Stokes shift ($\Delta \lambda ={\lambda }_{\mathrm{em},\max }-{\lambda }_{\mathrm{abs},\max }$)

The Stokes shift compares the lowest absorption-band maximum with the fluorescence emission spectrum maximum. Relaxation often makes $\Delta \lambda >0$. This comparison differs from using an arbitrary excitation wavelength. A spectral-density maximum can move when changing from per unit wavelength to per unit spectroscopic wavenumber; specify the spectral convention.^[ref]

An original illustrative energy budget makes the distinction concrete. Excitation at $450\mathrm{nm}$ and emission at $520\mathrm{nm}$ correspond to about $2.76$ and $2.38\mathrm{eV}$ per photon, using $E=hc/\lambda$ with Planck constant $h$ and speed of light $c$ in vacuum. The difference, about $0.37\mathrm{eV}$, must go elsewhere in the complete energy balance. These chosen wavelengths are not measured band maxima for a particular fluorophore.

Anti-Stokes emission has shorter wavelength than excitation. Additional energy, for example from a thermally populated level of molecular vibration or multiple excitation photons, makes it possible. The one-photon, initially relaxed drawing does not cover these cases.^[ref]