Advanced Chemistry - Mono Mole

April 30, 2024April 30, 2024

Normalisation constant for the wavefunctions of the quantum harmonic oscillator

The normalisation constant ensures that the wavefunctions of the quantum harmonic oscillator are properly scaled, thereby maintaining the probabilistic interpretation of quantum states.

To derive the normalisation constant for the wavefunctions of the quantum harmonic oscillator, we begin with the multiplication of eq33 by $e^{-y^2}$ and by itself (with a change of the dummy variable $t$ to $s$ and the dummy index $n$ to $m$ ) to give

$e^{-y^2}e^{-t^2+2ty}e^{-s^2+2sy}=e^{-y^2}\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\frac{H_m(y)}{m!}\frac{H_n(y)}{n!}s^mt^n\;\;\;\;\;\;\;\;41$

Integrating with respect to $y$ ,

$\int_{-\infty}^{\infty}e^{-y^2-t^2+2ty-s^2+2sy}dy=\int_{-\infty}^{\infty}e^{-y^2}\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\frac{H_m(y)}{m!}\frac{H_n(y)}{n!}s^mt^ndy$

Substituting the orthogonal relation $\int_{-\infty}^{\infty}e^{-y^2}H_m(y)H_n(y)dy=0$ when $m\neq n$ in the above equation and simplifying, we have

$e^{2st}\int_{-\infty}^{\infty}e^{-(y-s-t)^2}dy=\sum_{n=0}^{\infty}\frac{(st)^n}{n!n!}\int_{-\infty}^{\infty}e^{-y^2}\[H_n(y)\]^2dy\;\;\;\;\;\;\;\;42$

Question

Show that $I=\int_{-\infty}^{\infty}e^{-(y-s-t)^2}dy=\sqrt{\pi}$ .

Answer

Let $X=y-s-t$ and $dX=dy$ . So, $I=\int_{-\infty}^{\infty}e^{-X^2}dX$ with the limits unchanged. At the same time, let $I=\int_{-\infty}^{\infty}e^{-Y^2}dY$ , where we have changed the dummy variable $X$ to $Y$ .

$I^2=\int_{-\infty}^{\infty}e^{-X^2}dX\int_{-\infty}^{\infty}e^{-Y^2}dY=\int\int_{-\infty}^{\infty}e^{X^2+Y^2}dXdY$

In polar coordinates, $r^2=X^2+Y^2,X=rcos\theta,Y=rsin\theta$ . For infinitesimal changes, the change in area $dA$ can be approximated as a rectangle with sides $r$ and $rd\theta$ (see diagram below), with $dXdY=dA=rd\theta dr$ .

So,

$I^2=\int_{0}^{2\pi}d\theta\int_{0}^{\infty}re^{-r^2}dr=2\pi\int_{0}^{\infty}re^{-r^2}dr$

Let $u=r^2\;\Rightarrow\;du=2rdr$

$I^2=2\pi\int_{0}^{\infty}re^{-u}\frac{du}{2r}=\pi\int_{0}^{\infty}e^{-u}du=\pi$

$I=\int_{-\infty}^{\infty}e^{-(y-s-t)^2}dy=\sqrt{\pi}\;\;\;\;\;\;\;\;43$

Using eq43 and the Maclaurin series expansion of $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ , eq42 becomes

$\sqrt{\pi}\sum_{n=0}^{\infty}\frac{2^n(st)^n}{n!}=\sum_{n=0}^{\infty}\frac{(st)^n}{n!n!}\int_{-\infty}^{\infty}e^{-y^2}\[H_n(y)\]^2dy\;\;\;\;\;\;\;\;44$

Expanding eq44 and comparing the coefficients of $(st)^n$ ,

$n=0$	$\int_{-\infty}^{\infty}e^{-y^2}\[H_n(y)\]^2dy=\sqrt{\pi}$
$n=1$	$\int_{-\infty}^{\infty}e^{-y^2}\[H_n(y)\]^2dy=2\sqrt{\pi}$
$n=2$	$\int_{-\infty}^{\infty}e^{-y^2}\[H_n(y)\]^2dy=8\sqrt{\pi}$
$n=3$	$\int_{-\infty}^{\infty}e^{-y^2}\[H_n(y)\]^2dy=48\sqrt{\pi}$
$\vdots$	$\vdots$
$n=n$	$\int_{-\infty}^{\infty}e^{-y^2}\[H_n(y)\]^2dy=2^nn!\sqrt{\pi}$

In an earlier article, we have made a change of variable of ${x=y\biggr$\frac{\hbar}{m\omega}\biggr$^{\frac{1}{2}}}$ in deriving the un-normalised wavefunction of the quantum harmonic oscillator. Since $x$ is the variable representing position in the one-dimensional space that the oscillator is moving in, the normalisation of the wavefunction refers to an integral with respect to $x$ , i.e.

$N^2\int_{-\infty}^{\infty}e^{-\frac{m\omega}{\hbar}x^2}\biggr\[H_n\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$\biggr\]^2dx=1\;\;\;\;\;\;\;\;45$

Using ${dx=\biggr$\frac{\hbar}{m\omega}\biggr$^{\frac{1}{2}}}dy$ , eq45 is equivalent to

$N^2\biggr$\frac{\hbar}{m\omega}\biggr$^{\frac{1}{2}}\int_{-\infty}^{\infty}e^{-y^2}\[H_n(y)\]^2dy=1$

Therefore,

$N^2\biggr$\frac{\hbar}{m\omega}\biggr$^{\frac{1}{2}}2^nn!\sqrt{\pi}=1\;\;or\;\;N=\sqrt[4]{\frac{m\omega}{\pi\hbar}}\frac{1}{\sqrt{2^nn!}}$

The normalised wavefunction is then

$\psi_n(y)=\sqrt[4]{\frac{m\omega}{\pi\hbar}}\frac{1}{\sqrt{2^nn!}}H_n(y)e^{-\frac{y^2}{2}}\;\;\;n=0,1,2,\cdots\;\;\;\;\;\;\;\;45$

$\psi_n(x)=\sqrt[4]{\frac{m\omega}{\pi\hbar}}\frac{1}{\sqrt{2^nn!}}H_n\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$e^{-\frac{m\omega}{2\hbar}x^2}\;\;\;n=0,1,2,\cdots\;\;\;\;\;\;\;\;46$

Question

Show that $\psi_n(x)$ is either an even function or an odd function.

Answer

An even function is defined as $f(x)=f(-x)$ , while and odd function is when $-f(x)=f(-x)$ . So, $e^{-\frac{m\omega}{2\hbar}x^2}$ is an even function. On the other hand, $H_n\biggr$\sqrt{\frac{m\omega}{\hbar}}x\biggr$$ is an even function if $n$ is even and an odd function if $n$ is odd. Since the product of two even functions is an even function, while the product of an odd function and an even function is an odd function, $\psi_n(x)$ is an even function when $n$ is even and an odd function if $n$ is odd.

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April 30, 2024July 4, 2024

The Born-Oppenheimer approximation

The Born-Oppenheimer approximation assumes that the wavefunctions of atomic nuclei and electrons in a molecule can be treated separately, primarily because the nuclei are much heavier than the electrons.

We begin with the molecular Hamiltonian operator $\hat{H}$ :

$\hat{H}=-\sum_{\alpha}\frac{\hbar^2}{2m_{\alpha}}\nabla_{\alpha}^2-\sum_{i}\frac{\hbar^2}{2m_{e}}\nabla_{i}^2+\hat{V}_{NN}+\hat{V}_{Ne}+\hat{V}_{ee}\;\;\;\;\;\;\;\;51$

where

- $N$ and $e$ refer to nuclei and electrons respectively.
- The first and second terms are the kinetic energy operator of nuclei and electrons respectively.
- $\hat{V}$ is the potential energy of interaction between two particles.

For example, the Hamiltonian for $H_2$ (see diagram above) is:

$\hat{H}=-\sum_{\alpha}\frac{\hbar^2}{2m_{\alpha}}\nabla_{\alpha}^2-\sum_{\beta}\frac{\hbar^2}{2m_{\beta}}\nabla_{\beta}^2-\frac{\hbar^2}{2m_{e}}\nabla_{1}^2-\frac{\hbar^2}{2m_{e}}\nabla_{2}^2-\frac{e^2}{4\pi\varepsilon_0r_{1\alpha}}-\frac{e^2}{4\pi\varepsilon_0r_{1\beta}}-\frac{e^2}{4\pi\varepsilon_0r_{2\alpha}}-\frac{e^2}{4\pi\varepsilon_0r_{2\beta}}+\frac{e^2}{4\pi\varepsilon_0r_{12}}+\frac{e^2}{4\pi\varepsilon_0R}$

To solve the complicated Schrodinger equation $\hat{H}\psi=E\psi$ for a molecule, Max Born and J. Oppenheimer proposed separating the electronic and nuclear motions. The first assumption the two scientists made is that the molecular wavefunction $\psi$ is a product of the wavefunction of nuclear motion $\psi_N(q_{\alpha})$ and the wavefunction of electron motion $\psi_e(q_i,q_{\alpha})$ , i.e.

$\psi=\psi_N\psi_e\;\;\;\;\;\;\;\;52$

Question

Why is $\psi_e$ a function of nuclear coordinates $q_{\alpha}$ and electronic coordinates $q_i$ ?

Answer

The electron density of a molecule changes when the molecule vibrates and therefore the form of the wavefunction of electron motion changes based on the positions of the electrons and the nuclei.

With regard to eq51 and eq52, the Schrodinger equation is

$\biggr$-\sum_{\alpha}\frac{\hbar^2}{2m_{\alpha}}\nabla_{\alpha}^2-\sum_{i}\frac{\hbar^2}{2m_{e}}\nabla_{i}^2+\hat{V}_{NN}+\hat{V}_{Ne}+\hat{V}_{ee}\biggr$\psi_N\psi_e=E\psi_N\psi_e\;\;\;\;\;\;\;\;53$

Using the product rule, we have $\nabla_{\alpha}^2\psi_N\psi_e=\psi_e\nabla_{\alpha}^2\psi_N+2(\nabla_{\alpha}\psi_e)\nabla_{\alpha}\psi_N+\psi_N\nabla^2_{\alpha}\psi_e$ , which when substituted in eq53 gives

$-\sum_{\alpha}\frac{\hbar^2}{m_{\alpha}}(\nabla_{\alpha}\psi_e)\nabla_{\alpha}\psi_N-\sum_{\alpha}\frac{\hbar^2}{2m_{\alpha}}\psi_N\nabla^2_{\alpha}\psi_e-\sum_{\alpha}\frac{\hbar^2}{2m_{\alpha}}\psi_e\nabla^2_{\alpha}\psi_N-\\\psi_N\sum_{i}\frac{\hbar^2}{2m_{e}}\nabla_{i}^2\psi_e+(\hat{V}_{NN}+\hat{V}_{Ne}+\hat{V}_{ee})\psi_N\psi_e=E\psi_N\psi_e\;\;\;\;\;\;\;\;54$

The second assumption Born and Oppenheimer made was that $\psi_e$ is a slowly varying function of $q_{\alpha}$ . This is because the heavier nuclei are considered relatively stationary as they move much more slowly than electrons. Therefore, the change in the distribution of the electrons with respect to the positions of the nuclei $\nabla_{\alpha}\psi_e$ is very small and eq54 becomes

$-\psi_e\sum_{\alpha}\frac{\hbar^2}{2m_{\alpha}}\nabla^2_{\alpha}\psi_N-\psi_N\sum_{i}\frac{\hbar^2}{2m_{e}}\nabla_{i}^2\psi_e+(\hat{V}_{NN}+\hat{V}_{Ne}+\hat{V}_{ee})\psi_N\psi_e=E\psi_N\psi_e\;\;\;\;\;\;\;\;55$

Substituting the purely electronic Hamiltonian $\hat{H}_e=-\sum_{i}\frac{\hbar^2}{2m_{e}}\nabla_{i}^2+\hat{V}_{Ne}+\hat{V}_{ee}$ and its corresponding Schrodinger equation $\hat{H}_e\psi_e=E_e\psi_e$ in eq55, we have

$\biggr$-\sum_{\alpha}\frac{\hbar^2}{2m_{\alpha}}\nabla^2_{\alpha}+E_e+\hat{V}_{NN}\biggr$\psi_N=E\psi_N\;\;\;\;\;\;\;\;56$

Question

How do we interpret the term $E_e+\hat{V}_{NN}$ in eq56?

Answer

$E_e$ is the eigenvalue of the purely electronic Schrodinger equation $\hat{H}_e\psi_e=E_e\psi_e$ . Using the second assumption, we regard the nuclei as fixed while the electrons move. $q_{\alpha}$ in $\psi_e$ then becomes a constant for a particular nuclear configuation. In other words, $\psi_e$ is now a function of $q_i$ and at the same time parametrically dependent on $q_{\alpha}$ . For each electronic state , the purely electronic Schrodinger equation is solved repeatedly using different fixed values of $q_{\alpha}$ , giving a set of $\psi_e$ and a corresponding set of $E_e$ . Since $\hat{V}_{NN}=\sum_{\alpha}\sum_{\beta >\alpha}\frac{Z_{\alpha}Z_{\beta}e^2}{4\pi\epsilon_0r_{\alpha\beta}$ , we can then easily calculate a value $U=E_e+\hat{V}_{NN}$ for every $q_{\alpha}$ and obtain a curve (see diagram below). In effect, $U$ is a function of $q_{\alpha}$ , making it the potential energy for nuclear motion:

$\biggr\[-\sum_{\alpha}\frac{\hbar^2}{2m_{\alpha}}\nabla^2_{\alpha}+U(q_{\alpha})\biggr\]\psi_N=E\psi_N\;\;\;\;\;\;\;\;57$

We have therefore transformed eq53, the molecular Schrodinger equation, to eq57, which is an eigenvalue equation of a single variable of nuclear coordinates, with the eigenfunction $\psi_N$ as the wavefunction of nuclear motion, the eigenvalue $E$ representing the total energy of the molecule, and the term $U(q_{\alpha})$ as the potential energy for nuclear motion.

Question

How does the above molecular potential energy curve relate to the quadratic potential energy curve derived using the harmonic oscillator model?

Answer

The two curves coincide at low vibration energy levels, i.e around the equilibrium internuclear distance (see diagram below). Nuclear motion includes translational, rotational and vibrational motions. As translational and rotational motions are purely kinetic, $U(q_{\alpha})$ in eq57 is associated only with molecular vibrational motions. Consider a diatomic molecule with one of the nuclei at the origin and the other on the $x$ -axis. The potential energy term becomes $U(R)$ , where $R$ is the internuclear distance. Using a Taylor series, we can approximate the potential function by expanding it around its minimum at the equilibrium internuclear distance $R=R_e$ :

$U(R)=U(R_e)+U'(R_e)(R-R_e)+\frac{1}{2}U''(R_e)(R-R_e)^2+\cdots$

At $R_e$ , the slope of $U(R)$ is zero, i.e. $U'(R_e)=0$ . For small vibrations, where $R$ is close to $R_e$ , we ignore all the higher power terms, giving $U(R)=U(R_e)+\frac{1}{2}U''(R_e)(R-R_e)^2$ . The physical observables of a system, such as the transition frequencies $\Delta\nu$ of vibrational modes of a molecule, are not affected by the $U(R_e)$ term because it cancels out when we take the difference of two energy levels. Therefore, we can set $U(R_e)$ to zero, resulting in $U(R)=\frac{1}{2}U''(R_e)(R-R_e)^2$ , which is equivalent to the potential energy term of the harmonic oscillator (see eq49), where $U''(R_e)=k$ and $r=R-R_e$ .

In conclusion, the Born-Oppenheimer approximation offers a different approach, compared to the quantum harmonic oscillator, in deriving the Schrodinger equation for a diatomic molecule (see the Hamiltonian of eq57 vs eq48)) and the potential energy term for the vibrational motion of a diatomic molecule.

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April 30, 2024

Vibration of diatomic molecules

The vibration of a diatomic molecule can be modelled using the quantum harmonic oscillator.

The Schrodinger equation can be written as:

$\frac{p_1^2}{2m_1}\psi(x_1,x_2)+\frac{p_2^2}{2m_2}\psi(x_1,x_2)+\frac{1}{2}k(x_2-x_1-x_0)^2\psi(x_1,x_2)=E\psi(x_1,x_2)\;\;\;\;\;\;\;\;47$

To solve the equation, we need to make a change of coordinates to achieve a separation of variables. The new coordinate system consists of the variables $r=x_2-x_1-x_0$ and $s=\frac{m_1x_1+m_2x_2}{m_1+m_2}$ , where $s$ is the centre of mass of the molecule.

Question

What do the terms in eq47 mean? What is a separation of variables?

Answer

Eq47 is an eigenvalue equation, where the Hamiltonian $\hat{H}=\frac{\hat{p}_1^2}{2m_1}+\frac{\hat{p}_2^2}{2m_2}+\frac{1}{2}k(x_2-x_1-x_0)^2$ has two kinetic energy terms, one for each atom, and a potential energy term that describes the interaction between the atoms. According to the classical harmonic oscillator, the potential energy of the system is equal to $\frac{1}{2}kx^2$ , where $x$ is the displacement of the spring from its equilibrium position. For the diatomic molecule in the diagram above, $x=r=x_2-x_1-x_0$ .

To solve eq47, we need it to be of the form $\[\hat{H}_1(x_1)+\hat{H}_2(x_2)\]\psi(x_1,x_2)=(E_1+E_2)\psi(x_1,x_2)$ , where the Hamiltonian is separated into two, each being a function of only one variable. Unless it is transformed, the potential energy term in eq47 does not allow the separation to occur.

The derivatives of $r$ and $s$ with respect to time are $\dot r=\dot x_2-\dot x_1$ and $\dot s=\frac{m_1\dot x_1+m_2\dot x_2}{m_1+m_2}$ respectively. Substituting $\dot x_1=\dot x_2-\dot r$ and $\dot x_x=\dot x_1+\dot r$ in $\dot s$ and rearranging, we have $\dot x_2=\dot s+\frac{m_1\dot r}{m_1+m_2}$ and $\dot x_1=\dot s+\frac{m_2\dot r}{m_1+m_2}$ respectively. Using $p_i=m_i\dot x_i$ , the Hamiltonian of eq47 becomes

$\hat{H}=\frac{1}{2}(M\dot s^2+\mu\dot r^2)+\frac{1}{2}kr^2=\frac{p_s^2}{2M}+\frac{p_r^2}{2\mu}+\frac{1}{2}kr^2\;\;\;\;\;\;\;\;48$

where $p_s=M\dot s$ , $M=m_1+m_2$ , $p_r=\mu\dot r$ and $\mu=\frac{m_1m_2}{m_1+m_2}$ .

Since $M$ and $\dot s$ are the total mass of the molecule and the centre of mass coordinate of the molecule respectively, $\frac{p_s^2}{2M}$ corresponds to the kinetic energy of the translational motion of the molecule. The other kinetic energy energy term $\frac{p_r^2}{2\mu}$ must correspond to the internal motion (rotational and vibrational motion) of the molecule. However, there are only two degrees of freedom for a diatomic molecule in a one-dimensional space – a translational degree of freedom and a vibrational degree of freedom. Discarding the kinetic energy term for the translational motion, the Schrodinger equation describing the vibrational motion of the molecule is

$\biggr$\frac{p_r^2}{2\mu}+\frac{1}{2}kr^2\biggr$\psi(r)=E\psi(r)\;\;\;\;\;\;\;\;49$

Eq49 is equivalent to eq4 except for the substitution of $\mu$ , which is called the reduced mass. Its solutions are given by eq45 (with $m$ replaced with $\mu$ ). With reference to eq21, the vibrational energies of the diatomic molecule are:

$E_n\biggr$n+\frac{1}{2}\biggr$\hbar\omega\;\;\;\;n=0,1,2,\cdots\;\;\;\;\;\;\;\;50$

where ${\omega=\sqrt{\frac{k}{\mu}}}$ .

The potential energy curve of $\frac{1}{2}kr^2$ in eq49 for a typical diatomic molecule is represented by the dashed curve in the diagram below. The minimum potential energy of the molecule occurs when the atoms are at their equilibrium positions. In the next article, we will show that the actual potential energy curve has the shape of the solid curve, which correctly shows that the potential energy of the molecule increases rapidly as the bond length shortens and approaches a constant value representing the dissociation energy of the molecule as $r$ increases. Despite that, the two curves coincide at low vibrational energy levels,. This implies that the harmonic oscillator model of a diatomic molecule only serves as a good approximation at low vibrational energy levels.

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April 30, 2024May 9, 2024

Nuclear motion in diatomic molecules

Nuclear motion in diatomic molecules involves the rotation, vibration, and overall translation of nuclei, significantly impacting molecular behavior and spectroscopic transitions.

From eq57, the Schrodinger equation for nuclear motion in a diatomic-molecule is

$\biggr\[-\frac{\hbar^2}{2m_{\alpha}}\nabla^2_{\alpha}-\frac{\hbar^2}{2m_{\beta}}\nabla^2_{\beta}+U(R)\biggr\]\psi_N=E\psi_N\;\;\;\;\;\;\;\;60$

where $\alpha$ and $\beta$ refer to the nuclei and $R$ is the inter-nuclear distance.

The eigenfunctions and eigenvalues of eq60, which is a two-particle problem, are $E=E_{tr}+E_{int}$ and $\psi_N=\psi_{tr}\psi_{int}$ respectively, where the subscripts tr and int refer to translational motion and internal motion respectively. Since the operator $\frac{\hat{p}^2_{\mu}}{2\mu}$ in eq49 is equivalent to $-\frac{\hbar^2}{2\mu}\nabla^2$ (see this article for derivation), the Schrodinger equation for internal motion is

$\biggr\[-\frac{\hbar^2}{2\mu}\nabla^2+U(R)\biggr\]\psi_{int}=E_{int}\psi_{int}\;\;\;\;\;\;\;\;61$

The potential energy $U$ is spherically symmetric, as it depends on $R$ only. This implies that eq61 is a central force problem, with the following Schrodinger equation and eigenfunction:

$\biggr\[-\frac{\hbar^2}{2\mu}\biggr$\frac{2}{R}\frac{\partial}{\partial R}+\frac{\partial^2}{\partial R^2}\biggr$+\frac{\hat{L}^2}{2\mu R^2}+U(R)\biggr\]\psi_{int}=E_{int}\psi_{int}\;\;\;\;\;\;\;\;62$

$\psi_{int}=P(R)Y^M_J(\theta,\phi)\;\;\;\;\;\;\;\;63$

where $\hat{L}^2$ is the operator for the square of the magnitude of the orbital angular momentum of the molecule, $P(R)$ is a function of $R$ and $Y^M_J(\theta,\phi)$ is a spherical harmonic function with quantum numbers $J$ and $M$ .

The kinetic energy of internal motion is the sum of rotational kinetic energy and vibrational kinetic energy. This implies that the Hamiltonian of eq62 consists, in part, of two kinetic energy operators. If we assume that the diatomic molecule is a rigid rotor ( $R$ is a constant), whose rotational motion is independent of its vibrational motion, then there is no potential energy contribution in the Hamiltonian for the rotation of a rigid rotor because the bond length remains fixed at its equilibrium distance. In other words, rotation is characterised only by $\theta$ and $\phi$ , and the Hamiltonian for vibrational motion consists of both a kinetic energy operator and a potential energy function, as the vibrational motion of the molecule is described by a change in $R$ .

Since ${\biggr\[\frac{\hat{L}^2}{2\mu R^2}Y^M_J(\theta,\phi)\biggr\]P(R)=E_{rot}P(R)Y^M_J(\theta,\phi)}$ , where ${E_{rot}=\frac{J(J+1)\hbar^2}{2\mu R^2}}$ (see this article for derivation)

$\biggr\[-\frac{\hbar^2}{2\mu}\biggr$\frac{2}{R}\frac{\partial}{\partial R}+\frac{\partial^2}{\partial R^2}\biggr$+U(R)\biggr\]P(R)Y^M_J(\theta,\phi)=(E_{int}-E_{rot})P(R)Y^M_J(\theta,\phi)$

$Y^M_J(\theta,\phi)$ becomes a non-zero constant because the Hamiltonian in the above equation is a function of $R$ . So,

$\biggr\[-\frac{\hbar^2}{2\mu}\biggr$\frac{2}{R}\frac{\partial}{\partial R}+\frac{\partial^2}{\partial R^2}\biggr$+U(R)\biggr\]P(R)=(E_{int}-E_{rot})P(R)\;\;\;\;\;\;\;\;64$

To simplify eq64, we substitute ${P(R)=\frac{F(R)}{R}}$ , ${P'(R)=\frac{F'(R)}{R}-\frac{F(R)}{R^2}}$ and ${P''(R)=\frac{F''(R)}{R}-2\frac{F'(R)}{R^2}}+2\frac{F(R)}{R^3}$ in eq64 to give

$\biggr\[-\frac{\hbar^2}{2\mu}\frac{d^2}{dR^2}+U(R)\biggr\]F(R)=(E_{int}-E_{rot})F(R)\;\;\;\;\;\;\;\;65$

We can approximate $U(R)$ by expanding it in a Taylor series around the equilibrium internuclear distance $R=R_e$ :

$U(R)=U(R_e)+U'(R_e)(R-R_e)+\frac{1}{2}U''(R_e)(R-R_e)^2+\cdots\;\;\;\;\;\;\;\;66$

At $R_e$ , the slope of $U(R)$ is zero, i.e. $U'(R_e)=0$ . For small vibrations, where $R$ is close to $R_e$ , we ignore all the higher power terms, giving ${U(R)=U(R_e)+\frac{1}{2}U''(R_e)(R-R_e)^2}$ , which when substituted in eq65 gives

$\biggr\[-\frac{\hbar^2}{2\mu}\frac{d^2}{dR^2}+U(R_e)+\frac{1}{2}U''(R_e)(R-R_e)^2\biggr\]F(R)=(E_{int}-E_{rot})F(R)$

The molecule is in its most stable state when $R=R_e$ for a particular electronic configuration, with $U(R_e)$ being a minimum point on the potential energy curve. This implies that the electrons are in their lowest energy state for that configuration and $U(R_e)$ is called the equilibrium electronic energy $E_{elec}$ of the molecule.

Question

Elaborate on the concept of $E_{elec}$ .

Answer

$E_{elec}$ is the energy of the electrons that are interacting with the nuclei at their equilibrium positions. It is neither the purely electronic energy $E_{e}$ nor $E_{e}+\hat{V}_{NN}$ in eq56. It is also not a form of vibrational energy, which is characterised by a displacement of the nuclei from their equilibrium positions. Since $E_{elec}$ is a constant for a particular electronic state, it is subtracted out when we calculate vibrational transition energies for a given electronic state. This is why it is sometimes set to zero.

Letting $x=R-R_e$ and $k=U''(R_e)$ and therefore $F(R)=F(x+R_e)=S(x)$ , we have

$\biggr\[-\frac{\hbar^2}{2\mu}\frac{d^2}{dx^2}+\frac{1}{2}kx^2\biggr\]S(x)=E_{vib}S(x)\;\;\;\;\;\;\;\;67$

where $E_{vib}=E_{int}-E_{rot}-E_{elec}$ .

Question

Show that ${\frac{d^2}{dR^2}F(R)=\frac{d^2}{dx^2}S(x)}$ .

Answer

According to the chain rule, ${\frac{dS(x)}{dx}=\frac{dF(R)}{dx}=\frac{dF(R)}{dR}\frac{dR}{dx}}$ . Since $x=R-R_e$ , we have ${\frac{dR}{dx}=1}$ and ${\frac{dS(x)}{dx}=\frac{dF(R)}{dR}}$ . So, ${\frac{d^2S(x)}{dx^2}=\frac{d}{dx}F'(R)=\frac{dF'(R)}{dR}\frac{dR}{dx}=\frac{d^2F(R)}{dR^2}}$ .

Eq67 is the Schrodinger equation for a one-dimensional harmonic oscillator. Assuming that $S(x)$ behaves the same as a wave function of the harmonic oscillator, it is given by eq46. $E_{vib}$ is then represented by eq50. Therefore, $E$ and $\psi_N$ in eq60 are

$E=E_{tr}+E_{rot}+E_{vib}+E_{elec}\;\;\;\;\;\;\;\;68$

$\psi_N=\psi_{tr}\psi_{rot}\psi_{vib}\;\;\;\;\;\;\;\;69$

where $\psi_{vib}=\frac{S(x)}{R}$ .

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April 30, 2024

Vibration of polyatomic molecules

The vibration of polyatomic molecules involves stretching, bending and torsional motions.

The classical kinetic energy $T$ of a vibrating $N$ -nuclei molecule is

$T=\frac{1}{2}\sum_{\alpha=1}^Nm_{\alpha}\biggr\[\biggr$\frac{dx_{\alpha}}{dt}\biggr$^2+\biggr$\frac{dy_{\alpha}}{dt}\biggr$^2+\biggr$\frac{dz_{\alpha}}{dt}\biggr$^2\biggr\]\;\;\;\;\;\;\;\;85$

where $x_{\alpha}$ , $y_{\alpha}$ and $z_{\alpha}$ are the displacements of the $\alpha$ -th nucleus from its equilibrium position of $(x_{\alpha,e},y_{\alpha,e},z_{\alpha,e})$ (see diagram below).

Eq85 can be simplified by changing the Cartesian displacement coordinates $x_{\alpha},y_{\alpha},z_{\alpha}$ to mass-weighted displacement coordinates $q_1,\cdots,q_{3N}$ , where

$\begin{matrix} q_1=m_1^{1/2}x_1&q_2=m_1^{1/2}y_1&q_3=m_1^{1/2}z_1\\q_4=m_2^{1/2}x_2&\cdots &q_{3N}=m_N^{1/2}z_N\end{matrix}$

which gives

$T=\frac{1}{2}\sum_{i=1}^{3N}\dot q^2_i=\frac{1}{2}(\dot q_1\cdots \dot q_{3N})\begin{pmatrix}\dot q_1\\\vdots\\\dot q_{3N}\end{pmatrix}=\frac{1}{2}\boldsymbol{\mathit{\dot q}}^T\boldsymbol{\mathit{\dot q}}\;\;\;\;\;\;\;\;86$

where $\dot q_{i}=\frac{dq_i}{dt}$ and $\boldsymbol{\mathit{\dot q}}$ is a column vector with elements $\dot q_1,\cdots,\dot q_{3N}$ .

The potential energy $U$ of a vibrating $N$ -nuclei molecule is also a function of $x_{\alpha},y_{\alpha},z_{\alpha}$ and hence a function of $q_1,\cdots,q_{3N}$ . We can approximate the multivariable function $U$ by expanding it in a Maclaurin series around the equilibrium positions:

$U(q_1,\cdots,q_{3N})=U(q_{1,e=0},\cdots,q_{3N,e=0})+\sum_{i=1}^{3N}\frac{\partial U(q_{1,e=0},\cdots,q_{3N,e=0})}{\partial q_i}q_i+\frac{1}{2!}\sum_{i,j=1}^{3N}\frac{\partial^2 U(q_{1,e=0},\cdots,q_{3N,e=0})}{\partial q_i\partial q_j}q_iq_j+\cdots$

which can be abbreviated to

$U=U_e+\sum_{i=1}^{3N}\biggr$\frac{\partial U}{\partial q_i}\biggr$_eq_i+\frac{1}{2!}\sum_{i,j=1}^{3N}\biggr$\frac{\partial^2 U}{\partial q_i\partial q_j}\biggr$_eq_iq_j+\cdots$

When the nuclei are at their equilibrium positions, the slope of is zero, i.e. ${\biggr$\frac{\partial U}{\partial q_i}\biggr$_e=0}$ . For small vibrations, where $q_i$ is close to $q_{i,e}$ , we ignore all the higher power terms, giving

$U=U_e+\frac{1}{2}\sum_{i,j=1}^{3N}\biggr$\frac{\partial^2 U}{\partial q_i\partial q_j}\biggr$_eq_iq_j\;\;\;\;\;\;\;\;87$

or in matrix notation

$U=U_e+\frac{1}{2}\boldsymbol{\mathit{q}}^T\boldsymbol{\mathit{Uq}}\;\;\;\;\;\;\;\;88$

where $\boldsymbol{\mathit{q}}$ is the column vector with elements $q_1,\cdots,q_{3N}$ and $\boldsymbol{\mathit{U}}$ is a $3N\times 3N$ matrix, called the Hessian matrix, with elements $u_{ij}=\small{\biggr$\frac{\partial^2 U}{\partial q_i\partial q_j}\biggr$_e}$ .

Question

Show that $\boldsymbol{\mathit{q}}^T\boldsymbol{\mathit{Uq}}=\sum_{i,j=1}^{3N}\biggr$\frac{\partial^2 U}{\partial q_i\partial q_j}\biggr$_e$ .

Answer

$\begin{pmatrix}q_1&q_2&\cdots&q_{3N}\end{pmatrix}\begin{pmatrix}u_{11}&u_{12}&\cdots&u_{1j}\\u_{21}&u_{22}&\cdots&u_{2j}\\\vdots&\vdots&\ddots&\vdots\\u_{i1}&u_{i2}&\cdots&u_{ij}\end{pmatrix}\begin{pmatrix}q_1\\q_2\\\vdots\\q_{3N}\end{pmatrix}=\\\begin{pmatrix}q_1&q_2&\cdots&q_{3N}\end{pmatrix}\begin{pmatrix}\sum_{j=1}^{3N}u_{1j}q_j\\\sum_{j=1}^{3N}u_{2j}q_j\\\vdots\\\sum_{j=1}^{3N}u_{ij}q_j\end{pmatrix}=\sum_{i,j=1}^{3N}u_{ij}q_iq_j$

We want to derive an eigenvalue equation that resembles the harmonic oscillator equation so that we can easily solve for the vibrational energies of a polyatomic molecule. However, the Hamiltonian formed by combining eq86 and eq87 doesn’t resemble the Hamiltonian of a harmonic oscillator. Therefore, we need to further simply eq86 and eq87 with a change of variables.

With regard to eq88, the potential energy of a system is a real-valued physical property. This implies that $\boldsymbol{\mathit{U}}$ must consists of real-valued entries $u_{ij}$ . Since $u_{ij}=\small{\biggr$\frac{\partial^2 U}{\partial q_i\partial q_j}\biggr$_e=\biggr$\frac{\partial^2 U}{\partial q_j\partial q_i}\biggr$_e=u_{ji}}$ , the matrix $\boldsymbol{\mathit{U}}$ is both real and symmetric.

Question

i) Prove that eigenvectors of a real and symmetric matrix $O$ with distinct eigenvalues are orthogonal.

ii) What if some of the eigenvalues are degenerate?

Answer

i) Let $Ox=ax$ and $Oy=by$ , where $a\neq b$ . Multiply the first eigenvalue equation by $y^T$ , we have

$y^Tax=(y^TO)x=(O^Ty)^Tx=by^Tx$

where the 2^nd equality uses the identity mentioned in this article, while the 3^rd equality employs the property of a symmetrc matrix.

Since $a\neq b$ , we have $y^Tx=0$ , i.e. $y$ is orthogonal to $x$ .

ii) When some eigenvalues are degenerate, the corresponding eigenvectors belong to the same eigenspace. Since we can always select a set of linearly independent eigenvectors that span the eigenspace, these eigenvectors can be orthogonalised using the Gram-Schmidt process without changing the eigenvalue. So, for a real symmetric matrix, we can always find a basis of eigenvectors that are orthogonal to each other, regardless of whether the eigenvalues are distinct or degenerate.

Let the orthogonal eigenvectors $\boldsymbol{\mathit{l}}_m$ of $\boldsymbol{\mathit{U}}$ be the columns of the matrix $\boldsymbol{\mathit{L}}$ , where $\boldsymbol{\mathit{L}}^T\boldsymbol{\mathit{L=I}}$ . As shown in this article, we can express the eigenvalue problem $\boldsymbol{\mathit{Ul}}_m=\lambda_j\boldsymbol{\mathit{l}}_m$ as

$\boldsymbol{\mathit{UL=L\Lambda}}\;\;\;or\;\;\;\boldsymbol{\mathit{L}}^T\boldsymbol{\mathit{UL=\Lambda}}$

where $\boldsymbol{\mathit{\Lambda}}$ is the diagonal eigenvalue matrix with diagonal entries $\lambda_m$ .

Computing the secular equation $det(u_{ij}-\lambda_m\delta_{ij})$ allows us to find $\lambda_m$ , and solving the simultaneous equations of $((\boldsymbol{\mathit{U}}-\lambda_m \boldsymbol{\mathit{I}})\boldsymbol{\mathit{l}}_m=0$ , which is equivalent to $\sum_{j=1}^{3N}(u_{ij}-\lambda_m\delta_{ij})l_{jm}=0$ (where $i=1,\cdots,3N$ , enables us to find $\boldsymbol{\mathit{l}}_m$ .

Consider the following linear combinations of mass-weighted displacement coordinates

$Q_k=\sum_{j=1}^{3N}l_{jk}q_{j}=\boldsymbol{\mathit{L}}^T\boldsymbol{\mathit{q=Q}}\;\;\;\;\;\;k=1,\cdots,3N\;\;\;\;\;\;\;\;89$

where $\boldsymbol{\mathit{Q}}$ is the column vector with elements $Q_1,\cdots,Q_{3N}$ .

Since $l_{jk}$ are the components of the $k$ -th orthogonal eigenvector of $\boldsymbol{\mathit{U}}$ and each basis $q_j$ describes the displacement of a nucleus, $Q_k$ , known as the normal coordinates, is a set of orthogonal vectors that expresses $3N$ independent motions (called degrees of freedom) of the molecule as a whole.

Substituting $\boldsymbol{\mathit{q=LQ}}$ in eq88

$U=U_e+\frac{1}{2}(\boldsymbol{\mathit{LQ}})^T\boldsymbol{\mathit{ULQ}}=U_e+\frac{1}{2}\boldsymbol{\mathit{Q}}^T\boldsymbol{\mathit{L}}^T\boldsymbol{\mathit{ULQ}}=U_e+\frac{1}{2}\boldsymbol{\mathit{Q}}^T\boldsymbol{\mathit{\Lambda Q}}=U_e+\frac{1}{2}\sum_{j=1}^{3N}\lambda_iQ_j^2\;\;\;\;\;\;\;\;90$

Since $\boldsymbol{\mathit{q=LQ}}=\sum_{j=1}^{3N}l_{kj}Q_j=q_k$ , we have $\frac{dq_k}{dt}=\sum_{j=1}^{3N}l_{kj}\frac{dQ_j}{dt}$ or $\boldsymbol{\mathit{\dot q=L\dot Q}}$ and eq86 becomes

$T=\frac{1}{2}\boldsymbol{\mathit{\dot q}}^T\boldsymbol{\mathit{\dot q}}=\frac{1}{2}(\boldsymbol{\mathit{L\dot Q}})^T\boldsymbol{\mathit{L\dot Q}}=\frac{1}{2}\boldsymbol{\mathit{\dot Q}}^T\boldsymbol{\mathit{L}}^T\boldsymbol{\mathit{L\dot Q}}=\frac{1}{2}\boldsymbol{\mathit{\dot Q}}^T\boldsymbol{\mathit{\dot Q}}=\frac{1}{2}\sum_{j=1}^{3N}\dot Q_j^2\;\;\;\;\;\;\;\;91$

Combining eq90 and eq91, the classical total energy of the system is $E=\frac{1}{2}\sum_{j=1}^{3N}\dot Q_j^2+\frac{1}{2}\sum_{j=1}^{3N}\lambda_j Q_j^2$ , where we have set $U_e=0$ . This is because we are usually interested in computing transition energies and the eigenvalue corresponding to $U_e$ is subtracted out when we calculate vibrational transition energies for a given electronic state. The Hamiltonian is derived by replacing the classical observables by quantum-mechanical operators:

$\hat{H}=-\frac{\hbar^2}{2}\sum_{j=1}^{3N}\frac{\partial^2}{\partial Q^2_j}+\frac{1}{2}\sum_{j=1}^{3N}\lambda_j Q_j^2\;\;\;\;\;\;\;\;92$

Question

Show that the classical kinetic energy term $\frac{1}{2}\sum_{k=1}^{3N}\dot Q_k^2$ can be replaced by the quantum-mechanical operator $-\frac{\hbar^2}{2}\sum_{k=1}^{3N}\frac{\partial^2}{\partial Q^2_k}$ .

Answer

From eq89, $\dot Q_k=\sum_{j=1}^{3N}l_{kj}\dot q_j$ . Substituting the momentum operator $\hat{p}_1=m_1\dot x_1=\frac{\hbar}{i}\frac{\partial}{\partial x_1}$ in $\dot q_1=m_1^{1/2}\dot x_1$ , we have $\dot q_1=m_1^{-1/2}\frac{\hbar}{i}\frac{\partial}{\partial x_1}$ . Using the single-variable chain rule $\frac{\partial}{\partial x_1}=\frac{\partial}{\partial q_1}\frac{\partial q_1}{\partial x_1}$ , we have $\frac{\partial}{\partial x_1}=m_1^{1/2}\frac{\partial}{\partial q_1}$ , which when substituted in $\dot q_1$ gives $\dot q_1=\frac{\hbar}{i}\frac{\partial }{\partial q_1}$ . We can find similar expressions for $\dot q_2,\cdots,\dot q_{3N}$ . So, $\dot Q_k=\frac{\hbar}{i}\sum_{j=1}^{3N}l_{jk}\frac{\partial}{\partial q_j}$ . Since $U$ is a function of $Q_k$ , and $Q_k$ is a function of $q_j$ , the multi-variable chain rule gives $\frac{\partial}{\partial Q_k}=\sum_{j=1}^{3N}\frac{\partial}{\partial q_j}\frac{\partial q_j}{\partial Q_k}$ . Relabelling $q_k=\sum_{j=1}^{3N}l_{kj}Q_j$ as $q_j=\sum_{k=1}^{3N}l_{jk}Q_k$ , we have $\frac{\partial q_j}{\partial Q_k}=l_{jk}$ and therefore, $\frac{\partial}{\partial Q_k}=\sum_{j=1}^{3N}l_{jk}\frac{\partial}{\partial q_j}$ , which when substituted in $\dot Q_k$ gives $\dot Q_k=\frac{\hbar}{i}\frac{\partial}{\partial Q_k}$ . Substituting this last expression of $\dot Q_k$ in $\frac{1}{2}\sum_{k=1}^{3N}\dot Q_k^2$ gives $-\frac{\hbar^2}{2}\sum_{k=1}^{3N}\frac{\partial^2}{\partial Q_k^2}$ .

The kinetic energy term in eq92 involves a change in $Q_j$ , which is a function of $q_k$ and hence a function of the displacement coordinates $x_{\alpha}$ , $y_{\alpha}$ and $z_{\alpha}$ , which are relative to the equilibrium positions of the nuclei. In other words, $Q_j$ is a function of the displacements of all the nuclei from their respective equilibrium positions. For the molecule as a whole, all the equilibrium positions of the nuclei can be taken as the center of mass of the molecule. Since translational and rotational motions of the molecule involve no displacement relative to the molecule’s centre of mass, $\frac{\partial^2}{\partial Q_j^2}=0$ for such motions. Furthermore, translational and rotational motions are purely kinetic with no potential energy component. Therefore, we can rewrite eq92 as:

$\hat{H}=-\frac{\hbar^2}{2}\sum_{j=1}^m\frac{\partial^2}{\partial Q_j^2}+\frac{1}{2}\sum_{j=1}^m\lambda_jQ_j^2\;\;\;\;\;\;\;\;92$

where $m=3N-6$ and $m=3N-5$ for a non-linear molecule and a linear molecule respectively.

Each $m$ is called a normal mode and corresponds to a unique vibrational motion of the molecule. The Schrodinger equation is

$-\frac{\hbar^2}{2}\sum_{j=1}^m\frac{\partial^2}{\partial Q_j^2}\psi_{vib}+\frac{1}{2}\sum_{j=1}^m\lambda_jQ_j^2\psi_{vib}=E\psi_{vib}\;\;\;\;\;\;\;\;93$

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, $E=\sum_{j=1}^mE_j$ .

It follows that each $\psi_j(Q_j)$ describes a normal mode of the molecule. Comparing with eq4a, eq93 is the Schrodinger equation for a harmonic oscillator of unit mass and force constant $\lambda_j$ . This implies that (see eq46)

$\psi_j(Q_j)=N_jH_j\biggr$\sqrt{\frac{\omega_j}{\hbar}}Q_j\biggr$e^{-\frac{\omega_jQ^2_j}{2\hbar}}\;\;\;\;\;\;\;\;94$

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

Therefore, eq93 can be solved by computing $m$ individual Schrodinger equations. With reference to eq21, each solution gives

$E_j=\biggr$n_j+\frac{1}{2}\biggr$\hbar\omega_j\;\;\;\;\;\;n_j=0,1,2,\cdots\;\;\;\;\;\;\;\;95$

where $n_j$ is the vibrational quantum number of the $j$ -th normal mode of the molecule.

The total vibrational energy is

$E=\sum_{j=1}^m\biggr$n_j+\frac{1}{2}\biggr$\hbar\omega_j\;\;\;\;\;\;\;\;96$

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

$\vert 0,0,\cdots,0\rangle=\psi_0=N\prod_{j=1}^mH_j\biggr$\sqrt{\frac{\omega_j}{\hbar}}Q_j\biggr$e^{-\frac{\omega_jQ_j^2}{2\hbar}}=N\prod_{j=1}^me^{-\frac{\omega_jQ_j^2}{2\hbar}}\;\;\;\;\;\;\;\;97$

Eq97 is a Gaussian function that is symmetrical in $Q_j$ . The implication of this in group theory is that the ground state vibrational wavefunction of any polyatomic molecule is totally symmetric under all symmetry operations of the point group that the molecule belongs to. In fact, group theory provides an easy way to determine the vibrational modes of a molecule and to analyse the possible transitions between different vibrational states.

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April 30, 2024

Recurrence relations of the Hermite polynomials

The recurrence relations of the Hermite polynomials describe how each polynomial in the sequence can be obtained from its predecessors.

Expanding eq18 and assuming $n$ is an even number,

${H_n(y)=(-1)^{\frac{n}{2}}\frac{n!}{(\frac{n}{2})!}\biggr\[1-\frac{2n}{2!}y^2+\frac{2^2n(n-2)}{4!}y^4-\frac{2^3n(n-2)(n-4)}{6!}y^6+\cdots\biggr\]\;\;\;\;\;\;\;\;26}$

${H_{n-1}(y)=(-1)^{\frac{n-2}{2}}\frac{2(n-1)!}{(\frac{n-2}{2})!}\biggr\[y-\frac{2(n-2)}{3!}y^3+\frac{2^2(n-2)(n-4)}{5!}y^5-\frac{2^3(n-2)(n-4)(n-6)}{7!}y^7+\cdots\biggr\]\;\;\;\;\;\;\;\;27}$

Taking the derivative of eq26, we have

${\frac{dH_n(y)}{dy}=(-1)^{\frac{n}{2}-1}\frac{2(n-1)!}{(\frac{n-2}{2})!}\biggr\[2ny-\frac{2^2n(n-2)}{3!}y^3+\frac{2^3n(n-2)(n-4)}{5!}y^5-\cdots\biggr\]\;\;\;\;\;\;\;\;28}$

Substituting eq27 in eq28,

$\frac{dH_n(y)}{dy}=2nH_{n-1}(y)\;\;\;\;\;\;\;\;29$

If we repeat the steps from eq26 through eq29 on the assumption that $n$ is an odd number, we too end up with eq29. Therefore, $n$ in eq29 represents any number. Taking the derivative of eq29 again,

$\frac{d^2H_n(y)}{dy^2}=2n\frac{dH_{n-1}(y)}{dy}=4n(n-1)H_{n-2}(y)\;\;\;\;\;\;\;\;30$
Substitute 29 and 30 in eq23, we have

$H_n(y)-2yH_{n-1}(y)+2(n-1)H_{n-2}(y)=0\;\;\;\;\;\;\;\;31$

Replacing the dummy index $n$ in eq31 with $n+1$ gives

$H_{n+1}(y)-2yH_{n}(y)+2nH_{n-1}(y)=0\;\;\;\;\;\;\;\;32$

Eq29 and eq32 are the recurrence relations of the Hermite polynomials.

Question

Given $H_0(y)=1$ , show that eq32 can be used to generate the Hermite polynomials.

Answer

Substituting $n=0$ in eq32, we have $H_1(y)=2yH_0(y)=2y$ . Substituting $n=1$ in eq32, we have $H_2(y)=4y^2-2$ . Repeating the logic, we can generate the rest of the polynomials.

Question

Using eq32, ${y=\sqrt{\frac{m\omega}{\hbar}}x}$ and eq46, show that ${\langle\psi_k\vert x\vert\psi_n\rangle=\sqrt{\frac{\hbar(n+1)}{2m\omega}}\delta_{k,n+1}+\sqrt{\frac{n\hbar}{2m\omega}}\delta_{k,n-1}}$ .

Answer

Substituting eq32 in eq46, we have

$x\psi_n(x)=\sqrt{\frac{\hbar(n+1)}{2m\omega}}\psi_{n+1}(x)+\sqrt{\frac{n\hbar}{2m\omega}}\psi_{n-1}(x)$

Multiplying the above equation by $\psi^*_k(x)$ and integrating over all space,

$\langle\psi_k\vert x\vert\psi_n\rangle=\sqrt{\frac{\hbar(n+1)}{2m\omega}}\delta_{k,n+1}+\sqrt{\frac{n\hbar}{2m\omega}}\delta_{k,n-1}\;\;\;\;\;\;\;\;32a$

Next article: Taylor series of single-variable and multi-variable functions

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April 30, 2024May 17, 2024

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 $n,K,M$ 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,

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

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

$\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$

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

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

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

$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}$

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

$\vert A\vert^2=(n+1)(n+2)(n+3)\biggr$\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}$

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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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, $j-k=\pm 1$ .

Both 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

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

These two 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$_e=0}$ , while the normal mode of the molecule undergoing an antisymmetric stretch is IR active because it has a permanent electric dipole moment that varies with the displacement of the molecule.

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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

$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)$

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:

$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$

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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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,

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

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