elec­tric di­pole one, all else be­ing equal. Note how­ever that all else is def­i­nitely not equal, so these fac­tors should not be used even for ball­parks.

There are some of­fi­cial ball­parks for atomic nu­clei based on a more de­tailed analy­sis. These are called the Weis­skopf and Moszkowski es­ti­mates, chap­ter 14.20.4 and in par­tic­u­lar ad­den­dum {A.25.8}. But even there you should not be sur­prised if the ball­park is off by or­ders of mag­ni­tude. These es­ti­mates do hap­pen to work fine for the non­rel­a­tivis­tic hy­dro­gen atom, with ap­pro­pri­ate ad­just­ments, {A.25.8}.

The slow down fac­tors $T$$\raisebox{.5pt}{$/$}$$mc^2$ and $(R/\lambda)^2$ are of­ten quite com­pa­ra­ble. That makes the or­der of slow down of mag­netic di­pole tran­si­tions sim­i­lar to that of elec­tric quadru­pole tran­si­tions. To see the equiv­a­lence of the slow-down fac­tors, rewrite them as

\begin{displaymath}
\left(\frac{R}{\lambda}\right)^2 = \frac{1}{\hbar^2 c^2} R^...
...{T}{m c^2} = \frac{1}{\hbar^2 c^2} R^2 T \frac{\hbar^2}{m R^2}
\end{displaymath}

where the $2\pi$ in the wave length was again put back. For an atom, the en­ergy of the emit­ted pho­ton $\hbar\omega$ is of­ten com­pa­ra­ble to the ki­netic en­ergy $T$ of the outer elec­trons, and the fi­nal ra­tio in the equa­tions above is a rough es­ti­mate for that ki­netic en­ergy. It fol­lows that the two slow down fac­tors are com­pa­ra­ble. An­other way of look­ing at the sim­i­lar­ity be­tween mag­netic di­pole and elec­tric quadru­pole tran­si­tions will be given in {D.39}.

Note from the ta­ble that elec­tric quadru­pole and mag­netic di­pole tran­si­tions have the same par­ity. That means that they may com­pete di­rectly with each other on the same tran­si­tion, pro­vided that the atomic an­gu­lar mo­men­tum does not change more than one unit in that tran­si­tion.

For nu­clei, the pho­ton en­ergy tends to be sig­nif­i­cantly less than the nu­cleon ki­netic en­ergy. That is one rea­son that the Weis­skopf es­ti­mates have the elec­tric quadru­pole tran­si­tions a lot slower than mag­netic di­pole ones for typ­i­cal tran­si­tions. Also note that the ki­netic en­ergy es­ti­mate above does not in­clude the ef­fect of the ex­clu­sion prin­ci­ple. Ex­clu­sion raises the true ki­netic en­ergy if there are mul­ti­ple iden­ti­cal par­ti­cles in a given vol­ume.

There is an­other is­sue that should be men­tioned here. Mag­netic tran­si­tions have a ten­dency to un­der­per­form for sim­ple sys­tems like the hy­dro­gen atom. For these sys­tems, the mag­netic field has dif­fi­culty mak­ing ef­fec­tive use of spin in chang­ing the atomic or nu­clear struc­ture. That is dis­cussed in more de­tail in the next sub­sec­tion.

One very im­por­tant ad­di­tional prop­erty must still be men­tioned. The pho­ton can­not have zero net an­gu­lar mo­men­tum. Nor­mally it is cer­tainly pos­si­ble for a par­ti­cle with spin $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and or­bital an­gu­lar mo­men­tum quan­tum num­ber $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 to be in a state that has zero net an­gu­lar mo­men­tum, $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. How­ever, a pho­ton is not a nor­mal par­ti­cle; it is a rel­a­tivis­tic par­ti­cle with zero rest mass that can only move at the speed of light. It turns out that for a pho­ton, spin and or­bital an­gu­lar mo­men­tum are not in­de­pen­dent, but in­trin­si­cally linked. This lim­i­ta­tion pre­vents a state where the pho­ton has zero net an­gu­lar mo­men­tum, {A.21.3}.

There are some ef­fects in clas­si­cal physics that are re­lated to this lim­i­ta­tion. First of all, con­sider a pho­ton with def­i­nite lin­ear mo­men­tum. That cor­re­sponds to a light wave prop­a­gat­ing in a par­tic­u­lar di­rec­tion. Now lin­ear and an­gu­lar mo­men­tum do not com­mute, so such a pho­ton will not have def­i­nite an­gu­lar mo­men­tum. How­ever, the an­gu­lar mo­men­tum com­po­nent in the di­rec­tion of mo­tion is still well de­fined. The lim­i­ta­tion on pho­tons is in this case that the pho­ton must ei­ther have an­gu­lar mo­men­tum $\hbar$ or $\vphantom{0}\raisebox{1.5pt}{$-$}$$\hbar$ along the di­rec­tion of mo­tion. A nor­mal par­ti­cle of spin 1 could also have zero an­gu­lar mo­men­tum in the di­rec­tion of mo­tion, but a pho­ton can­not. The two states of def­i­nite an­gu­lar mo­men­tum in the di­rec­tion of mo­tion are called “right- and left-cir­cu­larly po­lar­ized” light, re­spec­tively.

Sec­ond, for the same type of pho­ton, there are two equiv­a­lent states that have def­i­nite di­rec­tions of the elec­tric and mag­netic fields. These states have un­cer­tainty in an­gu­lar mo­men­tum in the di­rec­tion of mo­tion. They are called “lin­early po­lar­ized” light. These states il­lus­trate that there can­not be an elec­tric or mag­netic field com­po­nent in the di­rec­tion of mo­tion. The elec­tric and mag­netic fields are nor­mal to the di­rec­tion of mo­tion, and to each other.

More gen­eral pho­tons of def­i­nite lin­ear mo­men­tum may have un­cer­tainty in both of the men­tioned prop­er­ties. But still there is zero prob­a­bil­ity for zero an­gu­lar mo­men­tum in the di­rec­tion of mo­tion, and zero prob­a­bil­ity for a field in the di­rec­tion of mo­tion.

Third, di­rectly re­lated to the pre­vi­ous case. Sup­pose you have a charge dis­tri­b­u­tion that is spher­i­cally sym­met­ric, but pul­sat­ing in the ra­dial di­rec­tion. You would ex­pect that you would get a fluc­tu­at­ing ra­dial elec­tri­cal field out­side this pul­sat­ing charge. But you do not, it does not ra­di­ate en­ergy. Such ra­di­a­tion would have the elec­tric field in the di­rec­tion of mo­tion, and that does not hap­pen. Now con­sider the tran­si­tion from the spher­i­cally sym­met­ric 2s state of a hy­dro­gen atom to the spher­i­cal sym­met­ric 1s state. Be­cause of the lack of spher­i­cally sym­met­ric ra­di­a­tion, you might guess that this tran­si­tion is in trou­ble. And it is; that is dis­cussed in the next sub­sec­tion.

In fact, the last ex­am­ple is di­rectly re­lated to the miss­ing state of zero an­gu­lar mo­men­tum of the pho­ton. Re­call from sec­tion 7.3 that an­gu­lar mo­men­tum is re­lated to an­gu­lar sym­me­try. In par­tic­u­lar, a state of zero an­gu­lar mo­men­tum (if ex­act to quan­tum ac­cu­racy) looks the same when seen from all di­rec­tions. The fact that there is no spher­i­cally sym­met­ric ra­di­a­tion is then just an­other way of say­ing that the pho­ton can­not have zero an­gu­lar mo­men­tum.


Key Points
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\end{picture}$
Nor­mal atomic tran­si­tions are called al­lowed or elec­tric di­pole ones. All oth­ers are called for­bid­den but can oc­cur just fine.

$\begin{picture}(15,5.5)(0,-3)
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In elec­tric di­pole tran­si­tions the emit­ted pho­ton has an­gu­lar mo­men­tum quan­tum num­ber $j_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and neg­a­tive par­ity $\pi_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$1.

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In the slower mag­netic di­pole tran­si­tions the pho­ton par­ity is pos­i­tive, $\pi_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

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Each higher mul­ti­pole or­der adds a unit to the pho­ton an­gu­lar mo­men­tum quan­tum num­ber $j_\gamma$ and flips over the par­ity $\pi_\gamma$.

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The higher the mul­ti­pole or­der, the slower the tran­si­tion will be.


7.4.4 Se­lec­tion rules

As dis­cussed, a given ex­cited atomic state may be able to tran­si­tion to a lower en­ergy state by emit­ting a pho­ton. But many tran­si­tions from a higher en­ergy state to a lower one sim­ply do not hap­pen. There are so-called se­lec­tion rules that pre­dict whether or not a given tran­si­tion process is pos­si­ble. This sub­sec­tion gives a brief in­tro­duc­tion to these rules.

The pri­mary con­sid­ered sys­tem will be the hy­dro­gen atom. How­ever, some gen­er­ally valid rules are given at the end. It will usu­ally be as­sumed that the ef­fect of the spin of the elec­tron on its mo­tion can be ig­nored. That is the same ap­prox­i­ma­tion as used in chap­ter 4.3, and it is quite ac­cu­rate. Ba­si­cally, the model sys­tem stud­ied is a spin­less charged elec­tron go­ing around a sta­tion­ary pro­ton. Spin will be tacked on af­ter the fact.

The se­lec­tion rules re­sult from the con­ser­va­tion laws and pho­ton prop­er­ties as dis­cussed in the pre­vi­ous two sub­sec­tions. Since the con­ser­va­tion laws are ap­plied to a spin­less elec­tron, the an­gu­lar mo­men­tum of the elec­tron is sim­ply its or­bital an­gu­lar mo­men­tum. That means that for the atomic states, the an­gu­lar mo­men­tum quan­tum num­ber $j$ be­comes the or­bital an­gu­lar mo­men­tum quan­tum num­ber $l$. For the emit­ted pho­ton, the true net an­gu­lar mo­men­tum quan­tum num­ber $\ell$ must be used.

Now sup­pose that the ini­tial high-en­ergy atomic state has an or­bital an­gu­lar mo­men­tum quan­tum num­ber $l_{\rm {H}}$ and that it emits a pho­ton with an­gu­lar mo­men­tum quan­tum num­ber $\ell$. The ques­tion is then what can be said about the or­bital an­gu­lar mo­men­tum $l_{\rm {L}}$ of the atomic state of lower en­ergy af­ter the tran­si­tion. The an­swer is given by sub­sec­tion 7.4.2 (7.11):

\begin{displaymath}
l_{\rm {L}} =
\vert l_{\rm {H}}{-}\ell\vert\mbox{, }
\ver...
...{, }
l_{\rm {H}}{+}\ell{-}1\mbox{, or }
l_{\rm {H}}{+}\ell %
\end{displaymath} (7.17)

That leads im­me­di­ately to a stun­ning con­clu­sion for the de­cay of the hy­dro­gen $\psi_{200}$ 2s state. This state has an­gu­lar mo­men­tum $l_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, as any s state. So the re­quire­ment above sim­pli­fies to $l_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell$. Now re­call from the pre­vi­ous sub­sec­tion that a pho­ton must have $\ell$ at least equal to 1. So $l_{\rm {L}}$ must be at least 1. But $l_{\rm {L}}$ can­not be at least 1. The only lower en­ergy state that ex­ists is the $\psi_{100}$ 1s ground state. It has $l_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. So the 2s state can­not de­cay!

Never say never, of course. It turns out that if left alone, the 2s state will even­tu­ally de­cay through the emis­sion of two pho­tons, rather than a sin­gle one. This takes for­ever on quan­tum scales; the 2s state sur­vives for about a tenth of a sec­ond rather than maybe a nanosec­ond for a nor­mal tran­si­tion. Also, to ac­tu­ally ob­serve the two-pho­ton emis­sion process, the atom must be in high vac­uum. Oth­er­wise the 2s state would be messed up by col­li­sions with other par­ti­cles long be­fore it could de­cay. Now you see why the in­tro­duc­tion to this sec­tion gave a 2p state, and not the seem­ingly more sim­ple 2s one, as a sim­ple ex­am­ple of an atomic state that de­cays by emit­ting a pho­ton.

Based on the pre­vi­ous sub­sec­tion, you might won­der why a sec­ond pho­ton can suc­ceed where a unit of pho­ton or­bital an­gu­lar mo­men­tum can­not. Af­ter all, pho­tons have only two in­de­pen­dent spin states, while a unit of or­bital an­gu­lar mo­men­tum has the full set of three. The ex­pla­na­tion is that in re­al­ity you can­not add a suit­able unit of or­bital an­gu­lar mo­men­tum to a pho­ton; the or­bital and spin an­gu­lar mo­men­tum of a pho­ton are in­trin­si­cally linked. But pho­tons do have com­plete sets of states with an­gu­lar mo­men­tum $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, {A.21.7}. For two pho­tons, these can com­bine into zero net an­gu­lar mo­men­tum.

It is cus­tom­ary to ex­plain pho­tons in terms of states of def­i­nite lin­ear mo­men­tum. That is in fact what was done in the fi­nal para­graphs of the pre­vi­ous sub­sec­tion. But it is sim­plis­tic. It is def­i­nitely im­pos­si­ble to un­der­stand how two pho­tons, each miss­ing the state of zero an­gu­lar mo­men­tum along their di­rec­tion of mo­tion, could com­bine into a state of zero net an­gu­lar mo­men­tum. In fact, they sim­ply can­not. Lin­ear and or­bital an­gu­lar mo­men­tum do not com­mute. But pho­tons do not have to be in quan­tum states of def­i­nite lin­ear mo­men­tum. They can be, and of­ten are, in quan­tum su­per­po­si­tions of such states. The states of def­i­nite an­gu­lar mo­men­tum are quan­tum su­per­po­si­tions of in­fi­nitely many states of lin­ear mo­men­tum in all di­rec­tions. To make sense out of that, you need to switch to a de­scrip­tion in terms of pho­ton states of def­i­nite an­gu­lar, rather than lin­ear, mo­men­tum. Those states are listed in {A.21.7}. Un­for­tu­nately, they are much more dif­fi­cult to de­scribe phys­i­cally than states of def­i­nite lin­ear mo­men­tum.

It should also be noted that if you in­clude rel­a­tivis­tic ef­fects, the 2s state can ac­tu­ally de­cay to the 2p state that has net an­gu­lar mo­men­tum (spin plus or­bital) $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. This 2p state has very slightly lower en­ergy than the 2s state due to a tiny rel­a­tivis­tic ef­fect called Lamb shift, {A.39.4}. But be­cause of the neg­li­gi­ble dif­fer­ence in en­ergy, such a tran­si­tion is even slower than two-pho­ton emis­sion. It takes over 100 years to have a 50/50 prob­a­bil­ity for the tran­si­tion.

Also, in­clud­ing rel­a­tivis­tic efects, a mag­netic di­pole tran­si­tion is pos­si­ble. An atomic state with net an­gu­lar mo­men­tum $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ (due to the spin) can de­cay to a state with again net an­gu­lar mo­men­tum $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ by emit­ting a pho­ton with an­gu­lar mo­men­tum $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, sub­sec­tion 7.4.2. A mag­netic ${\rm {M}}1$ tran­si­tion is needed in or­der that the par­ity stays the same. Un­for­tu­nately, in the non­rel­a­tivis­tic ap­prox­i­ma­tion an ${\rm {M}}1$ tran­si­tion does not change the or­bital mo­tion; it just flips over the spin. Also, with­out any en­ergy change the the­o­ret­i­cal tran­si­tion rate will be zero, sec­tion 7.6.1.

Rel­a­tivis­tic ef­fects re­move these ob­sta­cles. But since these ef­fects are very small, the one-pho­ton tran­si­tion does take sev­eral days, so it is again much slower than two-pho­ton emis­sion. In this case, it may be use­ful to think in terms of the com­plete atom, in­clud­ing the pro­ton spin. The elec­tron and pro­ton can com­bine their spins into a sin­glet state with zero net an­gu­lar mo­men­tum or a triplet state with one unit of net mo­men­tum. The pho­ton takes one unit of an­gu­lar mo­men­tum away, turn­ing a triplet state into a sin­glet state or vice-versa. If the atom ends up in a 1s triplet state, it will take an­other 10 mil­lion year or so to de­cay to the sin­glet state, the true ground state.

For ex­cited atomic states in gen­eral, dif­fer­ent types of tran­si­tions may be pos­si­ble. As dis­cussed in the pre­vi­ous sub­sec­tion, the nor­mal type is called an al­lowed, “elec­tric di­pole,” or $\rm {E}1$ tran­si­tion.

Yes, every one of these three names is con­fus­ing. Non­al­lowed tran­si­tions, called for­bid­den tran­si­tions, are per­fectly al­lowed and they do oc­cur. They are typ­i­cally just a lot slower. The atomic states be­tween which the tran­si­tions oc­cur do not have elec­tric di­pole mo­ments. And how many peo­ple re­ally know what an elec­tric di­pole is? And $\rm {E}1$ is just cryp­tic. $\rm {E}0$ would have been more in­tu­itive, as it in­di­cates the level to which the tran­si­tion is for­bid­den. Who cares about pho­ton an­gu­lar mo­men­tum?

The one good thing that can be said is that in the elec­tric di­pole ap­prox­i­ma­tion, the atom does in­deed re­spond to the elec­tric part of the elec­tro­mag­netic field. In such tran­si­tions the pho­ton comes out with one unit of an­gu­lar mo­men­tum, i.e. $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, and neg­a­tive par­ity. Then the se­lec­tion rules are:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{E1:}
\qquad l_{\rm{L}} = l_{\r...
...m_{l,\rm{H}}{\pm}1
\qquad m_{s,{\rm{L}}} = m_{s,{\rm{H}}}
$}
\end{displaymath} (7.18)

The first rule re­flects the pos­si­ble or­bital an­gu­lar mo­men­tum val­ues as given above. To be sure, these val­ues also al­low $l$ to stay the same. How­ever, since the par­ity of the pho­ton is neg­a­tive, par­ity con­ser­va­tion re­quires that the par­ity of the atom must change, sub­sec­tion 7.4.2. And that means that the or­bital an­gu­lar mo­men­tum quan­tum num­ber $l$ must change from odd to even or vice-versa. It can­not stay the same.

The sec­ond rule gives the pos­si­ble mag­netic quan­tum num­bers. Re­call that these are a di­rect mea­sure for the an­gu­lar mo­men­tum in the cho­sen $z$-​di­rec­tion. Since the pho­ton mo­men­tum $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, the pho­ton $z$ mo­men­tum $m_\gamma$ can be $\vphantom{0}\raisebox{1.5pt}{$-$}$1, 0, or 1. So the pho­ton can change the atomic $z$ mo­men­tum by up to one unit, as the se­lec­tion rule says. Note that while the pho­ton an­gu­lar mo­men­tum can­not be zero in the di­rec­tion of its mo­tion, the di­rec­tion of mo­tion is not nec­es­sar­ily the $z$-​di­rec­tion. In essence the pho­ton may be com­ing off side­ways. (The bet­ter way of think­ing about this is in terms of pho­ton states of def­i­nite an­gu­lar mo­men­tum. These can have the an­gu­lar mo­men­tum zero in the $z$-​di­rec­tion, while the di­rec­tion of pho­ton mo­tion is un­cer­tain.)

The fi­nal se­lec­tion rule says that the elec­tron spin in the $z$-​di­rec­tion does not change. That re­flects the fact that the elec­tron spin does not re­spond to an elec­tric field in a non­rel­a­tivis­tic ap­prox­i­ma­tion. (Of course, you might ar­gue that in a non­rel­a­tivis­tic ap­prox­i­ma­tion, the elec­tron should not have spin in the first place, chap­ter 12.12.)

Note that ig­nor­ing rel­a­tivis­tic ef­fects in tran­si­tions is a tricky busi­ness. Even a small ef­fect, given enough time to build up, might pro­duce a tran­si­tion where one was not pos­si­ble be­fore. In a more so­phis­ti­cated analy­sis of the hy­dro­gen atom, ad­den­dum {A.39}, there is a slight in­ter­ac­tion be­tween the or­bital an­gu­lar mo­men­tum of the elec­tron and its spin. That is known as spin-or­bit in­ter­ac­tion. Note that the s states have no or­bital an­gu­lar mo­men­tum for the spin to in­ter­act with.

As a re­sult of spin-or­bit in­ter­ac­tion the cor­rect en­ergy eigen­func­tions, ex­cept the s states, de­velop un­cer­tainty in the val­ues of both $m_l$ and $m_s$. In other words, the $z$ com­po­nents of both the or­bital and the spin an­gu­lar mo­menta have un­cer­tainty. That im­plies that the above rules are no longer re­ally right. The en­ergy eigen­func­tions do keep def­i­nite val­ues for $l$, rep­re­sent­ing the mag­ni­tude of or­bital an­gu­lar mo­men­tum, for $j$, rep­re­sent­ing the mag­ni­tude of net an­gu­lar mo­men­tum, or­bital plus spin, and $m_j$ rep­re­sent­ing the net an­gu­lar mo­men­tum in the $z$-​di­rec­tion. In those terms the mod­i­fied se­lec­tion rules be­come

\begin{displaymath}
\fbox{$\displaystyle
\mbox{E1$_{\rm so}$:}
\qquad l_{\rm{...
...\rm{L}}} = m_{j,{\rm{H}}} \mbox{ or } m_{j,{\rm{H}}} \pm 1
$}
\end{displaymath} (7.19)

The last two se­lec­tion rules above are a di­rect con­se­quence of an­gu­lar mo­men­tum con­ser­va­tion; since the pho­ton has $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, it can change each atomic quan­tum num­ber by at most one unit. In the first se­lec­tion rule, an­gu­lar mo­men­tum con­ser­va­tion could in prin­ci­ple al­low a change in $l$ by 2 units. A change in elec­tron spin could add to the pho­ton an­gu­lar mo­men­tum. But par­ity con­ser­va­tion re­quires that $l$ changes by an odd amount and 2 is not odd.

If the se­lec­tion rules are not sat­is­fied, the tran­si­tion is called for­bid­den. How­ever, the tran­si­tion may still oc­cur through a dif­fer­ent mech­a­nism. One pos­si­bil­ity is a slower mag­netic di­pole tran­si­tion, in which the elec­tron in­ter­acts with the mag­netic part of the elec­tro­mag­netic field. That in­ter­ac­tion oc­curs be­cause an elec­tron has spin and or­bital an­gu­lar mo­men­tum. A charged par­ti­cle with an­gu­lar mo­men­tum be­haves like a lit­tle elec­tro­mag­net and wants to align it­self with an am­bi­ent mag­netic field, chap­ter 13.4. The se­lec­tion rules in this case are

\begin{displaymath}
\fbox{$\displaystyle
\mbox{M1:}
\qquad l_{\rm{L}} = l_{\r...
...\rm{L}}} = m_{s,{\rm{H}}} \mbox{ or } m_{s,{\rm{H}}} \pm 1
$}
\end{displaymath} (7.20)

The rea­sons are sim­i­lar to the elec­tric di­pole case, tak­ing into ac­count that the pho­ton comes out with pos­i­tive par­ity rather than neg­a­tive. Also, the elec­tron spin def­i­nitely in­ter­acts with a mag­netic field. A more de­tailed analy­sis will show that ex­actly one of the two mag­netic quan­tum num­bers $m_l$ and $m_s$ must change, {D.39}.

It must be pointed out that an ${\rm {M}}1$ tran­si­tion is triv­ial for an hy­dro­gen atom in the non­rel­a­tivis­tic ap­prox­i­ma­tion. All the tran­si­tion does is change the di­rec­tion of the or­bital or spin an­gu­lar mo­men­tum vec­tor, {D.39}. Not only is this ho-hum, the rate of tran­si­tions will be van­ish­ingly small since it de­pends on the en­ergy re­lease in the tran­si­tion. The same prob­lem ex­ists more gen­er­ally for charged par­ti­cles in ra­dial po­ten­tials that only de­pend on po­si­tion, {A.25.8}.

Rel­a­tivis­tic ef­fects can change this. In par­tic­u­lar, in the pres­ence of spin-or­bit cou­pling, the se­lec­tion rules be­come

\begin{displaymath}
\fbox{$\displaystyle
\mbox{M1$_{\rm so}$:}
\qquad l_{\rm{...
...\rm{L}}} = m_{j,{\rm{H}}} \mbox{ or } m_{j,{\rm{H}}}{\pm}1
$}
\end{displaymath} (7.21)

In this case, it is less ob­vi­ous why $l$ could not change by 2 units. The ba­sic rea­son is that the mag­netic field wants to ro­tate the or­bital an­gu­lar mo­men­tum vec­tor, rather than change its mag­ni­tude, {D.39}. (Note how­ever that that de­riva­tion, and this book in gen­eral, uses a non­rel­a­tivis­tic Hamil­ton­ian for the in­ter­ac­tion be­tween the spin and the mag­netic field.) Sim­i­lar lim­i­ta­tions ap­ply for mag­netic tran­si­tions of higher mul­ti­pole or­der, {A.25.5} (A.175).

In higher-or­der mul­ti­pole tran­si­tions the pho­ton comes out with an­gu­lar mo­men­tum $\ell$ greater than 1. As the pre­vi­ous sub­sec­tion noted, this slows down the tran­si­tions. The fastest mul­ti­pole tran­si­tions are the elec­tric quadru­pole ones. In these tran­si­tions the emit­ted pho­ton has $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 and pos­i­tive par­ity. The se­lec­tion rules are then

\begin{displaymath}
\fbox{$\displaystyle
\mbox{E2:}
\quad\; l_{\rm{L}} = l_{\...
..._{l,\rm{H}}{\pm}2
\quad\; m_{s,{\rm{L}}} = m_{s,{\rm{H}}}
$}
\end{displaymath} (7.22)

In ad­di­tion $l_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is not pos­si­ble for such tran­si­tions. Nei­ther is $l_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $l_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 or vice-versa. In­clud­ing elec­tron spin, $j_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$ is not pos­si­ble. The rea­sons are sim­i­lar to the ones be­fore.

Mag­netic tran­si­tions at higher mul­ti­pole or­ders have sim­i­lar prob­lems as the mag­netic di­pole one. In par­tic­u­lar, con­sider the or­bital an­gu­lar mo­men­tum se­lec­tion rule (7.17) above. The low­est pos­si­ble mul­ti­pole or­der in the non­rel­a­tivis­tic case is

\begin{displaymath}
\ell_{\rm min} = \vert l_{\rm {H}}- l_{\rm {L}}\vert
\end{displaymath}

Be­cause of par­ity, that is al­ways an elec­tric mul­ti­pole tran­si­tion. (This ex­cludes the case that the or­bital an­gu­lar mo­menta are equal, in which case the low­est tran­si­tion is the al­ready dis­cussed triv­ial mag­netic di­pole one.)

The bot­tom line is that mag­netic tran­si­tions sim­ply can­not com­pete. Of course, con­ser­va­tion of net an­gu­lar mo­men­tum might for­bid the elec­tric tran­si­tion to a given fi­nal state. But in that case there will be an equiv­a­lent state that dif­fers only in spin to which the elec­tric tran­si­tion can pro­ceed just fine.

How­ever, for a multi-elec­tron atom or nu­cleus in an in­de­pen­dent-par­ti­cle model, that equiv­a­lent state might al­ready be oc­cu­pied by an­other par­ti­cle. Or there may be enough spin-or­bit in­ter­ac­tion to raise the en­ergy of the equiv­a­lent state to a level that tran­si­tion to it be­comes im­pos­si­ble. In that case, the low­est pos­si­ble tran­si­tion will be a mag­netic one.

Con­sider now more gen­eral sys­tems than hy­dro­gen atoms. Gen­eral se­lec­tion rules for elec­tric $\rm {E}\ell$ and mag­netic $\rm {M}\ell$ tran­si­tions are:

\begin{displaymath}
\fbox{$\displaystyle
\,\mbox{E$\ell$:} \quad
\vert j_{\rm...
...}
\qquad (\ell \mathrel{\raisebox{-1pt}{$\geqslant$}}1)
$} %
\end{displaymath} (7.23)


\begin{displaymath}
\fbox{$\displaystyle
\mbox{M$\ell$:} \quad
\vert j_{\rm{H...
...}
\qquad (\ell \mathrel{\raisebox{-1pt}{$\geqslant$}}1)
$} %
\end{displaymath} (7.24)

These rules rely only on the spin and par­ity of the emit­ted pho­ton. So they are quite gen­er­ally valid for one-pho­ton emis­sion.

If a nor­mal elec­tric di­pole tran­si­tion is pos­si­ble for an atomic or nu­clear state, it will most likely de­cay that way be­fore any other type of tran­si­tion can oc­cur. But if an elec­tric di­pole tran­si­tion is for­bid­den, other types of tran­si­tions may ap­pear in sig­nif­i­cant amounts. If both elec­tric quadru­pole and mag­netic di­pole tran­si­tions are pos­si­ble, they may be com­pet­i­tive. And elec­tric quadru­pole tran­si­tions can pro­duce two units of change in the atomic an­gu­lar mo­men­tum, rather than just one like the mag­netic di­pole ones.

Given the ini­tial state, of­ten the ques­tion is not what fi­nal states are pos­si­ble, but what tran­si­tion types are pos­si­ble given the fi­nal state. In that case, the gen­eral se­lec­tion rules can be writ­ten as

\begin{displaymath}
\fbox{$\displaystyle
\vert j_{\rm{H}} - j_{\rm{L}}\vert \m...
...x{: electric}\\ -1\mbox{: magnetic}
\end{array} \right.
$} %
\end{displaymath} (7.25)

Since tran­si­tion rates de­crease rapidly with in­creas­ing mul­ti­pole or­der $\ell$, nor­mally the low­est value of $\ell$ al­lowed will be the im­por­tant one. That is

\begin{displaymath}
\fbox{$\displaystyle
\ell_{\rm min} = \vert j_{\rm{H}} - j...
... $\vphantom0\raisebox{1.5pt}{$=$}${} 0 is not possible.}
$} %
\end{displaymath} (7.26)

If par­ity makes the cor­re­spond­ing tran­si­tion mag­netic, the next-higher or­der elec­tric tran­si­tion may well be of im­por­tance too.


Key Points
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Nor­mal atomic tran­si­tions are called elec­tric di­pole ones, or al­lowed ones, or $\rm {E}1$ ones. Un­for­tu­nately.

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The quan­tum num­bers of the ini­tial and fi­nal atomic states in tran­si­tions must sat­isfy cer­tain se­lec­tion rules in or­der for tran­si­tions of a given type to be pos­si­ble.

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If a tran­si­tion does not sat­isfy the rules of elec­tric di­pole tran­si­tions, it will have to pro­ceed by a slower mech­a­nism. That could be a mag­netic di­pole tran­si­tion or an elec­tric or mag­netic mul­ti­pole tran­si­tion.

$\begin{picture}(15,5.5)(0,-3)
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A state of zero an­gu­lar mo­men­tum can­not de­cay to an­other state of zero an­gu­lar mo­men­tum through any of these mech­a­nisms. For such tran­si­tions, two-pho­ton emis­sion is an op­tion.