The Theory Of Spectral Line Generation From Molecules

A molecular spectral emission line is formed when molecules change from from one energy state to another, lower, energy state, each releasing a photon in the process. The more molecules that do this the more intense the spectral line will be^2.1.

There are three different processes that cause energy level changes; electronic state changes, molecular vibration and angular momentum changes. Electronic level changes lead to spectral lines in the UV or visible regions and vibrational changes lead to spectral lines in the infra red region. However, the spectral lines of interest in millimetre wave radio astronomy are generally caused by changes in rotation energy levels of the molecules (ie. angular momentum changes). Most molecules will have either an electric or a magnetic dipole moment (or both). As the molecule changes its rotation rate (spontaneously or otherwise) the electric field caused by the dipole moment is changed and a photon is released. The photon carries away the energy difference between the 2 rotation states. The photon is detected on Earth and many such transitions will build up a spectral line at frequency $\nu = \frac{E}{h}$ where $E$ is the energy difference between the 2 rotation states and $h$ is Planck's constant.

Molecules themselves can be divided into several different groups, depending upon their construction. Which group a particular molecule falls into depends on the sizes of its principal moments of inertia, designated $I_a$, $I_b$ and $I_c$ and obeying $I_a \leq I_b \leq I_c$. The main types of astrophysical interest are:

 

Table 2.1: Types of Molecules

 

1.	linear molecules	$I_a=0$, $I_b=I_c$
2.	spherically symmetric tops	$I_a=I_b=I_c$
3a.	symmetric tops (oblate)	$I_a=I_b<I_c$
3b.	symmetric tops (prolate)	$I_a<I_b=I_c$
4.	asymmetric tops	$I_a<I_b<I_c$
5.	molecules with hindered internal motion

Spherically symmetric tops

(of which CH$_4$ is the classic example) are self explanatory, simply having all 3 principal moments of inertia identical.

Oblate and prolate symmetric tops

can be thought of as "coin shaped" and "pencil shaped" molecules respectively. Note that linear molecules are simply a special case of prolate symmetric tops. However, it is helpful to consider them separately as they have a much simpler structure - especially the diatomic linear molecules - than the other, more general, types.

Asymmetric tops

are those molecules with no 2 principal moments of inertia the same.

Molecules with internal motion

have a very much more complex spectrum and although they are usually asymmetric tops they are considered separately. This group includes not only internal rotators (eg. CH$_3$OH) but also other types of motion (eg. NH$_3$) involving quantum mechanical tunneling through potential barriers.

It has already been stated that spectral lines caused by rotation level changes in a molecule are caused by the loss of rotational energy. Classical mechanics states that rotational energy is given by $E_{rot}=\frac{1}{2}I \omega^2$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. $I$ and $\omega$ are related by $P=I\omega$ where $P$ is the angular momentum. This of course applies to each of the principal axes ie. $P_a=I_a\omega _a$, etc. and so:

$$E_{rot}=\frac{1}{2}I_a\omega_a^2+\frac{1}{2}I_b\omega_b^2+\frac{1}{2}I_c\omega_c^2$$ (2.1)

which can also be written:

$$E_{rot}=\frac{1}{2}\left(\frac{P_a^2}{I_a}\right)+\frac{1}{2}... ...P_b^2}{I_b}\right)+ \frac{1}{2}\left(\frac{P_c^2}{I_c}\right)$$ (2.2)

The components of $I$ are given by:

$$I_a=\sum_{i}m_i\left(y_i^2+z_i^2\right), \hspace{1cm} I_b=\su... ...\right), \hspace{1cm} I_c=\sum_{i}m_i\left(x_i^2+y_i^2\right).$$ (2.3)

where the $x_i$, $y_i$ & $z_i$ are aligned along the principal axes a, b & c respectively, and the origin is taken as the centre of mass of the molecule. The $m_i$'s are the masses of the individual atoms making up the molecule.

Similarly the components of $P$ are given by:

$$P_a=\sum_{i}m_i\left(y_i^2+z_i^2\right)^{\frac{1}{2}} \left(... ... \left(v_{x_i}^2+v_{z_i}^2\right)^{\frac{1}{2}}, \hspace{1cm}$$

$$P_c=\sum_{i}m_i\left(x_i^2+y_i^2\right)^{\frac{1}{2}} \left(v_{x_i}^2+v_{y_i}^2\right)^{\frac{1}{2}}.$$ (2.4)

where the $v$'s are the components of velocity. Also, by definition $P^2=P_a^2+P_b^2+P_c^2$. The commutation relations between P and its components can be summarised (see eg. Gordy & Cook [9], P13) by $\mathbf{P \times P=i \hbar P}$ (and similarly for the electronic spin vector $\mathbf{S}$ and the nuclear spin vector $\mathbf{I}$). From these 3 equations it can be shown (eg. King, Hainer & Cross [17]) that:

$$P^2\psi_{J,M}=\hbar^2J\left(J+1\right)\psi_{J,M}$$ (2.5)

$$P_z\psi_{J,M}=\hbar M \psi_{J,M}$$ (2.6)

where $\psi_{J,M}$ are the common eigenfunctions of $P^2$ and $P_z$ and $J\in \mathcal{Z}^++\left\{0\right\}$ and $M=J, J-1, \ldots, -J$. It can further be shown that pre multiplying both sides of equations 2.5 and 2.6 by $\psi^*_{J,M}$, the complex conjugate of $\psi_{J,M}$, yields:

$$\int \psi^*_{J,M}P^2\psi_{J,M}d\tau=\hbar^2J\left(J+1\right)$$ (2.7)

and

$$\int \psi^*_{J,M}P_c\psi_{J,M}d\tau=\hbar^2J\left(J+1\right)$$ (2.8)

this last step is possible since

$$\int \psi^*_{J,M}\psi_{J,M}d\tau=1$$ (2.9)

Next consider the Hamiltonian operator, this is related to the classical Hamiltonian $\frac{p^2_x}{2m}+V$ (in 1-D) where $p_x$ is the momentum and $V$ is the potential energy. For a molecule with no external forces acting only the kinetic energy contributes and so:

$$\mathcal{H}=\frac{1}{2}\left(\frac{P_a^2}{I_a}+\frac{P_b^2}{I_b}+\frac{P_c^2}{I_c}\right)$$ (2.10)

where the $P$'s are conjugate angular momentum operators. Now it is possible to work out the expected frequencies for different transitions since

$$E_J=\int \psi^*_{J,M}\mathcal{H}\psi_{J,M}d\tau$$ (2.11)

It will now be useful to consider the application of this equation for one molecule in detail. The simplest possible kind of molecule is one that contains just 2 atoms - ie. a diatomic linear molecule. From an astrophysical point of view probably the most interesting observable diatomic molecule is CO and its isotopes . Other molecules such as H$_2$ would be more interesting but are unfortunately unobservable as they have no dipole moment.

From the definition of a linear molecule $P_a=0$, $I_a=0$, $I_b=I_c=I$ and so $\mathcal{H}=\frac{P^2}{2I}$. Now using equations 2.7, 2.10 & 2.11

 

$$E_J = \int \psi^*_{J,M}\frac{1}{2}\frac{P^2}{I}\psi_{J,M}d\tau$$

$$= \frac{1}{2I}\int \psi^*_{J,M}P^2\psi_{J,M}d\tau$$

$$= \frac{1}{2I} \hbar^2 J(J+1)$$ (2.12)

It can be shown (eg. Gordy & Cook [9] section 2.6) that the selection rules for dipole transitions are $\Delta J=\pm 1$ so

$$h\nu =E_{J+1}-E_J=\frac{\hbar^2}{2I}\left[\left(J+1\right)\le... ...)-J\left(J+1 \right)\right]=\frac{\hbar^2}{2I}\left[J+1\right]$$ (2.13)

so

$$\nu=\frac{h}{4\pi^2I}\left(J+1\right)$$ (2.14)

This is conventionally written $\nu=2B\left(J+1\right)$ where $B=\frac{h}{8\pi^2I}$, the rotation constant. It can therefore be seen that the rotation spectrum will consist of lines at $\nu=2B,4B,6B,\ldots$. Consider for example CO, this has:

$m_C=2 \times 10^{-26}\:{\rm kg}$

$m_O=2.67 \times 10^{-26}\:{\rm kg}$

$d_{CO}=1.1282 \times 10^{10}\:{\rm m}$

The centre of mass is located on the CO axis $0.483 \times 10^{-10} {\rm m}$ from the O atom. Using equations 2.3, $I$ can next be calculated as $I=1.455 \times 10^{-46} {\rm kg}\, {\rm m}^2$. This then yields $B=57.68\:{\rm GHz}$ which would predict the $J=1 \rightarrow 0$transition at $\nu=115.35\:{\rm GHz}$ compared to the measured value of $\nu=115.27\:{\rm GHz}$. The slight difference here is due to ignored minor effects, for example, as the molecule rotates centrifugal effects will cause the bond length to increase slightly, thus altering the rotation constant.

Unfortunately not all diatomic molecules are as simple as CO. If the molecule has an unbalanced electronic angular momentum this will lead to a splitting of the energy levels due to the coupling of the angular momentum vectors. An unbalanced electronic angular momentum is caused by there being an unpaired electron present. Such molecules have historically been difficult to study as they are generally unstable and therefore short lived on Earth. However, in the interstellar medium several are observable, with CN being one of the prime examples.  

Before continuing it will be instructive to review some notation. Unfortunately this notation is not necessarily consistent with that used in atomic spectroscopy. The total angular momentum of a molecule is made up of several components, namely the angular momentum caused by the rotation of the whole molecule, the orbital angular momentum of the electrons, the electron spin angular momentum and the nucleus spin angular momentum. Each of these is quantised and assigned a quantum number as follows:

 

In addition the value K occurs in many formulae and is the component of total angular momentum along the principal axis of the molecule.

The vector $\mathbf{N}$, which represents the total angular momentum minus the electron spin angular momentum is made up out of $\mathbf{\Lambda}$ and $\mathbf{O}$ (ie. $\mathbf{N=\Lambda+O}$) where $\mathbf{\Lambda}$ is the vector component of $\mathbf{L}$directed along the internuclear axis and $\mathbf{O}$ is end over end rotation angular momentum. The component of $\mathbf{L}$ (orbital electron angular momentum) along the internuclear axis is usually given as $\Lambda=\vert\mathbf{\Lambda}\vert=0,1,2,\ldots,L$. The values of $\Lambda$ are termed the different electronic states for the molecule and are assigned the letters $\Sigma, \Pi, \Delta, \Phi, \ldots$ for $\Lambda=0,1,2,3,\ldots$respectively (eg. a $\Pi$ state has $\Lambda=1$). To confuse things even more, for values of $\Lambda \geq1$ the electronic spin angular momentum interacts with the spin magnetic moment. This necessitates another quantum number to measure the component of $\mathbf{S}$ along the internuclear axis - this is denoted $\Sigma$ (not the same $\Sigma$ as for $\Lambda=0$! For $\Lambda=0,\:\: \Sigma$ is undefined). It takes the values $\Sigma=S,S-1,\ldots,-S$. The angular momentum along the internuclear axis is then $\Omega=\vert\Lambda+\Sigma\vert$. So $\Omega$ represents the total angular momentum due to the electrons for a non-rotating molecule. Note that in the above diagram $\mathbf{F}$ is only used for molecules in which there is coupling with the nuclear spin, otherwise $\mathbf{F=J}$ in which case $\mathbf{J}$ is the quantum number quoted for total angular momentum. Finally, the nomenclatures are often combined in the format $^{2S+1}(letter for \Lambda)_{\Lambda+\Sigma}$ where $2S+1$ is the multiplicity (eg. for $\Lambda=2, \:\:S=1\: :\: ^3\Delta_1,\:\:^3\Delta_2,\:\:^3\Delta_3$ are possible) and ' $letter for \Lambda$' refers to the letters used to represent the various values of $\Lambda$ as listed above. Most diatomic molecules in their lowest state are in the $^1\Sigma$ state. However, CN in its lowest state is in the $^2\Sigma$ state. 

The various angular momentum vectors present in a $^2\Sigma$ molecule can be represented as in figure 2.2 (from Gordy & Cook [9]). This is referred to as Hund's case (b). This is an idealised case where the molecule has no orbital momentum but does possess electronic spin. This causes the spin $\mathbf{S}$ to be coupled with $\mathbf{N}$. The dotted circles represent the movement of the angular momentum vectors around a given axis. For example the $\mathbf{L}$ vector precesses about the internuclear axis and the $\mathbf{S}$ and $\mathbf{N}$ vectors both precess about the $\mathbf{J}$ axis. Note also that the entire molecule precesses about the $\mathbf{N}$axis.

Hund's case (b)   To calculate the effect that this has on the spectrum of the molecule the Hamiltonian for the magnetic interaction must be considered. As the field generated by the rotation of the molecule is proportional to both $\mathbf{N}$ and $\mathbf{S}$, this will be given by

$$\mathcal{H}=-\gamma \mathbf{S}\cdotp \mathbf{N}$$ (2.15)

where $\gamma$ is the constant of proportionality (it is labeled negative here so that $\gamma$ will be positive). Now considering the vectors $\mathbf{J}$, $\mathbf{N}$ and $\mathbf{S}$ together with the cosine law, the situation in figure 2.2 can be represented by

$$\mathbf{J^2=N^2+S^2-2\vert S\vert\vert N\vert}\cos(\mathbf{S,N})$$ (2.16)

Rearranging this and substituting into 2.16 to replace the dot product yields

$$\mathcal{H}=\frac{\gamma}{2}\left(\mathbf{J^2-S^2-N^2}\right)$$ (2.17)

It can be shown that this leads to

$$\frac{E}{h}=\frac{\gamma}{2}\left[J(J+1)-S(S+1)-N(N+1)\right]$$ (2.18)

with $J=N+S,\:N+S-1,\ldots,\:\vert N-S\vert$ and $N=\mathcal{Z}^++\{0\}$. So for a $^2\Sigma$ molecule (eg. CN) where $S=\frac{1}{2}$ the possible values of $J$ are $J=N+\frac{1}{2},\:N-\frac{1}{2}$. Note that as there is only one unpaired electron $J=N$ is not a possibility (as it is for a $^3\Sigma$ state). It can therefore be seen that $\frac{E}{h}=\frac{\gamma N}{2},\: -\frac{\gamma}{2} (N+1)$. So combining with equation 2.13 gives the two split energy levels as:

$$\frac{E}{h} = BN(N+1)+\frac{\gamma N}{2}$$ (2.19)

$$BN(N+1)-\frac{\gamma}{2}(N+1)$$ (2.20)

Figure 2.3 shows how these split energy levels can be represented.

For some molecules such as CN, however, this is still not the full story. There is the effect of nuclear spin which leads to the hyperfine splitting of the energy levels. Called hyperfine because the splitting generally has an very small effect it is nonetheless an important process that can be very useful in the analysis of molecules such as CN because it splits what would be just one line for a particular transition into several sub-lines of different intensities. Since these sublines have different intensities they also have different optical depths and it is thus possible to gain more information about the structure of a cloud than would be possible if the line were not split.

Since the quantum mechanical description of hyperfine splitting is rather complex it will not be described here.

Energy levels  

Transition types