Advanced Chemistry - Mono Mole

April 30, 2024August 17, 2025

Anharmonicity

Anharmonicity refers to the deviation of a system’s vibrational motion from the harmonic oscillator model, resulting in changes to the energy levels.

According to the harmonic oscillator model, the potential energy curve of a diatomic molecule is a quadratic function and there are infinite vibrational energy levels that are equally spaced. In reality, the potential energy curve resembles the one derived using the Born-Oppenheimer approximation (see diagram below). Furthermore, the energy levels are finite and the spacing between adjacent levels decreases as $n$ increases.

To understand how anharmonicity results in the converging energy levels, we refer to eq67, which was derived using the Born-Oppenheimer approximation by neglecting some terms in eq66. The inclusion of these terms via the perturbation theory will show the changes in the energy levels.

Consider the following:

Description	Formula	Reference equation
Zeroth-order wavefunction	$\psi^{(0)}_{nJM}=\frac{S_n(x)}{R}Y_J^M(\theta,\phi)$	70
Zeroth-order energy	$E_n^{(0)}=\biggr$n+\frac{1}{2}\biggr$\hbar\omega$	50
1^st-order perturbation Hamiltonian	$\hat{H}^{(1)}=\frac{1}{6}U'''(R_e)x^3+\frac{1}{24}U^{iv}(R_e)x^4$	71
1^st-order perturbation energy	$E_n^{(1)}=\langle\psi_{nJM}^{(0)}\vert\hat{H}^{(1)}\vert\psi_{nJM}^{(0)}\rangle$	72

where $n,J,M$ are quantum numbers and $x=R-R_e$ .

Substituting eq70 in eq72 and noting that $S(x)$ is real,

$E_n^{(1)}=\int_0^{\infty}\frac{S_n(x)^2}{R^2}\biggr\[\frac{1}{6}U'''(R_e)x^3+\frac{1}{24}U^{iv}(R_e)x^4\biggr\]R^2dR\int_0^{2\pi}\int_0^{\pi}[Y_J^M(\theta,\phi)]^2sin\theta d\theta d\phi$

For a set of orthonormal spherical harmonic functions, ${\int_0^{2\pi}\int_0^{\pi}[Y_J^M(\theta,\phi)]^2sin\theta d\theta d\phi=1}$ . Furthermore, ${\frac{dR}{dx}=\frac{d}{dx}(x+R_e)=1\;\;\Rightarrow\;\;dR=dx}$ and so,

$E_n^{(1)}=\int_{-R_e}^{\infty}S_n(x)^2\biggr\[\frac{1}{6}U'''(R_e)x^3+\frac{1}{24}U^{iv}(R_e)x^4\biggr\]dx\;\;\;\;\;\;\;\;73$

If we assume that the probability of a nucleus of a vibrating molecule undergoing a displacement of more than $\vert R_e\vert$ is very small, then $S(x)\approx 0$ for $x < -R_e$ , and we can express the lower limit of the integral in eq73 as $-\infty$ without any major error:

$E_n^{(1)}=\frac{U'''(R_e)}{6}\int_{-\infty}^{\infty}S_n(x)^2x^3dx+\frac{U^{iv}(R_e)}{24}\int_{-\infty}^{\infty}S_n(x)^2x^4dx\;\;\;\;\;\;\;\;74$

As mentioned in this article, $S(x)$ is either an even function or an odd funcion, which makes $S(x)^2$ an even function. Therefore, the integrand $S(x)^2x^3$ is an odd function, whose integral over all space is zero. Eq74 becomes

$E_n^{(1)}=\frac{U^{iv}(R_e)}{24}\langle S_n\vert x^4\vert S_n\rangle\;\;\;\;\;\;\;\;75$

Question

$R$ , $S$ , and $T$ are linear operators with matrix representations $\boldsymbol{\mathit{R}}$ , $\boldsymbol{\mathit{S}}$ and $\boldsymbol{\mathit{T}}$ respectively. The matrix elements of $\boldsymbol{\mathit{R}}$ , $\boldsymbol{\mathit{S}}$ and $\boldsymbol{\mathit{T}}$ are $R_{mn}=\langle f_m\vert R\vert f_n\rangle$ , $S_{mn}=\langle f_m\vert S\vert f_n\rangle$ and $T_{mn}=\langle f_m\vert T\vert f_n\rangle$ respectively, where $\{f_i\}$ is a complete orthonormal basis set. Show that $\boldsymbol{\mathit{R}}=\boldsymbol{\mathit{S}}\boldsymbol{\mathit{T}}$ if $R=ST$ .

Answer

Since $\{f_i\}$ is a complete orthonormal basis set, ${\sum_i\vert f_i\rangle\langle f_i\vert=I}$ , where $I$ is the identity matrix. So,

$\begin{align}R_{mn}&=\langle f_m\vert R\vert f_n\rangle=\langle f_m\vert ST\vert f_n\rangle=\langle f_m\vert S\sum_i\vert f_i\rangle\langle f_i\vert T\vert f_n\rangle\\&=\sum_i\langle f_m\vert S\vert f_i\rangle\langle f_i\vert T\vert f_n\rangle=\sum_iS_{mi}T_{in}\end{align}$

which is the definition of the matrix multiplication of $\boldsymbol{\mathit{R}}=\boldsymbol{\mathit{S}}\boldsymbol{\mathit{T}}$ .

Consider the integral $\langle S_{n'}\vert x^2\vert S_n\rangle$ . As $x^2=xx$ , we have

$\langle S_{n'}\vert x^2\vert S_n\rangle=\sum_i\langle S_{n'}\vert x\vert S_i\rangle\langle S_{i}\vert x\vert S_n\rangle\;\;\;\;\;\;\;\;76$

Replacing $m$ in eq32a with $\mu$ and substituting the resultant equation in eq76

$\begin{align}\langle S_{n'}\vert x^2\vert S_n\rangle=&\frac{\hbar}{2\mu\omega}\biggr\[\sum_i\sqrt{(i+1)(n+1)}\delta_{n',i+1}\delta_{i,n+1}+\sum_i\sqrt{n(i+1)}\delta_{n',i+1}\delta_{i,n-1}\\&+\sum_i\sqrt{i(n+1)}\delta_{n',i-1}\delta_{i,n+1}+\sum_i\sqrt{in}\delta_{n',i-1}\delta_{i,n-1}\biggr\]\;\;\;\;\;\;\;\;77\end{align}$

All the terms in the first summation vanish except for when $i=n'-1$ and $i=n+1$ . Using $i=n+1$ , the 1^st summation becomes $\sqrt{(n+2)(n+1)}\delta_{n',n+2}$ . Applying the same logic to the remaining summations, eq77 becomes

$\langle S_{n'}\vert x^2\vert S_n\rangle=\frac{\hbar}{2\mu\omega}\biggr\[\sqrt{(n+2)(n+1)}\delta_{n',n+2}+(2n+1)\delta_{n',n}+\sqrt{n(n-1)}\delta_{n',n-2}\biggr\]\;\;\;\;\;\;\;\;78$

Let’s now consider the integral $\langle S_{n'}\vert x^4\vert S_n\rangle$ . As $x^4=x^2x^2$ , we have

$\langle S_{n'}\vert x^4\vert S_n\rangle=\sum_i\langle S_{n'}\vert x^2\vert S_i\rangle\langle S_{i}\vert x^2\vert S_n\rangle\;\;\;\;\;\;\;\;79$

Substituting eq78 in eq79 and employing the same logic as before, we have

$\begin{align}\langle S_{n'}\vert x^4\vert S_n\rangle=&\biggr$\frac{\hbar}{2\mu\omega}\biggr$^2\biggr\[\sqrt{(n+4)(n+3)(n+2)(n+1)}\delta_{n',n+4}\\&+(2n+1)\sqrt{(n+2)(n+1)}\delta_{n',n+2}+n(n-1)\delta_{n',n}\\&+(2n+5)\sqrt{(n+2)(n+1)}\delta_{n',n+2}+(2n+1)^2\delta_{n',n}\\&+(2n-3)\sqrt{n(n-1)}\delta_{n',n-2}+(n+2)(n+1)\delta_{n',n}\\&+(2n+1)\sqrt{n(n-1)}\delta_{n',n-2}+\sqrt{n(n-1)(n-2)(n-3)}\delta_{n',n-4}\biggr\]\end{align}$

If $n'=n$ , the above equation becomes

$\langle S_{n}\vert x^4\vert S_n\rangle=\biggr$\frac{\hbar}{2\mu\omega}\biggr$^2(6n^2+6n+3)\;\;\;\;\;\;\;\;80$

Substitute eq80 in eq75,

$E_n^{(1)}=\frac{U^{iv}(R_e)}{8}\biggr$\frac{\hbar}{2\mu\omega}\biggr$^2(2n^2+2n+1)\;\;\;\;\;\;\;\;81$

With reference to eq74, the $x^3$ term disappeared for the computation of the first-order energy correction $E_n^{(1)}$ . So, we need to account for it by calculating the second-order energy correction $E_n^{(2)}$ . From eq270a,

$E_n^{(2)}=\sum_{m\neq n}\frac{\vert\langle S_m^{(0)}\vert\hat{H}^{(1)}\vert S_n^{(0)}\rangle\vert^2}{E_n^{(0)}-E_m^{(0)}}\;\;\;\;\;\;\;\;82$

Substituting eq50 and $H^{(1)}=\frac{1}{6}U'''(R_e)x^3$ in eq82

$E_n^{(2)}=\frac{U'''(R_e)}{6}\sum_{m\neq n}\frac{\vert\langle S_m^{(0)}\vert x^3\vert S_n^{(0)}\rangle\vert^2}{(n-m)\hbar\omega}$

As $x^3=x^2x$ , we have $\langle S_{m}^{(0)}\vert x^3\vert S_n^{(0)}\rangle=\sum_i\langle S_{m}\vert x^2\vert S_i\rangle\langle S_{i}\vert x\vert S_n\rangle$ and

$E_n^{(2)}=\frac{U'''(R_e)}{6}\sum_{m\neq n}\frac{\vert\sum_i\langle S_{m}\vert x^2\vert S_i\rangle\langle S_{i}\vert x\vert S_n\rangle\vert^2}{(n-m)\hbar\omega}\;\;\;\;\;\;\;\;83$

Substituting eq32a (where $m=\mu$ ) and eq78 in eq83, and using the same logic in deriving eq78,

$E_n^{(2)}=\frac{U'''(R_e)}{6}\sum_{m\neq n}\frac{\vert A\vert^2}{(n-m)\hbar\omega}\;\;\;\;\;\;\;\;83$

where

$\begin{align}A=&\frac{\hbar}{2\mu\omega}\sqrt{\frac{(n+1)(n+2)(n+3)\hbar}{2\mu\omega}}\delta_{m,n+3}+3\biggr\[\frac{(n+1)\hbar}{2\mu\omega}\biggr\]^{\frac{3}{2}}\delta_{m,n+1}\\&+3\biggr$\frac{n\hbar}{2\mu\omega}\biggr$^{\frac{3}{2}}\delta_{m,n-1}+\frac{\hbar}{2\mu\omega}\sqrt{\frac{n(n-1)(n-2)\hbar}{2\mu\omega}}\delta_{m,n-3}\end{align}$

Since $\delta_{x,y}\delta_{x,y}=\delta_{x,y}$ and $\delta_{x,y}\delta_{x,z}=0$

$\begin{align}\vert A\vert^2 = & (n+1)(n+2)(n+3)\biggr$\frac{\hbar}{2\mu\omega}\biggr$^3\delta_{m,n+3}+9\biggr\[\frac{(n+1)\hbar}{2\mu\omega}\biggr\]^3\delta_{m,n+1}\\&+9\biggr$\frac{n\hbar}{2\mu\omega}\biggr$^3\delta_{m,n-1}+n(n-1)(n-2)\biggr$\frac{\hbar}{2\mu\omega}\biggr$^3\delta_{m,n-3}\end{align}$

Substituting the above equation in eq83 and expanding the summation gives

$E_n^{(2)}=-\frac{5\hbar^2U'''(R_e)}{8\mu^3\omega^3}\biggr$n^2+n+\frac{11}{30}\biggr$\;\;\;\;\;\;\;\;84$

With reference to eq266, $E_n\approx E_n^{(0)}+\lambda E_n^{(1)}+\lambda^2E_n^{(2)}$ . Since $E_n^{(2)}$ is negative, $\Delta E_n$ becomes smaller as $n$ increases.

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April 30, 2024October 8, 2025

Vibrational selection rules for molecules

Vibrational selection rules for molecules define the transition probabilities between different vibrational states during spectroscopic transitions.

According to the time-dependent perturbation theory, the transition probability between the orthogonal vibrational states $\psi_j=H_j\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$e^{-\frac{m\omega x^2}{2\hbar}}$ and $\psi_k=H_k\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$e^{-\frac{m\omega x^2}{2\hbar}}$ within a given electronic state of a diatomic molecule being observed by infrared (IR) spectroscopy is proportional to $\vert\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle\vert^2$ , where $\boldsymbol{\mathit{\mu}}$ is the operator for the molecule’s electric dipole moment. Since the dipole moment is dependent on the bond length of the molecule, we can express it as a Taylor series about the equilibrium position of the molecule:

$\boldsymbol{\mathit{\mu}}=\boldsymbol{\mathit{\mu}}_e+\biggr$\frac{d\boldsymbol{\mathit{\mu}}}{dx}\biggr$_ex+\frac{1}{2}\biggr$\frac{d^2\boldsymbol{\mathit{\mu}}}{dx^2}\biggr$_ex^2+\cdots\;\;\;\;\;\;\;\;98$

where $x$ is the displacement of the molecule from its equilibrium position.

Multiplying eq98 on the left and right by $\psi_j$ and $\psi_k$ respectively and integrating over all space,

$\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle=\boldsymbol{\mathit{\mu}}_e\langle\psi_j\vert\psi_k\rangle+\biggr$\frac{d\boldsymbol{\mathit{\mu}}}{dx}\biggr$_e\langle\psi_j\vert x\vert\psi_k\rangle+\frac{1}{2}\biggr$\frac{d^2\boldsymbol{\mathit{\mu}}}{dx^2}\biggr$_e\langle\psi_j\vert x^2\vert\psi_k\rangle+\cdots$

Noting that $\langle\psi_j\vert\psi_k\rangle=0$ and ignoring the higher power terms, we have

$\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle=\biggr$\frac{d\boldsymbol{\mathit{\mu}}}{dx}\biggr$_e\langle\psi_j\vert x\vert\psi_k\rangle\;\;\;\;\;\;\;\;99$

With reference to eq32a,

$\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle=\biggr$\frac{d\boldsymbol{\mathit{\mu}}}{dx}\biggr$_e\biggr\[\sqrt{\frac{\hbar(k+1)}{2m\omega}}\delta_{j,k+1}+\sqrt{\frac{k\hbar}{2m\omega}}\delta_{j,k-1}\biggr\]\;\;\;\;\;\;\;\;100$

For a non-zero transition probability, $\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle\neq 0$ . This is only possible when

1) $\biggr$\frac{d\boldsymbol{\mathit{\mu}}}{dx}\biggr$_e\neq 0$ , and

2) $j=k\pm 1$ , or equivalently, $j-k=\pm 1$ .

Furthermore, the total angular momentum of the system must be conserved during a photon-mediated transition. All three conditions must be met for transition to occur.

The first condition states that the electric dipole moment of a diatomic molecule must vary with the displacement of the molecule. As for the second condition, since $j-k$ is the change in vibrational quantum number, which is denoted by $\Delta n$ , transition between vibrational states for a diatomic molecule occurs when

$\Delta n=\pm 1\;\;\;\;\;\;\;\;101$

Lastly, vibrational transitions in IR spectroscopy involves molecules absorbing photons, each carrying an intrinsic spin angular momentum of $\hbar$ . Allowed transitions require the conservation of total angular momentum after absorption (see below for details).

These three conditions are the selection rules that govern the transition between vibrational energy levels of a diatomic molecule. It follows that a homonuclear diatomic molecule does not show an IR vibrational spectrum because its electric dipole moment doesn’t vary with the displacement of the molecule.

Question

i) Why do homonuclear diatomic molecules have zero electric dipole moment?

ii) If a homonuclear diatomic molecule has zero electric dipole moment, does it mean that the molecule can only remain in the vibrational ground state?

Answer

i) An electric dipole moment of two charges $+q$ and $-q$ , separated by a distance $\boldsymbol{\mathit{R}}$ , is defined as a vector $\boldsymbol{\mathit{\mu}}$ with magnitude $\mu=qR$ . The direction of $\boldsymbol{\mathit{\mu}}$ points from the positive charge to the negative charge. It’s important to note that this definition follows the chemistry convention; in physics, the convention is that the arrow goes from negative to positive.

In a system of multiple electric charges $q_i$ , such as chlorine gas, we can consider the protons of the two atoms as having an effective charge and an effective position. The same principle applies to the electrons. Since the effective positions of negative and positive charges coincide, $\boldsymbol{\mathit{R}}=0$ and $\boldsymbol{\mathit{\mu}}=0$ .

ii) No, a homonuclear diatomic molecule having a zero electric dipole moment does not mean that the molecule cannot undergo vibrational transitions. The selection rules that are described above pertain only to transitions observed via electric-dipole spectroscopy, such as IR spectroscopy. Other methods, such as Raman spectroscopy, can be used to observe these transitions.

For a polyatomic molecule, the electric dipole moment can be expressed as a multi-variable Taylor series about the equilibrium position

$\boldsymbol{\mathit{\mu}}=\boldsymbol{\mathit{\mu}}_e+\sum_{i=1}^m\biggr$\frac{\partial\boldsymbol{\mathit{\mu}}}{\partial Q_i}\biggr$_eQ_i\;\;\;\;\;\;\;\;102$

where $Q_i$ are the normal modes of the molecule, and where we have ignored the higher order terms.

For the sole excitation of a single normal mode $i$ , we multiply eq102 on the left and right by $\psi_j=\vert n_a,n_b,\cdots,n'_i,\cdots,n_m\rangle$ and $\psi_k=\vert n_a,n_b,\cdots,n_i,\cdots,n_m\rangle$ respectively and integrate over all space to give,

$\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle=\sum_{i=1}^m\biggr$\frac{\partial\boldsymbol{\mathit{\mu}}}{\partial Q_i}\biggr$_e\langle\psi_j\vert Q_i\vert\psi_k\rangle=\biggr$\frac{\partial\boldsymbol{\mathit{\mu}}}{\partial Q_i}\biggr$_e\langle\psi_j\vert Q_i\vert\psi_k\rangle\;\;\;\;\;\;\;\;103$

where the second equality is due to the fact that ${\biggr$\frac{\partial\boldsymbol{\mathit{\mu}}}{\partial Q_i}\biggr$_e=0}$ for all the other non-excited normal modes, which remain at their equilibrium positions.

Expanding eq103,

$\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle=\biggr$\frac{\partial\boldsymbol{\mathit{\mu}}}{\partial Q_i}\biggr$_e\int\cdots\int\psi^*_{n_a}\cdots\psi^*_{n'_i}\cdots\psi^*_{n_m}Q_i\psi_{n_a}\cdots\psi_{n_i}\cdots\psi_{n_m}dQ_a\cdots dQ_m$

$\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle=\biggr$\frac{\partial\boldsymbol{\mathit{\mu}}}{\partial Q_i}\biggr$_e\int\psi^*_{n'_i}Q_i\psi_{n_i}dQ_i\;\;\;\;\;\;\;\;104$

Substituting eq32a with $Q_i$ in place of $x$ in eq104,

$\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle=\biggr$\frac{\partial\boldsymbol{\mathit{\mu}}}{\partial Q_i}\biggr$_e\biggr\[\sqrt{\frac{\hbar(n_i+1)}{2m\omega}}\delta_{n'_i,n_i+1}+\sqrt{\frac{n_i\hbar}{2m\omega}}\delta_{n'_i,n_i-1}\biggr\]$

Consequently, we end up with the same two selection rules. For example, the normal mode of the symmetric stretch of $CO_2$ is IR inactive because the net electric dipole moment is always zero and hence ${\biggr$\frac{\partial\boldsymbol{\mathit{\mu}}}{\partial Q_i}\biggr$_e=0}$ . In contrast, the normal mode of the antisymmetric stretch is IR active because it produces an electric dipole moment that varies with displacement. However, the absorption peak for the asymmetric stretch does not correspond to a pure vibrational transition, but a rovibrational transition, such as $\vert n=0,J=0\rangle\rightarrow\vert n=1,J=\pm 1\rangle$ , in which the total angular momentum, including that of the photon, is conserved.

Question

What are the components of the rotational quantum number $J$ ?

Answer

The quantum number $J$ represents the magnitude of the coupled total angular momentum vector $\boldsymbol{\mathit{J=R+L+G+S}}$ , where

$\boldsymbol{\mathit{R}}$ is the nuclear rotation angular momentum vector (end-over-end rotation).
$\boldsymbol{\mathit{L}}$ is the electronic orbital angular momentum.
$\boldsymbol{\mathit{G}}$ is the vibrational angular momentum.
$\boldsymbol{\mathit{S}}$ is the electronic spin angular momentum vector.

The nuclear spin momentum is excluded because it couples only weakly to the other components. As the asymmetric stretch of CO₂ involves a closed shell molecule, $\boldsymbol{\mathit{L=S=0}}$ . Furthermore, this vibrational mode does not produce vibrational angular momentum ( $\boldsymbol{\mathit{G=0}}$ ). Therefore, the absorption of the photon’s angular momentum must be accounted for by $\Delta\boldsymbol{\mathit{R=\pm1}}$ , resulting in $J=0\rightarrow J=\pm1$ .

The degenerate bending modes are also IR active but differ from the asymmetric stretch in an important way: they can involve vibrational angular momentum. Because the two bending modes occur in perpendicular planes and are degenerate, their linear combination can produce a circular (or elliptical) motion of the O atoms about the C atom, effectively giving rise to vibrational angular momentum. This allows the molecule to absorb the photon’s angular momentum along an alternate pathway without requiring a change in the rotational quantum number $J$ . In other words, $\Delta\boldsymbol{\mathit{G}}=\pm\boldsymbol{1}$ after photon absorption, but the vector addition of $\boldsymbol{R}$ and $\boldsymbol{G}$ can result in the magnitude of $\boldsymbol{J}$ being unchanged (possible when $\boldsymbol{R\neq 0}$ ), enabling $\vert n=0,J\rangle\rightarrow\vert n=1,J\rangle$ transitions to appear in the IR spectrum as the Q-branch.

Next article: Fourier series

Previous article: Anharmonicity

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April 30, 2024December 12, 2024

Fourier series

A Fourier series is a mathematical tool that represents a function, which is periodic on a specific interval, as a linear combination of cosine and sine waves.

Consider the trigonometric functions ${g(x)=cos\biggr$\frac{2\pi nx}{L}\biggr$}$ and ${h(x)=sin\biggr$\frac{2\pi nx}{L}\biggr$}$ that are periodic on the interval $0\leq x\leq L$ :

$g(x)=cos\biggr\[\frac{2\pi n(x+L)}{L}\biggr\]=cos\biggr$\frac{2\pi nx}{L}+2\pi n\biggr$=cos\biggr$\frac{2\pi nx}{L}\biggr$$

$h(x)=sin\biggr\[\frac{2\pi n(x+L)}{L}\biggr\]=sin\biggr$\frac{2\pi nx}{L}+2\pi n\biggr$=sin\biggr$\frac{2\pi nx}{L}\biggr$$

Question

i) What is the period of $h(x)=sin(bx)$ ?

ii) Show that the sum of two periodic functions produces another periodic function if the ratio of the periods of the component functions is a rational number.

Answer

i) The period is the interval after which the range of the function repeats itself. The sine wave completes one full cycle when $bx=2\pi$ , which happens when ${x=\frac{2\pi}{b}}$ . Hence, the period of $h(x)$ is ${\frac{2\pi}{b}}$ .

ii) Let $p(x)=cos(ax)$ and $q(x)=sin(bx)$ with periods ${\frac{2\pi}{a}}$ and ${\frac{2\pi}{b}}$ , respectively, where $a,b\neq 0$ . For all $x$ , the output of the function $f(x)=Acos(ax)+Bsin(bx)$ repeats itself when

$\begin{align}f(x)&=Acos\biggr\[a\biggr$x+m\frac{2\pi }{a}\biggr$\biggr\]+Bsin\biggr\[b\biggr$x+n\frac{2\pi }{b}\biggr$\biggr\]\\&=Acos(ax+m2\pi)+Bsin(bx+n2\pi)\\&=Acos(ax)+Bsin(bx)\end{align}$

where $m,n\in \mathbb{Z}$ .

In other words, the period of $f(x)$ is ${T=m\frac{2\pi}{a}=n\frac{2\pi}{b}}$ , which implies that ${\frac{m}{n}=\frac{a}{b}}$ . Therefore, ${\frac{a}{b}}$ and hence ${\frac{2\pi}{b}/\frac{2\pi}{a}}$ must be a rational number.

The linear combination ${f(x)=Ag(x)+Bh(x)=Acos\biggr$\frac{2\pi nx}{L}\biggr$+Bsin\biggr$\frac{2\pi nx}{L}\biggr$}$ is a periodic function since the ratio of the periods of the cosine and sine functions is a rational number. It follows that ${f(x)+Ccos\biggr$\frac{2\pi nx}{L}\biggr$}$ is also a periodic function. Therefore, we can express a function that is periodic on the interval $0\leq x\leq L$ as an infinite sum of cosine and sine functions:

$\begin{align}f(x)&=\sum_{n=0}^{\infty}\biggr\[a_ncos\biggr$\frac{2\pi nx}{L}\biggr$+b_nsin\biggr$\frac{2\pi nx}{L}\biggr$\biggr\]\\&=a_0+\sum_{n=1}^{\infty}\biggr\[a_ncos\biggr$\frac{2\pi nx}{L}\biggr$+b_nsin\biggr$\frac{2\pi nx}{L}\biggr$\biggr\]\;\;\;\;\;\;\;\;105\end{align}$

Eq105 is known as a Fourier series.

To find $a_0$ , we integrate eq105 with respect to $x$ over the interval $L$ :

$\int_0^Lf(x)dx=\int_0^L\biggr\{a_0+\sum_{n=1}^{\infty}\biggr\[a_ncos\biggr$\frac{2\pi nx}{L}\biggr$+b_nsin\biggr$\frac{2\pi nx}{L}\biggr$\biggr\]\biggr\}dx$

The cosine and sine functions have the period of $L/n$ . As we are integrating over $n$ multiples of $L/n$ , we have ${a_n\int_0^Lcos\biggr$\frac{2\pi nx}{L}\biggr$dx=b_n\int_0^Lsin\biggr$\frac{2\pi nx}{L}\biggr$dx=0}$ . Hence, ${\int_0^Lf(x)dx=a_0L}$ or

$a_0=\frac{1}{L}\int_0^Lf(x)dx\;\;\;\;\;\;\;\;106$

To find $a_n$ , we multiply eq105 by ${cos\biggr$\frac{2\pi mx}{L}\biggr$}$ and integrate with respect to $x$ over the period $L$ :

$\int_0^Lf(x)cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\int_0^L\biggr\{a_0+\sum_{n=1}^{\infty}\biggr\[a_ncos\biggr$\frac{2\pi nx}{L}\biggr$+b_nsin\biggr$\frac{2\pi nx}{L}\biggr$\biggr\]\biggr\}cos\biggr$\frac{2\pi mx}{L}\biggr$dx$

Since $m$ is an integer, ${a_0\int_0^Lcos\biggr$\frac{2\pi mx}{L}\biggr$dx=\frac{a_0L}{2\pi m}sin2\pi m=0}$ . Using the trigonometric identity $2sinAcosB=sin(A+B)+sin(A-B)$ , we have ${b_n\int_0^Lsin\biggr$\frac{2\pi nx}{L}\biggr$cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\frac{b_n}{2}\int_0^L\biggr\{sin\biggr\[(n+m)\frac{2\pi x}{L}\biggr\]+sin\biggr\[(n-m)\frac{2\pi x}{L}\biggr\]\biggr\}dx=0}$ because we are again integrating over an integral number of complete periods for either sine function. Therefore,

$\int_0^Lf(x)cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\sum_{n=1}^{\infty}a_n\int_0^Lcos\biggr$\frac{2\pi nx}{L}\biggr$cos\biggr$\frac{2\pi mx}{L}\biggr$dx$

Using the trigonometric identity of $2cosAcosB=cos(A+B)+cos(A-B)$ ,

$\int_0^Lf(x)cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\sum_{n=1}^{\infty}\frac{a_n}{2}\int_0^L\biggr\{cos\biggr\[(n+m)\frac{2\pi x}{L}\biggr\]+cos\biggr\[(n-m)\frac{2\pi x}{L}\biggr\]\biggr\}dx\;\;\;\;\;\;\;\;107$

Since we are again integrating over an integral number of complete periods for either cosine function, all the terms in the summation equals zero except when $n=m$ . We have ${\int_0^Lf(x)cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\frac{a_m}{2}\int_0^Ldx=\frac{a_m}{2}L}$ or

$a_n=\frac{2}{L}\int_0^Lf(x)cos\biggr$\frac{2\pi nx}{L}\biggr$dx\;\;\;\;\;\;n=1,2,\cdots\;\;\;\;\;\;\;\;108$

where we have relabelled the index from $m$ to $n$ .

To find $b_n$ , we multiply eq105 by ${sin\biggr$\frac{2\pi mx}{L}\biggr$}$ and integrate with respect to $x$ over the period $L$ . Using the same logic as the derivation of eq108 and the trigonometric identity $2sinAsinB=cos(A-B)-cos(A+B)$ , we have

$b_n=\frac{2}{L}\int_0^Lf(x)sin\biggr$\frac{2\pi nx}{L}\biggr$dx\;\;\;\;\;\;n=1,2,\cdots\;\;\;\;\;\;\;\;109$

Question

i) Can a Fourier series be defined for a function that is periodic on the interval ${-\frac{L}{2}\leq x\leq \frac{L}{2}}$ instead of $0\leq x\leq L$ ?

ii) Can we represent a periodic function of period $L$ with a Fourier series on the interval $0\leq x\leq 2L$ instead of $0\leq x\leq L$ ?

Answer

i) Yes, the representation is the same as eq105 but has slightly different formulae for the coefficients. Repeating the above steps in determining the coefficients, we have

$a_0=\frac{1}{L}\int_{-L/2}^{L/2}f(x)dx\;\;\;\;\;\;\;\;109a$

$a_n=\frac{2}{L}\int_{-L/2}^{L/2}f(x)cos\biggr$\frac{2\pi nx}{L}\biggr$dx\;\;\;\;\;\;\;\;109b$

$b_n=\frac{2}{L}\int_{L/2}^{L/2}f(x)sin\biggr$\frac{2\pi nx}{L}\biggr$dx\;\;\;\;\;\;\;\;109c$

ii) Yes. However, the Fourier series representation is $f(x)=a_0+\sum_{n=1}^{\infty}\biggr\[a_ncos\biggr$\frac{\pi nx}{L}\biggr$+b_nsin\biggr$\frac{\pi nx}{L}\biggr$\biggr\]$ , which is slightly different from eq105. The corresponding coefficients are ${a_0=\frac{1}{2L}\int_0^{2L}f(x)dx}$ , ${a_n=\frac{1}{L}\int_0^{2L}f(x)cos\biggr$\frac{\pi nx}{L}\biggr$dx}$ and ${b_n=\frac{1}{L}\int_0^{2L}f(x)sin\biggr$\frac{\pi nx}{L}\biggr$dx}$ . The function in this representation now repeats every $2L$ instead of every $L$ .

The fourier series can also be expressed in the complex form. Substituting the identities ${cosy=\frac{e^{iy}+e^{-iy}}{2}}$ and ${siny=\frac{e^{iy}-e^{-iy}}{2i}}$ , where ${y=\frac{2\pi nx}{L}}$ in eq105, we have

$f(x)=a_0+\sum_{n=1}^{\infty}\biggr\[\biggr$\frac{a_n-ib_n}{2}\biggr$e^{iy}+\biggr$\frac{a_n+ib_n}{2}\biggr$e^{-iy}\biggr\]\;\;\;\;\;\;\;\;110$

Let ${c_n=\frac{a_n-ib_n}{2}}$ and ${c_{-n}=\frac{a_{-n}-ib_{-n}}{2}}$ , where $n=1,2,\cdots$ . From eq108, $a_{-n}=\frac{2}{L}\int_0^Lf(x)cos(-y)dx=\frac{2}{L}\int_0^Lf(x)cos(y)dx=a_n$ . Similarly, from eq109, $b_{-n}=-b_n$ . Therefore, ${c_{-n}=\frac{a_{n}+ib_{n}}{2}}$ and eq110 becomes

$f(x)=c_0+\sum_{n=1}^{\infty}c_ne^{i\frac{2\pi nx}{L}}+\sum_{n=1}^{\infty}c_{-n}e^{-i\frac{2\pi nx}{L}}\;\;\;\;\;\;\;\;111$

where we have relabelled the constant $a_0$ as $c_0$ without loss of generality (see Q&A below).

If we let the dummy index $n=-k$ for the second summation and subsequently relabel $k$ as $n$ , we have

$f(x)=c_0+\sum_{n=1}^{\infty}c_ne^{i\frac{2\pi nx}{L}}+\sum_{k=-1}^{-\infty}c_{k}e^{i\frac{2\pi kx}{L}}=c_0+\sum_{n=1}^{\infty}c_ne^{i\frac{2\pi nx}{L}}+\sum_{n=-1}^{-\infty}c_{n}e^{i\frac{2\pi nx}{L}}$

Therefore, $f(x)$ can be expressed as

$f(x)=\sum_{n=-\infty}^{\infty}c_{n}e^{i\frac{2\pi nx}{L}}\;\;\;\;\;\;\;\;112$

To find $c_n$ , we substitute eq108 and eq109 in ${c_n=\frac{a_n-ib_n}{2}}$ to give

$c_n=\frac{1}{L}\int_0^Lf(x)e^{-i\frac{2\pi nx}{L}}dx\;\;\;\;\;\;\;\;113$

Question

i) Show that $c_0=a_0$ .

ii) What is the corresponding equation for eq113 if $f(x)$ is periodic on the interval ${-\frac{L}{2}\leq x\leq \frac{L}{2}}$ instead of $0\leq x\leq L$ ?

Answer

i) From eq113, $c_0=\frac{1}{L}\int_0^Lf(x)dx$ . Comparing this equation with eq106, $c_0=a_0$ . This implies that eq113 is valid for all $n$ , including $n=0$ even though the derivation involves ${c_n=\frac{a_n-ib_n}{2}}$ , which is for $n=1,2,\cdots$ .

ii) Repeating the above steps for the derivation of the complex form of a Fourier series using eq109a, eq109b and eq109c, we have

$c_n=\frac{1}{L}\int_{-L/2}^{L/2}f(x)e^{-i\frac{2\pi nx}{L}}dx\;\;\;\;\;\;\;\;114$

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April 30, 2024December 7, 2024

Fourier transform

The Fourier transform is a mathematical operation that produces a function in one domain by integrating a function from another domain.

As mentioned in the previous article, a Fourier series represents a function $f(x)$ within a specific interval, for example ${-\frac{L}{2}\leq x\leq \frac{L}{2}}$ . Outside this interval, the function is assumed to repeat itself. Therefore, a non-periodic function cannot be accurately represented by a Fourier series. If we want a mathematical tool to represent both periodic and non-periodic functions, we need a more generalised version of the Fourier series that is defined on the interval ${-\frac{L}{2}\leq x\leq \frac{L}{2}}$ , as $L\rightarrow\infty$ . To express this mathematically, let ${k_n=\frac{2\pi n}{L}}$ and substitute eq113 for this new interval in eq112 to give

$f(x)=\sum_{n=-\infty}^{\infty}\biggr\[\frac{1}{L}\int_{-\infty}^{\infty}f(x)e^{-ik_nx}dx\biggr\]e^{ik_nx}\;\;\;\;\;\;\;\;114$

Since ${\Delta k_n=\frac{2\pi \Delta n}{L}}$ and $\Delta n=1$ ,

$f(x)=\sum_{n=-\infty}^{\infty}\biggr\[\frac{1}{L}\int_{-\infty}^{\infty}f(x)e^{-ik_nx}dx\biggr\]e^{ik_nx}\Delta n=\sum_{n=-\infty}^{\infty}\biggr\[\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ik_nx}dx\biggr\]e^{ik_nx}\Delta k_n$

As ${\lim_{L\rightarrow\infty}\Delta k_n=\lim_{L\rightarrow\infty}\frac{2\pi\Delta n}{L}\rightarrow 0}$ , the discrete sum in the above equation becomes a continuous integral:

$f(x)=\sum_{n=-\infty}^{\infty}\biggr\[\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\biggr\]e^{ikx}dk\;\;\;\;\;\;\;\;115$

where we have replaced the discrete variable $k_n$ with the continuous variable $k$ .

Question

Show that $\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$ in eq115 is equal to the function $g(k)$ .

Answer

The integral $\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$ is essentially a function of $k$ , as there is a corresponding integral value (output) for every value of $k$ (input).

Therefore,

$f(x)=\int_{-\infty}^{\infty}g(k)e^{ikx}dk\;\;\;where\;\;\;g(k)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\;\;\;\;\;\;\;\;117$

$g(k)$ is known as the Fourier transform of $f(x)$ , and vice versa. The two equations in eq117 are regarded as a Fourier transform pair, which allow one function to be transformed into the other function. We can also say that $g(k)$ is the inverse Fourier transform of $f(x)$ , and vice versa.

Question

i) Show that $h(x)=f(x)g(x)$ is an even function if both $f(x)$ and $g(x)$ are even functions.

ii) Show that $h(x)=f(x)g(x)$ is an odd function if $f(x)$ is an even function and $g(x)$ is an odd function or vice versa.

iii) If $f(x)$ is an odd function, show that $\int_{-\infty}^{\infty}f(x)dx=0$ , assuming that the integral is carried out over a symmetric interval.

iv) Is $\int_{-\infty}^{\infty}q(x)dx=0$ if $q(x)$ is an even function?

Answer

i) An even function and an odd function are defined as $f(-x)=f(x)$ and $f(-x)=-f(x)$ respectively. So, $h(-x)=f(-x)g(-x)=f(x)g(x)=h(x)$ .

ii) $h(-x)=f(-x)g(-x)=-f(x)g(x)=-h(x)$ .

iii) Using the Cauchy principal value, ${\int_{-\infty}^{\infty}f(x)dx=\lim_{a\rightarrow\infty,b\rightarrow 0}\biggr\[\int_{-a}^bf(x)dx+\int_b^af(x)dx\biggr\]=\int_{-\infty}^0f(x)dx+\int_0^{\infty}f(x)dx}$ . Let $x=-x$ , $dx=-dx$ for $\int_{-\infty}^0f(x)dx$ . So,

$\int_{-\infty}^{\infty}f(x)dx=-\int_{\infty}^{0}f(-x)dx+\int_{0}^{\infty}f(x)dx=0$

iv) No. With $x=-x$ , $dx=-dx$ for $\int_{-\infty}^0f(x)dx$ ,

$\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^0f(x)dx+\int_0^{\infty}f(x)dx=2\int_{0}^{\infty}f(x)dx$

Special forms of eq117 arise according to whether $f(x)$ is even or odd. From eq117,

$g(k)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)coskxdx-\frac{i}{2\pi}\int_{-\infty}^{\infty}f(x)sinkxdx$

$coskx$ is an even function, while $sinkx$ is an odd function. So, if $f(x)$ is even, the first integrand is an even function, while the second integrand is an odd function. As shown in parts iii and iv the Q&A above,

$g(k)=\frac{1}{\pi}\int_{0}^{\infty}f(x)coskxdx\;\;\;\;\;\;\;\;118$

It follows that $g(k)$ is also an even function because $g(-k)=\frac{1}{\pi}\int_{0}^{\infty}f(x)cos(-kx)dx=\frac{1}{\pi}\int_{0}^{\infty}f(x)coskxdx=g(k)$ . From eq117,

$\begin{align}f(x)&=\int_{-\infty}^{\infty}g(k)e^{ikx}dx\\&=\int_{-\infty}^{\infty}g(k)coskxdk+i\int_{-\infty}^{\infty}g(k)sinkxdk\\&=2\int_{0}^{\infty}g(k)coskxdk\;\;\;\;\;\;\;\;119\end{align}$
Eq118 and eq119 are known as a pair of Fourier cosine transforms, which has applications in Fourier-transform infrared spectroscopy.

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April 30, 2024December 3, 2024

Fourier-transform infrared spectroscopy

Fourier-transform infrared spectroscopy (FT-IR) is the study of the interaction of infrared radiation with matter, enabling the identification of chemical substances or functional groups using an interferometer to measure the entire infrared (IR) spectrum simultaneously.

The diagram below outlines a typical FT-IR spectrometer, which utilises a heated rod, wire or coil as the radiation source. The materials used for fabricating these components can be rare-earth oxides (for the Nernst Glower), silicon carbide (for the Globar) or nichrome (for the Nichrome Coil). All of these materials produce continuous radiations when heated to specific temperatures within the range of 1000^oC to 1800^oC.

A beam of broad-spectrum infrared light produced by the source (beam A) is collimated and then equally split into two separate beams, B and C, by a beam splitter (BS1). The beam splitter is a sheet of glass that is partially coated with a thin layer of metal, making it half-transmitting and half-reflecting. Beam B is reflected by a moving mirror towards a second beam splitter (BS2), where it is again split into two beams (beam D and a balanced output). The moving mirror is driven by a motor and moves in the direction indicated in the above diagram. Beam C, on the other hand, is reflected by a fixed mirror towards BS2*, where it is similarly split into two beams (beam D and a balanced output). Beam D, which is a superposition of two beams, then passes through the sample cell, beyond which lies the photon detector.

* Beam C appears to cross beam A in the diagram above. However, it is redirected by mirrors (not shown for simplicity) before and after the apparent crossing point, so that it avoids beam A but subsequently interferes with beam B at BS2.

The two beams forming beam D interfere with each other based on the principle of superposition of waves. Let’s suppose that beam A is a monochromatic light of wavelength $\lambda$ . If the path difference $\delta$ between the two beams is $\delta=2x$ , constructive interference and destructive interference occur when $\delta=n\lambda$ and ${\delta=\biggr$n+\frac{1}{2}\biggr$\lambda}$ , respectively.

Question

i) Prove the identity ${cosa+cosb=2cos\biggr$\frac{a-b}{2}\biggr$cos\biggr$\frac{a+b}{2}\biggr$}$ .

ii) Show that the intensity $I$ of the superposition of two waves of the same frequency is $I=2kA^2(1+cos\delta)$ , where $k$ is a constant, $A$ is the common maximum amplitude of the two waves, and $\delta$ is the phase difference of the two waves.

Answer

i) Let ${\frac{a+b}{2}=x}$ and ${\frac{a-b}{2}=y}$ . Then, $a=x+y$ and $b=x-y$ . Substituting $a$ and $b$ in the equation $cos(a+b)+cos(a-b)=2cosacosb$ gives $cos(2x)+cos(2y)=2cos(x+y)cos(x-y)$ . Substituting $a=2x$ and $b=2y$ in this equation completes the proof.

ii) Consider the travelling waves $y_1=Acos(kx-\omega t+\phi)$ and $y_2=Acos(kx-\omega t)$ , both of which have the same amplitude $A$ , the same frequency $\omega$ , but different phases. The superposition of the two waves is $y=A[cos(kx-\omega t+\phi)+cos(kx-\omega t)]$ . Substituting the trigonometric identity in part i) in $y$ gives ${y=2Acos\biggr$\frac{\phi}{2}\biggr$cos\biggr$kx-\omega t+\frac{\phi}{2}\biggr$}$ , which is the resultant travelling wave whose amplitude is ${2Acos\biggr$\frac{\phi}{2}\biggr$}$ . As the intensity of a wave is proportional to the square of its amplitude, ${I=4kA^2cos^2\biggr$\frac{\phi}{2}\biggr$=2kA^2\biggr$1+2cos^2\frac{\phi}{2}-1\biggr$=2kA^2(1+cos\phi)}$ , where $k$ is the proportionality constant. Since the phase constant $\phi$ can be any value in radians, we can express the intensity of the superposition of the two waves as

$I=2kA^2\biggr$1+cos\frac{2\pi\delta}{\lambda}\biggr$\;\;\;\;\;\;\;\;120$

where $\lambda$ is the common wavelength of the two waves and $\delta$ is the path difference of the two waves.

According to eq120, the intensity of the supposed monochromatic beam D is $I_{mono,d}=2kA^2\biggr$1+cos\frac{2\pi\delta}{\lambda}\biggr$$ , which is an even function of $\delta$ (see above diagram). We can also express the equation as

$I_{mono,D}(\delta)=\frac{1}{2}B_D[1+cos(2\pi\delta\tilde{\nu})]\;\;\;\;\;\;\;\;121$

where $\tilde{\nu}$ is the wavenumber and $B_D$ is $I_{mono,d}$ at $\delta=0$ .

When the monochromatic wave passes through the sample, it is partially absorbed. This absorption results in a reduced intensity for the exiting wave (beam E), which remains an even function of $\delta=0$ . Beam E ends its journey at the photodetector, which operates based on the photoelectric effect. This effect involves the interaction of IR radiation with a semiconductor, promoting electrons to the conduction band and generating a small current that is proportional to the intensity of the radiation and independent of the wavenumber. When a broad spectrum of light hits the photodetector, each photon with a specific wavenumber interacts individually with the detector material. The total current is then the sum of the contributions from all the photons. Since beam E actually contains a continuous range of wavenumbers, it produces a signal at the detector that corresponds to the sum of the intensities at each individual wavenumber:

$I_E(\delta)=\frac{1}{2}\int_{0}^{\infty}B_E(\tilde\nu)[1+cos(2\pi\delta\tilde{\nu})]d\tilde{\nu}\;\;\;\;\;\;\;\;122$

where $B_E(\tilde\nu)$ is the intensity of beam E when $\delta=0$ .

Question

Is $I_E(\delta)$ an even function of $\delta$ ?

Answer

Yes, $I_E(\delta)$ remains an even function of $\delta$ because we are essentially summing different even functions of the reduced intensities of monochromatic waves in eq122.

Substituting $I_E(0)=\int_{0}^{\infty}B_E(\tilde\nu)d\tilde{\nu}$ in eq122, we have $I_E(\delta)=\frac{1}{2}I_E(0)+\frac{1}{2}\int_{0}^{\infty}B_E(\tilde\nu)cos(2\pi\delta\tilde{\nu})d\tilde{\nu}$ or

$F(\delta)=\frac{1}{2}\int_{0}^{\infty}B_E(\tilde\nu)cos(2\pi\delta\tilde{\nu})d\tilde{\nu}\;\;\;\;\;\;\;\;123$

where the interferogram function $F(\delta)$ is equal to $I_E(\delta)-\frac{1}{2}I_E(0)$ .

Comparing with eq118 and eq119, eq123 is a Fourier cosine transform with the following inverse Fourier cosine transform:

$B_E(\tilde\mu)=\frac{4}{\pi}\int_{0}^{\infty}F(\delta)cos(2\pi\delta\tilde{\nu})d\delta\;\;\;\;\;\;\;\;124$

Since $I_E(\delta)$ and $I_E(0)$ are known, $F(\delta)$ is also known. Using eq124, the computer is able to compute $B_E(\tilde\nu)$ , which is the intensity of the signal as a function of wavenumber. In other words, $B_E(\tilde\nu)$ is the IR spectrum we commonly see.

As mentioned above, when beam E hits the photodetector, each photon with a specific wavenumber interacts individually with the detector material. Ideally, we would want an equal number of photons of each wavenumber of a broad spectrum of light to reach the sample at any given time. However, the types of IR source used do not emit an equal number of photons at each wavenumber. Instead, the distribution of photons over different wavenumbers follows a curve known as the blackbody radiation curve. To resolve this issue, a background spectrum is first produced without the sample (or with just the solvent if the sample is aqueous). A second spectrum of the sample is then collected and the final spectrum is generated by the computer using a ratio of the sample spectrum against the background spectrum.

Finally, let’s look at a frequent application of FT-IR spectroscopy. According to an earlier article, a normal mode of vibration involves the oscillation of most, if not all, atoms in a molecule. However, for certain normal modes, the vibrational amplitudes of most atoms may be minimal, localising the normal mode within two or three atoms. This localisation allows for the identification of specific chemical moieties in a polyatomic molecule. This is why the FT-IR spectrometer is particularly useful for identifying the functional groups of organic polyatomic molecules.

In a typical FT-IR experiment involving organic compounds, the most intense absorption transitions occur from the molecule’s ground state to the fundamental states. For example, transitions may occur from $\psi_j=\vert 0,0,\cdots,0,\cdots,0\rangle$ to $\psi_j=\vert 0,0,\cdots,1,\cdots,0\rangle$ . Some characteristic absorption ranges of such transitions for the functional groups of organic molecules include:

Bond	Functional group	Absorption wavenumber/ cm^-1	Appearance of peak
$C-O$	Alcohols, ethers, esters	1040-1300	Strong
$C=C$	Aromatic compounds, alkenes	1500-1680	Weak unless conjugated
$C=O$	Amides Ketones & aldehydes Esters	1640-1690 1670-1740 1715-1750	Strong Strong Strong
$C\equiv C$	Alkynes	2150-2250	Weak unless conjugated
$C\equiv N$	Nitriles	2200-2250	Weak
$C-H$	Alkanes Alkenes & arenes	2850-2950 3000-3100	Strong Weak
$N-H$	Amines, amides	3300-3500	Weak
$O-H$	Carboxylic acids H-bonded alcohols Non H-bonded alcohols	2500-3000 3200-3600 3580-3650	Strong and broad Strong and broad Strong

For example, the spectra of butanone and butan-2-ol are:

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Group theory and infra-red spectroscopy

Group theory allows us to determine the symmetries of a molecule, enabling an efficient way to obtain relevant information in infra-red (IR) spectroscopy.

The objective in this article is to use group theory to work out the symmetries of the normal modes of a molecule and then ascertain whether they are IR-active. As an example, let’s consider $H_{2}O$ , which belongs to the $C_{2v}$ point group. Here’s the corresponding character table:

Instead of using Cartesian displacement vectors to form a basis for generating representations of the $C_{2v}$ point group, as we did in the previous article, we shall use mass-weighted Cartesian coordinate unit vectors $q_i$ to form a basis (see diagram above). Since a symmetry operation of $C_{2v}$ transforms $H_2O$ 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 (see eq55 to eq61). Therefore, $q_i$ centred on every atom of $H_2O$ generate a representation $\Gamma_{3N}$ of the $C_{2v}$ point group. To find the matrices of the representation, we arrange the 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,

$q_1,q_2,q_3,q_4,q_5,q_6,q_7,q_8,q_9\)\xrightarrow{C_2}\(-q_1,-q_2,q_3,-q_7,-q_8,q_9,-q_4,-q_5,q_6$

$\begin{pmatrix}-q_1\\-q_2\\q_3\\-q_7\\-q_8\\q_9\\-q_4\\-q_5\\q_6\end{pmatrix}=\begin{pmatrix}-1&0&0&0&0&0&0&0&0\\0&-1&0&0&0&0&0&0&0\\0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&-1&0&0\\0&0&0&0&0&0&0&-1&0\\0&0&0&0&0&0&0&0&1\\0&0&0&-1&0&0&0&0&0\\0&0&0&0&-1&0&0&0&0\\0&0&0&0&0&1&0&0&0\end{pmatrix}\begin{pmatrix}q_1\\q_2\\q_3\\q_4\\q_5\\q_6\\q_7\\q_8\\q_9\end{pmatrix}$

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 $\Gamma_{3N}=3A_1\oplus A_2\oplus 3B_1\oplus 2B_2$ .

Applying the projection operator on each element of the $3N$ basis set for all irreducible representations, we have

$A_1$	$\hat{P}^{A_1}q_3=q_3$	$B_1$	$\hat{P}^{B_1}q_1=q_1$
	$\hat{P}^{A_1}q_6=\frac{1}{2}$q_6+q_9$$		$\hat{P}^{B_1}q_6=\frac{1}{2}$q_6-q_9$$
	$\hat{P}^{A_1}q_4=\frac{1}{2}$q_4-q_7$$		$\hat{P}^{B_1}q_4=\frac{1}{2}$q_4+q_7$$
	$\hat{P}^{A_1}q_9=\frac{1}{2}$q_6+q_9$$		$\hat{P}^{B_1}q_9=\frac{1}{2}$q_9-q_6$$
	$\hat{P}^{A_1}q_7=\frac{1}{2}$q_7-q_4$$		$\hat{P}^{B_1}q_7=\frac{1}{2}$q_7+q_4$$
$A_2$	$\hat{P}^{A_2}q_5=\frac{1}{2}$q_5-q_8$$	$B_2$	$\hat{P}^{B_2}q_2=q_2$
	$\hat{P}^{A_2}q_8=\frac{1}{2}$q_8-q_5$$		$\hat{P}^{B_2}q_5=\frac{1}{2}$q_5+q_8$$
	$\hat{P}^{A_2}q_8=\frac{1}{2}$q_8-q_5$$		$\hat{P}^{B_2}q_8=\frac{1}{2}$q_8+q_5$$

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.

We have shown in an earlier article that

1. the number of orthogonal or orthonormal basis functions of a representation corresponds to the dimension of the representation.
2. if $f_1,f_2,\cdots,f_i$ is a basis of a representation $\Gamma$ of a group, then any linear combination of $f_1,f_2,\cdots,f_i$ is a basis of a representation $\Gamma '$ that is equivalent to $\Gamma$ .

This implies that we can always select a set of nine orthonormal SALC for a block diagonal representation that is equivalent to $\Gamma_{3N}$ . Since none of the nine SALCs can be expressed as a linear combination of the others, each SALC, known as normal coordinates $Q_j$ , must describe one of nine independent motions of the molecule.

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.

Question

Does the basis function $x$ in the character table of the $C_{2v}$ point group correspond to the normal coordinate $Aq_1+Bq_4+Cq_7$ for $H_2O$ ?

Answer

Yes. In fact, there is a correspondence between the basis functions $x,y,z,\boldsymbol{\mathit{R}}_x,\boldsymbol{\mathit{R}}_y,\boldsymbol{\mathit{R}}_z$ and six of the normal coordinates of $H_2O$ . The three normal coordinates describing the vibrational motions of $H_2O$ are known as normal modes.

To verify whether the three normal modes are IR-active, we refer to the Schrodinger equation for vibration motion (see eq93):

$-\frac{\hbar^2}{2}\sum_{j=1}^m\frac{\partial^2}{\partial Q_j^2}\psi_{vib}+\frac{1}{2}\sum_{j=1}^m\lambda_jQ_j^2\psi_{vib}=E\psi_{vib}$

where the separation of variables technique allows us to approximate the total vibrational wavefunction as $\psi_{vib}=\prod_{j=1}^m\psi_j(Q_j)$ and hence, $E=\sum_{j=1}^mE_j$ .

Each $\psi_j(Q_j)$ describes a normal mode of the molecule and has the formula (see eq94):

$\psi_j(Q_j)=N_jH_j\biggr$\sqrt{\frac{\omega_j}{\hbar}}Q_j\biggr$e^{-\frac{\omega_jQ^2_j}{2\hbar}}$

where $N_j=\sqrt[4]{\frac{\omega_j}{\pi\hbar}}\frac{1}{\sqrt{2^nn!}}$ is the normalisation constant for the Hermite polynomials $H_j$ .

The total vibrational energy (see eq96) is

$E=\sum_{j=1}^m\biggr$n_j+\frac{1}{2}\biggr$\hbar\omega_j$

A vibrational state of a polyatomic molecule $\vert n_1,n_2,\cdots,n_m\rangle$ is characterised by $m$ quantum numbers. Hence, the vibrational ground state of $H_2O$ , where $m=3N-6=3$ , is $\vert 0,0,0\rangle$ or

$\vert 0,0,0\rangle=\psi_0=N\prod_{j=1}^3H_j\biggr$\sqrt{\frac{\omega_j}{\hbar}}Q_j\biggr$e^{-\frac{\omega_jQ_j^2}{2\hbar}}=N\prod_{j=1}^3e^{-\frac{\omega_jQ_j^2}{2\hbar}}$

Many IR-spectroscopy experiments are conducted at room temperature, where most molecules are in their vibrational ground state. According to the time-dependent perturbation theory, the transition probability between orthogonal vibrational states within a given electronic state of a molecule is proportional to

$\vert\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle\vert^2=\vert\langle\psi_j\vert\mu_x\vert\psi_k\rangle\vert^2+\vert\langle\psi_j\vert\mu_y\vert\psi_k\rangle\vert^2+\vert\langle\psi_j\vert\mu_z\vert\psi_k\rangle\vert^2\;\;\;\;\;\;\;\;104$

where $\psi_j$ and $\psi_k$ are the initial and final states respectively and $\boldsymbol{\mathit{\mu}}=\boldsymbol{\mathit{i}}\mu_x+\boldsymbol{\mathit{j}}\mu_y+\boldsymbol{\mathit{k}}\mu_z$ is the operator for the molecule’s electric dipole moment.

In other words, no transition between states occurs when $\langle\psi_j\vert\mu_x\vert\psi_k\rangle=\langle\psi_j\vert\mu_y\vert\psi_k\rangle=\langle\psi_j\vert\mu_z\vert\psi_k\rangle=0$ . Since the objective is to ascertain the IR activity of each of the three normal modes of $H_2O$ , it is suffice to study the fundamental transitions, where $\psi_j=\psi_0=\vert 0,0,0\rangle$ and $\psi_k=\vert 1,0,0\rangle$ or $\vert 0,1,0\rangle$ or $\vert 0,0,1\rangle$ .

The next step involves determining which irreducible representation of the $C_{2v}$ point group $\psi_0$ , $\psi_k$ , $\mu_x$ , $\mu_y$ and $\mu_z$ belong to. Since the components $\mu_x$ , $\mu_y$ and $\mu_z$ represent the electric dipole moment along the $x$ , $y$ and $z$ directions respectively, they transform in the same way as the basis functions $x$ , $y$ and $z$ respectively (see character table above).

Question

Show that $R_{\alpha}Q_j=\pm Q_j$ , where $R_{\alpha}$ represents the symmetry operations of a point group and $j=1,\cdots,3N-6$ .

Answer

In general, if $R_{\alpha}f(x)=f(x)$ , then the function $f(x)$ is invariant under the symmetry operation $R_{\alpha}$ . Every value of $x$ is mapped into $x$ by $R_{\alpha}$ and we can say that $R_{\alpha}x=x$ if $R_{\alpha}f(x)=f(x)$ . The converse is also true, i.e. if the variable $x$ of the function $f(x)$ transforms according to $R_{\alpha}x=x$ , then $R_{\alpha}f(x)=f(x)$ . From this article, the potential term of the vibrational Hamiltonian for $H_2O$ is ${U(Q_j)=U_e+\frac{1}{2}\sum_{j=1}^{3N-6}\lambda_jQ^2_j}$ . Since the function $U(Q_j)$ is invariant under any symmetry operation, ${U(Q_j)=U_e+\frac{1}{2}\sum_{j=1}^{3N-6}\lambda_jQ^2_j=U_e+\frac{1}{2}\sum_{j=1}^{3N-6}\lambda_j(R_{\alpha}Q_j)^2}$ or simply

$\sum_{j=1}^{3N-6}\lambda_j(R_{\alpha}Q_j)^2=\sum_{j=1}^{3N-6}\lambda_jQ_j^2\;\;\;\;\;\;\;\;105$

If each $\lambda_j$ is distinct, then $(R_{\alpha}Q_j)^2=Q_j^2$ or

$R_{\alpha}Q_j=\pm Q_j\;\;\;\;\;\;\;\;106$

If $\lambda_j$ is degenerate, eq55 states that ${R_{\alpha}Q_j=\sum_{m=1}^{k}a_{mj}Q_m$ and eq105 becomes

$\lambda_1(a_{11}Q_1+a_{21}Q_2)^2+\lambda_1(a_{12}Q_1+a_{22}Q_2)^2+\lambda_3(R_{\alpha}Q_3)^2+\cdots=\\\lambda_1Q_1^2+\lambda_1Q^2_2+\lambda_3Q_3^2+\cdots\;\;\;\;\;\;\;\;107$

where we have assumed that $Q_1$ and $Q_2$ are degenerate.

As described in this article, $Q_j$ forms a set of orthonormal eigenvectors of the vibrational Hamiltonian. So, $R_{\alpha}Q_1=a_{11}Q_1+a_{21}Q_2$ and $R_{\alpha}Q_2=a_{12}Q_1+a_{22}Q_2$ are orthonormal vectors. The matrix formed by using these vectors as its columns is an orthogonal matrix $M$ , which is defined as $M^TM=I$ . This is because the dot product of different columns will be zero, and the dot product of a column with itself will be 1. Since $M^T=M^{-1}$ if $M$ is an orthogonal matrix (see property 11 of this link for proof), $M^TM=I=MM^T$ or

$\begin{pmatrix}a_{11}&a_{21}\\a_{12}&a_{22}\end{pmatrix}\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}\begin{pmatrix}a_{11}&a_{21}\\a_{12}&a_{22}\end{pmatrix}$

which implies that $a_{11}^2+a_{21}^2=a_{12}^2+a_{22}^2=a_{11}^2+a_{12}^2=a_{21}^2+a_{22}^2=1$ and $a_{11}a_{21}+a_{12}a_{22}=a_{11}a_{12}+a_{21}a_{22}=0$ .

The first two terms on LHS of eq107 then becomes $\lambda_1Q_1^2+\lambda_1Q^2_2$ . This implies that $(R_{\alpha}Q_j)^2$ must also be equal to $Q_j^2$ in eq105 if $\lambda_i$ is degenerate. Therefore, $R_{\alpha}Q_j=\pm Q_j$ regardless of whether $\lambda_i$ is non-degenerate or degenerate.

As mentioned in the above Q&A, if the variable $x$ of the function $f(x)$ transforms according to $R_{\alpha}x=x$ , then $R_{\alpha}f(x)=f(x)$ . It follows that if the variable $x$ of the function $f(x)$ transforms according to $R_{\alpha}x=-x$ , then $R_{\alpha}f(x)=f(-x)$ , in which case $R_{\alpha}$ is the reflection operator about the vertical axis of the graph of $f(x)$ against $x$ . Therefore, the ground state vibrational wave function $\psi_0$ of $H_2O$ transforms according to the totally symmetric irreducible representation $A_1$ of the $C_{2v}$ point group because

$R_{\alpha}\psi_0(Q_j)=\psi_0(\pm Q_j)=N\prod_{j=1}^3e^{-\frac{\omega_j(\pm Q_j)^2}{2\hbar}}=\psi_0$

Next, let’s analyse the symmetries of $\psi_k=\vert 1,0,0\rangle$ , $\psi_k=\vert 0,1,0\rangle$ and $\psi_k=\vert 0,0,1\rangle$ . These three wave functions (see this article) have the same form of:

$\psi_k=NQ_3\prod_{j=1}^3e^{-\frac{\omega_j Q_j^2}{2\hbar}}\;\;\;\;\;\;\;\;108$

Eq108 is a product of two functions $Q_3$ and ${\prod_{j=1}^3e^{-\frac{\omega_j Q_j^2}{2\hbar}}}$ . Since ${\prod_{j=1}^3e^{-\frac{\omega_j Q_j^2}{2\hbar}}}$ is totally symmetric, the theory of direct product representation states that $\psi_k$ must transform according to the irreducible representation of the $C_{2v}$ point group that $Q_3$ belongs to.

With reference to the $C_{2v}$ character table above, the theory of direct product representation again finds that the functions $\psi_0\mu_x$ , $\psi_0\mu_y$ and $\psi_0\mu_z$ in eq104 transform according to the irreducible representations of $B_1$ , $B_2$ and $A_1$ respectively. Therefore, the theory of vanishing integrals states that

	$\psi_k\in \Gamma_{A_1}$	$\psi_k\in \Gamma_{B_1}$
$\langle\psi_0\vert\mu_x\vert\psi_k\rangle$	Zero	Not zero
$\langle\psi_0\vert\mu_y\vert\psi_k\rangle$	Zero	Zero
$\langle\psi_0\vert\mu_z\vert\psi_k\rangle$	Not zero	Zero

Since $\vert\langle\psi_j\vert\boldsymbol{\mathit{\mu}}\vert\psi_k\rangle\vert^2\neq 0$ in eq104 for each of the three normal modes, all three normal modes of $H_2O$ are IR-active.

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Vector subspace and eigenspace

A vector subspace is a subset of a vector space, and it is a vector space by itself.

More formally, the vector subspace $W=\{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}},\boldsymbol{\mathit{v_3}}\}$ of the vector space $V=\left \{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}}, \cdots,\boldsymbol{\mathit{v_k}}\right \}$ satisfies the following conditions:

1) Commutative and associative addition for all elements of the closed set.

$\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}}=\boldsymbol{\mathit{v_{2}}}+\boldsymbol{\mathit{v_{1}}}$

$(\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}})+\boldsymbol{\mathit{v_{3}}}=\boldsymbol{\mathit{v_{1}}}+(\boldsymbol{\mathit{v_{2}}}+\boldsymbol{\mathit{v_{3}}})$

2) Associativity and distributivity of scalar multiplication for all elements of the closed set

$c_1(c_2\boldsymbol{\mathit{v_{1}}})=(c_1c_2)\boldsymbol{\mathit{v_{1}}}$

$c_1(\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}})=c_1\boldsymbol{\mathit{v_{1}}}+c_1\boldsymbol{\mathit{v_{2}}}$

$(c_1+c_2)\boldsymbol{\mathit{v_{1}}}=c_1\boldsymbol{\mathit{v_{1}}}+c_2\boldsymbol{\mathit{v_{1}}}$

where $c_1$ and $c_2$ are scalars.

3) Scalar multiplication identity.

$\boldsymbol{\mathit{1}}\boldsymbol{\mathit{v_{1}}}=\boldsymbol{\mathit{v_{1}}}$

4) Additive inverse.

$\boldsymbol{\mathit{v_{1}}}+(-\boldsymbol{\mathit{v_{1}}})=0$

5) Existence of null vector , such that

$\boldsymbol{\mathit{0}}+\boldsymbol{\mathit{v_{1}}}=\boldsymbol{\mathit{v_{1}}}$

6) Closed under addition: the sum of any two or more vectors in $W$ is another vector in $W$ .

7) Closed under scalar multiplication: the product of any vector in $W$ with a scalar is another vector in $W$ .

For example, in the $\mathbb{R}^{3}$ space, a plane through the origin is a subspace, as is a line through the origin. Hence, $\mathbb{R}^{2}$ and $\mathbb{R}$ are subspaces of $\mathbb{R}^{3}$ . The entire $\mathbb{R}^{3}$ space and the single point at the origin are also subspaces of the $\mathbb{R}^{3}$ space. This implies that a subspace contains a set of orthonormal basis vectors.

An eigenspace is the set of all eigenvectors associated with a particular eigenvalue. In other words, it is a vector subspace formed by eigenvectors corresponding to the same eigenvalue. Consider the eigenvalue equation $Ax_i=\lambda_ix_i$ , where $i=1,2,3$ . If the eigenvalues of $x_1$ , $x_2$ and $x_3$ are $\lambda_1=1$ , $\lambda_2=2$ and $\lambda_3=2$ respectively, then $J=\{x_1\}$ and $K=\{x_2,x_3\}$ are the eigenspaces of the operator $A$ . Since an eigenspace is a vector subspace, it must contain a set of orthonormal basis vectors.

Question

Show that all orthonormal basis eigenvectors $\boldsymbol{\mathit{v_{k}}}$ in an eigenspace $S=\{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}},\cdots,\boldsymbol{\mathit{v_k}}\}$ are linearly independent of one another.

Answer

A set of eigenvectors is linearly independent if the only solution to eq1 is when $c_k=0$ for all $k$ . Taking the dot product of eq1 with $\boldsymbol{\mathit{v_{i}}}$ gives

$c_1\boldsymbol{\mathit{v_{1}}}\cdot\boldsymbol{\mathit{v_{i}}}+c_2\boldsymbol{\mathit{v_{2}}}\cdot\boldsymbol{\mathit{v_{i}}}+\cdots +c_i\boldsymbol{\mathit{v_{i}}}\cdot\boldsymbol{\mathit{v_{i}}}+\cdots +c_k\boldsymbol{\mathit{v_{k}}}\cdot\boldsymbol{\mathit{v_{i}}}=0$

which reduces to $c_i=0$ because $\boldsymbol{\mathit{v_{i}}}\cdot\boldsymbol{\mathit{v_{i}}}\neq 0$ . Since $i$ is arbitrary, we conclude that $c_1=c_2=\cdots =c_k=0$ and that the set $\{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}},\cdots,\boldsymbol{\mathit{v_k}}\}$ is linearly independent.

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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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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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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$$

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 $E$ .

$A_1$	$\hat{P}^{A_1}q_3=q_3$	$B_1$	$\hat{P}^{B_1}q_1=q_1$
	$\hat{P}^{A_1}q_6=\frac{1}{2}\(q_6+q_9\)$		$\hat{P}^{B_1}q_6=\frac{1}{2}\(q_6-q_9\)$
	$\hat{P}^{A_1}q_4=\frac{1}{2}\(q_4-q_7\)$		$\hat{P}^{B_1}q_4=\frac{1}{2}\(q_4+q_7\)$
	$\hat{P}^{A_1}q_9=\frac{1}{2}\(q_6+q_9\)$		$\hat{P}^{B_1}q_9=\frac{1}{2}\(q_9-q_6\)$
	$\hat{P}^{A_1}q_7=\frac{1}{2}\(q_7-q_4\)$		$\hat{P}^{B_1}q_7=\frac{1}{2}\(q_7+q_4\)$
$A_2$	$\hat{P}^{A_2}q_5=\frac{1}{2}\(q_5-q_8\)$	$B_2$	$\hat{P}^{B_2}q_2=q_2$
	$\hat{P}^{A_2}q_8=\frac{1}{2}\(q_8-q_5\)$		$\hat{P}^{B_2}q_5=\frac{1}{2}\(q_5+q_8\)$
	$\hat{P}^{A_2}q_8=\frac{1}{2}\(q_8-q_5\)$		$\hat{P}^{B_2}q_8=\frac{1}{2}\(q_8+q_5\)$

$A_1$	$\hat{P}^{A_1}z_1=z_1$	$B_1$	$\hat{P}^{B_1}x_1=x_1$
	$\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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