What happens to the $2\pi$ factor when calculating Raman-shifts in units of wavenumbers?

Question

So from the classical theory, you find a formula for a dipole in a planar electromagnetic wave, where there will be two cosine terms with a frequency (actually angular velocity in $[rad/s]$, as the argument of a trigonometric function should be dimensionless and it is multiplied by $t[s]$)

$\omega_i \pm \omega_n [rad/s]$

corresponding with the Stokes & anti-Stokes Raman shifts. ($i$ for impinging and $n$ for denoting the $n$th normal vibrational mode)

The wikipedia page states that

$\Delta \omega_{\pm} = \frac{1}{\lambda_i}\pm \frac{1}{\lambda_n}$

Now here is something that alarms me: this is not the same $\omega$ (though it is proportional to it), it can't be. The units are all wrong. On top of that, even if you would just accept that assignment to $\Delta \omega$, then it's not a wavenumber, as the wavenumber $k$ is defined as

$k=\frac{2\pi}{\lambda} [rad/m]$

I understand the cm$^{-1}$ part (magnitudes are then usually between 0-2000, which is perfect), but what happened to the $2\pi [rad]$ factor? Is the "Raman shift" actually in terms of "inverse wavelength" ? Do you have to multiply $\Delta \omega$ by $2\pi [rad]$ to get the $\text{actual}$ wavenumber?

Extra remark: nowhere in the wikipedia-article is implied what $\omega$ should be (which is something I (possibly wrongly) assume), though I find it a bit confusing to use the symbol commonly known as the angular velocity ($\propto$ frequence) for denoting a difference in wavenumbers (usually $k$)

Answer 1

There are two definitions for the wavenumber, one is $k=\frac{2\pi}{\lambda}$, the number of wavelengths per $2\pi$ units of distance (i.e. on the circle), the other one is $k=\frac{1}{\lambda}$, which is the number of wavelengths per unit distance. The formula for the Raman shift is refering to the latter definition.