3 ABSORPTION, EMISSION, AND DISPERSION OF LIGHT
3.14 REFRACTIVE INDEX
† Note that, if the 3S1/2(F¼1)23S1/2(F¼2) splitting were very small compared to the Doppler width, we would haveS(1)(n)ffiS(2)(n);S(n) and
s(n)ffil2A21
8p g2
g1
3 8þ5
8
S(n)¼l2A21
8p g2
g1
S(n): (3:13:13) We would in effect be justified in ignoring the energy difference between theF¼1 andF¼2 levels.
The same sort of arguments as used in Section 3.5 for broadband radiation can be made to show that, if theradiation has a spectral width dn large compared to the 3S1/2(F¼1)2 3S1/2(F¼2) separation, then the absorption is also given approximately by (3.13.13). In other words, if either the spectral width of the transition or the spectral width of the radiation is large com- pared to the 3S1/2hyperfine splitting, we can treat 3S1/2as a single “unresolved” level. In particular, this approximation can be made if the radiation is in the form of a pulse of durationtp(1=dn) that is short compared to 1/(2p1772 MHz)¼90 ps.
Such considerations can also be applied, of course, to excited states and can be used to justify our treatment of 3P3/2as a single unresolved level in the calculation ofs(n). †
As in the case of spontaneous emission and absorption, this result of the classical oscillator model must be modified to include the oscillator strengthf:
n2(v)1¼Ne2f me0
v20v2
(v20v2)2þ4b2v2: (3:14:4) Unlike absorption, the refractive index is usually attributable tononresonanttransitions, that is, transitions such thatjv20v2j bv. In this case
n2(v)1Ne2f me0
1
v20v2: (3:14:5) In this nonresonant situation, however, no one transition is necessarily dominant, and so we must add the contributions of all transitions connected to the ground state in which the atoms are presumed (for now) to reside. Thus, if the transitions from the ground state have oscillator strengthsfjand transition frequenciesvj, the refractive index at the radi- ation frequencyvis given by the formula
n2(v)1¼Ne2 me0
X
j
fj
v2jv2: (3:14:6)
This result applies when there is one type of atom or molecule in the medium; more generally we simply add the contributions of the different species. In a gas, further- more, the densityNis generally sufficiently low thatn(v) 1 and thereforen221¼ (n21)(nþ1)2(n21). Thus, for a gas consisting of a single type of atom or molecule with number densityN, the formula for the refractive index is approximately
n(v)¼1þ Ne2 2me0
X
j
fj
v2j v2: (3:14:7) It is interesting to relate this result to a formula that is often used in tabulations of the refractive index of gases. For this purpose we first rewrite (3.14.7) in terms of radiation wavelengthl(¼2pc/v) and transition wavelengthslj (¼ 2pc/vj):
n(l)¼1þ Ne2 8p2e0mc2
X
j
l2jfj
1l2j=l2: (3:14:8) As noted in the Introduction, electronic resonances in molecules (and in many atoms) tend to lie in the ultraviolet, in which caseljlfor optical wavelengthsl. In this case we can approximate (1l2j=l2)1 by the first two terms of its binomial series expansion, 1þl2j=l2:
n(l)1A1(1þB1=l2), (3:14:9)
where
A1¼ Ne2 8p2e0mc2
X
j
fjl2j (3:14:10)
and
B1¼ P
jfjl4j P
jfjl2j : (3:14:11)
An empirical relation of the form (3.14.9) was proposed by Cauchy in 1830, before the electromagnetic theory of light. Our derivation ofCauchy’s formulagives explicit expressions for the coefficientsA1andB1. Unfortunately, it is difficult to calculate the numerical values ofA1andB1for a given atom or molecule because we require the tran- sition wavelengths and the oscillator strengths of all transitions connected to the ground state, including transitions to “continuum” states in which the electrons are unbound, that is, in which an atom is ionized.
For a gas at STP [P¼760 Torr, T¼273K, and, from Eq. (3.8.20), N¼ 2.691025 m23],
A1¼1:21010X
j
fjl2j: (3:14:12)
Consider as an example a gas of ground-state helium atoms, for which the 58.4-nm transition from the ground state to the first excited state has an oscillator strength of about 0.28. If we include only the contribution of this transition to the summations in (3.14.10) and (3.14.11), we obtain A1¼1.21025 and B1¼3.410215m2, in contrast to the tabulated (measured) values of 3.481025 and 2.310215m2, respectively.13 Adding the contributions of transitions from the ground state to higher-energy bound states does not change these results very significantly because the oscillator strengths for these transitions are considerably smaller than that for the transition to the first excited state. We conclude that transitions to the conti- nuum are mainly responsible for the discrepancy between our simple calculation and the measured values of the Cauchy constants for helium.
Cauchy’s formula correctly accounts for the fact that most transparent materials we encounter daily (e.g., water, air, glass) have refractive indices greater than unity at visible wavelengths. According to our analysis, this is a consequence of these materials having resonance wavelengths lj that are small compared to optical wavelengths (which lie roughly between 400 and 700 nm). It also follows from (3.14.9) that dn/dl,0, which is also a familiar feature of refractive indices in the visible: A glass prism, for instance, causes violet to be dispersed more than red when it separates white light into its spectral components. In fact, the increase ofn(l) with decreasingl(dn/ dl,0) is sufficiently ubiquitous that it is called “normal dispersion.” An example of normal dispersion appears in Fig. 3.21.
13M. Born and E. Wolf,Principles of Optics, 7th ed., Cambridge University Press, Cambridge, 1999, p. 101.
3.14 REFRACTIVE INDEX 125
It is also interesting to consider the case in which the frequency of the radiation is much greater than the resonance frequencies of the medium. The simplest example occurs for free electrons, for which there is no binding force. The resonance frequencies vjare then zero, and the dispersion formula (3.14.6) reduces to [see Eq. (3.14.16)]
n(v)¼ 1 Ne2 e0mv2 1=2
¼ 1v2p v2
!1=2
, (3:14:13) whereNis now the density of free electrons (Problem 3.20) and
vp¼ Ne
2
me0 1=2
(3:14:14) is called theplasma frequency. In some cases this result is known to be fairly accurate; it is applicable, for instance, to the upper atmosphere, where ultraviolet solar radiation pro- duces free electrons by photoionization (Problem 3.20). Another example is the refrac- tion of X rays by glass. In this case the resonance frequencies of the medium are not zero but are much less than X-ray frequencies. In both examples the refractive index is less than one.
Ifv,vp, the refractive index (3.14.13) is a pure imaginary number. In this case the free-electron gas will not support a propagating electromagnetic wave, and an incident wave is instead reflected. This applies, for instance, to the propagation of radio waves in Earth’s atmosphere. High-frequency FM radio waves are not reflected by the ionosphere (v.vp), whereas the lower frequency AM waves are. AM radio broadcasts, therefore, reach more distant points on Earth’s surface (Problem 3.20).
† Consider the limit of Eq. (3.14.6) in which the field frequencyvis very large compared to any of the transition frequenciesvj:
n2(v)1¼ Ne2 e0mv2
X
j
fj: (3:14:15)
3.65
n( ) – 1 vs.
(Å) 3.70 × 10–5
3.60 3.55 3.50
2000 6000 10,000
λ λ λ
Figure 3.21 Refractive index of helium at standard temperature and pressure.
In this limit the transition frequenciesvjare effectively zero, that is, the atom behaves as though its energy levels form a continuum, as is the case for unbound electrons. It is then plausible that in this limit the refractive index should be identical to that ofNfree electrons per unit volume, which is given by Eq. (3.14.13). This implies that the oscillator strengthsfjmust obey theelectric dipole sum rule
X
j
fj¼1, (3:14:16)
which in fact may be derived using quantum mechanics. Our less rigorous “derivation” of this sum rule is based on the assumption that each of theNatoms per unit volume has one bound elec- tron. In the case ofZelectrons per atom Eq. (3.14.15) should reduce ton2(v)21 for the case of NZfree electrons per unit volume, so that the sum rule for aZ-electron atom is
X
j
fj¼Z: (3:14:17)
Table 3.2 lists oscillator strengths for the hydrogen atom for “allowed” electric dipole tran- sitions, i.e., transitions for whichD‘¼+1.14The sum of the oscillator strengths of allthe bound – bound 1s2nptransitions is 0.565, so that the electric dipole sum rule implies that the sum of the oscillator strengths for transitions from 1sto continuum states is 0.435. For the 2s state the corresponding bound – bound and bound – free contributions to thefsums are 0.649 and 0.351. Table 3.2 exemplifies the fact, noted above for helium, that transitions to the first excited state tend to be stronger than transitions to higher-energy bound states. Note also that the transitions to continuum states contribute significantly to the sum of the oscillator strengths.
ForZ-electron atoms the transitions to continuum states make a contributionZto the sum of the oscillator strengths for all the electrons. Bound – bound transitions of single electrons, such as those for the valence electrons of the alkali atoms, have oscillator strengths comparable in
magnitude to those of hydrogen. †
We have assumed in our discussion of the refractive index that the N atoms per unit volume are all in the ground state with high probability, but it is straightforward to deal with the more general situation where there areNi atoms per unit volume in TABLE 3.2 Oscillator Strengths for Some Bound – Bound Transitions of Hydrogen
Initial state 1s 2s 2p 3s 3p 3d
Final state np np ns nd np ns nd np nf
n¼1 — — 20.139 — — 20.026 — — —
n¼2 0.4162 — — — 20.041 20.145 — 20.417 —
n¼3 0.0791 0.435 0.0134 0.696 — — — — —
n¼4 0.0290 0.103 0.0030 0.122 0.484 0.032 0.619 0.011 1.018
n¼5 0.0139 0.042 0.0012 0.044 0.121 0.007 0.139 0.002 0.156
14Selection rules and many other fundamental aspects of atomic theory are discussed in R. D. Cowan,The Theory of Atomic Structure and SpectraUniversity of California Press, Berkeley, CA, 1981. Useful tabula- tions of oscillator strengths and formulas for atomic and molecular transitions may be found inAllen’s Astrophysical Quantitiesand other sources (see footnote 5).
3.14 REFRACTIVE INDEX 127
energy levelEi. Equation (3.14.7), for instance, generalizes to n(v)¼1þ e2
2me0 X
i
X
j
Nifij
v2jiv2, (3:14:18) where vji¼(EjEi)=h and fij is the oscillator strength for the i!j transition. In particular, the contribution to the index from the 1!2 transition is
n(v)12¼1þ e2 2me0
N1f12 v221v2þ
N2f21 v221v2
¼1þ e2 2me0
f12
v221v2 N1 g1 g2
N2
, (3:14:19) where we have used Eq. (3.7.26) to relatef21tof12.
† The sum rule (3.14.17) may be written more generally as X
j
fij¼X
j.i
fijþX
j,i
fij¼X
j.i
fijX
j,i
jfijj ¼Z (3:14:20) for any leveliof an atom. In other words, the sum over oscillator strengths in the electric dipole sum rule must, in the case of excited states, include both the oscillator strengths for absorption (positive) and for emission (negative).
The sum rule for oscillator strengths played an important role in the formulation of quantum theory in the 1920s. It was already known, based on the physical argument we have used in going from (3.14.15) to (3.14.16), before some of the most important features of quantum mechanics (e.g., before the Schro¨dinger equation).
Historically, downward transitions associated with stimulated emission were referred to in terms of “negative oscillators.” In the case of the refractive index, the term proportional toN2 in Eq. (3.14.19) corresponds to such a negative oscillator. The contribution of negative oscillators to the refractive index was studied experimentally by R. Ladenburg and H. Kopfermann around 1928. They measured the variation of refractive index with electric current of a discharge tube filled with neon. According to our theory, forv,v21in (3.14.19), with neither 1 nor 2 the ground level,n(v)12should initially increase with increasing current because electron – atom collisions produce atoms in excited level 1. With further increase of the current, however, the rate of growth ofn(v)12with current decreases because excited level 2 has appreciable population N2 and acts as a negative oscillator. This sort of behavior was observed by Ladenburg and Kopfermann, thus confirming the role of negative oscillators. † The atoms or molecules of a medium do not form a continuum but have empty space between them. As a result, there is a difference between the “mean” field and the actual field acting on a given atom. In many cases the only practical consequence of this differ- ence is that the relation between the refractive index and the polarizabilityabecomes
n2(v)1
n2(v)þ2¼Na(v)
3e0 : (3:14:21)
The origin of this “Lorentz – Lorenz relation” is discussed in many textbooks on electro- magnetism. Note that when the refractive index is close to unity, so thatn2(v)þ23, the Lorentz – Lorenz relation reduces to the relation between n and a assumed in writing (3.14.3).