= Thermal Raman line ratio
{c}
{title2=$I_{\rm AS}/I_{\rm S}$}
Assume one harmonic <normal mode> at <thermal equilibrium>, weak spontaneous excitation far from electronic resonances, the same collection <solid angle> and <wave polarization>, and matched response after instrumental correction. The <Boltzmann distribution> gives
$$
r=\exp\left(-\frac{h\nu_v}{k_BT}\right),\qquad
P_v=(1-r)r^v,\qquad
\overline v=\frac{r}{1-r}.
$$
Here $k_B$ is the <Boltzmann constant>, $T>0$ is <temperature>, and $P_v$ is the probability of vibrational <quantum number> $v$. Thus $P_1/P_0=r$. Linear coupling to $Q$ has squared oscillator matrix elements proportional to $v+1$ for <Stokes Raman scattering> and $v$ for <Anti-Stokes Raman scattering>. Summing over initial levels gives strengths proportional to $\overline v+1$ and $\overline v$ respectively; their ratio is $r$, even when levels above $v=1$ are occupied.
Let $I_{\rm S}$ and $I_{\rm AS}$ denote line-integrated <radiant intensities>, meaning <power> per collection <solid angle>. The off-resonant <electric dipole radiation> law adds the fourth power of emitted <frequency>:
$$
\frac{I_{\rm AS}}{I_{\rm S}}
=\left(\frac{\nu_0+\nu_v}{\nu_0-\nu_v}\right)^4
\exp\left(-\frac{h\nu_v}{k_BT}\right).
$$
The <photon> rate divides each line's <power> by its own $h\nu_s$. Consequently, the ideal response-corrected anti-Stokes/Stokes count-rate ratio has the third power of the same <frequency> ratio. Peak heights alone do not give line-integrated <radiant intensities> if linewidths differ. Uncorrected detector counts can also include unequal detector response and unequal <optical filter> transmission. Resonant enhancement, overlapping <spectral lines>, nonthermal populations and local optical heating can invalidate a <temperature> inferred from this simple formula.
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