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Femtosecond stimulated Raman spectroscopy, or FSRS, measures a Raman spectrum through gain induced in a short broadband probe by a narrowband Raman pump. At a molecular vibration frequency matching the pump–probe frequency difference, Stokes Raman scattering transfers energy from the pump into the probe. The narrow pump and spectrograph supply spectral selectivity while the short probe samples the system in time; the probe duration alone is not the spectral linewidth.^[ref]

The2003 apparatus of David W. McCamant, Philipp Kukura and Richard A. Mathies filtered a Ti:sapphire amplifier output to make a narrowband pump near $793\mathrm{nm}$ and generated a broadband probe in sapphire. Overlapped beams traversed the sample; a spectrograph dispersed probe and reference spectra onto a dual-diode array. The paper demonstrates gain bands in cyclohexane.^[ref]

After dark correction, let $S$ and $R$ denote probe and reference counts. The Raman pump-on pump-off gain ratio isFor an original chosen spectral channel, take $(S,R{)}_{\mathrm{off}}=(10000,10000)$ and $(S,R{)}_{\mathrm{on}}=(11440,11000)$. The normalized ratio is $1.04$, a four-percent Raman gain; the unnormalized probe ratio $1.144$ would mistake a source fluctuation for additional gain. This assumes a linear detector and a reference that tracks that fluctuation.

A second original calculation uses an $800\mathrm{nm}$ pump and a $1000\mathrm{c}{\mathrm{m}}^{-1}$ vibration. The Stokes probe channel is at wavenumber $12500-1000=11500\mathrm{c}{\mathrm{m}}^{-1}$, or about $869.6\mathrm{nm}$. A broadband probe can sample many such differences simultaneously. Spatial and temporal overlap, wavelength calibration, probe chirp and non-Raman nonlinear backgrounds still require characterization. A short pulse label does not by itself establish the time resolution of a changing molecular response.

The gross requirement is an anisotropic polarizability tensor, so that changing molecular orientation changes the induced response. A permanent electric dipole moment is not required; a homonuclear diatomic can therefore have a rotational Raman scattering spectrum.

For the inelastic lines of a linear rigid rotor, the detailed rule is $\Delta J=\pm 2$, with $J\ge 0$. The $\Delta J=0$ contribution is unshifted Rayleigh scattering, not a rotational Stokes Raman scattering line. Stokes Raman scattering has $\Delta J=+2$ because the molecule gains rotational energy.

The Stokes vibrational band has O, Q and S branches, with $\Delta N=-2,0,+2$, respectively. Here $N$ denotes end-over-end rotational angular momentum; treating electronic fine structure as unresolved gives the usual elementary rotor calculation. The absolute incident wavenumber isand energy conservation for Stokes Raman scattering givesFor the actual ${}^{16}{\mathrm{O}}_2$ ground state, only odd $N$ have nonzero nuclear-spin statistical weight. Thus the lowest initial levels are $N=3$ for O and $N=1$ for Q and S. Ignoring the small change in $B$ upon vibration, the lines are:

The odd-$N$ restriction follows from the zero nuclear spin and exchange symmetry of ${}^{16}{\mathrm{O}}_2$; the electronic spin splits each such $N$ into several $J$ levels without changing this simplified branch-centre calculation.^[ref]

If the phrase simplest approximation is instead taken to discard nuclear-spin statistics as well, a fictitious unrestricted rotor gives O $2\to 0$ at $7826.9{\mathrm{cm}}^{-1}$ and S $0\to 2$ at $7810.1{\mathrm{cm}}^{-1}$, with Q still at $7818.5{\mathrm{cm}}^{-1}$. Those even-$N$ lines are not physical ground-state ${}^{16}{\mathrm{O}}_2$ lines.

Raman scattering is inelastic scattering of light in which the outgoing photon has a different frequency because matter gains or loses internal energy. Molecular vibration is a particularly useful recipient: a visible photon can reveal an internal energy spacing normally associated with infrared frequencies. Rayleigh scattering supplies the much stronger, approximately unchanged-frequency reference. The distinction concerns the transferred energy, rather than merely a change of direction or the colour of the illuminating beam.

The elementary bookkeeping is conservation of energy:Here $h$ is the Planck constant, ${\nu }_0$ and ${\nu }_s$ are the incident and scattered photon frequencies, and $E_i,E_f$ are initial and final internal molecular energies. This expression neglects translational recoil and Doppler broadening; those can matter in sufficiently precise gas measurements. A positive internal energy change gives Stokes Raman scattering, and a negative one gives Anti-Stokes Raman scattering. The energy difference determines a possible line position, but it does not by itself establish that a transition is optically allowed or predict its strength.

The spontaneous molecular process developed here has a connection to the Raman transition used in atomic measurement methods: two photon interactions transfer an internal energy difference. Coherently driving that process with two applied light fields is a different experimental regime from collecting weak spontaneous Raman scattering beside an illuminating beam.

The signed Raman shift subtracts scattered from incident spectroscopic wavenumber:Here $c$ is the speed of light, and both wavelengths are in vacuum. Positive shifts denote Stokes Raman scattering; negative shifts denote Anti-Stokes Raman scattering. Some plots reverse the horizontal direction or report separate unsigned magnitudes, so check the convention. Spectroscopic wavenumber has no $2\pi$ factor, unlike the common angular wavenumber $k=2\pi /\lambda$. A shift of $1000\mathrm{c}{\mathrm{m}}^{-1}$ means $100000{\mathrm{m}}^{-1}$, not a frequency of $1000\mathrm{Hz}$.

For a one-quantum change of molecular vibration with ordinary frequency ${\nu }_v$, the magnitude is ${\tilde{\nu }}_v={\nu }_v/c$. This makes Raman spectroscopy attractive: the same molecular energy spacing can be read using different visible excitation wavelengths. Equal and opposite Raman shifts are not equal and opposite changes measured in nanometres, because wavelength is the reciprocal of spectroscopic wavenumber.

A Raman spectrum records the strength of Raman scattering as a function of scattered optical frequency or, usually, Raman shift relative to the excitation. For vacuum wavelengths in nanometres, a common wavenumber convention isHere the numerical wavelengths are in nanometres. Stokes Raman scattering has positive shift under this convention; Anti-Stokes Raman scattering has negative shift. The strong unshifted Rayleigh scattering line is not a vibrational Raman band.

For an original chosen excitation ${\lambda }_0=500\mathrm{nm}$ and a $1500\mathrm{c}{\mathrm{m}}^{-1}$ vibration, the Stokes wavenumber is $20000-1500=18500\mathrm{c}{\mathrm{m}}^{-1}$, giving ${\lambda }_s\approx 540.54\mathrm{nm}$. The corresponding anti-Stokes wavelength is $10^7/21500\approx 465.12\mathrm{nm}$. Changing the excitation changes these optical wavelengths while the same assumed transition retains its Raman shift. The One-mode Raman spectrum example gives an existing level diagram and a separate numerical spectrum.

The vertical quantity must be stated too: detector counts per spectral bin, calibrated scattered power density and a stimulated gain ratio are different observables. Femtosecond stimulated Raman spectroscopy can plot the Raman pump-on pump-off gain ratio rather than spontaneous photon counts. Its primary apparatus disperses probe and reference beams onto an array and calibrates shifts with known solvent lines.^[ref]

Spectral-density conversion between wavelength and Raman shift also changes the density's units: rebadging the horizontal axis alone need not preserve a spectral integral. Background, excitation rejection, instrument response and spectral resolution affect observed band heights and widths. An integrated band area is not determined by peak height alone. A chemical inference should compare calibrated positions and responses under specified conditions, rather than identify a material from one unqualified peak.

The simplest off-resonant explanation is a vibrating polarizability. Let $Q(t)=Q_0\cos (2\pi {\nu }_{v}t)$ be a small normal mode coordinate of molecular vibration, and use one scalar component of polarizability, $\alpha (Q)\simeq {\alpha }_0+{\alpha }_0^{\prime }Q$. For incident electric field $E(t)=E_0\cos (2\pi {\nu }_{0}t)$, the induced electric dipole moment isThe product-to-sum formula produces the two displaced frequencies. A nonzero derivative ${\alpha }_0^{\prime }$ gives vibrational Raman scattering in this linear model. Real polarizability is generally a tensor, and the selected components depend on incident and observed wave polarization. This condition differs from vibrational infrared activity, which requires a changing molecular electric dipole moment without an applied optical electric field. A vibration can therefore appear in one measurement and be absent in the other.

This classical modulation argument explains line locations but supplies equal sideband wave amplitudes; it does not explain their thermal asymmetry or the nonzero Stokes Raman scattering from a vibrational ground state. Those require the quantum harmonic oscillator population and transition strengths. The far-off-resonance drawing uses dashed virtual intermediate levels: they represent terms in a scattering probability amplitude, not a long-lived occupied electronic energy eigenstate. Near an electronic resonance, a more detailed treatment of the intermediate response is needed.

In Stokes Raman scattering, matter gains internal energy and the photon loses the same amount. For a quantum harmonic oscillator representing one molecular vibration, $E_v=h{\nu }_v(v+1/2)$ with $v=0,1,\dots$. The transition $v\to v+1$ gives ${\nu }_s={\nu }_0-{\nu }_v$, requiring ${\nu }_0>{\nu }_v$ for an outgoing photon. Even a molecule initially in its vibrational ground state can undergo this process. The shared name does not turn it into the absorption/emission-band Stokes shift used for fluorescence.

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 givesHere $k_B$ 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 $I_{\mathrm{S}}$ and $I_{\mathrm{AS}}$ 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: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.