A.24 Quan­tum spon­ta­neous emis­sion

Chap­ter 7.8 ex­plained the gen­eral in­ter­ac­tion be­tween atoms and elec­tro­mag­netic fields. How­ever, spon­ta­neous emis­sion of ra­di­a­tion was found us­ing a dirty trick due to Ein­stein. He peeked at the so­lu­tion for black­body ra­di­a­tion. This ad­den­dum will give a proper quan­tum de­scrip­tion. Warn­ing: while this ad­den­dum tries to be rea­son­ably self-con­tained, to re­ally ap­pre­ci­ate the de­tails you may have to read some other ad­denda too.

The prob­lem with the de­scrip­tions of emis­sion and ab­sorp­tion of ra­di­a­tion in chap­ter 7.7 and 7.8 is that they as­sume that the elec­tro­mag­netic field is given. The elec­tro­mag­netic field is not given; it changes by one pho­ton. That is rather im­por­tant for spon­ta­neous emis­sion, where it changes from no pho­tons to one pho­ton. To ac­count for that cor­rectly re­quires that the elec­tro­mag­netic field is prop­erly quan­tized. That is done in this note.

To keep it sim­ple, it will be as­sumed that the atom is a hy­dro­gen one. Then there is just one elec­tron to worry about. (The gen­eral analy­sis can be found in {A.25}). The hy­dro­gen atom is ini­tially in some high en­ergy state $\psi_{\rm {H}}$. Then it emits a pho­ton and tran­si­tions to a lower en­ergy state $\psi_{\rm {L}}$. The emit­ted pho­ton comes out in a state with en­ergy

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

Re­call that the pho­ton en­ergy is given in terms of its fre­quency $\omega$ by the Planck-Ein­stein re­la­tion. This pho­ton en­ergy is ap­prox­i­mately the dif­fer­ence be­tween the atomic en­er­gies. It does not have to be ex­act; there can be some en­ergy slop, chap­ter 7.6.1.

Only a sin­gle pho­ton en­ergy state needs to be con­sid­ered at a time. At the end of the story, the re­sults can be summed over all pos­si­ble pho­ton states. To al­low for stim­u­lated emis­sion, it will be as­sumed that ini­tially there may al­ready be $i$ pre­ex­ist­ing pho­tons present. For spon­ta­neous emis­sion, $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The ini­tial sys­tem state will be in­di­cated as:

\begin{displaymath}
\psi_1 = \psi_{\rm {H}} {\left\vert i\right\rangle}
\end{displaymath}

Here the so-called Fock space ket ${\left\vert i\right\rangle}$ is sim­ply a con­cise way of in­di­cat­ing that there are $i$ pho­tons in the con­sid­ered pho­ton quan­tum state.

In the fi­nal state the atom has de­cayed to a lower en­ergy state $\psi_{\rm {L}}$. In do­ing so it has re­leased 1 more pho­ton into the con­sid­ered pho­ton state. So the fi­nal wave func­tion is

\begin{displaymath}
\psi_2 = \psi_{\rm {L}} {\left\vert i{+}1\right\rangle}
\end{displaymath}

The key to the emis­sion process is now the set of Hamil­ton­ian co­ef­fi­cients, chap­ter 7.6,

\begin{displaymath}
\langle E_1\rangle = \langle\psi_1\vert H \psi_1\rangle
\q...
...qquad
\langle E_2\rangle = \langle\psi_2\vert H \psi_2\rangle
\end{displaymath}

Here $H$ is the Hamil­ton­ian. All that re­ally needs to be done in this note is to iden­tify these co­ef­fi­cients, and in par­tic­u­lar the so-called ma­trix el­e­ment $H_{21}$. With the ma­trix el­e­ment known, Fermi’s golden rule can be used to find the pre­cise tran­si­tion rate, chap­ter 7.6.1.

To iden­tify the Hamil­ton­ian co­ef­fi­cients, first the Hamil­ton­ian must be iden­ti­fied. Re­call that the Hamil­ton­ian is the op­er­a­tor of the to­tal en­ergy of the sys­tem. It will take the form

\begin{displaymath}
H = H_{\rm {atom}} + H_\gamma + H_{\rm {atom,\gamma}}
\end{displaymath}

The first term in the right hand side is the in­her­ent en­ergy of the hy­dro­gen atom. This Hamil­ton­ian was writ­ten down way back in chap­ter 4.3. How­ever, its pre­cise form is of no in­ter­est here. The sec­ond term in the right hand side is the en­ergy in the elec­tro­mag­netic field. Elec­tro­mag­netic fields too have in­her­ent en­ergy, about $\hbar\omega$ per pho­ton in fact. The third term is the en­ergy of the in­ter­ac­tion be­tween the atomic elec­tron and the elec­tro­mag­netic field.

Un­like the first term in the Hamil­ton­ian, the other two are in­her­ently rel­a­tivis­tic: the num­ber of pho­tons is hardly a con­served quan­tity. Pho­tons are read­ily cre­ated or ab­sorbed by a charged par­ti­cle, like the elec­tron here. And it turns out that Hamil­to­ni­ans for pho­tons are in­trin­si­cally linked to op­er­a­tors that an­ni­hi­late and cre­ate pho­tons. Math­e­mat­i­cally, at least. These op­er­a­tors are de­fined by the re­la­tions

\begin{displaymath}
\widehat a{\left\vert i\right\rangle} = \sqrt{i} {\left\ver...
...t i{-}1\right\rangle} = \sqrt{i} {\left\vert i\right\rangle} %
\end{displaymath} (A.166)

for any num­ber of pho­tons $i$. In words, the an­ni­hi­la­tion op­er­a­tor $\widehat a$ takes a state of $i$ pho­tons and turns it into a state with one less pho­ton. The cre­ation op­er­a­tor $\widehat a^\dagger $ puts the pho­ton back in. The scalar fac­tors $\sqrt{i}$ are a mat­ter of con­ve­nience. If you did not put them in here, you would have to do it else­where.

The Hamil­ton­ian that de­scribes the in­her­ent en­ergy in the elec­tro­mag­netic field turns out to be, {A.23},

\begin{displaymath}
H_\gamma = {\textstyle\frac{1}{2}}\hbar\omega(\widehat a^\dagger \widehat a+\widehat a\widehat a^\dagger )
\end{displaymath}

As a san­ity check, this Hamil­ton­ian can be ap­plied on a state of $i$ pho­tons. Us­ing the de­f­i­n­i­tions of the an­ni­hi­la­tion and cre­ation op­er­a­tors given above,

\begin{displaymath}
H_\gamma {\left\vert i\right\rangle}
= {\textstyle\frac{1}...
...r\omega(i+{\textstyle\frac{1}{2}}) {\left\vert i\right\rangle}
\end{displaymath}

The fac­tor in front of the fi­nal ket is the en­ergy eigen­value. So the en­ergy in the field in­creases by one unit $\hbar\omega$ for each pho­ton added, ex­actly as it should. The ad­di­tional half pho­ton is the ground state en­ergy of the elec­tro­mag­netic field. Even in its ground state, the elec­tro­mag­netic field has some en­ergy left. That is much like a one-di­men­sion­al har­monic os­cil­la­tor still has ${\textstyle\frac{1}{2}}\hbar\omega$ of en­ergy left in its ground state, chap­ter 4.1.

Fi­nally the en­ergy of the in­ter­ac­tion be­tween the elec­tron and elec­tro­mag­netic field is needed. This third part of the to­tal Hamil­ton­ian is the messi­est. To keep it as sim­ple as pos­si­ble, it will as­sumed that the tran­si­tion is of the nor­mal elec­tric di­pole type. In such tran­si­tions the elec­tron in­ter­acts only with the elec­tric part of the elec­tro­mag­netic field. In ad­di­tion, just like in the analy­sis of chap­ter 7.7.1 us­ing a clas­si­cal elec­tro­mag­netic field, it will be as­sumed that the elec­tric field is in the $z$-​di­rec­tion and prop­a­gates in the $y$-​di­rec­tion. (The gen­eral mul­ti­pole analy­sis can be found in {A.25}).

Now re­call that in quan­tum me­chan­ics, ob­serv­able prop­er­ties of par­ti­cles are the eigen­val­ues of Her­mit­ian op­er­a­tors, chap­ter 3.3. For ex­am­ple, the ob­serv­able val­ues of lin­ear mo­men­tum of an elec­tron in the $y$-​di­rec­tion are the eigen­val­ues of the lin­ear mo­men­tum op­er­a­tor ${\widehat p}_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\partial$$\raisebox{.5pt}{$/$}$${\rm i}\partial{y}$. This op­er­a­tor acts on the elec­tron wave func­tion.

Sim­i­larly, the elec­tric field ${\cal E}_z$ that the elec­tron in­ter­acts with is an ob­serv­able prop­erty of the cor­re­spond­ing pho­tons. So the ob­serv­able val­ues of the elec­tric field must be the eigen­val­ues of a Her­mit­ian elec­tric field op­er­a­tor $\skew4\widehat{\cal E}_z$. And this op­er­a­tor acts on pho­ton wave func­tions.

In the analy­sis us­ing a clas­si­cal elec­tro­mag­netic field, the en­ergy of in­ter­ac­tion be­tween the elec­tron and the elec­tro­mag­netic field was taken to be ap­prox­i­mately $e{\cal E}_zz$. That is sim­i­lar to the $mgh$ po­ten­tial of a par­ti­cle due to grav­ity. The elec­tron elec­tric charge $\vphantom0\raisebox{1.5pt}{$-$}$$e$ takes the place of the mass $m$, the elec­tric field ${\cal E}_z$ that of the ac­cel­er­a­tion of grav­ity $g$, and $z$ that of the height $h$. Us­ing the quan­tized elec­tric field, there is no given clas­si­cal field ${\cal E}_z$, and in­stead the op­er­a­tor $\skew4\widehat{\cal E}_z$ must be used:

\begin{displaymath}
H_{\rm {atom,\gamma}} = e \skew4\widehat{\cal E}_z z
\end{displaymath}

The op­er­a­tor $\skew4\widehat{\cal E}_z$ acts on the ket part of the com­bined atom-pho­ton wave func­tion. (And, al­though you may not think of it that way, the fac­tor $z$ is re­ally an op­er­a­tor that acts on the elec­tron wave func­tion part. That is true even in the analy­sis us­ing the clas­si­cal field.)

The elec­tric field op­er­a­tor $\skew4\widehat{\cal E}_z$ can be iden­ti­fied from the ap­pro­pri­ate pho­ton wave func­tion. The pho­ton wave func­tion here is as­sumed to have its lin­ear mo­men­tum in the $y$-​di­rec­tion and its un­ob­serv­able elec­tric field in the $z$-​di­rec­tion. The cor­re­spond­ing nor­mal­ized wave func­tion and un­ob­serv­able elec­tric field were given in {A.21.6} (A.95):

\begin{displaymath}
\skew3\vec A_\gamma^{\rm n}= \frac{\varepsilon_k}{{\rm i}kc...
... \varepsilon_k = \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}}
\end{displaymath}

Here $\epsilon_0$ is the per­mit­tiv­ity of space. Also ${\cal V}$ is the vol­ume of the large pe­ri­odic box in which the en­tire sys­tem will be as­sumed to be lo­cated. In truly in­fi­nite space the analy­sis would be ex­tremely messy, lit­tered with ugly delta func­tions.

The rules to get the op­er­a­tor of the ob­serv­able elec­tric field were dis­cussed in ad­den­dum {A.23}. First the un­ob­serv­able elec­tric field above is mul­ti­plied by the an­ni­hi­la­tion op­er­a­tor, then the Her­mit­ian con­ju­gate of that prod­uct is added, and the sum is di­vided by $\sqrt{2}$:

\begin{displaymath}
\skew4\widehat{\cal E}_z = \frac{\varepsilon_k}{\sqrt{2}}(\widehat ae^{{\rm i}ky} + \widehat a^\dagger e^{-{\rm i}ky})
\end{displaymath}

(Note that for the usual Schrö­din­ger ap­proach fol­lowed here, time de­pen­dence is de­scribed by the wave func­tion. Most sources switch here to a Heisen­berg ap­proach where the time-de­pen­dence is pushed into the op­er­a­tors. There is how­ever no par­tic­u­lar need to do so.)

In the elec­tric di­pole ap­prox­i­ma­tion, it is as­sumed that the atom is so small com­pared to the wave length of the pho­ton that $ky$ can be as­sumed to be zero. There­fore

\begin{displaymath}
\skew4\widehat{\cal E}_z = \frac{\varepsilon_k}{\sqrt{2}}(\widehat a+ \widehat a^\dagger )
\end{displaymath}

The com­bined Hamil­ton­ian is then

\begin{displaymath}
H = H_{\rm {atom}} + H_\gamma + e\frac{\varepsilon_k}{\sqrt{2}}(\widehat a+\widehat a^\dagger )z
\end{displaymath}

with the first two terms as de­scribed ear­lier.

Next the Hamil­ton­ian ma­trix co­ef­fi­cients are needed. The first one is

\begin{displaymath}
\langle E_1\rangle = \langle \psi_1\vert H\psi_1\rangle
= ...
...t a^\dagger )z
\Big)\psi_{\rm {H}}{\left\vert i\right\rangle}
\end{displaymath}

Now the atomic part of the Hamil­ton­ian pro­duces a mere fac­tor $E_{\rm {H}}$ when it acts on the atomic part of the right hand wave func­tion. Fur­ther, as dis­cussed above, the elec­tro­mag­netic Hamil­ton­ian pro­duces a fac­tor $(i+\frac12)\hbar\omega$ when it acts on the right hand ket. Fi­nally the in­ter­ac­tion part of the Hamil­ton­ian does not pro­duce a con­tri­bu­tion. One way to see that is from the atomic in­ner prod­uct. The atomic in­ner prod­uct is zero be­cause neg­a­tive val­ues of $z$ in­te­grate away against pos­i­tive ones. An­other way to see it is from the elec­tro­mag­netic in­ner prod­uct. The op­er­a­tors $\widehat a$ and $\widehat a^\dagger $ pro­duce states ${\left\vert i{-}1\right\rangle}$ re­spec­tively ${\left\vert i{+}1\right\rangle}$ when they act on the right hand ket. And those are or­thog­o­nal to the left hand ket; in­ner prod­ucts be­tween kets with dif­fer­ent num­bers of pho­tons are zero. Kets are by de­f­i­n­i­tion or­tho­nor­mal.

All to­gether then

\begin{displaymath}
\langle E_1\rangle = E_{\rm {H}} + (i+{\textstyle\frac{1}{2}})\hbar\omega
\end{displaymath}

The same way

\begin{displaymath}
\langle E_2\rangle = E_{\rm {L}} + (i+1+{\textstyle\frac{1}{2}})\hbar\omega
\end{displaymath}

Fi­nally the ma­trix el­e­ment:

\begin{displaymath}
H_{21} = \langle \psi_2\vert H\psi_1\rangle
= {\left\langl...
...t a^\dagger )z
\Big)\psi_{\rm {H}}{\left\vert i\right\rangle}
\end{displaymath}

In this case the atomic part of the Hamil­ton­ian pro­duces zero. The rea­son is that this Hamil­ton­ian pro­duces a sim­ple scalar fac­tor $E_{\rm {H}}$ when it acts on the right hand state. It leaves the state $\psi_{\rm {H}}$ it­self un­changed. And this state pro­duces zero in an in­ner prod­uct with the atomic state $\psi_{\rm {L}}$; en­ergy eigen­states are or­tho­nor­mal. Sim­i­larly, the elec­tro­mag­netic Hamil­ton­ian pro­duces zero. It leaves the ket ${\left\vert i\right\rangle}$ in the right hand wave wave func­tion un­changed, and that is or­thog­o­nal to the left hand ${\left\langle i+1\hspace{0.3pt}\right\vert}$. How­ever, in this case the in­ter­ac­tion Hamil­ton­ian pro­duces a nonzero con­tri­bu­tion:

\begin{displaymath}
H_{21} = \frac{\varepsilon_k\sqrt{i+1}}{\sqrt{2}}
\big\langle \psi_{\rm {L}}\big\vert e z \psi_{\rm {H}}\big\rangle
\end{displaymath}

The rea­son is that the cre­ation op­er­a­tor $\widehat a^\dagger $ act­ing on the right hand ket pro­duces a mul­ti­ple $\sqrt{i+1}$ times the left hand ket. The re­main­ing in­ner prod­uct $\big\langle\psi_{\rm {L}}\big\vert ez\psi_{\rm {H}}\big\rangle$ is called the atomic ma­trix el­e­ment, as it only de­pends on what the atomic states are.

The task laid out in chap­ter 7.6.1 has been ac­com­plished: the rel­a­tivis­tic ma­trix el­e­ment has been found. A fi­nal ex­pres­sion for the spon­ta­neous emis­sion rate can now be de­ter­mined.

Be­fore do­ing so, how­ever, it is good to first com­pare the ob­tained re­sult with that of chap­ter 7.7.1. That sec­tion used a clas­si­cal given elec­tro­mag­netic field, not a quan­tized one. So the com­par­i­son will show up the ef­fect of the quan­ti­za­tion of the elec­tro­mag­netic field. The sec­tion de­fined a mod­i­fied ma­trix el­e­ment

\begin{displaymath}
\overline{H}_{21}
= H_{21} e^{{\rm i}(\langle E_2\rangle-\langle E_1\rangle)t/\hbar}
\end{displaymath}

This ma­trix el­e­ment de­ter­mined the en­tire evo­lu­tion of the sys­tem. For the quan­tized elec­tric field dis­cussed here, this co­ef­fi­cient works out to be
\begin{displaymath}
\overline{H}_{21} =
\frac{\varepsilon_k\sqrt{i+1}}{\sqrt{2...
... \varepsilon_k = \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}}
\end{displaymath} (A.167)

where $\omega_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(E_{\rm {H}}-E_{\rm {L}})$$\raisebox{.5pt}{$/$}$$\hbar$.

That is es­sen­tially the same form as for the clas­si­cal field. Re­call that the sec­ond term in (7.44) for the clas­si­cal field can be ig­nored. The first term is the same as above, within a con­stant. To see the real dif­fer­ence in the con­stants, note that the tran­si­tion prob­a­bil­ity is pro­por­tional to the square mag­ni­tude of the ma­trix el­e­ment. The square mag­ni­tudes are:

\begin{displaymath}
\mbox{quantized: } \vert\overline{H}_{21}^2\vert =
\frac{(...
...e \psi_{\rm {L}}\big\vert e z \psi_{\rm {H}}\big\rangle\vert^2
\end{displaymath}

Now if there is a large num­ber $i$ of pho­tons in the state, the two ex­pres­sions are ap­prox­i­mately the same. The elec­tro­mag­netic en­ergy of the wave ac­cord­ing to clas­si­cal physics, $\epsilon_0{\cal E}_{\rm {f}}^2{\cal V}$$\raisebox{.5pt}{$/$}$​2, {A.23}, is then ap­prox­i­mately the num­ber of pho­tons $i$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $i+1$ times $\hbar\omega$.

But for spon­ta­neous emis­sion there is a big dif­fer­ence. In that case, clas­si­cal physics would take the ini­tial elec­tro­mag­netic field ${\cal E}_{\rm {f}}$ to be zero. And that then im­plies that the atom stays in the ex­cited state $\psi_{\rm {H}}$ for al­ways. There is no elec­tro­mag­netic field to move it out of the state. So there is no spon­ta­neous emis­sion.

In­stead quan­tum me­chan­ics takes the ini­tial field to have $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 pho­tons. But note the square ma­trix el­e­ment above. It is not zero! The ma­trix el­e­ment is as if there is still a full pho­ton left in the elec­tro­mag­netic field. So spon­ta­neous emis­sion can and does oc­cur in the quan­tized elec­tro­mag­netic field. Also, as noted in chap­ter 7.8, one full pho­ton is ex­actly what is needed to ex­plain spon­ta­neous emis­sion. Ein­stein’s $A$ co­ef­fi­cient has been found us­ing pure quan­tum analy­sis. With­out peek­ing at the black body spec­trum.

That can also be seen with­out de­tour­ing through the messy analy­sis of chap­ter 7.7 and 7.8. To find the spon­ta­neous emis­sion rate di­rectly, the ma­trix el­e­ment above can be plugged into Fermi’s Golden Rule (7.38) of chap­ter 7.6.1. The den­sity of states needed in it was given ear­lier in chap­ter 6.3 (6.7) and 6.19. Do note that these modes in­clude all di­rec­tions of the elec­tric field, not just the $z$-​di­rec­tion. To ac­count for that, you need to av­er­age the square atomic ma­trix el­e­ment over all three Carte­sian di­rec­tions. That pro­duces the spon­ta­neous tran­si­tion rate

\begin{displaymath}
\frac{\omega^3}{\pi\hbar c^3\epsilon_0}
\frac{\vert\langle...
...t\langle\psi_{\rm {L}}\vert ez\psi_{\rm {H}}\rangle\vert^2}{3}
\end{displaymath}

The above re­sult is the same as Ein­stein’s, (7.47) and (7.48). (To see why a sim­ple av­er­age works in the fi­nal term, first note that it is ob­vi­ously the right av­er­age for pho­tons with ax­ial lin­ear mo­menta and fields. Then note that the av­er­age is in­de­pen­dent of the an­gu­lar ori­en­ta­tion of the axis sys­tem in which the pho­tons are de­scribed. So it also works for pho­tons that are ax­ial in any ro­tated co­or­di­nate sys­tem. To ver­ify that the av­er­age is in­de­pen­dent of an­gu­lar ori­en­ta­tion does not re­ally re­quire lin­ear al­ge­bra; it suf­fices to show that it is true for ro­ta­tion about one axis, say the $z$-​axis.)

Some ad­di­tional ob­ser­va­tions may be in­ter­est­ing. You might think of the spon­ta­neous emis­sion as caused by ex­ci­ta­tion from the ground state elec­tro­mag­netic field. But as seen ear­lier, the ac­tual en­ergy of the ground state is half a pho­ton, not one pho­ton. And the zero level of en­ergy should not af­fect the dy­nam­ics any­way. Ac­cord­ing to the analy­sis here, spon­ta­neous emis­sion is a twi­light ef­fect, chap­ter 5.3. The Hamil­ton­ian co­ef­fi­cient $H_{21}$ is the en­ergy if the atom is not ex­cited and there is a pho­ton if the atom is ex­cited and there is no pho­ton. In quan­tum me­chan­ics, the twi­light term al­lows the ex­cited atom to in­ter­act with the pho­ton that would be there if it was not ex­cited. Sic.