Appendix: A Properties of the CCS radical

CCS is a linear radical whose electronic ground state is the $^{3}{\Sigma}$configuration (Saito et al. 1987), i.e., the projection of the orbital angular momentum onto the molecular axis vanishes, ${\Lambda}$ = 0, and spin angular momentum S = 1. Since the electronic orbital angular momentum $\vec{L}$ is far more strongly coupled to the molecular axis than the spin (at least for the slow molecular rotations considered here), the intrinsic magnetic field due to the molecular rotation is expected to couple the spin magnetic moment to the molecular axis. The limiting case is usually described in terms of Hund's case b (e.g., Townes & Schawlow 1975). In the case of CCS, however, the dipole-dipole interaction between the two unpaired electrons is unusually large. For small rotational and projected orbital angular momenta, $\vec{R}$ and $\vec{\Lambda}$, i.e., small $\vec{N}$(= $\vec{R}$ + $\vec{\Lambda}$), the intrinsic magnetic field associated with the molecular rotation becomes so small that the spin-spin interaction dominates the coupling of the electronic spin onto the molecular axis. This case is rather described by Hund's case a, where the total angular momentum $\vec{J}$ results from the coupling of $\vec{\Sigma}$ and $\vec{N}$; the angular momentum $\vec{\Sigma}$ is the projection of $\vec{S}$ onto the molecular axis, and, correspondingly, $\vec{\Omega}$ is the projection of $\vec{J}$.

The energy associated with the magnetic dipole interaction of a paramagnetic molecule with an external magnetic field $\vec{B}$ is given by

$${\Delta}{E}_{B} = - \vec{\mu}_{\rm JN} \vec{B},$$

where $\vec{\mu}_{\rm JN}$ is the magnetic moment of the CCS radical in state J_N, associated with the total angular momentum operator $\vec{J}$,

$$\vec{\mu}_{\rm JN} = - g_{\rm JN}~ \frac{\mu_{\rm B}}{\hbar}~{\vec{J}}.$$ (A.1)

Generally, $\vec{\mu}_{\rm JN}$ results from individual contributions (e.g., Judd 1975),

$$\vec{\mu}_{\rm JN} = - \frac{\mu_{\rm B}}{\hbar} (g_{L} {\ve... ...\frac{\mu_{\rm n}}{\hbar} (g_{I} {\vec{I}} + g_{R} {\vec{R}}),$$

where $\mu_{\rm B}$ is the Bohr magneton, $\mu_{\rm n}$ is the nuclear magneton, $\vec{I}$ is the nuclear spin, $\vec{L}$ is the electronic orbital and $\vec{N}$ the total orbital angular momentum. The g-factors are defined as usual. For CCS in its ground state configuration the terms involving $\vec{I}$ and $\vec{N}$ can be neglected since $\mu_{\rm n} \ll \mu_{\rm B}$. Therefore, the magnetic moment $\vec{\mu}_{\rm JN}$ can be expressed by

$${\vec{\mu}}_{\rm JN} = - \frac{\mu_{\rm B}}{\hbar} (g_{L} {\vec{L}} + g_{S} {\vec{S}})$$ (A.2)

to an excellent approximation. The J_N = 0₁ level has a vanishing magnetic moment because of J = 0. The rotational ground state can thus not split into magnetic sublevels due to an external magnetic field. Taking the quantization axis along the direction of $\vec{B}$ the splitting energy of other rotational levels J_N due to the Zeeman effect is given by

$${\Delta}{E}_{B} = g_{\rm JN} {\mu}_{\rm B} {M}_{\rm JN} B,$$ (A.3)

where $M_{\rm JN}$ is the corresponding magnetic quantum number and B is the magnetic field strength.

A.1 The g $_{\mathsfsl{JN}}$-factor

Since the lower rotational states can be approximated by Hund's case a, we designate eigenvectors by $\vert\Sigma \Omega S J$>. Because of S=1 and ${\Lambda}$ = 0 in the J = 1 state, $\Sigma =$ +1, 0, -1, and ${\Omega} =$ $\vert\Lambda + {\Sigma}\vert =$ 0, 1. Hence, there are three corresponding eigenvectors: |1111>, |0011> and |-1111>.

For any linear molecule in the $^{3}{\Sigma}$ configuration, the Hamiltonian is given by (Gordy & Cook 1984)

$$\vec{H} = h ( B_0 {\vec{N}}^{2} \hspace{-0.5ex}+ \hspace{-0.3... ...e{-0.5ex}+ \hspace{-0.5ex} \hfill{} {\gamma} \vec{N} \vec{S}),$$

where (in the case of CCS) B₀ (=6477.74952 MHz) is the rotation constant, D₀ (= 1.72704 kHz) is the centrifugal stretching, $\lambda_0$ (= 97193.79 MHz) is the spin-spin coupling constant, $\lambda_1$ (= 27.14 kHz) is the centrifugal correction, and $\gamma$ (= -14.645 MHz) is the spin-rotation coupling constant (Saito et al. 1987).

In order to express the above Hamiltonian in a form that is suitable for case a eigenvectors, we notice that in case b

$$\vec{J} = \vec{N} + \vec{S},$$

which is equivalent to

$$\vec{N} \vec{S} = ( \vec{J} - \vec{S} ) \vec{S} = \vec{J} \vec{S} - \vec{S}^{2}$$

so that

$$\vec{H} = h {B}_{0} (\vec{J}^{2} + \vec{S}^{2} - 2 \vec{J} \v... ... {D}_{0} (\vec{J}^{2} + \vec{S}^{2} - 2 \vec{J} \vec{S})^{2} +$$

$$\frac{2\,h}{3} ({\lambda}_{0} + {\lambda}_{1} (\vec{J}^{2} + ... ...} - \vec{S}^{2}) + h {\gamma} (\vec{J} \vec{S} - \vec{S}^{2}).$$

The operator product $\vec{J} \vec{S}$ will be expressed in terms of the ladder operators in order to apply eigenvalue equations later on,

$$\vec{J} \vec{S} = \frac{1}{2} (\vec{J}_{+} \vec{S}_{-} + \vec{J}_{-} \vec{S}_{+}) + \vec{J}_{Z} \vec{S}_{Z}$$ (A.4)

using

$$\vec{J}_{\pm} = \vec{J}_{X}\;{\pm}\;{i} \vec{J}_{Y} ,$$

$$\vec{S}_{\pm} = \vec{S}_{X}\;{\pm}\;{i} \vec{S}_{Y}.$$

The state vectors for $\vec{H}$ can now be formed from linear combinations of the eigenvectors,

$$\vert{\psi}_{n}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} = {\... ...{\Sigma}{\Omega} S J{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}.$$ (A.5)

For the solution of Schrödinger's equation,

$$\vec{H}\vert\psi_{n}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}... ...E}_{n} \vert\psi_{n}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }},$$ (A.6)

the following eigenvalue equations have to be considered (e.g., Hougen 1970),

$$\vec{J}^{2}\,\vert{\Sigma}{\Omega} S J{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} \hbar^{2}~{J}\,(J + 1)~ \vert{\Sigma}{\Omega}SJ{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$$ = (A.7)

$$\vec{S}^{2}\,\vert{\Sigma}{\Omega} S J{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} \hbar^{2}~{S}\,(S + 1)~ \vert{\Sigma}{\Omega} S J{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$$ = (A.8)

$$\vec{J}_{Z}\,\vert{\Sigma}{\Omega}S J{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} \hbar~{\Omega}~\vert{\Sigma}{\Omega} S J{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$$ = (A.9)

$$\vec{S}_{Z}\,\vert{\Sigma}{\Omega}S J{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} \hbar~{\Sigma}~\vert{\Sigma}{\Omega} S J{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$$ = (A.10)

and

$$\vec{S}_{+}\,\vert{\Sigma}{\Omega} S J \hbar \sqrt{({S} - {\Sigma}) ({S} + {\Sigma} + 1)} ~ \vert{\Sigma + 1} \, {\Omega} S J$$ = (A.11)

$$\vec{J}_{+}\,\vert{\Sigma}{\Omega} S J \hbar \sqrt{({J} + {\Omega}) ({J} - {\Omega} + 1)} ~ \vert{\Sigma} \, {\Omega - 1} \, S J$$ = (A.12)

$$\vec{S}_{-}\,\vert{\Sigma}{\Omega} S J \hbar \sqrt{({S} + {\Sigma}) ({S} - {\Sigma} + 1)} ~ \vert{\Sigma -1} \, {\Omega} S J$$ = (A.13)

$$\vec{J}_{-}\,\vert{\Sigma}{\Omega} S J \hbar \sqrt{({J} - {\Omega}) ({J} + {\Omega} + 1)} ~ \vert{\Sigma} \, {\Omega +1} \, S J.$$ = (A.14)

With those equations we obtain the matrix elements listed in Table A.1. The solutions to Schrödinger's equation in matrix form

$$$<$\psi_{m}\vert\vec{H}\vert\psi_{n}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} - {E}_{n} {\delta}_{mn} = 0$$ (A.15)

can be found from the secular equation

$${\rm det} ({H}_{mn} - {E}_{n} {\delta}_{mn}) = 0.$$

The solutions are the energy eigenvalues which correspond to three energy levels in the J = 1 state, N = 0, 1, 2, as shown in Fig. 2 of Saito et al. (1987),

$$\begin{array}{crl} E_{0} = & h (-104.964\,{\rm GHz})& \\ E_{... ... GHz})& \\ E_{2} = & h ( +79.583\,{\rm GHz})&. \\ \end{array}$$

The smallest of those three energy eigenvalues corresponds to the energy level J_N = 1₀.

 

Table A.1: Matrix elements

<1111 | $\vec{H}$ | $1111{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$	=	$h ( 2 {B}_{0} - 8 {D}_{0} + \frac{2}{3} {\lambda}_{0} + \frac{4}{3} {\lambda}_{1} - {\gamma}) = h (77.766026\,{\rm GHz})$
<1111 | $\vec{H}$ | $0011{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$	=	$h ( -2 {B}_{0} + 12 {D}_{0} + \frac{8}{3} {\lambda}_{1} + {\gamma}) = h (-12.970087\,{\rm GHz})$
<1111 | $\vec{H}$ | $\mbox{--1111}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$	=	$h (-4 {D}_{0}) = h (-7\,{\rm kHz})$
<0011 |$\vec{H}$ | $1111{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$	=	$h (-2 {B}_{0} + 12 {D}_{0} - \frac{4}{3} {\lambda}_{1} + {\gamma}) = h (-12.970160\,{\rm GHz})$
<0011 |$\vec{H}$ | $0011{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$	=	$h (4 {B}_{0} - 24 {D}_{0} - \frac{4}{3} {\lambda}_{0} - \frac{16}{3} {\lambda}_{1} - 2 {\gamma}) = h (-103.651618\,{\rm GHz})$
<0011 |$\vec{H}$ | $\mbox{--1111}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$	=	$h (-2 {B}_{0} + 12 {D}_{0} - \frac{4}{3} {\lambda}_{1} + {\gamma}) = h (-12.970160\,{\rm GHz})$
< $\mbox{--1111}$ |$\vec{H}$ | $1111{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$	=	$h (-4 {D}_{0}) = h (-7\,{\rm kHz})$
< $\mbox{--1111}$ |$\vec{H}$ | $0011{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$	=	$h ( -2 {B}_{0} + 12 {D}_{0} + \frac{8}{3} {\lambda}_{1} + {\gamma}) = h (-12.970087\,{\rm GHz})$
< $\mbox{--1111}$ | $\vec{H}$ | $\mbox{--1111}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$	=	$h ( 2 {B}_{0} - 8 {D}_{0} + \frac{2}{3} {\lambda}_{0} + \frac{4}{3} {\lambda}_{1} - {\gamma}) = h (77.766026\,{\rm GHz})$

In order to find the wavefunction for this energy eigenstate, the coefficients c $^{0}_{\rm i}$ from the linear combination (Eq. (A.5)) must be determined. Rewriting Eq. (A.5) as (omitting the superscript 0 in the following)

$$\vert{\psi}_{0}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} = \l... ...array}{lll} c_{-1} \\ c_{0} \\ c_{+1} \\ \end{array}\right)$$

Eq. (A.15) yields

$$\begin{array}{rcrcrl} 182.730054 c_{-1} & - & 12.970087 c_{0}... ... 12.970087 c_{0} & + & 182.730054 c_{+1} & = 0. \\ \end{array}$$

where the matrix elements in Table 1 were used. Both the first and third equation give c_-1 = c₊₁, and consequently

$$\frac{c_{+1}}{c_{0}} = 0.07098.$$

Hence the above system of equations is redundant, and a normalization is introduced,

c²_-1 + c²₀ + c²₊₁ = 1, (A.16)

which allows to determine the individual coefficient to

$$c_{0} = \left(1 + 2 \left( \frac{c_{+1}}{c_0} \right)^2 \right)^{-\frac{1}{2}} = 0.99500$$

$$c_{+1} = c_{-1} = \left(\frac{1 - c^{2}_{0}}{2} \right)^{\frac{1}{2}} = 0.07062.$$

The wavefunction for the J_N = 1₀ level can now be given explicitely,

$$\vert{\psi}_{0}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} =$$

$$0.07062\,\vert 1111{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} ... ...\,\vert\mbox{--1111}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}.$$

In order to find the $g_{\rm JN}$-factor for the J_N = 1₀ level, Eqs. (A.1) and (A.2) were multiplied by $\vec{J}$ to obtain

$$\frac{{\mu}_{\rm B}}{\hbar} ({g}_{L} \vec{J} \vec{L} + {g}_{S... ...ec{S}) = {g}_{\rm JN} \frac{{\mu}_{\rm B}}{\hbar} \vec{J}^{2}.$$

The corresponding expectation value in state $\vert{\psi}_{0}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$ is found by using Eqs. (A.7) and (A.16),

$${g}_{L} \, <{\psi}_{0}\vert \vec{J} \vec{L} \vert{\psi}_{0}{\... ...{S} \vert{\psi}_{0}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }} =$$

$${g}_{\rm JN} \, <{\psi}_{0}\vert \vec{J}^{2} \vert{\psi}_{0}{... ...hspace{-0.1ex}$>$ }} = {g}_{\rm JN} \hbar^{2} \, {J}\,(J + 1).$$

The evaluation of the expectation value for $\vec{J} \vec{L}$ in state $\vert{\psi}_{0}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$ requires eigenvalue equations for $\vec{L_X}$, $\vec{L_Y}$and $\vec{L_Z}$. In analogy to the eigenvalue equations for the spin angular momentum, corresponding equations for the electronic orbital angular momentum can be found substituting $\vec{S}$ by $\vec{L}$, S by L and ${\Sigma}$ by ${\Lambda}$ in Eqs. (A.7) through (A.14). (Hougen 1970). Obviously, the operators $\vec{L_{\pm}}$ and $\vec{L_Z}$ have non-vanishing matrix elements only if the selection rule ${\Delta}{\Lambda}$ = $\pm$1 is satisfied, and ${\Lambda}$ $\neq$ 0, respectively. Neither of these requirements is fulfilled for the expectation value of $\vec{J} \vec{L}$ in state $\vert{\psi}_{0}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$, so that < ${\psi}_{0}\vert \vec{J} \vec{L} \vert{\psi}_{0}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}$ = 0, hence

$${g}_{\rm JN} = \frac{{g}_{S}}{{J}~({J}\,+\,1)}~ \frac{1}{{\h... ...c{S} \vert{\psi}_{0}{\makebox[1.5ex][l]{\hspace{-0.1ex}$>$ }}.$$ (A.17)

Using Eq. (A.4) and the eigenvalue Eqs. (A.9)-(A.14),

$$<{\psi}_{0}\vert \vec{J} \vec{S} \vert{\psi}_{0}{\makebox[1.5... ...\hbar^{2} \, ( 2 c^{2}_{+1} + 4 c_{0} c_{+1} ) = \, \hbar^{2}.$$

From Eq. (A.17) and recalling that g_S = 2.0023, the $g_{\rm JN}$-factor for J_N = 1₀ level is finally determined to be

g₁₀ = 0.29.