Determining Population Inversion with Level Degeneracy

Welcome back. The preceding lesson established the induced transition rates and, crucially, the Einstein-coefficient relation

$$g_1{B}_{12}=g_2{B}_{21}.$$

That relation means that comparing the total populations $N_1$ and $N_2$ is not always enough to tell whether a resonant field will be attenuated or amplified. In this lesson, the goal is to make the degeneracy-corrected inversion test automatic: given populations and degeneracies, classify the transition as absorptive, transparent, or inverted.

The result is local to a specified optical transition. A medium may be inverted on one transition while being absorptive on another, and inversion alone is not yet the same thing as a laser oscillating; cavity feedback and losses enter later. But inversion is the essential material condition for stimulated optical gain.

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Population lives in levels; transitions begin in substates

A reported level population such as $N_2$ is usually the total number density in an energy manifold. If the upper manifold contains $g_2$ degenerate substates, and those substates are equally populated, then the number density in any one upper substate is

$$\frac{N_2}{g_2}.$$

Likewise, the population per lower substate is

$$\frac{N_1}{g_1}.$$

Degeneracy can arise from, for example, unresolved angular-momentum projections, spin states, or closely spaced Stark or Zeeman sublevels. The key point is statistical: a manifold with more substates can hold a larger total population without any individual substate being unusually occupied.

This is why the phrase population inversion has a precise meaning:

The upper laser level must have a greater population per available state than the lower laser level.

The Oxford lecture notes give the compact statement of this criterion and contrast it with the thermal Boltzmann distribution.

A useful visual distinction appears in the four-level optical-gain diagram below. Its high-excitation panel labels gain by comparing $n_3$ and $n_4$; that simple comparison presumes equal effective degeneracies. When the degeneracies differ, replace it with the population-per-substate comparison developed here.

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Deriving the correct inversion criterion

For radiation resonant with the transition, the ensemble rates are

$$R_{\mathrm{abs}}=N_1{B}_{12}\rho ,$$ $$R_{\mathrm{st}}=N_2{B}_{21}\rho .$$

Here $\rho$ is the appropriate resonant spectral energy density. For the present question, its precise magnitude does not matter: it multiplies both rates and cannot change which one is larger.

Optical gain requires stimulated emission to exceed absorption:

$$R_{\mathrm{st}}>R_{\mathrm{abs}}.$$

Substituting the rates gives

$$N_2{B}_{21}\rho >N_1{B}_{12}\rho .$$

Using

$$B_{12}=\frac{g_2}{g_1}{B}_{21},$$

and cancelling the positive factors $B_{21}$ and $\rho$, gives

$$N_2>\frac{g_2}{g_1}{N}_1.$$

Dividing each side by $g_2$ produces the form to remember:

$$\boxed{\frac{N_2}{g_2}>\frac{N_1}{g_1}}$$

This is the population-inversion condition for the transition from level 2 to level 1.

The complete classification is:

Degeneracy-corrected population comparison	Induced response at the transition	Classification
$\frac{N_2}{g_2}>\frac{N_1}{g_1}$	Stimulated emission exceeds absorption	Population inversion; positive stimulated gain
$\frac{N_2}{g_2}=\frac{N_1}{g_1}$	Stimulated emission balances absorption	Transparency
$\frac{N_2}{g_2}<\frac{N_1}{g_1}$	Absorption exceeds stimulated emission	Absorption

It is useful to define a degeneracy-corrected inversion density,

$$\Delta N_{\mathrm{inv}}=N_2-\frac{g_2}{g_1}{N}_1.$$

Its sign alone gives the answer:

$$\Delta N_{\mathrm{inv}}\{\begin{array}{ll}>0,& \text{inverted},\\ =0,& \text{transparent},\\ <0,& \text{absorptive}.\end{array}$$

Later, when gain is expressed in terms of stimulated-emission and absorption cross sections, this same quantity controls the sign of the small-signal gain coefficient.

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Three examples that expose the role of degeneracy

Example 1: equal degeneracies

Suppose

$$g_1=g_2=2,$$ $$N_1=7.0\times 10^{18}{\mathrm{m}}^{-3},N_2=9.0\times 10^{18}{\mathrm{m}}^{-3}.$$

Because the degeneracies are equal, the normalization changes neither comparison:

$$\frac{N_1}{g_1}=3.5\times 10^{18}{\mathrm{m}}^{-3},$$ $$\frac{N_2}{g_2}=4.5\times 10^{18}{\mathrm{m}}^{-3}.$$

Therefore,

$$\frac{N_2}{g_2}>\frac{N_1}{g_1}.$$

The transition is inverted. In this special case, the familiar shorthand $N_2>N_1$ happens to be valid.

Example 2: more total population in the upper level, but no inversion

Now take

$$g_1=2,g_2=4,$$ $$N_1=8.0\times 10^{18}{\mathrm{m}}^{-3},N_2=1.4\times 10^{19}{\mathrm{m}}^{-3}.$$

At first glance, the upper level looks favorable because

$$N_2>N_1.$$

But there are twice as many upper substates. The relevant populations are

$$\frac{N_1}{g_1}=4.0\times 10^{18}{\mathrm{m}}^{-3},$$ $$\frac{N_2}{g_2}=3.5\times 10^{18}{\mathrm{m}}^{-3}.$$

Thus,

$$\frac{N_2}{g_2}<\frac{N_1}{g_1}.$$

The medium is still absorptive on this transition. The upper manifold contains more atoms in total, but not enough atoms per substate to overcome absorption.

Equivalently, the threshold upper-level population would have been

$$N_{2,\mathrm{trans}}=\frac{g_2}{g_1}{N}_1,$$ $$N_{2,\mathrm{trans}}=\frac{4}{2}\left(8.0\times 10^{18}\right),$$ $$N_{2,\mathrm{trans}}=1.6\times 10^{19}{\mathrm{m}}^{-3}.$$

The actual population $1.4\times 10^{19}{\mathrm{m}}^{-3}$ lies below that transparency threshold.

Example 3: fewer total upper-level atoms, yet inversion

Finally, reverse the degeneracy advantage:

$$g_1=6,g_2=2,$$ $$N_1=1.2\times 10^{19}{\mathrm{m}}^{-3},N_2=5.0\times 10^{18}{\mathrm{m}}^{-3}.$$

This time,

$$N_2<N_1.$$

Nevertheless,

$$\frac{N_1}{g_1}=2.0\times 10^{18}{\mathrm{m}}^{-3},$$ $$\frac{N_2}{g_2}=2.5\times 10^{18}{\mathrm{m}}^{-3}.$$

Hence,

$$\frac{N_2}{g_2}>\frac{N_1}{g_1}.$$

The transition is inverted even though fewer atoms occupy the upper manifold in total. Its smaller degeneracy means that each upper substate is, on average, more populated than each lower substate.

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A reliable procedure for any population data

When given a candidate laser transition, use this short procedure.

1. Identify the two manifolds participating in the optical transition.
   Label the lower laser level $1$ and upper laser level $2$. Ignore populations in other levels unless they affect the stated $N_1$ or $N_2$.

2. Confirm what the quoted populations mean.
   Usually $N_1$ and $N_2$ are totals over all substates in the respective manifolds.

3. Identify the corresponding degeneracies.
   Use $g_1$ and $g_2$ for those same manifolds. Do not mix a population summed over several substates with a degeneracy referring to only one.

4. Compute population per substate.

   $$n_1^{\mathrm{sub}}=\frac{N_1}{g_1},n_2^{\mathrm{sub}}=\frac{N_2}{g_2}.$$

5. Compare the two quantities.
   The sign of

   $$n_2^{\mathrm{sub}}-n_1^{\mathrm{sub}}$$

   classifies the transition as gain, transparency, or loss.

Two compact alternatives are often faster:

$$\frac{N_2}{N_1}>\frac{g_2}{g_1}$$

for inversion, or

$$N_2>\frac{g_2}{g_1}{N}_1.$$

These forms are equivalent, but the population-per-substate form is usually safest conceptually.

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Thermal equilibrium and the need for pumping

At positive temperature, thermal equilibrium imposes

$$\frac{N_2/g_2}{N_1/g_1}=\exp \left(-\frac{E_2-E_1}{k_{\mathrm{B}}T}\right).$$

For $E_2>E_1$, the exponent is negative, so

$$\exp \left(-\frac{E_2-E_1}{k_{\mathrm{B}}T}\right)<1.$$

Therefore,

$$\frac{N_2}{g_2}<\frac{N_1}{g_1}$$

in ordinary thermal equilibrium. A material at equilibrium is absorptive on an upward optical transition, even if a large upper-level degeneracy makes its total $N_2$ appear substantial.

A laser must therefore be driven out of thermal equilibrium by pumping and relaxation processes. The central engineering task of a laser medium is not merely to put population into an excited manifold, but to create and maintain

$$\frac{N_2}{g_2}>\frac{N_1}{g_1}.$$

One final qualification matters in real media. The simple division by $g$ assumes the degenerate substates are equally populated, as occurs when collisions, relaxation, or averaging over an isotropic ensemble redistribute population efficiently. Strongly polarized pumping, magnetic fields, or slow sublevel mixing can produce unequal magnetic-sublevel populations. Then the relevant absorption and emission rates must be calculated by resolving the individual substates and allowed polarization-dependent transitions, rather than by applying a single degeneracy factor.

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Takeaways

Population inversion is a statement about a specified transition, not simply about whether an excited level holds more particles overall. The correct test is

$$\boxed{\frac{N_2}{g_2}>\frac{N_1}{g_1}}$$

where $N_i$ is the total population of level $i$, and $g_i$ is the number of relevant degenerate substates.

Remember the three outcomes:

• Upper population per substate larger: inversion and stimulated gain.
• Equal populations per substate: transparency.
• Lower population per substate larger: absorption.

Next, we will turn from population balance to spectral structure: homogeneous and inhomogeneous broadening, and how different physical mechanisms shape a laser transition’s linewidth.