Electronic States in Selected Polyatomic Molecules

Abstract

Polyatomic molecules from bonds between pairs of atoms.

Appendix 9.1 Ammonia Molecule in Electric Field and the Ammonia Maser

According to Fig. 9.1 the Ammonia molecule can be found in two equivalent configurations, depending on the position of the N atom above (state $|1>$) or below (state $|2>$) the xy plane of the H atoms. By considering the molecule in its ground electronic state and neglecting all other degrees of freedom, let us discuss the problem of the position of the N atom along the z direction perpendicular to the xy plane, therefore involving the vibrational motion in which N oscillates against the three coplanar H atoms (for details on the vibrational motions see Sects. 10.3 and 10.6).

The potential energy V(z), that in the framework of the Born-Oppenheimer separation (Sect. 7.1) controls the nuclear motions and that is the counterpart of the energy $E(R_{AB})$ in diatomic molecules, has the shape sketched below

The distance of the N atom from the xy plane corresponding to the minima in V(z) is $z_o= 0.38$ Å, while the height of the potential energy for $z=0$ is $V_o \simeq 25 \mathrm {meV}$. In the state $|1>$ the molecule has an electric dipole moment ${\varvec{\mu }_{e}}$ along the negative z direction, while in the state $|2>$ the dipole moment is parallel to the reference z-axis. Within each state the N atoms vibrate around $+z_o$ or $-z_o$. As for any molecular oscillator the ground state has a zero-point energy different from zero, that we label $E_o$ (correspondent to the two levels A and B sketched in Fig. 9.1c). The vibrational eigenfunction in the ground state is a Gaussian one, centered at $\pm z_o$ (see Sect. 10.3). The effective mass of the molecular oscillator is $\mu = 3M_HM_N/(3M_H+M_N)$.

Thus the system is formally similar to the H$_2^+$ molecule discussed at Sect. 8.1, the $|1>, |2>$ states corresponding to the electron hydrogenic states $1s_A$ and $1s_B$, while the vibration zero-point energy corresponds to $-R_Hhc$. Therefore, the generic state of the system is written

$$\begin{aligned} |\psi>= c_1|1>+ c_2|2> \end{aligned}$$

(A.9.1.1)

with coefficients $c_i$ obeying to Eqs. (8.13). Here $H_{12}=-A$ is the probability amplitude that because of the quantum tunneling the N atom jumps from $|1>$ to $|2>$ and vice versa, in spite of the fact that $E_o\ll V_o$. Two stationary states are generated, say $|g>$ and $|u>$, with eigenvalues $E_o-A$ and $E_o+A$, respectively. The correspondent eigenfunctions are linear combinations of the Gaussian functions describing the oscillator in its ground state (see Sect. 10.3):

The degeneracy of the original states is thus removed and the vibrational levels are in form of doublets (inversion doublets). For the ground-state the splitting $E_g-E_u=2A$ corresponds to $0.793\,\mathrm{cm}^{-1}$, while it increases in the excited vibrational states, owing to the increase of $H_{12}$. For the first excited state $2A'=36.5\,\mathrm{cm}^{-1}$ and for the second excited state $2A''=312.5\,\mathrm{cm}^{-1}$. It can be remarked that the vibrational frequency (see Sect. 10.3) of N around the minimum in one of the wells is about $950\,\mathrm{cm}^{-1}$.

The inversion splitting are drastically reduced in the deuterated Ammonia molecule ND$_3$ where for the ground-state $2A=0.053\,\mathrm{cm}^{-1}$. Thus the tunneling frequency, besides being strongly dependent on the height of the effective potential barrier $V_o$, is very sensitive to the reduced mass $\mu$. For instance, in the AsH$_3$ molecule, the time required for a complete tunneling cycle of the As atom is estimated to be about two years. These marked dependences on $V_o$ and $\mu$ explains why in most molecules the inversion doublet is too small to be observed.

In NH$_3$ the so-called inversion spectrum was first observed (Cleeton and Williams, 1934) as a direct absorption peak at a wavelength around 1.25 cm, by means of microwave techniques. This experiment opened the field presently known as microwave spectroscopy.

The typical experimental setup is schematically shown below

Finally it should be remarked that the rotational motions of the molecule (Sect. 10.2), as well as the magnetic and quadrupolar interactions (Chap. 5), in general cause fine and hyperfine structures in the inversion spectra.

As already mentioned the $|g>$ and $|u>$ states of the inversion doublet in NH$_3$ have been used in the first experiment (Townes and collaborators) of microwave amplification by stimulated emission of radiation (see Problem 1.24). The maser action requires that the statistical population $N_u$ is maintained larger than $N_g$ while a certain number of transitions from $|u>$ to $|g>$ take place.

Now we are going to discuss how the Ammonia molecule behaves in a static electric field. Then we show how by applying an electric field gradient (quadrupolar electric lens) one can select the Ammonia molecule in the upper energy state.

In the presence of a field $\mathcal {E}$ along z the eigenvalue for the states $|1>$ and $|2>$ become

$$ H_{11}=E_o + \mu _e\mathcal {E}\,\,\, \mathrm {and}\,\,\, H_{22}=E_o - \mu _e\mathcal {E} $$

The rate of exchange can be assumed approximately the same as in absence of the field, namely $H_{12}=-A$. The analogous of Eqs. (8.13) for the coefficients $c_i$ in Eq. (A.9.1.1) are then modified in

$$\begin{aligned} i\hbar \frac{dc_1}{dt}&=(E_o + \mu _e\mathcal {E})c_1 - Ac_2 \end{aligned}$$

(A.9.1.2)

$$\begin{aligned}&\qquad \qquad \,\, i\hbar \frac{dc_2}{dt}=(E_o - \mu _e\mathcal {E})c_2 - Ac_1 \end{aligned}$$

(A.9.1.3)

The solutions of these equations must be of the form $c_i= a_i exp(-i E t/\hbar )$, with E the unknown eigenvalue. The resulting equations for $a_i$ are

$$\begin{aligned} (E-E_o - \mu _e\mathcal {E})a_1 + Aa_2=0 \nonumber \\ Aa_1 + (E-E_o + \mu _e\mathcal {E})a_2=0 \nonumber \end{aligned}$$

and the solubility condition yields

$$\begin{aligned} E_{\pm }= \frac{H_{11}+H_{22}}{2}\pm \sqrt{\frac{(H_{11}-H_{22})^2}{4}+A^2}= E_o \pm \sqrt{A^2 + \mu _e^2\mathcal {E}^2} \end{aligned}$$

(A.9.1.4)

(representing a particular case of the perturbation effects described in Appendix 1.2 (Eq. (A.1.2.4))). When the perturbation is not too strong compared to the inversion splitting, Eq. (A.9.1.4) can be approximated in the form

$$\begin{aligned} E_{\pm }= E_o \pm A \pm \frac{\mu _e^2\mathcal {E}^2}{2A}. \end{aligned}$$

(A.9.1.5)

$E_{\pm }$ are reported below as a function of the field.

Equation (A.9.1.5) can be read in terms of induced dipole moments $\mu _{ind}^{\pm }=-dE_{\pm }/d\mathcal {E}= \mp \mu _e^2\mathcal {E}/A$. Therefore, if a collimated beam of molecules passes in a region with an electric field gradient across the beam itself, molecules in the $|u>$ and $|g>$ states will be deflected along opposite directions (this effect is analogous to the one observed in the Rabi experiment at Sect. 6.2). In particular, the molecules in the $|g>$ state will be deflected towards the region of stronger $\mathcal {E}^2$, owing to the force $-\mathbf {\nabla }[-(\mu _e\mathcal {E})^2/2A]$.

In practice, to obtain a beam with molecules in the upper energy state one uses quadrupole electric lenses, providing a radial gradient of $\mathcal {E}^2$. The square of the electric field varies across the beam. Passing through the lens the beam is enriched in molecules in the excited state and once they enter the microwave cavity the maser action becomes possible. The experimental setup of the Ammonia maser is sketched in the following figure.

The basic principles outlined above for the Ammonia maser are also at work in other type of atomic or solid-state masers. In the Hydrogen or Cesium atomic maser the stimulated transition involves the hyperfine atomic levels (see Chap. 5). For the line at 1420 MHz, for instance, the selection of the atoms in the upper hyperfine state with $F=1$ is obtained by a magnetic multipolar lens. Then the atomic beam enters a microwave cavity tuned at the resonance frequency. The resolution (ratio between the linewidth and 1420 MHz) can be improved up to $10^{-10}$, since the atoms can be kept in the cavity up to a time of the order of a second. The experimental value of the frequency of the $F=1\rightarrow 0$ transition in Hydrogen is presently known to be ($1420405751.781 \pm 0.016\,\mathrm{Hz}$), while for $^{133}$Cesium the $F=4\rightarrow 3$ transition is estimated 9192631770 Hz, which is the frequency used to calibrate the unit of time (see Sect. 5.2).

Solid state masers are usually based on crystals with a certain number of paramagnetic transition ions, kept in a magnetic field and at low temperature, in order to increase the spin-lattice relaxation time $T_1$ and to reduce the linewidth associated with the life-time broadening (see Chap. 6) (as well as to reduce the spontaneous emission acting against the population inversion). A typical solid state maser involves ruby, a single crystal of Al$_2$O$_3$ with diluted Cr$^{3+}$ ions (electronic configuration $3d^3$). The crystal field removes the degeneracy of the 3d levels (details will be given at Sect. 13.3) and the magnetic field causes the splitting of the $M_J=\pm 3/2 , \pm 1/2$ levels. The population inversion between these levels is obtained by microwave irradiation of proper polarization.

Here we have presented only a few aspects of the operational principles of masers, which nowadays have a wide range of applications, due to their resolution (which can be increased up to $10^{-12}$) and sensitivity (it can be recalled that maser signals reflected on the surface of Venus have been detected).

Problems

Problem 9.1

Under certain circumstances the cyclobutadiene molecule can be formed in a configuration of four C atoms at the vertices of a square. In the MO.LCAO picture of delocalized $2p_z$ electrons derive the eigenvalues and the spin molteplicity of the ground state within the same approximations used for C$_6$H$_6$.

Solution: The secular equation is

$$ \left| \begin{array}{cccc} \alpha - E &{} \beta &{} 0 &{} \beta \\ \beta &{} \alpha - E &{} \beta &{} 0\\ 0 &{} \beta &{} \alpha - E &{} \beta \\ \beta &{} 0 &{} \beta &{} \alpha - E \end{array} \right| = 0. $$

By setting $\alpha - E = x$, one has $x^4 -4\beta ^2 x^2 = 0$ and then $E_1 = -2\vert \beta \vert +\alpha \,, \qquad E_{2,3}= \alpha \,,\qquad E_4 = 2\vert \beta \vert + \alpha \,.$

Ground state:    $4\alpha -4|\beta |$. Since the Hund rules hold also in molecules (see Sect. 8.2), the ground-state is a triplet.

Problem 9.2

Refer to the C$_3$H$_3$ molecule, with carbon atoms at the vertices of an equilateral triangle. Repeat the treatment given for C$_6$H$_6$, deriving eigenfunctions and the energy of the ground-state. Then release the assumption of zero overlap integral among orbitals centered at different sites and repeat the derivation. Estimate, for the ground-state configuration, the average electronic charge per C atom .

Solution: For $S_{ij}=0$ for $i\ne j$, the secular equation is

$$\left| \begin{array}{ccc} E_0-E &{} \beta &{} \beta \\ \beta &{} E_0-E &{} \beta \\ \beta &{} \beta &{} E_0-E \end{array} \right| = 0 $$

so that

$$ E_I= E_0 + 2\beta \,\,\,\qquad E_{\textit{II},\textit{III}}= E_0 - \beta $$

and the ground-state energy is

$$ E_g= 3E_0 + 4\beta - \beta = 3E_0 + 3\beta $$

The eigenfunctions turn out

$$ \phi _I= \frac{1}{\sqrt{3}}[ \phi _1 + \phi _2+\phi _3]\equiv \frac{A}{\sqrt{3}} $$

$$ \phi _{\textit{II}}= \frac{1}{\sqrt{2}}[\phi _1 -\phi _3]\equiv \frac{B}{\sqrt{2}} $$

$$ \phi _{\textit{III}}=\frac{1}{\sqrt{6}}[-\phi _1 + 2\phi _2 -\phi _3]\equiv \frac{C}{\sqrt{6}} $$

The total amount of electronic charge on a given atom (e.g. atom 1) is given by the sum of the squares of the coefficient pertaining to $\phi _1$ in $\phi _{I, \textit{II}, \textit{III}}$:

$$ q= 2\left( \frac{1}{\sqrt{3}}\right) ^2 + \frac{1}{2}\left[ \left( \frac{1}{\sqrt{2}}\right) ^2 + \left( \frac{1}{\sqrt{6}}\right) ^2\right] =1 $$

(having taken the average of the two degenerate states).

For $S_{ij}\equiv S\ne 0$, the secular equation becomes

$$\left| \begin{array}{ccc} E_0-E\,\, &{} \beta - SE\,\, &{} \beta - SE\,\, \\ \beta - SE\,\, &{} E_0-E\,\, &{} \beta - SE\,\, \\ \beta - SE\,\, &{} \beta - SE\,\, &{} E_0-E\,\, \end{array} \right| = 0 $$

and the eigenvalues are

$$ E_I= \frac{E_0 + 2\beta }{1+ 2S} \,\,\, E_{\textit{II},\textit{III}}= \frac{E_0 - \beta }{1- S} $$

The ground-state energy is

$$ E_g= 2E_I + E_{\textit{II}}= 3 \frac{E_0 + \beta (1-2S)}{(1+2S)(1-S)} $$

with normalized eigenfunctions

$$ \phi _I= \frac{A}{\sqrt{3(1+2S)}} $$

$$ \phi _{\textit{II}}= \frac{B}{\sqrt{2(1-S)}} $$

$$ \phi _{\textit{III}}= \frac{C}{\sqrt{6(1-S)}} $$

Again, by estimating the squares of the coefficients the charge at a given atom turns out

$$ q'= \frac{1}{(1-S)(1+2S)}. $$

The charge in the region “in between” two atoms (e.g. atoms 1 and 2) is obtained by evaluating the sum of the coefficients $c_1c_2$ (for $\phi _1$ and $\phi _2$) in $\phi _{I, \textit{II}, \textit{III}}$, multiplied by the overlap integral. Thus

$$ q''= \frac{S(1-2S)}{(1-S)(1+2S)}. $$

Problem 9.3

Estimate the order of magnitude of the diamagnetic contributions to the susceptibility in benzene, for magnetic field perpendicular to the molecular plane.

Solution: The diamagnetic susceptibility (per molecule) can approximately be

$$\chi _{\psi } = - \frac{n_{\psi } e^2}{4mc^2} <r^2>_{\psi },$$

where $n_{\psi }$ is the number of electrons in a molecular state $\psi$ and $<r^2>_{\psi }$ is the mean square distance.

In benzene there are 12 1s electrons of C, with $<r^2>_{1s}\simeq \frac{3a_0^2}{Z^2}$ ($Z=6$). Then there are 24 electrons in $\sigma$ bonds for which, approximately,

$$<r^2>_{\sigma }\, \simeq \int _ {-\frac{L}{2}} ^{\frac{L}{2}} \frac{dx}{L} x^2 = \frac{L^2}{12},$$

the length of the $\sigma$ bond being $L = 1.4\,{\AA }$.

Finally there are 6 electrons in the delocalized bond $\pi _z$, where one can assume $<r^2>_{\pi _z} \simeq L^2$. The diamagnetic correction at the center of the molecule is in large part due to the delocalized electrons and from that value of $<r^2>_{\pi _z}$ one can crudely estimate $\chi _{\pi } \approx -0.49 \cdot 10^{-28}$  cm$^3\,$. The experimental values for the single-molecule susceptibility are $\chi _{dia}^{\perp }= -1.52 \cdot 10^{-28}$  cm$^3\,$ and $\chi _{dia}^{\parallel }= -0.62 \cdot 10^{-28}$  cm$^3$, for magnetic field perpendicular and parallel to the plane of the molecule.