Sub­sec­tions


7.7 Ab­sorp­tion and Stim­u­lated Emis­sion

This sec­tion will ad­dress the ba­sic physic of ab­sorp­tion and emis­sion of ra­di­a­tion by a gas of atoms in an elec­tro­mag­netic field. The next sec­tion will give prac­ti­cal for­mu­lae.

Fig­ure 7.9: Emis­sion and ab­sorp­tion of ra­di­a­tion by an atom.
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... \put(199,-5){\makebox(0,0){$E_\gamma=\hbar\omega$}}
\end{picture}
\end{figure}

Fig­ure 7.9 shows the three dif­fer­ent processes of in­ter­est. The pre­vi­ous sec­tions al­ready dis­cussed the process of spon­ta­neous emis­sion. Here an atom in a state $\psi_{\rm {H}}$ of high en­ergy emits a pho­ton of elec­tro­mag­netic ra­di­a­tion and re­turns to an atomic state $\psi_{\rm {L}}$ of lower en­ergy. For ex­am­ple, for a hy­dro­gen atom the ex­cited state $\psi_{\rm {H}}$ might be the $\psi_{210}$ 2p$_z$ state, and the lower en­ergy state $\psi_{\rm {L}}$ the $\psi_{100}$ 1s ground state, as de­fined in chap­ter 4.3.

To a su­perb ap­prox­i­ma­tion, the pho­ton car­ries off the dif­fer­ence in en­ergy be­tween the atomic states. In view of the Planck-Ein­stein re­la­tion, that means that its fre­quency $\omega$ is given by

\begin{displaymath}
\hbar\omega = E_{\rm {H}} - E_{\rm {L}}
\end{displaymath}

Un­for­tu­nately, the dis­cus­sion of spon­ta­neous emis­sion in the pre­vi­ous sec­tions had to re­main in­com­plete. Non­rel­a­tivis­tic quan­tum me­chan­ics as cov­ered in this book can­not ac­com­mo­date the cre­ation of new par­ti­cles like the pho­ton in this case. The num­ber of par­ti­cles has to stay the same.

The sec­ond process of in­ter­est in fig­ure 7.9 is ab­sorp­tion. Here an atom in a low en­ergy state $\psi_{\rm {L}}$ in­ter­acts with an ex­ter­nal elec­tro­mag­netic field. The atom picks up a pho­ton from the field, which al­lows it to en­ter an ex­cited en­ergy state $\psi_{\rm {H}}$. Un­like spon­ta­neous emis­sion, this process can rea­son­ably be de­scribed us­ing non­rel­a­tivis­tic quan­tum me­chan­ics. The trick is to ig­nore the pho­ton ab­sorbed from the elec­tro­mag­netic field. In that case, the elec­tro­mag­netic field can be ap­prox­i­mated as a known one, us­ing clas­si­cal elec­tro­mag­net­ics. Af­ter all, if the field has many pho­tons, one more or less is not go­ing to make a dif­fer­ence.

The third process is stim­u­lated emis­sion. In this case an atom in an ex­cited state $\psi_{\rm {H}}$ in­ter­acts with an elec­tro­mag­netic field. And now the atom does not do the log­i­cal thing; it does not pick up a pho­ton to go to a still more ex­cited state. In­stead it uses the pres­ence of the elec­tro­mag­netic field as an ex­cuse to dump a pho­ton and re­turn to a lower en­ergy state $\psi_{\rm {L}}$.

This process is the op­er­at­ing prin­ci­ple of lasers. Sup­pose that you bring a large num­ber of atoms into a rel­a­tively sta­ble ex­cited state. Then sup­pose that one of the atoms per­forms a spon­ta­neous emis­sion. The pho­ton re­leased by that atom can stim­u­late an­other ex­cited atom to re­lease a pho­ton too. Then there are two co­her­ent pho­tons, which can go on to stim­u­late still more ex­cited atoms to re­lease still more pho­tons. And so on in an avalanche ef­fect. It can pro­duce a run­away process of pho­ton re­lease in which a macro­scopic amount of mono­chro­matic, co­her­ent light is cre­ated.

Masers work on the same prin­ci­ple, but the ra­di­a­tion is of much lower en­ergy than vis­i­ble light. It is there­fore usu­ally re­ferred to as mi­crowaves in­stead of light. The am­mo­nia mol­e­cule is one pos­si­ble source of such low en­ergy ra­di­a­tion, chap­ter 5.3.

The analy­sis in this sec­tion will il­lu­mi­nate some of the de­tails of stim­u­lated emis­sion. For ex­am­ple, it turns out that pho­ton ab­sorp­tion by the lower en­ergy atoms, fig­ure 7.9(b), com­petes on a per­fectly equal foot­ing with stim­u­lated emis­sion, fig­ure 7.9(c). If you have a 50/50 mix­ture of atoms in the ex­cited state $\psi_{\rm {H}}$ and the lower en­ergy state $\psi_{\rm {L}}$, just as many pho­tons will be cre­ated by stim­u­lated emis­sion as will be ab­sorbed. So no net light will be pro­duced. To get a laser to work, you must ini­tially have a “pop­u­la­tion in­ver­sion;” you must have more ex­cited atoms than lower en­ergy ones.

(Note that the lower en­ergy state is not nec­es­sar­ily the same as the ground state. All else be­ing the same, it ob­vi­ously helps to have the lower en­ergy state it­self de­cay rapidly to a state of still lower en­ergy. To a con­sid­er­able ex­tent, you can pick and choose de­cay rates, be­cause de­cay rates can vary greatly de­pend­ing on the amount to which they are for­bid­den, sec­tion 7.4.)


Key Points
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An elec­tro­mag­netic field can cause atoms to ab­sorb pho­tons.

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How­ever, it can also cause ex­cited atoms to re­lease pho­tons. That is called stim­u­lated emis­sion.

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In lasers and masers, an avalanche ef­fect of stim­u­lated emis­sion pro­duces co­her­ent, mono­chro­matic light.


7.7.1 The Hamil­ton­ian

To de­scribe the ef­fect of an elec­tro­mag­netic field on an atom us­ing quan­tum me­chan­ics, as al­ways the Hamil­ton­ian op­er­a­tor is needed.

The atom will be taken to be a hy­dro­gen atom for sim­plic­ity. Since the pro­ton is heavy, the elec­tro­mag­netic field in­ter­acts mainly with the elec­tron. The pro­ton will be as­sumed to be at rest.

It is also nec­es­sary to sim­plify the elec­tro­mag­netic field. That can be done by de­com­pos­ing the field into sep­a­rate “plane waves.” The to­tal in­ter­ac­tion can usu­ally be ob­tained by sim­ply sum­ming the ef­fects pro­duced by the sep­a­rate waves.

A sin­gle plane wave has an elec­tric field $\skew3\vec{\cal E}$ and a mag­netic field $\skew2\vec{\cal B}$ that can be writ­ten in the form, (13.10):

\begin{displaymath}
\skew3\vec{\cal E}= {\hat k}{\cal E}_{\rm {f}} \cos\Big(\om...
...frac1c {\cal E}_{\rm {f}} \cos\Big(\omega(t - y/c)-\alpha\Big)
\end{displaymath}

For con­ve­nience the $y$-​axis was taken in the di­rec­tion of prop­a­ga­tion of the wave. Also the $z$-​axis was taken in the di­rec­tion of the elec­tric field. Since there is just a sin­gle fre­quency $\omega$, the wave is mono­chro­matic; it is a sin­gle color. And be­cause of the di­rec­tion of the elec­tric field, the wave is said to be po­lar­ized in the $z$-​di­rec­tion. Note that the elec­tric and mag­netic fields for plane waves are nor­mal to the di­rec­tion of prop­a­ga­tion and to each other. The con­stant $c$ is the speed of light, ${\cal E}_{\rm {f}}$ the am­pli­tude of the elec­tric field, and $\alpha$ is some unim­por­tant phase an­gle.

For­tu­nately, the ex­pres­sion for the wave can be greatly sim­pli­fied. The elec­tron re­acts pri­mar­ily to the elec­tric field, pro­vided that its ki­netic en­ergy is small com­pared to its rest mass en­ergy. That is cer­tainly true for the elec­tron in a hy­dro­gen atom and for the outer elec­trons of atoms in gen­eral. There­fore the mag­netic field can be ig­nored. (The er­ror made in do­ing so is de­scribed more pre­cisely in {D.39}.) Also, the wave length of the elec­tro­mag­netic wave is usu­ally much larger than the size of the atom. For ex­am­ple, the Ly­man-tran­si­tion wave lengths are of the or­der of a thou­sand Å, while the atom is about one Å. So, as far as the light wave is con­cerned, the atom is just a tiny speck at the ori­gin. That means that $y$ can be put to zero in the ex­pres­sion for the plane wave. Then the wave sim­pli­fies to just:

\begin{displaymath}
\skew3\vec{\cal E}= {\hat k}{\cal E}_{\rm {f}} \cos(\omega t-\alpha)
\end{displaymath} (7.40)

This may not be ap­plic­a­ble to highly en­er­getic ra­di­a­tion like X-rays.

Now the ques­tion is how this field changes the Hamil­ton­ian of the elec­tron. Ig­nor­ing the time de­pen­dence of the elec­tric field, that is easy. The Hamil­ton­ian is

\begin{displaymath}
H = H_{\rm {atom}} + e {\cal E}_{\rm {f}} \cos(\omega t-\alpha) z %
\end{displaymath} (7.41)

where $H_{\rm {atom}}$ is the Hamil­ton­ian of the hy­dro­gen atom with­out the ex­ter­nal elec­tro­mag­netic field. The ex­pres­sion for $H_{\rm {atom}}$ was given in chap­ter 4.3, but it is not of any in­ter­est here.

The in­ter­est­ing term is the sec­ond one, the per­tur­ba­tion caused by the elec­tro­mag­netic field. In this term $z$ is the $z$-​po­si­tion of the elec­tron. It is just like the $mgh$ po­ten­tial en­ergy of grav­ity, with the charge $e$ play­ing the part of the mass $m$, the elec­tric field strength ${\cal E}_{\rm {f}}\cos({\omega}t-\alpha)$ that of the grav­ity strength $g$, and $z$ that of the height $h$.

To be sure, the elec­tric field is time de­pen­dent. The above per­tur­ba­tion po­ten­tial re­ally as­sumes that “the elec­tron moves so fast that the field seems steady to it.” In­deed, if an elec­tron speed is ball­parked from its ki­netic en­ergy, the elec­tron does seem to travel through the atom rel­a­tively fast com­pared to the fre­quency of the elec­tric field. Of course, it is much bet­ter to write the cor­rect un­steady Hamil­ton­ian and then show it works out pretty much the same as the quasi-steady one above. That is done in {D.39}.


Key Points
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An ap­prox­i­mate Hamil­ton­ian was writ­ten down for the in­ter­ac­tion of an atom with an elec­tro­mag­netic wave.

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By ap­prox­i­ma­tion the atom sees a uni­form, quasi-steady elec­tric field.


7.7.2 The two-state model

The big ques­tion is how the elec­tro­mag­netic field af­fects tran­si­tions be­tween a typ­i­cal atomic state $\psi_{\rm {L}}$ of lower en­ergy and one of higher en­ergy $\psi_{\rm {H}}$.

The an­swer de­pends crit­i­cally on var­i­ous Hamil­ton­ian co­ef­fi­cients. In par­tic­u­lar, the ex­pec­ta­tion val­ues of the en­er­gies of the two states are needed. They are

\begin{displaymath}
E_{\rm {L}} = \langle\psi_{\rm {L}}\vert H\vert\psi_{\rm {L...
... {H}} = \langle\psi_{\rm {H}}\vert H\vert\psi_{\rm {H}}\rangle
\end{displaymath}

Here the Hamil­ton­ian to use is (7.41) of the pre­vi­ous sub­sec­tion; it in­cludes the elec­tric field. But it can be seen that the en­er­gies are un­af­fected by the elec­tric field. They are the un­per­turbed atomic en­er­gies of the states. That fol­lows from sym­me­try; if you write out the in­ner prod­ucts above us­ing (7.41), the square wave func­tion is the same at any two po­si­tions ${\skew0\vec r}$ and $\vphantom{0}\raisebox{1.5pt}{$-$}$${\skew0\vec r}$, but $z$ in the elec­tric field term changes sign. So in­te­gra­tion val­ues pair­wise can­cel each other.

Note how­ever that the two en­er­gies are now ex­pec­ta­tion val­ues of en­ergy; due to the elec­tric field the atomic states de­velop un­cer­tainty in en­ergy. That is why they are no longer sta­tion­ary states.

The other key Hamil­ton­ian co­ef­fi­cient is

\begin{displaymath}
H_{\rm {HL}} = \langle\psi_{\rm {H}}\vert H\vert\psi_{\rm {L}}\rangle
\end{displaymath}

Plug­ging in the Hamil­ton­ian (7.41), it is seen that the atomic part $H_{\rm {atom}}$ does not con­tribute. The states $\psi_{\rm {H}}$ and $\psi_{\rm {L}}$ are or­thog­o­nal, and the atomic Hamil­ton­ian just mul­ti­plies $\psi_{\rm {L}}$ by $E_{\rm {L}}$. But the elec­tric field gives

\begin{displaymath}
H_{\rm {HL}}
= {\cal E}_{\rm {f}} \langle\psi_{\rm {H}}\ve...
...e^{{\rm i}(\omega t-\alpha)}+e^{-{\rm i}(\omega t-\alpha)}}{2}
\end{displaymath}

Here the co­sine in (7.41) was taken apart into two ex­po­nen­tials us­ing the Euler for­mula (2.5).

The next ques­tion is what these co­effients mean for the tran­si­tions be­tween two atomic states $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$. First, since the atomic states are com­plete, the wave func­tion can al­ways be writ­ten as

\begin{displaymath}
\Psi = c_{\rm {L}} \psi_{\rm {L}} + c_{\rm {H}} \psi_{\rm {H}} + \ldots
\end{displaymath}

where the dots stand for other atomic states. The co­ef­fi­cients $c_{\rm {L}}$ and $c_{\rm {H}}$ are the key, be­cause their square mag­ni­tudes give the prob­a­bil­i­ties of the states $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$. So they de­ter­mine whether tran­si­tions oc­cur be­tween them.

Evo­lu­tion equa­tions for these co­ef­fi­cients fol­low from the Schrö­din­ger equa­tion. The way to find them was de­scribed in sec­tion 7.6, with ad­di­tional ma­nip­u­la­tions in de­riva­tion {D.38}. The re­sult­ing evo­lu­tion equa­tions are:

\begin{displaymath}
\fbox{$\displaystyle
{\rm i}\hbar \dot {\bar c}_{\rm{L}}
...
...rm{H}}
= \overline{H}_{\rm{HL}}\bar c_{\rm{L}} + \ldots
$} %
\end{displaymath} (7.42)

where the dots rep­re­sent terms in­volv­ing states other than $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$. These equa­tions use the mod­i­fied co­ef­fi­cients
\begin{displaymath}
\bar c_{\rm {L}} = c_{\rm {L}} e^{{\rm i}E_{\rm {L}} t/\hba...
... \bar c_{\rm {H}} = c_{\rm {H}} e^{{\rm i}E_{\rm {H}} t/\hbar}
\end{displaymath} (7.43)

The mod­i­fied co­ef­fi­cients have the same square mag­ni­tudes as the orig­i­nal ones and the same val­ues at time zero. That makes them fully equiv­a­lent to the orig­i­nal ones. The mod­i­fied Hamil­ton­ian co­ef­fi­cient in the evo­lu­tion equa­tions is
\begin{displaymath}
\overline{H}_{\rm {HL}} = \overline{H}_{\rm {LH}}^* =
{\te...
...t\psi_{\rm {L}}\rangle
e^{{\rm i}(\omega_0+\omega)t-\alpha} %
\end{displaymath} (7.44)

where $\omega_0$ is the fre­quency of a pho­ton that has the ex­act en­ergy $E_{\rm {H}}-E_{\rm {L}}$.

Note that this mod­i­fied Hamil­ton­ian co­ef­fi­cient is re­spon­si­ble for the in­ter­ac­tion be­tween the states $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$. If this Hamil­ton­ian co­ef­fi­cient is zero, the elec­tro­mag­netic wave can­not cause tran­si­tions be­tween the two states. At least not within the ap­prox­i­ma­tions made.

Whether this hap­pens de­pends on whether the in­ner prod­uct $\langle\psi_{\rm {H}}\vert ez\vert\psi_{\rm {L}}\rangle$ is zero. This in­ner prod­uct is called the “atomic ma­trix el­e­ment” be­cause it de­pends only on the atomic states, not on the strength and fre­quency of the elec­tric wave.

How­ever, it does de­pend on the di­rec­tion of the elec­tric field. The as­sumed plane wave had its elec­tric field in the $z$-​di­rec­tion. Dif­fer­ent waves can have their elec­tric fields in other di­rec­tions. There­fore, waves can cause tran­si­tions as long as there is at least one nonzero atomic ma­trix el­e­ment of the form $\langle\psi_{\rm {L}}\vert er_i\vert\psi_{\rm {H}}\rangle$, with $r_i$ equal to $x$, $y$, or $z$. If there is such a nonzero ma­trix el­e­ment, the tran­si­tion is called al­lowed. Con­versely, if all three ma­trix el­e­ments are zero, then tran­si­tions be­tween the states $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$ are called for­bid­den.

Note how­ever that so-called for­bid­den tran­si­tions of­ten oc­cur just fine. The de­riva­tion in the pre­vi­ous sub­sec­tion made sev­eral ap­prox­i­ma­tions, in­clud­ing that the mag­netic field can be ig­nored and that the elec­tric field is in­de­pen­dent of po­si­tion. If these ig­nored ef­fects are cor­rected for, many for­bid­den tran­si­tions turn out to be pos­si­ble af­ter all; they are just much slower.

The ap­prox­i­ma­tions made to ar­rive at the atomic ma­trix el­e­ment $\langle\psi_{\rm {H}}\vert ez\vert\psi_{\rm {L}}\rangle$ are known as the elec­tric di­pole ap­prox­i­ma­tion. The cor­re­spond­ing tran­si­tions are called elec­tric di­pole tran­si­tions. If you want to know where the term comes from, why? Any­way, in that case note first that if the elec­tron charge dis­tri­b­u­tion is sym­met­ric around the pro­ton, the ex­pec­ta­tion value of $ez$ will be zero by sym­me­try. Neg­a­tive $z$ val­ues will can­cel pos­i­tive ones. But the elec­tron charge dis­tri­b­u­tion might get some­what shifted to the pos­i­tive $z$ side, say. The to­tal atom is then still elec­tri­cally neu­tral, but it be­haves a bit like a com­bi­na­tion of a neg­a­tive charge at a pos­i­tive value of $z$ and an equal and op­po­site pos­i­tive charge at a neg­a­tive value of $z$. Such a com­bi­na­tion of two op­po­site charges is called a di­pole in clas­si­cal elec­tro­mag­net­ics, chap­ter 13.3. So in quan­tum me­chan­ics the op­er­a­tor $ez$ gives the di­pole strength in the $z$-​di­rec­tion. And if the above atomic ma­trix el­e­ment is nonzero, it can be seen that non­triv­ial com­bi­na­tions of $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$ have a nonzero ex­pec­ta­tion di­pole strength. So the name elec­tric di­pole tran­si­tions is jus­ti­fied, es­pe­cially since ba­sic elec­tric tran­si­tions would be un­der­stand­able by far too many non­ex­perts.

Al­lowed and for­bid­den tran­si­tions were dis­cussed ear­lier in sec­tion 7.4. How­ever, that was based on as­sumed prop­er­ties of the emit­ted pho­ton. The al­lowed atomic ma­trix el­e­ments above, and sim­i­lar for­bid­den ones, make it pos­si­ble to check the var­i­ous most im­por­tant re­sults di­rectly from the gov­ern­ing equa­tions. That is done in de­riva­tion {D.39}.

There is an­other re­quire­ment to get a de­cent tran­si­tion prob­a­bil­ity. The ex­po­nen­tials in the mod­i­fied Hamil­ton­ian co­ef­fi­cient (7.44) must not os­cil­late too rapidly in time. Oth­er­wise op­po­site val­ues of the ex­po­nen­tials will av­er­age away against each other. So no sig­nif­i­cant tran­si­tion prob­a­bil­ity can build up. (This is sim­i­lar to the can­ce­la­tion that gives rise to the adi­a­batic the­o­rem, {D.34}.) Now un­der real-life con­di­tions, the sec­ond ex­po­nen­tial in (7.44) will al­ways os­cil­late rapidly. Nor­mal elec­tro­mag­netic fre­quen­cies are very high. There­fore the sec­ond term in (7.44) can nor­mally be ig­nored.

And in or­der for the first ex­po­nen­tial not too os­cil­late too rapidly re­quires a pretty good match be­tween the fre­quen­cies $\omega$ and $\omega_0$. Re­call that $\omega$ is the fre­quency of the elec­tro­mag­netic wave, while $\omega_0$ is the fre­quency of a pho­ton whose en­ergy is the dif­fer­ence be­tween the atomic en­er­gies $E_{\rm {H}}$ and $E_{\rm {L}}$. If the elec­tric field does not match the fre­quency of that pho­ton, it will not do much. Us­ing the Planck-Ein­stein re­la­tion, that means that

\begin{displaymath}
\omega \approx \omega_0 \equiv (E_{\rm {H}}-E_{\rm {L}})/\hbar
\end{displaymath}

One con­se­quence is that in tran­si­tions be­tween two atomic states $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$, other states usu­ally do not need to be con­sid­ered. Un­less an other state matches ei­ther the en­ergy $E_{\rm {H}}$ or $E_{\rm {L}}$, it will give rise to rapidly os­cil­lat­ing ex­po­nen­tials that can be ig­nored.

In ad­di­tion, the in­ter­est is of­ten in the so-called col­li­sion-dom­i­nated regime in which the atom evolves for only a short time be­fore be­ing dis­turbed by col­li­sions with its sur­round­ings. In that case, the short evo­lu­tion time pre­vents non­triv­ial in­ter­ac­tions be­tween dif­fer­ent tran­si­tion processes to build up. Tran­si­tion rates for the in­di­vid­ual tran­si­tion processes can be found sep­a­rately and sim­ply added to­gether.

The ob­tained evo­lu­tion equa­tions (7.42) can ex­plain why ab­sorp­tion and stim­u­lated emis­sion com­pete on an equal foot­ing in the op­er­a­tion of lasers. The rea­son is that the equa­tions have a re­mark­able sym­me­try: for every so­lu­tion $\bar{c}_{\rm {L}}$, $\bar{c}_{\rm {H}}$ there is a sec­ond so­lu­tion $\bar{c}_{\rm {L,2}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar{c}_{\rm {H}}^{\,*}$, $\bar{c}_{\rm {H,2}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\bar{c}_{\rm {L}}^{\,*}$ that has the prob­a­bil­i­ties of the low and high en­ergy states ex­actly re­versed. It means that

An elec­tro­mag­netic field that takes an atom out of the low en­ergy state $\psi_{\rm {L}}$ to­wards the high en­ergy state $\psi_{\rm {H}}$ will equally take that atom out of the high en­ergy state $\psi_{\rm {H}}$ to­wards the low en­ergy state $\psi_{\rm {L}}$.
It is a con­se­quence of the Her­mit­ian na­ture of the Hamil­ton­ian; it would not ap­ply if ${\overline{H}}_{\rm {LH}}$ was not equal to ${\overline{H}}_{\rm {HL}}^*$.


Key Points
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The gov­ern­ing evo­lu­tion equa­tions for the prob­a­bil­i­ties of two atomic states $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$ in an elec­tro­mag­netic wave have been found.

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The equa­tions have a sym­me­try prop­erty that makes elec­tro­mag­netic waves equally ef­fec­tive for ab­sorp­tion and stim­u­lated emis­sion.

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...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
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Nor­mally the elec­tro­mag­netic field has no sig­nif­i­cant ef­fect on tran­si­tions be­tween the states un­less its fre­quency $\omega$ closely matches the fre­quency $\omega_0$ of a pho­ton with en­ergy $E_{\rm {H}}-E_{\rm {L}}$.

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The gov­ern­ing equa­tions can ex­plain why some tran­si­tions are al­lowed and oth­ers are for­bid­den. The key are so-called atomic ma­trix el­e­ments.