# Clarification of term symbol explanation of laser crystal doping from textbook

I am currently studying the textbook Diode Lasers and Photonic Integrated Circuits, second edition, by Coldren, Corzine, and Mashanovitch. Chapter 1.2 ENERGY LEVELS AND BANDS IN SOLIDS says the following:

In gas and solid-state lasers, the energy levels of the active atomic species are only perturbed slightly by the surrounding gas or solid host atoms, and they remain effectively as sharp as the original levels in the isolated atom. For example, lasers operating at the $$1.06 \ \text{\mum}$$ wavelength transition in Nd-doped YAG, use the $$^4\text{F}_{3/2}$$ level of the Nd atom for the upper laser state #2 and the $$^4\text{I}_{11/2}$$ level for the lower laser state #1.

I don't understand what is meant in this part:

For example, lasers operating at the $$1.06 \ \text{\mum}$$ wavelength transition in Nd-doped YAG, use the $$^4\text{F}_{3/2}$$ level of the Nd atom for the upper laser state #2 and the $$^4\text{I}_{11/2}$$ level for the lower laser state #1.

I understand that these are term symbols. However, these are usually taught in the quantum mechanics part of physical chemistry textbooks, rather than physics-focused quantum mechanics textbooks. I have only just begun slowly studying physical chemistry, so I do not yet have any exposure to term symbols. Furthermore, after skimming through the textbook, this looks like it might be the only section (or at least one of the only few sections) that use term symbols. Therefore, I do not want to get hung-up on this part, and would rather keep progressing and then learn the details of term symbols when I get to them in my physical chemistry textbook. So I was wondering if people here with physical chemistry understanding would please take the time to explain what this is saying in a way that does not require understanding of the details of term symbols, so that I may understand "the gist" of this part of the textbook and move on.

• All this is really saying is that the Nd ions in the Nd-doped YAG matrix have extremely similar environments in the crystal. So their interactions with the pump beam are almost the same and a narrow laser emission profile occurs. This is what happens in standard green laser pointers: the Nd-doped YAG emits at 1064 nm and is frequency doubled to 532 nm. In a Nd-doped glass laser medium, the Nd ions are in slightly different environments, because glass is not ordered, so the energy levels are perturbed and emission from the ensemble of perturbed Nd ions results in broader emission profile.
– Ed V
Jan 30 '21 at 17:00
• @EdV Oh, so it's referring to the environments of the Nd ions themselves relative to each other? And, in the glass case, all of the Nd ions are not in extremely similar environments, so the energy levels are perturbed? Jan 30 '21 at 17:17
• Minor fix: in green laser pointers, Nd-doped yttrium orthovanadate is used rather than Nd-doped YAG.
– Ed V
Jan 30 '21 at 17:18
• Correct! Glass is just not the nice ordered crystal environment and this has consequences.
– Ed V
Jan 30 '21 at 17:20
• @EdV Ok, this makes sense to me. Thanks for taking the time to clarify. Jan 30 '21 at 17:21

## 2 Answers

The answer provided by orthocresol provides a very nice explanation of term symbols, so my answer will focus on the laser specifics aspect of the question. To begin, figure 1 shows two laser rods: ruby crystal on the left and Nd-doped phosphate glass on the right. Both laser rods can be optically pumped to achieve the necessary population inversions. The neodymium lasers, i.e., Nd-doped glass or YAG (yttrium aluminum garnet) are 4 level systems, while the ruby laser is a three level system and is not discussed further.

Source: my photo of two of my laser rods.

Side note: The Nd-glass rod is polished on the ends and on bands around the ends. That is where the o-rings would seal the rod in the flashlamp cavity. The ruby rod is only polished on the ends.

Verdeyen 1 shows the usual emission manifolds for the Nd-YAG laser and a typical glass Nd-glass laser.

Source: 1, Fig. 10-5(a) and 10-5(b).

This is what Verdeyen 1 has to say about the emission manifolds:

Only the lowest level in each manifold is shown in Fig. 10-5 (a) and only a few of the important ones are shown in Fig. 10-5(b). For instance, the $$\mathrm{^{4}I_{9/2}}$$ manifold consists of five states extending from $$\pu{0}$$ to $$\pu{857 cm^{-1}}$$, the $$\mathrm{^{4}I_{11/2}}$$ manifold consists of six states extending from $$\pu{2001}$$ to $$\pu{2518 cm^{-1}}$$, and so on. (See Ref. 6 for more details.) One consequence of this large number of states in a given manifold is that the fluorescence line shape is inhomogeneously broadened (even in YAG) and that one would be hard pressed to describe the fluorescent line shape functions, shown in Fig. 10-5 (e) and (f) by a simple mathematical expression.

Source: 1, p. 261. Reference 6 in the quotation is 3 in the references below.

Yariv 2 provides a slightly more detailed emission manifold figure:

Source: 2, Fig. 10.11. Reference 8 in the figure caption is 4 in the references below.

The emission profiles mentioned in the first quotation are shown below:

Source: 1, Fig. 10-5(e) and 10-5(f).

Verdeyen 1 says

The line width of the transitions around $$\pu{1.06 \mu m}$$ for YAG is much smaller than that for glass $$\mathrm{(\Delta \lambda_{YAG} \sim 7 \overset{\lower.5em\circ}{\mathrm{A}} \ll \Delta \lambda_{glass} \sim 300 \overset{\lower.5em\circ}{\mathrm{A}})}$$ but is still asymmetric. The asymmetry is due to the fact that there are two different transitions present in the smooth curve of Fig. 10-5(e) - a strong one at $$\pu{1.0641 \mu m}$$ and a weaker one at $$\pu{1.0645 \mu m}$$ (in about a 3:1 mix). Nevertheless, the huge difference in the line shapes points out the role of the atoms’ local environment. YAG, being a crystalline substance with a well-defined structure, presents a nearly identical environment for each Nd atom and its line width is narrow. Glass, being an amorphous material, presents a different site at each location; consequently, its line shape is much broader.

Source: 1, p. 261.

The transmittance spectrum of a Nd-silicate glass is shown below:

Source: 2, Fig. 10.13. Reference 8 in the figure caption is 4 in the references below.

The strong absorbance just past $$\pu{800 nm}$$ is especially useful for laser diode pumping at $$\pu{808 nm}$$. Inexpensive green laser pointers use yttrium orthovanadate crystals, similar to YAG, excited by IR laser diodes with emission wavelength near $$\pu{800 nm}$$. The resulting IR lasing is near $$\pu{1064 nm}$$, which then gets frequency-doubled to the green $$\pu{532 nm}$$ output. It is important to filter out residual IR, to avoid inadvertent eye damage. Not every inexpensive green laser pointer does this properly.

Also note the strong absorbance just below $$\pu{600 nm}$$: this absorbs the famous sodium D lines extremely well, making “didymium” glass useful for blocking the over-whelming sodium emission that occurs in glass blowing.

One more thing: the OP’s question looks reasonably easy on the surface, but it is easy to find contradictory information. For example, Lakshman and Kumar 5 provide the following branching ratios table:

So they claim that the major lasing line is to the $$\mathrm{^{4}I_{9/2}}$$ manifold, instead of the the $$\mathrm{^{4}I_{11/2}}$$ manifold. N.B. SASP glass is sodium acetate and sodium hexametaphosphate glass, fused and then quick quenched.

References:

1. Joseph T. Verdeyen, Laser Electronics, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1981.

2. Amnon Yariv, Quantum Electronics, 2nd Ed., John Wiley & Sons, New York, 1975.

3. H.G. Danielmey, “Progress in YAG Lasers”, in Lasers, Vol. 4, ed. A.K. Levine and A.J. DeMaria, New York: Marcel Dekker, Inc., 1976, p. 8.

4. E. Snitzer and C.G. Young, “Glass Lasers”, in Lasers, Vol. 2, ed. A.K. Levine, New York: Marcel Dekker, Inc., 1968, p. 191.

5. S.V.J. Lakshman and A. Suresh Kumar, “Lifetimes of laser lines of $$\ce{Pr^{3+}}$$, $$\ce{Nd^{3+}}$$, $$\ce{Er^{3+}}$$ and $$\ce{Tm^{3+}}$$ in acetophosphate glass”, J. Phys. Chem. Solids 1988, 49, 807-811; doi 10.1016/0022-3697(88)90032-7.

• Since I never dared to open the Surelite NdYAG's this might be a naïve question: why does the rod next to «Nd-glass» looks like a polished cap, while the lower end of the rod next to «phosphate glass» still is less glossy (like the ruby rod along all of its mantle)? Oct 17 '21 at 0:00
• The ruby rod is from a 1970s ruby laser. The ends are polished, but the rest is unpolished. It was in a silver-plated water-cooled brass elliptical housing, at one foci and a Xe flashlamp was at the other foci. The ends of the ruby rod has o-rings on them, so the design was not good for several reasons: the ends after the o-rings were not optically pumped, so they were actually detrimental. And the rod was really heavily doped: I have three other ruby rods that are lighter doped: one is basically pink rather than pigeon blood red. The Nd-glass rod is polished on the ends and on bands around
– Ed V
Oct 17 '21 at 0:08
• the ends. That is where the o-rings would seal the rod in the flashlamp cavity. I never used it: it was a gift from a colleague at Corning. The ruby rod in my photo was used, but not by me. It is beautifully dichroic: rotating it about its axis is quite neat. I used to pass it around class (a grad class), telling students to be extremely careful not to drop it. Looking through it is wonderful: the most intense red I will ever see, most likely. By the way, my photography skills are poor, so maybe your question was caused by my bad photo?
– Ed V
Oct 17 '21 at 0:13
• @Buttonwood At CSI XXV in Toronto on June, 1987, I saw 1981 Physics co-Nobelist Arthur Schawlow give a plenary lecture. He brought along his ruby laser in a toy pistol housing and fired it at the projection screen. The red was so intense it appeared to me to be white in the center and it visually seemed to ‘peel off’ the screen. Unforgettable!
– Ed V
Oct 17 '21 at 0:21
• I think it might be worth to add this as a side note to the answer (because comments are prone to be removed at some point) that the upper part of the middle glass rod was polished intentionally on this face and the small portion of the mantle. Otherwise on might get (erroneously) get the impression this were a cap added to the rod. The observation about the «laser pistol» reminds me that some researchers are good to explain their point convincingly to the point, accessible even to the non specialist. Oct 17 '21 at 16:28

I notice that Ed V has addressed the chemistry aspects in the comments. To address the title question about the term symbols themselves: if you have a physics / quantum mechanics background, then actually term symbols are not all that complicated.

When two sources of angular momentum (with quantum numbers $$j_1$$ and $$j_2$$) are coupled, the resulting total angular momentum quantum number has the permitted values

$$j \in \{ j_1 + j_2, j_1 + j_2 - 1, \cdots |j_1 - j_2| \}$$

and the corresponding projection quantum number is

$$m \in \{j, j-1, \cdots, -j\}$$

There are two ways of representing the eigenstates, either the "coupled" representation $$|j m j_1j_2\rangle$$ which uses eigenstates of $$J^2, J_z, J_1^2, J_2^2$$; or the "uncoupled" representation $$|j_1m_1j_2m_2\rangle$$ with eigenstates of $$J_1^2, J_{1z}, J_2^2, J_{2z}$$. All of this is described in any physics textbook (indeed, IMO, it is explained far better in physics textbooks than in chemistry ones).

The term symbols are just shorthand ways of writing the coupled eigenstates $$|jmj_1j_2\rangle$$, where we don't care about the value of $$m$$ (because the $$2j+1$$ states of different $$m$$ are degenerate in the absence of a magnetic field):

$${}^{2j_2 + 1}(j_1)_j$$

In the case of atoms / molecules $$j_2$$ refers to the spin angular momentum and $$j_1$$ to the orbital angular momentum, so it's common to label them as $$s$$ and $$l$$ respectively. Also, chemists often prefer to use capital letters, because the small letters usually refer to single electrons and atoms have many electrons:

$${}^{2S + 1}(L)_J$$

In particular, letters are used in the middle instead of numbers: $$l=0$$ is labelled as S, $$l=1$$ is P, and so on. The full sequence is SPDFGHIK....

So $$^4\mathrm{F}_{3/2}$$ is just refers to the "state" $$S = 3/2$$, $$L = 3$$, and $$J = 3/2$$. Actually, this is not a state: it is a collection of four states with $$m = 3/2, 1/2, -1/2, -3/2$$. Hence, "term" symbol, instead of "state" symbol.

Likewise, $$^4\mathrm{I}_{11/2}$$ means a collection of 12 states with $$S = 3/2$$, $$L = 7$$, and $$J = 11/2$$.