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
\setlength{\unitlength}{1pt}
\begin{picture}(...
...71,0){\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
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Atoms can transition to a lower electronic energy level while emitting a photon of electromagnetic radiation.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
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
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
In a transition, the difference in atomic energy levels gives the energy, and so the frequency, of the emitted photon.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Unstable states have some uncertainty in energy, but it is usually very small. For extremely unstable particles, the uncertainty can be a lot.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
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
\setlength{\unitlength}{1pt}
\begin{picture}(...
...ut(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.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...\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_\ga...
...} \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...
...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 $\vphantom0\raisebox{1.5pt}{$-$}$$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_\gamm...
...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 $\vphantom0\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. Parity can have only two values, 1 or $\vphantom0\raisebox{1.5pt}{$-$}$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 $\vphantom0\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. In other words, they change by a trivial factor 1. That makes the parity 1, or even, or positive. The spherical harmonics for odd $l$ change sign when ${\skew0\vec r}$ is replaced by $\vphantom0\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. In other words, they get multiplied by a factor $\vphantom0\raisebox{1.5pt}{$-$}$1. That makes the parity $\vphantom0\raisebox{1.5pt}{$-$}$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 $\vphantom0\raisebox{1.5pt}{$-$}$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. $\vphantom0\raisebox{1.5pt}{$-$}$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
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The rules for combining angular momenta and parities were discussed.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
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 $\vphantom0\raisebox{1.5pt}{$-$}$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...
...ant$}}1$}}
\\ \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 $\vphantom0\raisebox{1.5pt}{$-$}$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 $\vphantom0\raisebox{1.5pt}{$-$}$$\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
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Normal atomic transitions are called allowed or electric dipole ones. All others are called forbidden but can occur just fine.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
In electric dipole transitions the emitted photon has angular momentum quantum number $j_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and negative parity $\pi_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
In the slower magnetic dipole transitions the photon parity is positive, $\pi_\gamma$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Each higher multipole order adds a unit to the photon angular momentum quantum number $j_\gamma$ and flips over the parity $\pi_\gamma$.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
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{, }
\...
...}
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) $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. This 2p state has very slightly lower energy than the 2s state due to a tiny relativistic effect called Lamb shift, {A.38.4}. 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ (due to the spin) can decay to a state with again net angular momentum $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ 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_...
...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 $\vphantom0\raisebox{1.5pt}{$-$}$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.38}, 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_{\...
...m{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_...
...m{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_{\...
...m{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...
...,\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$:} \quad
\vert j_{...
... \qquad (\ell \mathrel{\raisebox{-1pt}{$\geqslant$}}1)
$} %
\end{displaymath} (7.23)


\begin{displaymath}
\fbox{$\displaystyle
\mbox{M$\ell$:} \quad
\vert j_{\r...
... \qquad (\ell \mathrel{\raisebox{-1pt}{$\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 ...
...lectric}\\ -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{$\displaystyle
\ell_{\rm min} = \vert j_{\rm{H}} -...
...\vphantom0\raisebox{1.5pt}{$=$}${} 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
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Normal atomic transitions are called electric dipole ones, or allowed ones, or $\rm {E}1$ ones. Unfortunately.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
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.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
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.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
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.