Advanced Chemistry - Mono Mole

February 22, 2024May 11, 2024

Reduction of the two-particle problem to one-particle problems

The reduction of a two-particle problem to one-particle problems simplifies the Schrodinger equation by separating it into two one-particle equations. This process involves treating the two particles as if they were a single effective particle, utilising the concepts of center of mass and reduced mass $\mu$ .

With reference to the above diagram, we define the relative coordinates $x,y,z$ of the system as

$x=x_2-x_1\;\;\;y=y_2-y_1\;\;\;z=z_2-z_1\;\;\;\;\;\;\;\;302$

The coordinates $X,Y,Z$ of the center of mass of the two particles of masses $m_1$ and $m_2$ are

$X=\frac{m_1x_1+m_2x_2}{m_1+m_2}\;\;\;Y=\frac{m_1y_1+m_2y_2}{m_1+m_2}\;\;\;Z=\frac{m_1z_1+m_2z_2}{m_1+m_2}\;\;\;\;\;\;\;\;303$

Question

How is centre of mass of the diatomic molecule derived?

Answer

The centre of mass can be derived via the principle of moments. For example, $m_1(X-x_1)=m_2(x_2-X)$ , which rearranges to $X=\frac{m_1x_1+m_2x_2}{m_1+m_2}$ .

Substitute eq302 in eq303, we have

$\begin{matrix}x_1=X-\frac{m_2}{m_1+m_2}x&y_1=Y-\frac{m_2}{m_1+m_2}y&z_1=Z-\frac{m_2}{m_1+m_2}z\\\\x_2=X+\frac{m_1}{m_1+m_2}x&y_2=Y+\frac{m_1}{m_1+m_2}y&z_2=Z+\frac{m_1}{m_1+m_2}z\end{matrix}\;\;\;\;\;\;303a$

The magnitudes of the linear momentum vector of the particles with mass $m_1$ and $m_2$ are

$p_1^2=p_{x_1}^2+p_{y_1}^2+p_{z_1}^2=m_1^2\left[\left(\frac{dx_1}{dt}\right)^2+\left(\frac{dy_1}{dt}\right)^2+\left(\frac{dz_1}{dt}\right)^2\right]\;\;\;\;\;\;\;\;304$

$p_2^2=p_{x_2}^2+p_{y_2}^2+p_{z_2}^2=m_2^2\left[\left(\frac{dx_2}{dt}\right)^2+\left(\frac{dy_2}{dt}\right)^2+\left(\frac{dz_2}{dt}\right)^2\right]\;\;\;\;\;\;\;\;305$

Substituting eq303a in eq304 and eq305 and summing the results, we have

$\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}=\frac{1}{2}M\left[\left(\frac{dX}{dt}\right)^2+\left(\frac{dY}{dt}\right)^2+\left(\frac{dZ}{dt}\right)^2\right]+\frac{1}{2}\mu\left[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2\right]\;\;\;\;\;\;\;\;306$

where $M=m_1+m_2$ and $\mu=\frac{m_1m_2}{m_1+m_2}$ .

Question

Explain in detail the concept of reduced mass.

Answer

The reduced mass of two particles is a quantity that simplifies the description of the two-body problem, making it equivalent to a one-body problem, which consists of a fictitious particle of mass $\mu$ . Such a problem, e.g. one in the field of molecular vibrations, is easier to solve.

Eq306, which represents the total kinetic energy $T$ of the two particles, can be rewritten as

$T=\frac{p_M^2}{2M}+\frac{p_\mu^2}{2\mu}$

$p_M^2$ is expressed in center-of-mass coordinates and $M$ is the total mass of the system. Therefore, $\frac{p_M^2}{2M}$ is the kinetic energy of the translational motion of the system, and consequently, $\frac{p_\mu^2}{2\mu}$ must describe the kinetic energy of the rotational motion and vibrational motion (collectively known as internal motion) of the system.

If the potential energy $V$ of the system is a function of the relative coordinates of the two particles, the Schrodinger equation is

$\left[\frac{\hat{p}_M^2}{2M}+\frac{\hat{p}_\mu^2}{2\mu}+V(x,y,z)\right]\psi=E\psi\;\;\;\;\;\;\;\;307$

Since translational motion is independent from rotational and vibrational motions, $E=E_M+E_{\mu}$ , where $E_M$ and $E_{\mu}$ are the translational energy of the system and the internal motion energy of the system respectively. This implies that $\psi=\psi_M\psi_{\mu}$ . Noting that translational energy is purely kinetic, we can separate eq307 into two one-particle problems:

$\frac{\hat{p}_M^2}{2M}\psi_M=E_M\psi_M\;\;\;\;\;\;\;\;308$

$\left[\frac{\hat{p}_\mu^2}{2\mu}+V(x,y,z)\right]\psi_{\mu}(x,y,z)=E_{\mu}\psi_{\mu}(x,y,z)\;\;\;\;\;\;\;\;309$

Eq308 is a Schrodinger equation of a fictitious particle of mass $M$ that is subject to a zero-potential field, while eq309 is a Schrodinger equation of another fictitious particle of mass $\mu$ that is under the influence of the potential $V$ . Instead of relative cartesian coordinates $x,y,z$ , the two equations can also be expressed in spherical coordinates $R,\theta,\phi$ of one particle relative to the other (see diagram below):

$\frac{\hat{p}_M^2}{2M}\psi_M=E_M\psi_M\;\;\;\;\;\;\;\;310$

$\left[\frac{\hat{p}_\mu^2}{2\mu}+V(R)\right]\psi_{\mu}(R,\theta,\phi)=E_{\mu}\psi_{\mu}(R,\theta,\phi)\;\;\;\;\;\;\;\;311$

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February 22, 2024August 29, 2024

Central force problem

The central force problem involves solving the Schrodinger equation for a particle moving under the influence of a central potential, which is a spherically symmetric function.

The Hamiltonian of a particle subject to a central force is

$\hat{H} =-\frac{\hbar^2}{2m}\nabla^2+V(r)\;\;\;\;\;\;\;\;300$

where $\nabla^2=\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\left [ \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\left ( sin\theta\frac{\partial}{\partial\theta}\right )\right ]$ (see this article for derivation) and $V$ is a function of $r$ only since it is a spherically symmetric function.

Substituting eq49 and eq50 in eq300 gives

$\hat{H} =-\frac{\hbar^2}{2m}\left(\frac{2}{r}\frac{\partial}{\partial r}+\frac{\partial^2}{\partial r^2}\right)+\frac{\hat{L}^2}{2mr^2}+V(r)$

The eigenfunctions of $\hat{L}^2$ , which is the operator for the square of the magnitude of the orbital angular momentum of the particle, are the spherical harmonics $Y_l^m(\theta,\phi)$ , which are independent of $r$ (for molecules, we use $J$ and $M$ instead of the quantum numbers $l$ and $m$ ). Therefore, the Hamiltonian is separable and the solution to the Schrodinger equation is of the form

$\psi=R(r)Y_l^m(\theta,\phi)\;\;\;\;\;\;\;\;301a$

In conclusion, the Schrodinger equation of a particle subject to a central force separates into radial and angular parts. This is possible because of the spherically symmetric potential. The eigenfunction is the product of two functions, each independent of the other’s coordinates. The central force problem arises when solving the Schrodinger equations for the hydrogen atom and the nuclear motion of a diatomic molecule.

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December 30, 2023November 20, 2024

Determining the degrees of freedom of a molecule using group theory

Group theory allows us to determine the symmetries of the degrees of freedom of a molecule in a simple way, especially the vibrational modes of the molecule.

Consider $H_{2}O$ , which belongs to the $C_{2v}$ point group with the following character table:

Let’s form a set of basis with 9 unit displacement vectors representing instantaneous motions of the atoms (see diagram above). Since a symmetry operation of $C_{2v}$ transforms $H_{2}O$ into an indistinguishable copy of itself, it transforms each of the elements of the basis set into a linear combination of elements of the set. Therefore, orthogonal unit vectors centred on every atom of $H_{2}O$ generate a representation $\Gamma_{3N}$ of the $C_{2v}$ point group. To find the matrices of the representation, we arrange the unit vectors into a row vector and apply the symmetry operations of the $C_{2v}$ point group on it to obtain the transformed vector. The matrices of the representation are then constructed by inspection. For example,

$x_1,y_1,z_1,x_2,y_2,z_2,x_3,y_3,z_3\)\xrightarrow{C_2}\(-x_1,-y_1,z_1,-x_3,-y_3,z_3,-x_2,-y_2,z_2$

$\begin{pmatrix}-x_1\\-y_1\\z_1\\-x_3\\-y_3\\z_3\\-x_2\\-y_2\\z_2\end{pmatrix}=\begin{pmatrix}-1&0&0&0&0&0&0&0&0\\0&-1&0&0&0&0&0&0&0\\0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&-1&0&0\\0&0&0&0&0&0&0&-1&0\\0&0&0&0&0&0&0&0&1\\0&0&0&-1&0&0&0&0&0\\0&0&0&0&-1&0&0&0&0\\0&0&0&0&0&1&0&0&0\end{pmatrix}\begin{pmatrix}x_1\\y_1\\z_1\\x_2\\y_2\\z_2\\x_3\\y_3\\z_3\end{pmatrix}$

The traces of the matrices are

Using eq27a and the character table of the $C_{2v}$ point group, we have

$A_1$	$\frac{1}{4}\[1\cdot 9+1\cdot$-1$+1\cdot 3+1\cdot 1\]=3$
$A_2$	$\frac{1}{4}\[1\cdot 9+1\cdot$-1$-1\cdot 3-1\cdot 1\]=1$
$B_1$	$\frac{1}{4}\[1\cdot 9-1\cdot$-1$+1\cdot 3-1\cdot 1\]=3$
$B_2$	$\frac{1}{4}\[1\cdot 9-1\cdot$-1$-1\cdot 3+1\cdot 1\]=2$

which implies that the decomposition of the reducible representation is $\Gamma_{3N}=3A_1\oplus A_2\oplus 3B_1\oplus 2B_2$ .

A quick method to assign the irreducible representations from the decomposition of $\Gamma_{3N}$ to the nine degrees of freedom of $H_2O$ is to refer to the character table, where the basis functions $x$ , $y$ and $z$ represent the three independent translational motion and $\boldsymbol{\mathit{R}}_x$ , $\boldsymbol{\mathit{R}}_y$ and $\boldsymbol{\mathit{R}}_z$ represent the three independent rotational motion. Deducting the irreducible representations associated with these six basis functions from the direct sum of $3A_1\oplus A_2\oplus 3B_1\oplus 2B_2$ , we are left with $A_1$ , $A_1$ and $B_1$ . These remaining three irreducible representations correspond to the vibrational degrees of freedom.

To understand how the quick method works, we apply the projection operator on each element of the $3N$ basis set for all irreducible representations to give

$A_1$	$\hat{P}^{A_1}z_1=z_1$	$B_1$	$\hat{P}^{B_1}x_1=x_1$
	$\hat{P}^{A_1}z_2=\frac{1}{2}$z_2+z_3$$		$\hat{P}^{B_1}z_2=\frac{1}{2}$z_2-z_3$$
	$\hat{P}^{A_1}x_2=\frac{1}{2}$x_2-x_3$$		$\hat{P}^{B_1}x_2=\frac{1}{2}$x_2+x_3$$
	$\hat{P}^{A_1}z_3=\frac{1}{2}$z_2+z_3$$		$\hat{P}^{B_1}z_3=\frac{1}{2}$z_3-z_2$$
	$\hat{P}^{A_1}x_3=\frac{1}{2}$x_3-x_2$$		$\hat{P}^{B_1}x_3=\frac{1}{2}$x_3+x_2$$
$A_2$	$\hat{P}^{A_2}y_2=\frac{1}{2}$y_2-y_3$$	$B_2$	$\hat{P}^{B_2}y_1=y_1$
	$\hat{P}^{A_2}y_3=\frac{1}{2}$y_3-y_2$$		$\hat{P}^{B_2}y_2=\frac{1}{2}$y_2+y_3$$
	$\hat{P}^{A_2}y_3=\frac{1}{2}$y_3-y_2$$		$\hat{P}^{B_2}y_3=\frac{1}{2}$y_3+y_2$$

Each equation in the above table is a basis of the irreducible representation it belongs to. Since any linear combination of bases that transform according to a one-dimensional irreducible representation is also a basis of that irreducible representation, we can generate symmetry-adapted linear combinations (SALC) from the above basis vectors that belong to their respective irreducible representations, e.g. $\hat{P}^{A_1}z_1+\hat{P}^{A_1}z_2+\hat{P}^{A_1}z_3=z_1+z_2+z_3$ belongs to $\Gamma_{A_1}$ .

Question

Show that $z_1+z_2+z_3$ is a basis of $\Gamma_{A_1}$ and that it describes the translational motion of $H_2O$ in the $z$ -direction.

Answer

Let $\hat{R}_{\alpha}$ be the $\alpha$ -th symmetry operation of the $C_{2v}$ point group. With reference to the subspace of $\Gamma_{A_1}$ , we have $\hat{R}_{\alpha}z_1=\chi_{\alpha}z_1$ and $\hat{R}_{\alpha}\frac{1}{2}$z_2+z_3$=\chi_{\alpha}\frac{1}{2}$z_2+z_3$$ because $z_1$ and $\frac{1}{2}$z_2+z_3$$ are bases of $\Gamma_{A_1}$ .

$z_2+z_3$ and $z_1+z_2+z_3$ are also bases of $\Gamma_{A_1}$ , as $\hat{R}_{\alpha}$z_2+z_3$=\hat{R}_{\alpha}\biggr\[\frac{1}{2}$z_2+z_3$+\frac{1}{2}$z_2+z_3$\biggr\]=\chi_{\alpha}$z_2+z_3$$ and $\hat{R}_{\alpha}$z_1+z_2+z_3$=\hat{R}_{\alpha}z_1+\hat{R}_{\alpha}$z_2+z_3$=\chi_{\alpha}$z_1+z_2+z_3$$ .

In an earlier article, we explained that the displacement of $H_2O$ in the $z$ -direction is described in terms of unit instantaneous displacements vectors that are centred on the atoms, or equivalently, a single instantaneous displacement vector on the centre of mass. Using the second description, we then showed that the translational motion of $H_2O$ in the $z$ -direction transforms according to $\Gamma_{A_1}$ . Since the two descriptions are equivalent (see diagram below), the three $z$ -bases in the first description must transform according to the subspace of $\Gamma_{A_1}$ . This is only possible if they are expressed as the SALC $T_z=z_1+z_2+z_3$ .

Consequently, the SALCs describing the three translational degrees of freedom are:

$\boldsymbol{\mathit{\Gamma}}$	SALC	Description
$A_1$	$T_z=z_1+z_2+z_3$	Translation in the $z$ -direction
$B_1$	$T_x=x_1+x_2+x_3$	Translation in the $x$ -direction
$B_2$	$T_y=y_1+y_2+y_3$	Translation in the $y$ -direction

Using the logic mentioned the above Q&A, the rotational motion are given by:

$\boldsymbol{\mathit{\Gamma}}$	SALC	Description
$A_2$	$R_z=\hat{P}^{A_2}y_2-\hat{P}^{A_2}y_3=y_2-y_3$	Rotation around the centre of mass with reference to the $z$ -axis
$B_1$	$\begin{align}R_y&=a\hat{P}^{B_1}x_1+b\hat{P}^{B_1}x_2+c\hat{P}^{B_1}z_2\\&=ax_1+\biggr$\frac{bx_2}{2}+\frac{cz_2}{2}\biggr$+\biggr$\frac{bx_3}{2}-\frac{cz_3}{2}\biggr$\end{align}$	Rotation around the centre of mass with reference to the $y$ -axis
$B_2$	$\begin{align}R_x&=-a\hat{P}^{B_2}y_1+d\hat{P}^{B_2}y_2\\&=-ay_1+\frac{dy_2}{2}+\frac{dy_3}{2}\end{align}$	Rotation around the centre of mass with reference to the $x$ -axis

where $a,b,c,d$ are positive coefficients and each displacement vector is tangent to the relevant circular path (see diagram below).

Why do the remaining three irreducible representations of $A_1$ , $A_1$ and $B_1$ correspond to vibrational degrees of freedom? We have proven, in an earlier article, that if the $3N$ basis vectors transform according to a reducible representation $\Gamma_{3N}$ of the $C_{2v}$ point group, then any linear combination (SALC) of the vectors is also a basis of a reducible representation $\Gamma_{3N}^{'}$ that is equivalent to $\Gamma_{3N}$ . We have also shown, in the same article, that the number of linearly independent basis functions of a representation corresponds to the dimension of the representation. Hence, we need to select nine linearly independent SALCs consisting of six SALCs listed in the two tables above and three remaining SALCs, each of which belonging to $A_1$ , $A_1$ and $B_1$ respectively. Since none of the nine SALCs can be expressed as a linear combination of the others, the remaining three SALCs must describe the independent vibrational degrees of freedom of $H_2O$ . The possible vibrational modes, noting that two are totally-symmetric ( $A_1$ ) and one is antisymmetric ( $B_1$ ) with respect to rotation about the principal axis, are

The SALCs corresponding to these modes have the form:

$\boldsymbol{\mathit{\Gamma}}$	SALC	Description
$A_1$	$\begin{align}Q_s&=-e\hat{P}^{A_1}z_1+f$\hat{P}^{A_1}z_2+\hat{P}^{A_1}z_3$-\hat{P}^{A_1}x_2\\&=-ez_1-\biggr$\frac{x_2}{2}-fz_2\biggr$+\biggr$\frac{x_3}{2}+fz_3\biggr$\end{align}$	Vibration: symmetric stretching
$A_1$	$\begin{align}Q_b&=-e\hat{P}^{A_1}z_1+f\hat{P}^{A_1}z_2+\hat{P}^{A_1}x_2\\&=-ez_1-\biggr$\frac{x_2}{2}+\frac{fz_2}{2}\biggr$-\biggr$\frac{x_3}{2}-\frac{fz_3}{2}\biggr$\end{align}$	Vibration: bending
$B_1$	$\begin{align}Q_a&=-\hat{P}^{B_1}x_1+\hat{P}^{B_1}x_3+\hat{P}^{B_1}z_3\\&=-x_1+\biggr$\frac{x_2}{2}-\frac{z_2}{2}\biggr$+\biggr$\frac{x_3}{2}+\frac{z_3}{2}\biggr$\end{align}$	Vibration: antisymmetric stretching

where $e,f$ are positive coefficients.

To determine the coefficients, we note that orthogonal vectors are linearly independent. If we let $a=d=1$ and solve for $b,c,e,f$ , a possible set of orthogonal SALC vectors are:

$\boldsymbol{\mathit{\Gamma}}$	SALC
$A_1$	$T_z=z_1+z_2+z_3$
	$Q_s=-\frac{z_1}{\sqrt{3}}-\biggr$\frac{x_2}{2}-\frac{z_2}{2\sqrt{3}}\biggr$+\biggr$\frac{x_3}{2}+\frac{z_3}{2\sqrt{3}}\biggr$$
	$Q_b=-\frac{z_1}{\sqrt{3}}+\biggr$\frac{x_2}{2}+\frac{z_2}{2\sqrt{3}}\biggr$-\biggr$\frac{x_3}{2}-\frac{z_3}{2\sqrt{3}}\biggr$$
$A_2$	$R_z=y_2-y_3$
$B_1$	$T_x=x_1+x_2+x_3$
	$R_y=-x_1+\biggr$\frac{x_2}{2}+\frac{3z_2}{2}\biggr$+\biggr$\frac{x_3}{2}-\frac{3z_3}{2}\biggr$$
	$Q_a=-x_1+\biggr$\frac{x_2}{2}-\frac{z_2}{2}\biggr$+\biggr$\frac{x_3}{2}+\frac{z_3}{2}\biggr$$
$B_2$	$T_y=y_1+y_2+y_3$
$B_2$	$R_x=-y_1+\frac{y_2}{2}+\frac{y_3}{2}$

which can further be normalised to give orthonormal SALC vectors.

The method that we have used so far does not take into account the absolute masses of the atoms, only that mass of $O$ is different from mass of $H$ . Furthermore, the restriction of motion due to the presence and stiffness of chemical bonds is also not considered. Therefore, the centre of mass does not correspond to the actual centre of mass of the water molecule. It is along the $-z_1$ -axis but not beyond a line joining the centres of the hydrogen atoms. The exact position of the centre of mass of the water molecule affects the values of the coefficients of the SALC vectors, but is inconsequential when the objective is determine which irreducible representation a particular degree of freedom belongs to.

Question

Determine the symmetries of the vibrational modes of $NH_3$ .

Answer

There are a total of 12 degrees of freedom. Using a basis set of 3 orthogonal displacement vectors on each atom, we have

which decomposes to $\Gamma_{3N}=3A_1\oplus A_2\oplus 4E$ .

Deducting the irreducible representations in the $C_{3v}$ character table for translational and rotational motion, the symmetries of the 6 vibrational modes of $NH_3$ are $A_1$ , $A_1$ , $E$ and $E$ .

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December 30, 2023

Degrees of freedom

The degrees of freedom of a chemical species are the number of independent displacements of the species.

An atom has three degrees of freedom because its instantaneous displacement is defined by three orthogonal directions in the $\mathbb{R}^3$ space. A molecule of $N$ atoms is characterised by $3N$ degrees of freedom, as every atom can be displaced from its equilibrium position in three orthogonal directions. Each of the $3N$ motion is independent and cannot be expressed as a combination of the others. However, the presence of chemical bonds restrict some of these displacements, resulting in three groups of independent motion: translation, rotation and vibration.

To understand how the restriction of molecular motion by chemical bonds results in three groups of motion, each of which is independent of the others, we consider a diatomic molecule (see diagram above), where the red, purple and brown arrows represent instantaneous displacements of the atoms.

Firstly, we would expect the molecule to have six degrees of freedom since $N=2$ . Next, the motion of the two atoms (red arrows) in the same direction and with the same magnitude, which is equivalent to the motion of the centre of mass of the molecule in the positive $x$ -direction, is a translational motion. Similarly, the centre of mass of the molecule can move independently in the $z$ -direction and $y$ -direction, resulting in another two degrees of translational freedom.

The rotational motion of the molecule around the $y$ -axis passing through its centre of mass is described by the two brown arrows that are tangent to the circle of motion. The rotation about the $z$ -axis contributes to another degree of freedom. However, the molecule spinning around the $x$ -axis is not considered a degree of freedom, as no atomic displacement has occurred. In fact, all linear molecules have two degrees of rotational freedom. One common property of translational motion and rotational motion is that they are both rigid transformations, meaning that the bond length doesn’t change throughout either motion.

When the atoms are displaced in opposite directions (purple arrows), they oscillate about their equilibrium positions because they are restricted by the chemical bond. This is known as vibrational motion.

Combining two degrees of rotational freedom with three degrees of translational freedom, all linear molecules have $3N-5$ degrees of vibrational freedom. Since the centre of mass moves during translation, while the position of the centre of mass of a molecule does not change during rotational motion and vibrational motion, translational motion is independent of rotational motion and vibrational motion. Furthermore, rotational motion is independent of vibrational motion because rotational motion is a rigid transformation, whereas vibrational motion is not. In general, the three types of motion are independent of one another and all possible motion of atoms of a molecule are a combination of them.

Question

What if the brown arrows represent linear motions, instead of tangents to the circle of motion?

Answer

If that’s the case, the atoms move linearly in the directions of the arrows until they are restricted by the chemical bond. Such a motion is a combination of vibrational and rotational motions and not an independent motion. Therefore, it is not a degree of freedom.

Since an arrow on an atom can be expressed as a linear combination of three orthogonal vectors centred on the atom, the degrees of freedom of a molecule with $N$ atoms are often expressed in terms of $3N$ orthogonal unit displacement vectors.

For example, when the motion of the water molecule in the above diagram is described only by the three $z$ -unit vectors, and the magnitudes of the displacement of all atoms are equal, the molecule is translating in the $z$ -direction. Similarly, if the only unit vectors describing the molecule’s motion are $-y_2$ and $y_3$ , and the atoms have equal displacement magnitude, the molecule is rotating around the $z_1$ -axis. We will show, in the next article, that the translational motion of $H_2O$ in the $z$ -direction and the rotational motion of $H_2O$ around the $z_1$ -axis can be expressed as the symmetry-adapted linear combinations $z_1+z_2+z_3$ and $-y_2+y_3$ respectively.

$-y_2$ and $y_3$ can also describe a twisting motion of the molecule (like a gym glider machine). However, such a motion, as explained in the Q&A above, is not a degree of freedom of the molecule. Unlike a linear molecule, a non-linear molecule has three rotational degrees of freedom, and hence, $3N-6$ degrees of vibrational freedom. Group theory allows us to determine which irreducible representation of a point group a particular degree of freedom belongs to, and consequently provides a simple way to predict spectroscopic properties the molecule.

Next article: Determining the degrees of freedom of a molecule using group theory

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December 7, 2023

Set and subset

A set is collection of different mathematical objects called elements, which can be numbers, people, colours, matrices, etc. Examples of sets are {4, 20, 83, 1059, …} and {red, blue, cyan, grey}, where the former is an infinite set and the latter is a finite set. Two sets $A$ and $B$ are equal if they have the same elements, e.g. $A=\{3,10,17\}$ and $B=\{17,3,10\}$ . A set with no element, denoted by $\varnothing$ , is called an empty set. An element $k$ in a set $A$ is denoted by $k\in A$ .

A subset is an equal or smaller collection of elements of a particular set. With reference to the above diagram, the set $A=\{7,13,58\}$ is a subset of the set $B=\{1,5,7,13,58,82,296,941\}$ , which is mathematically denoted by $A\subseteq B$ . Similarly, $\varnothing\subseteq B$ and $B\subseteq B$ .

Question

Is $\{3,10,17,3\}$ a set.

Answer

No, because a set is defined as a collection of different elements.

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December 7, 2023January 5, 2025

Group theory (overview)

Group theory is the study of mathematical structures called groups, which have properties that are used to classify molecules by symmetry.

The classification of a molecule according to its symmetry in turn allows us to analyse its molecular properties. In the field of quantum chemistry, group theory simplifies calculations, e.g. in determining if overlap integrals are zero and investigating the degree of degeneracy of a system. Group theory is also useful in predicting spectroscopic transitions, especially in electronic and vibrational transitions.

To fully understand the theory, we need to have some knowledge of abstract algebra and apply it to derive the great orthogonality theorem.

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December 7, 2023January 5, 2025

Binary operation

A binary operation is a rule that combines two elements to form another element.

The most basic binary operations are the addition and multiplication of numbers in a set. For example, $1+2$ for $A=\{1,2,3\}$ is a binary operation. In general, a binary operation is denoted by $a*b$ , where $a$ and $b$ are the operands and $*$ is the chosen binary operation.

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Group

A group is a set, together with a chosen binary operation, that satisfies the properties of closure, identity, inverse and associativity as follows:

1. Closure: if $a$ and $b$ are elements of the group $G$ , then $c=a*b$ is also an element of $G$ .
2. Identity: there exists an element called the identity, which is denoted by $e$ , such that $e*k=k*e=k$ for $k\in G$ .
3. Inverse: every element $a$ in $G$ has an inverse $a^{-1}$ , which is itself an element of $G$ , such that $a^{-1}*a=a*a^{-1}=e$ .
4. Associativity: $(a*b)*c=a*(b*c)$ for all $a,b,c\in G$ .

Consider the set $A=\{0,1,2,3\}$ under the binary operation of addition mod 4. It is a group because it satisfies the four properties mentioned above. For example, the element 0 is the identity, since $a+0=0+a=a$ , where $a\in A$ . As for inverses, $0^{-1}=0$ , because $0^{-1}+0$ gives the identity. Similarly, $1^{-1}=3$ , $2^{-1}=2$ and $3^{-1}=1$ . The closure and associativity properties can be easily verified.

Question

What is addition mod 4 for the set $A=\{0,1,2,3\}$ ?

Answer

It means that if the sum of two elements of the set is greater than 3, then we divide the result by 4 (called the modulus) and take the remainder as the answer. For example, $0+3=3$ , $0+4=0$ and $2+3=1$ .

Similarly, the set $B=\{1,3,5,7\}$ under the binary operation of multiplication mod 8 is a group. In this case, if the product of two elements of $B$ is greater than 7, we divide the result by 8 and take the remainder as the answer. All possible binary operations and answers can be presented in the form of a multiplication table as follows:

where the left operand and right operand of the binary operation $a*b$ are listed in the top row and the leftmost column respectively.

Clearly, the multiplication of integers is associative. From the multiplication table, we can validate the closure property of $B$ , authenticate that 1 is the identity element and determine that the inverse of an element is the element itself. The associativity property is also easily verified separately.

Question

Is the set $A=\{0\}$ under the binary operation of multiplication a group?

Answer

Yes, it is a trivial group with only one element.

If the results of all binary operations of a group are independent of the order of the operands (i.e., $a*b=b*a$ ), the group is said to be Abelian. Both groups $A$ and $B$ mentioned above are examples of Abelian groups.

The order of a group $G$ , denoted by $\vert G\vert$ , is the number of elements in $G$ . For example, $\vert G\vert=4$ for both $A$ and $B$ . The order of an element $g$ of a group $G$ under the binary operation of multiplication, denoted by $\vert g\vert$ , is the smallest value of $n$ such that $g^n=e$ . For example, the order of the all elements in the group $B$ is 2.

Finally, some useful properties of a group can be inferred from the main properties of closure, identity, inverses and associativity. Two such inferred properties are:

Cancellation

(Statement) If $ab=ac$ , then $b=c$ for all $a$ in the group $G$ . In other words, $a$ is “cancellable”.

According to the inverses property of a group, if $a\in G$ , there exists a unique $a^{-1}$ , where $a^{-1}\in G$ . The proof for the cancellation theorem then involves left-multiplying $ab=ac$ by $a^{-1}$ and using the associativity and identity properties:

$a^{-1}(ab)=a^{-1}(ac)\implies(a^{-1}a)b=(a^{-1}a)c\implies eb=ec\implies b=c$

Question

Proof the uniqueness of inverses in a group, i.e. for each $a\in G$ , there is a unique element $b$ such that $ab=ba=e$ .

Answer

Suppose $a$ has two inverses $b$ and $c$ . Then $ab=e=ac$ or $ab=ac$ . By cancellation, $b=c$ .

Rearrangement theorem

(Statement) Let $G=\{e,a,b,\cdots,n\}$ and $k\in G$ . Suppose we multiply each element of $G$ by $k$ to form a new group $G'$ , where $G'=Gk=\{ek,ak,bk,\cdots,nk\}$ . Then, $G'$ contains each element of $G$ once and only once. In other words, $G'=G$ .

This theorem is proven by contradiction. Suppose two elements of $G'$ are equal, where $ak=bk$ . By cancellation, we have $a=b$ , which contradicts the definition that a group is a set, which is a collection of different elements. However, if $k\neq e$ , the order of elements in $G'$ may not be the same as that in $G$ , and thus the term ‘rearrangement’.

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December 7, 2023

Molecular point group

Molecular point groups are point groups to which molecules are assigned according to their inherent symmetries. They are used for analysing and predicting chemical properties of the assigned molecules.

The laborious way to identify the point group of a molecule is to visually determine all the symmetry operations associated with the molecule and compare them with the sets of symmetry operations of all point groups. A less arduous method involves finding key symmetry elements of a molecule and matching them sequentially to the characteristic symmetry operations of a point group. For example, cyclohexane in its twisted boat form (see diagram above) has two $C_2$ axes perpendicular to the main $C_2$ axis, while the dihedral groups $D_n,D_{nh}$ and $D_{nd}$ all have in common, the elements $C_n$ and $nC_2'\perp C_n$ . Furthermore, the molecule does not have any mirror plane and hence belongs to the point group $D_2$ . Similarly, a molecule with at least one $C_{\infty}$ axis belongs to either $C_{\infty},D_{\infty h}$ or $SO(3)$ , while a molecule with two or more $C_n$ , where $n > 2$ , belongs to one of the cubic or icosahedral point groups. We can summarise such an identification logic in the following flow chart:

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Similarity transformation

A similarity transformation of a matrix $A$ to a matrix $A'$ is expressed as $A'=P^{-1}AP$ , where

1. $P$ is an invertible matrix called the change of basis matrix.
2. $A$ is a linear transformation matrix with respect to the basis $\boldsymbol{\mathit{e}}$ .
3. $A'$ is the transformed representation of $A$ , such that $A'$ performs the same linear transformation as $A$ but with respect to another basis $\boldsymbol{\mathit{e}}'$ .

Let a vector be $\boldsymbol{\mathit{x}}$ with respect to the basis $\boldsymbol{\mathit{e}}$ and $\boldsymbol{\mathit{x}}'$ with respect to the basis $\boldsymbol{\mathit{e}}'$ . Let another vector be $\boldsymbol{\mathit{y}}$ with respect to the basis $\boldsymbol{\mathit{e}}$ and $\boldsymbol{\mathit{y}}'$ with respect to the basis $\boldsymbol{\mathit{e}}'$ . Consider the transformation of these vectors as follows:

$\boldsymbol{\mathit{x}}=P\boldsymbol{\mathit{x}}'\;\;\;\;\;\;\boldsymbol{\mathit{y}}=P\boldsymbol{\mathit{y}}'\;\;\;\;\;\;\boldsymbol{\mathit{x}}=A\boldsymbol{\mathit{y}}$

where the first two equations describe change of basis transformations and the last equation is a linear transformation of $\boldsymbol{\mathit{y}}$ to $\boldsymbol{\mathit{x}}$ in the same basis $\boldsymbol{\mathit{e}}$ .

Combining the three equation, we have $P\boldsymbol{\mathit{x}}'=A\boldsymbol{\mathit{y}}=AP\boldsymbol{\mathit{y}}'$ . If $P$ is invertible, we can multiply $P\boldsymbol{\mathit{x}}'=AP\boldsymbol{\mathit{y}}'$ by $P^{-1}$ on the left to give $\boldsymbol{\mathit{x}}'=A'\boldsymbol{\mathit{y}}'$ , where $A'=P^{-1}AP$ . Comparing $\boldsymbol{\mathit{x}}=A\boldsymbol{\mathit{y}}$ and $\boldsymbol{\mathit{x}}'=A'\boldsymbol{\mathit{y}}'$ , $A'$ is the transformed representation of $A$ , where $A'$ performs the same linear transformation as $A$ but with respect to another basis $\boldsymbol{\mathit{e}}'$ . We say that $A'$ is similar to $A$ because $A'$ has properties that are similar to $A$ . For example, the trace of $A'$ , which is defined as $Tr(A')=\sum_ia_{ii}'$ , is the same as the trace of $A$ .

To show that $Tr(A')=Tr(A)$ , we have,

$Tr(A')=Tr(G^{-1}AG)=Tr(GG^{-1}A)=Tr(EA)=Tr(A)$

where $E$ is the identity matrix and where we have used the identity $Tr(XY)=Tr(YX)$ for the second equality.

Question

Proof that $Tr(XY)=Tr(YX)$ .

Answer

$Tr(XY)=(XY)_{11}+(XY)_{22}+\cdots +(XY)_{nn}$

$=\begin{matrix}x_{11}y_{11}+&x_{12}y_{21}+&\cdots&x_{1k}y_{k1}\\+x_{21}y_{12}+&x_{22}y_{22}+&\cdots&x_{2k}y_{k2}\\\vdots&\vdots&\vdots&\vdots\\+x_{n1}y_{1n}+&x_{n2}y_{2n}+&\cdots&x_{nk}y_{kn} \end{matrix}=\begin{matrix}x_{11}y_{11}+&x_{21}y_{12}+&\cdots&x_{n1}y_{1n}\\+x_{12}y_{21}+&x_{22}y_{22}+&\cdots&x_{n2}y_{2n}\\\vdots&\vdots&\vdots&\vdots\\+x_{1k}y_{k1}+&x_{2k}y_{k2}+&\cdots&x_{nk}y_{kn} \end{matrix}$

$=\begin{matrix}y_{11}x_{11}+&y_{12}x_{21}+&\cdots&y_{1n}x_{n1}\\+y_{21}x_{12}+&y_{22}x_{22}+&\cdots&y_{2n}x_{n2}\\\vdots&\vdots&\vdots&\vdots\\+y_{k1}x_{1k}+&y_{k2}x_{2k}+&\cdots&y_{kn}x_{nk} \end{matrix}=(YX)_{11}+(YX)_{22}+\cdots+(YX)_{nn}\\=Tr(YX)$

The common properties of similar matrices are useful for explaining certain group theory concepts, such as why there are exactly 32 crystallographic point groups.

One of the most common applications of similarity transformations is to transform a matrix to a diagonal matrix $D$ . Consider the eigenvalue problem $A\boldsymbol{\mathit{u}}_j=\lambda_j\boldsymbol{\mathit{u}}_j$ and let the eigenvectors of $A$ be the columns of $P$ :

$P=(\boldsymbol{\mathit{u}}_1,\boldsymbol{\mathit{u}}_2,\cdots ,\boldsymbol{\mathit{u}}_n)=\begin{pmatrix}u_{11} & u_{21} & \cdots & u_{n1} \\u_{12} & u_{22} & \cdots & u_{n2}\\ \vdots & \vdots & \ddots & \vdots \\u_{1n} & u_{2n} & \cdots & u_{nn} \end{pmatrix}$

Since $A=\begin{pmatrix}u_{j1}\\ u_{j2}\\ \vdots\\ u_{jn}\end{pmatrix}=\lambda_j\begin{pmatrix}u_{j1}\\ u_{j2}\\ \vdots\\ u_{jn} \end{pmatrix}$ , where $A=\begin{pmatrix}a_{11} & a_{12} & \cdots & a_{1n} \\a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots \\a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}$ , we have

$AP=\begin{pmatrix}\lambda_1u_{11} &\lambda_2u_{21} & \cdots &\lambda_nu_{n1} \\\lambda_1u_{12} &\lambda_2u_{22} & \cdots &\lambda_nu_{n2}\\ \vdots & \vdots & \ddots & \vdots \\\lambda_1u_{1n} &\lambda_2u_{2n} & \cdots &\lambda_nu_{nn} \end{pmatrix}=P\begin{pmatrix}\lambda_1 &0 & \cdots &0 \\0 &\lambda_2 & \cdots &0\\ \vdots & \vdots & \ddots & \vdots \\\0 &0 & \cdots &\lambda_n \end{pmatrix}=PD$

If the eigenvectors of $A$ are linearly independent, then $P$ is non-singular (i.e. invertible). This allows us to multiply $AP=PD$ on the left by $P^{-1}$ to give $D=P^{-1}AP$ , with the diagonal entries of $D$ being the eigenvalues of $A$ .

Question

Why is $P$ non-singular if the eigenvectors of $A$ are linearly independent?

Answer

The eigenvectors $\boldsymbol{\mathit{u}}_1,\boldsymbol{\mathit{u}}_2,\cdots,\boldsymbol{\mathit{u}}_n$ are linearly independent if the only solution to $x_1\boldsymbol{\mathit{u}}_1+x_2\boldsymbol{\mathit{u}}_2+\cdots+x_n\boldsymbol{\mathit{u}}_n=\boldsymbol{\mathit{0}}$ is when $x_n=0$ for all $n$ . In other words,

$(\boldsymbol{\mathit{u}}_1,\boldsymbol{\mathit{u}}_2,\cdots,\boldsymbol{\mathit{u}}_n)\begin{pmatrix}x_1 \\x_2 \\\vdots \\ x_n \end{pmatrix}=\boldsymbol{\mathit{0}}$

$\begin{pmatrix}u_{11} & u_{21} & \cdots & u_{n1} \\u_{12} & u_{22} & \cdots & u_{n2}\\ \vdots & \vdots & \ddots & \vdots \\u_{1n} & u_{2n} & \cdots & u_{nn} \end{pmatrix}\begin{pmatrix}x_1 \\x_2 \\\vdots \\ x_n \end{pmatrix}=\begin{pmatrix}0 \\0 \\\vdots \\ 0 \end{pmatrix}$

or simply $Px=0$ .

We need to show that the only solution to $Px=0$ is $x=0$ and this is possible if $P$ is invertible such that

$Px=0\implies P^{-1}Px=P^{-1}0\implies x=0$

Using the same eigenvalue problem, we can show that Hermitian operators $\hat{H}$ are diagonalisable, i.e. $D=P^{-1}\hat{H}P$ (see this article).

Question

Show that a Hermitian matrix $H$ can be diagonalised by $U$ , i.e. $D=U^{-1}HU$ , where $U$ is a unitary matrix, and that $D$ is also Hermitian.

Answer

A unitary matrix has the property: $U^{\dagger}U=UU^{\dagger}=I$ . If a complete set of orthonormal eigenvectors of $H$ are the columns of $P$ , we have $HP=PD$ and $D=P^{-1}HP$ because orthonormal eigenvectors are linearly independent. The remaining step is to show that $P^{\dagger}P=PP^{\dagger}=I$ .

$P^{\dagger}=\begin{pmatrix}p_{11} & p_{21} & \cdots & p_{n1} \\p_{12} & p_{22} & \cdots & p_{n2}\\ \vdots & \vdots & \ddots & \vdots \\p_{1n} & p_{2n} & \cdots & p_{nn} \end{pmatrix}^{\dagger}=\begin{pmatrix}p_{11}^* & p_{12}^* & \cdots & p_{1n}^* \\p_{21}^* & p_{22}^* & \cdots & p_{2n}^*\\ \vdots & \vdots & \ddots & \vdots \\p_{n1}^* & p_{n2}^* & \cdots & p_{nn}^* \end{pmatrix}$

$P^{\dagger}P=\begin{pmatrix}p_{11}^* & p_{12}^* & \cdots & p_{1n}^* \\p_{21}^* & p_{22}^* & \cdots & p_{2n}^*\\ \vdots & \vdots & \ddots & \vdots \\p_{n1}^* & p_{n2}^* & \cdots & p_{nn}^* \end{pmatrix}\begin{pmatrix}p_{11} & p_{21} & \cdots & p_{n1} \\p_{12} & p_{22} & \cdots & p_{n2}\\ \vdots & \vdots & \ddots & \vdots \\p_{1n} & p_{2n} & \cdots & p_{nn} \end{pmatrix}\\=(\langle\boldsymbol{\mathit{p}}_i\vert\boldsymbol{\mathit{p}}_j\rangle)_{ij}\delta_{ij}=I$

For example, $\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$ . Since $P$ is non-singular, multiplying $P^{-1}$ on the right of $P^{\dagger}P=I$ gives $P^{\dagger}=P^{-1}$ . So, $I=PP^{-1}=PP^{\dagger}$ .

To show that $D$ is also Hermitian, we have

$D^{\dagger}=(U^{\dagger}HU)^{\dagger}=U^{\dagger}H^{\dagger}U=U^{\dagger}HU=D$ .

As mentioned above, $D=P^{-1}AP\implies AP=PD$ . The $i$ -th column $c_i$ of $AP$ is $Ac_i$ , while the $i$ -th column of $PD$ is $D_{ii}c_i$ . So, $Ac_i=D_{ii}c_i$ , where $c_i$ is an eigenvector of $A$ and $D_{ii}$ is the corresponding eigenvalue. Therefore, the order of the columns of the change of basis matrix corresponds to the order of the diagonal entries in the diagonal matrix.

$A_1$	$\hat{P}^{A_1}z_1=z_1$	$B_1$	$\hat{P}^{B_1}x_1=x_1$
	$\hat{P}^{A_1}z_2=\frac{1}{2}\(z_2+z_3\)$		$\hat{P}^{B_1}z_2=\frac{1}{2}\(z_2-z_3\)$
	$\hat{P}^{A_1}x_2=\frac{1}{2}\(x_2-x_3\)$		$\hat{P}^{B_1}x_2=\frac{1}{2}\(x_2+x_3\)$
	$\hat{P}^{A_1}z_3=\frac{1}{2}\(z_2+z_3\)$		$\hat{P}^{B_1}z_3=\frac{1}{2}\(z_3-z_2\)$
	$\hat{P}^{A_1}x_3=\frac{1}{2}\(x_3-x_2\)$		$\hat{P}^{B_1}x_3=\frac{1}{2}\(x_3+x_2\)$
$A_2$	$\hat{P}^{A_2}y_2=\frac{1}{2}\(y_2-y_3\)$	$B_2$	$\hat{P}^{B_2}y_1=y_1$
	$\hat{P}^{A_2}y_3=\frac{1}{2}\(y_3-y_2\)$		$\hat{P}^{B_2}y_2=\frac{1}{2}\(y_2+y_3\)$
	$\hat{P}^{A_2}y_3=\frac{1}{2}\(y_3-y_2\)$		$\hat{P}^{B_2}y_3=\frac{1}{2}\(y_3+y_2\)$

$A_1$	$\frac{1}{4}\[1\cdot 9+1\cdot\(-1\)+1\cdot 3+1\cdot 1\]=3$
$A_2$	$\frac{1}{4}\[1\cdot 9+1\cdot\(-1\)-1\cdot 3-1\cdot 1\]=1$
$B_1$	$\frac{1}{4}\[1\cdot 9-1\cdot\(-1\)+1\cdot 3-1\cdot 1\]=3$
$B_2$	$\frac{1}{4}\[1\cdot 9-1\cdot\(-1\)-1\cdot 3+1\cdot 1\]=2$

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