Just to add to gigacyan's answer, the harmonic oscillator Hamiltonian may be written in terms of raising and lowering operators:
\begin{eqnarray}
\hat{H}\psi&=&-\frac{1}{2m}\frac{\partial^2\psi}{\partial x^2}+\frac{1}{2}m\omega^2x^2\psi\nonumber\\
&=&(a^{\dagger}a+\frac{1}{2})\omega\psi
\end{eqnarray}
where
\begin{equation}
a=\sqrt{\frac{m\omega}{2}}\hat{x}+\frac{i\hat{p}}{\sqrt{2m\omega}}
\end{equation}
and
\begin{equation}
a^{\dagger}=\sqrt{\frac{m\omega}{2}}\hat{x}-\frac{i\hat{p}}{\sqrt{2m\omega}}
\end{equation}
so
\begin{equation}
\hat{x}=\frac{1}{\sqrt{m\omega}}(a+a^{\dagger})
\end{equation}
In other words, the $\hat{x}$ contains a single raising and lowering operator, so facilitates only transitions with $\Delta\nu=\pm1$.
In general, normalizing means making a range go from $0$ to $1$.
Transmittance is the ratio of incident light to transmitted light intensity. It should always be between $0$ and $1$. If the intensity of incident light varies, perhaps because the power source isn't perfect or any other such cause, it changes the transmitted intensity. It can appear that the transmittance $> 1$.
One way of doing a measurement is to measure the intensity of input light once, and then the transmitted spectrum.
However, if the input intensity varies, it is better to continually measure it together with the transmitted intensity. Then dividing removes the apparent variability and "normalizes" the transmittance.
The article you reference mentions something like this:
Light source spectral power variation, interference frin ges due to reflected laser light, and wavelength dependence of the optical components can all cause background variation. For high accuracy applications, the SRM user is advised to normalize the signal to the light source spectrum.
Here is another random article that describes the same thing. Z-Scan Measurements of Optical Nonlinearities
In many practical cases where consider able laser power fluctuations
may occur during the scan, a reference detector can be used to monitor
and normalize the transmittance
Best Answer
There are several different issues conflated together here: selection rules, separation between energy levels, and energy level population (which you didn't mention).
For vibrational spectroscopy, in the approximation that a vibrational mode behaves like a quantum harmonic oscillator, the energy levels are equally spaced and the selection rule is $\Delta n=\pm 1$, where $n$ is the quantum number. The reason for this is explained here. So you expect to see (and do see) an absorption transition from $n=0$ to $n=1$. You might also expect to see a transition from $n=1$ to $n=2$ etc. This would occur at the same frequency since the gap between successive energy levels is the same. However, it relies on there being a thermal equilibrium population of molecules already in the $n=1$ state. For most molecules, at normal temperatures, the population of $n=1$ and higher levels (determined by the Boltzmann factor) is rather low. At elevated temperatures, you might see such transitions; also the frequency won't be exactly at the same frequency as the $n=0\rightarrow 1$ transition, because of anharmonicity effects.
For a linear rotor, the quantum levels are at $BJ(J+1)$ where $B$ is a constant and $J$ is the quantum number. These are not evenly spaced. The selection rule is $\Delta J=\pm 1$ (angular momentum conservation). So you expect to see (and do see) transitions between successive levels: $J=0\rightarrow 1$, $J=1\rightarrow 2$ etc. In this case, at normal temperatures, the spacing between rotational levels is typically small compared with the available thermal energy. So those higher states are populated, at least for $J$ not too high. The frequencies are not all the same, but the energy level spacings change linearly with $J$: $$\Delta E_{J\rightarrow J+1}=B(J+1)(J+2)-BJ(J+1)=2B(J+1).$$ So you see a spectrum with equally spaced lines for $J=0,1,2\ldots$ (in this rigid rotor approximation).
EDIT following OP comment.
And so on. Hence the lines in the spectrum are equally spaced, $2B$ apart (in energy units) or $2B/h$ in frequency units. You can also see a diagram of this in the Linear Molecules section of the Rotational Spectroscopy Wikipedia page (reproduced below under the terms of the CC BY-SA 3.0 licence). The diagram shows the link between the energy levels and the lines in the spectrum (the only difference is that the transitions on the energy level diagram on that page are drawn for emission lines, $J\leftarrow J+1$, but exactly the same frequencies occur for the corresponding absorption lines $J\rightarrow J+1$).