Advanced Chemistry - Mono Mole

October 27, 2025

Rotation operator

A rotation operator $\hat{D}_k(\alpha)$ is a unitary operator that rotates quantum states by an angle $\alpha$ about an axis in a Hilbert space.

Mathematically, it is given by:

$\hat{D}_k(\alpha)=e^{-i\alpha\hat{J}_k}\;\;\;\;\;\;\;\;100$

where $\hat{J}_k$ , the angular momentum operator, is the generator of rotations about axis $k$ .

To derive eq100, we begin by noting that the probability outcome of a quantum mechanical measurement is described by the Born interpretation:

$\vert\langle\phi\vert\psi\rangle\vert^2$

where $\phi$ and $\psi$ are wavefunctions representing quantum states.

Since rotations are symmetry operations that do not change physical probabilities, we require:

$\vert\langle\hat{D}_k\phi\vert\hat{D}_k\psi\rangle\vert^2=\vert\langle\phi'\vert\psi'\rangle\vert^2=\vert\langle\phi\vert\psi\rangle\vert^2$

Using the matrix identity $(ABC\cdots)^{\dagger}=\cdots C^{\dagger}B^{\dagger}A^{\dagger}$ (see property 13 of this article for proof), we have $\vert\langle\hat{D}_k\phi\vert\hat{D}_k\psi\rangle\vert^2=\vert\langle\phi\vert\hat{D}_k^{\dagger}\hat{D}_k\vert\psi\rangle\vert^2=\vert\langle\phi\vert\psi\rangle\vert^2$ . Therefore, $\hat{D}_k$ is unitary, where $\hat{D}_k^{\dagger}\hat{D}_k=\hat{I}$ .

Let’s approximate $\hat{D}_k$ as a power series of a small change in $\alpha$ :

$\hat{D}_k(\delta\alpha)\approx\hat{D}_k(0)+\delta\alpha\hat{G}_k+(\delta\alpha)^2\hat{O}_k+\cdots\;\;\;\;\;\;\;\;101$

where $\hat{G}_k$ and $\hat{O}_k$ are first and second order rotation generator matrices respectively.

Since a rotation by zero angle must do nothing, $\hat{D}_k(0)=\hat{I}$ , and eq101 (ignoring higher order terms) becomes

$\hat{D}_k(\delta\alpha)\approx\hat{I}+\delta\alpha\hat{G}_k\;\;\;\;\;\;\;\;102$

To preserve unitarity, $(\hat{I}+\delta\alpha\hat{G}_k)^{\dagger}(\hat{I}+\delta\alpha\hat{G}_k)=\hat{I}$ . Expanding this equation and ignoring higher-order terms gives $\hat{I}+\delta\alpha\hat{G}_k^{\dagger}+\delta\alpha\hat{G}_k=\hat{I}$ or $\hat{G}_k^{\dagger}=-\hat{G}_k$ . This implies that $\hat{G}_k$ is not Hermitian, which contradicts the postulate that all physical observables in quantum mechanics are represented by Hermitian operators. It follows that eq102 must have the following form:

$\hat{D}_k(\delta\alpha)=\hat{I}-i\delta\alpha\hat{G}_k\;\;\;\;\;\;\;\;103$

For a quantum mechanical operator $\hat{A}$ to represent the same physical observable in the passively rotated frame, its expectation value must be the same whether calculated in the initial frame or with respect to the new frame:

$\langle\psi'\vert\hat{A}'\vert\psi'\rangle=\langle\psi\vert\hat{A}\vert\psi\rangle\;\;\;\;\;\;\;\;104$

Substituting $\vert\psi'\rangle=\hat{D}_k\vert\psi\rangle$ into eq104 gives $\langle\psi\vert\hat{D}_k^{\dagger}\hat{A}'\hat{D}_k\vert\psi\rangle=\langle\psi\vert\hat{A}\vert\psi\rangle$ , which means that $\hat{A}$ and $\hat{A}'$ are related by the similarity transformation $\hat{D}_k^{\dagger}\hat{A}'\hat{D}_k=\hat{A}$ or equivalently,

$\hat{A}'=\hat{D}_k\hat{A}\hat{D}_k^{\dagger}\;\;\;\;\;\;\;\;105$

where $\hat{D}_k^{\dagger}=\hat{D}_k^{-1}$ because $\hat{D}_k^{\dagger}\hat{D}_k=\hat{I}$ .

Substituting eq103 into eq105 yields:

$\hat{\boldsymbol{\mathit{r}}}'=(\hat{I}-i\delta\alpha\hat{G}_k)\hat{\boldsymbol{\mathit{r}}}(\hat{I}+i\delta\alpha\hat{G}_k)\;\;\;\;\;\;\;\;106$

where we have let the operator be the position operator $\hat{\boldsymbol{\mathit{r}}}$ .

Expanding eq106 and ignoring higher order terms results in $\hat{\boldsymbol{\mathit{r}}}'=\hat{\boldsymbol{\mathit{r}}}+i\boldsymbol{\mathit{r}}\hat{G}_k\delta\alpha-i\hat{G}_k\hat{\boldsymbol{\mathit{r}}}\delta\alpha$ , or equivalently,

$\delta\hat{\boldsymbol{\mathit{r}}}=-i\[\hat{G}_k,\hat{\boldsymbol{\mathit{r}}}\]\delta\alpha\;\;\;\;\;\;\;\;107$

where $\delta\hat{\boldsymbol{\mathit{r}}}=\hat{\boldsymbol{\mathit{r}}}'-\hat{\boldsymbol{\mathit{r}}}$ .

To determine the nature of $\hat{G}_k$ , we refer to the active classical rotation of the vector $\vec{r}$ by an infinitesimal angle $\delta\alpha$ about an arbitrary axis represented by the unit vector $\boldsymbol{\mathit{k}}$ (see diagram above), with the infinitesimal change in $\vec{r}$ given by:

$\delta\vec{r}=\delta\alpha(\boldsymbol{\mathit{k}}\times\vec{r})\;\;\;\;\;\;\;\;108$

where the cross product results in a vector with a magnitude proportional to $\delta\alpha$ and a direction tangential to the path of rotation.

Using the methodology of replacing classical variables with their quantum mechanical analogues to derive quantum mechanical expressions, eq108 becomes:

$\delta\hat{\boldsymbol{\mathit{r}}}=-\delta\alpha(\boldsymbol{\mathit{k}}\times\hat{\boldsymbol{\mathit{r}}})\;\;\;\;\;\;\;\;109$

where we have added a minus sign because $\delta\vec{r}$ in eq108 is defined by an active rotation about the axis, while $\delta\hat{\boldsymbol{\mathit{r}}}$ in eq107 describes a passive rotation about the axis.

Question

Why is an active rotation and a passive rotation related by a negative sign in eq109?

Answer

An active rotation refers an anticlockwise rotation of a position vector in a fixed coordinate system by the angle $\theta$ (see this article for details). In a passive rotation, the position vector remains stationary while the coordinate system rotates around it. To produce the same effect on the coordinates, a passive rotation corresponds to an anticlockwise rotation of the coordinate system by $-\theta$ . In other words, a passive rotation of the coordinate system by $-\theta$ has the same result as an active rotation of the vector by $\theta$ . Therefore, $\delta\vec{r}=\delta\alpha(\boldsymbol{\mathit{k}}\times\vec{r})=\delta\alpha\vert\boldsymbol{\mathit{k}}\vert\vert\vec{r}\vert sin\theta=-\delta\hat{\boldsymbol{\mathit{r}}}$ .

Substituting $\hat{\boldsymbol{\mathit{r}}}=\hat{x}\boldsymbol{\mathit{i}}+\hat{y}\boldsymbol{\mathit{j}}+\hat{z}\boldsymbol{\mathit{k}}$ into eq109 and expanding it gives:

$\delta\hat{x}\boldsymbol{\mathit{i}}+\delta\hat{y}\boldsymbol{\mathit{j}}+\delta\hat{z}\boldsymbol{\mathit{k}}=-\delta\alpha\hat{x}(\boldsymbol{\mathit{k}}\times\boldsymbol{\mathit{i}})-\delta\alpha\hat{y}(\boldsymbol{\mathit{k}}\times\boldsymbol{\mathit{j}})-\delta\alpha\hat{z}(\boldsymbol{\mathit{k}}\times\boldsymbol{\mathit{k}})$

Since $\boldsymbol{\mathit{k}}\times\boldsymbol{\mathit{i}}=\boldsymbol{\mathit{j}}$ , $\boldsymbol{\mathit{k}}\times\boldsymbol{\mathit{j}}=-\boldsymbol{\mathit{i}}$ and $\boldsymbol{\mathit{k}}\times\boldsymbol{\mathit{k}}=\boldsymbol{\mathit{0}}$ , we have:

$\delta\hat{x}\boldsymbol{\mathit{i}}+\delta\hat{y}\boldsymbol{\mathit{j}}+\delta\hat{z}\boldsymbol{\mathit{k}}=\delta\alpha\hat{y}\boldsymbol{\mathit{i}}-\delta\alpha\hat{x}\boldsymbol{\mathit{j}}\;\;\;\;\;\;\;\;110$

Substituting $\hat{\boldsymbol{\mathit{r}}}=\hat{x}\boldsymbol{\mathit{i}}+\hat{y}\boldsymbol{\mathit{j}}+\hat{z}\boldsymbol{\mathit{k}}$ into 107 and setting $\boldsymbol{\mathit{k}}$ to be the $z$ -axis for simplicity gives:

$\delta\hat{x}\boldsymbol{\mathit{i}}+\delta\hat{y}\boldsymbol{\mathit{j}}+\delta\hat{z}\boldsymbol{\mathit{k}}=-i\[\hat{G}_z,\hat{x}\]\delta\alpha\boldsymbol{\mathit{i}}-i\[\hat{G}_z,\hat{y}\]\delta\alpha\boldsymbol{\mathit{j}}-i\[\hat{G}_z,\hat{z}\]\delta\alpha\boldsymbol{\mathit{k}}\;\;\;\;\;\;\;\;111$

Comparing eq110 with eq111 yields:

$\[\hat{G}_z,\hat{x}\]=i\hat{y}$

$\[\hat{G}_z,\hat{y}\]=-i\hat{x}$

$\[\hat{G}_z,\hat{z}\]=0$

Comparing these three commutation relations with the total angular commutation relations of $\[\hat{J}_z,\hat{x}\]=i\hbar\hat{y}$ , $\[\hat{J}_z,\hat{y}\]=-i\hbar\hat{x}$ and $\[\hat{J}_z,\hat{z}\]=0$ , we find that $\hat{G}_k=\hat{J}_k/\hbar$ . In other words, $\hat{G}_k$ is the $k$ -th component of the total angular momentum operator, expressed in units of $\hbar$ .

A finite rotation by an angle $\alpha$ can be constructed by apply $N$ successive infinitesimal rotations, each by an angle $\delta\alpha=\alpha/N$ . From eq103,

$\hat{D}_k(\delta\alpha)=\lim_{N\rightarrow\infty}\biggr\[\hat{I}-i\frac{\alpha}{N\hbar}\hat{J}_k\biggr\]^N\;\;\;\;\;\;\;\;112$

Question

Show that $e^x=\lim_{n\rightarrow\infty}\biggr$1+\frac{x}{n}\biggr$^n$ .

Answer

Let $f(n)=\biggr$1+\frac{x}{n}\biggr$^n$ . So, $lnf(n)=nln\biggr$1+\frac{x}{n}\biggr$$ . Using the Taylor series $ln(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+\cdots$ for small $y$ , we have

$lnf(n)=n\biggr\[\frac{x}{n}-\frac{1}{2}\biggr$\frac{x}{n}\biggr$^2+\cdots\biggr\]=x-\frac{x^2}{2n}+\cdots$

As $n\rightarrow\infty$ , we find that $lnf(n)\rightarrow x$ , or equivalently, $f(n)\rightarrow e^x$ . So, $e^x=\lim_{n\rightarrow\infty}\biggr$1+\frac{x}{n}\biggr$^n$ . This makes eq112 the definition of the matrix exponential function.

Therefore, eq112 becomes:

$\hat{D}_k(\delta\alpha)=e^{-i\frac{\alpha}{\hbar}\hat{J}_k}\;\;\;\;\;\;\;\;113$

Eq113 is usually written as:

$\hat{D}_k(\delta\alpha)=e^{-i\alpha\hat{J}_k}\;\;\;\;\;\;\;\;114$

where the eigenvalues of $\hat{J}_k$ and hence $\hat{D}_k$ are expressed in units of $\hbar$

Since $\hat{J}_k$ is a finite-dimensional linear operator acting on a finite-dimensional vector space, $e^{-i\alpha\hat{J}_k}$ can be represented by an $n\times n$ matrix in both group theory and quantum mechanical calculations. In general, the rotation operator in an Euler angle system, which is defined by three successive rotations by the angles $\phi$ , $\theta$ and $\chi$ , is given by:

$\hat{D}(\phi,\theta,\chi)=e^{-i\phi\hat{J}_z}e^{-i\theta\hat{J}_y}e^{-i\chi\hat{J}_z}\;\;\;\;\;\;\;\;115$

where $0\leq\phi\leq2\pi$ , $0\leq\theta\leq \pi$ and $0\leq\chi\leq2\pi$ .

The application of eq115 involves the rotation operator transforming a quantum state $\vert J,K\rangle$ with total angular momentum projection along the molecular axis into a linear combination of states $\vert J,M_J\rangle$ with the projection along the lab $c$ -axis:

$\hat{D}(\phi,\theta,\chi)\vert J,K\rangle=\sum_{M_J}D^J_{M_J,K}(\phi,\theta,\chi)\vert J,M_J\rangle$

where $D^J_{M_J,K}(\phi,\theta,\chi)$ , the entries of the Wigner D-matrix, are the coefficients of $\vert J,M_J\rangle$ .

Question

Why is the transformed state a linear combination of $\vert J,M_J\rangle$ ?

If eq114 describes an operator that rotates quantum states by an angle $\alpha$ about an axis $k$ in a Hilbert space, how do three consecutive rotations described by eq115 ensure the complete transformation from the molecular coordinate system to the lab coordinate system?

Answer

In quantum mechanics, the states $\vert J,M_J\rangle$ form a complete orthonormal basis in the Hilbert space corresponding to a fixed $J$ . Therefore, any state with total angular momentum $J$ , including the rotated state $\hat{D}(\phi,\theta,\chi)\vert J,K\rangle$ , can be expressed as a linear combination of the basis states $\vert J,M_J\rangle$ .

In the diagram above, $a,b,c$ represent the molecular axes, while $x,y,z$ represent the lab-frame axes. Without assuming any specific rotation convention, the diagram illustrates that any target orientation can be achieved through three consecutive rotations. The first rotation about $c$ brings $a$ into the $xy$ -plane. The second rotation, about $a$ , aligns $c$ with $z$ . Finally, the last rotation about $c$ aligns the remaining axes of the two coordinate systems.

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October 26, 2025November 1, 2025

SO(3) Group

The SO(3) group, or special orthogonal group in three dimensions, is an infinite group of all 3D rotations.

Each element of the group corresponds to a rotation operator, characterised by a unit rotation axis $\boldsymbol{\mathit{k}}$ and an angle $\alpha$ . Since there are infinitely many possible rotation angles and axes, SO(3) contains infinitely many elements. These elements satisfy the properties of a group. For example, the binary operation of two rotation operators, each with a specific rotation angle about an axis, results in another rotation operator, demonstrating the closure property.

Consider the rotation of the spherical harmonics around the $z$ -axis. When the rotation operator $\hat{C}_{\alpha}$ acts on $Y^{m_l}_l(\theta,\phi)=P(\theta)e^{im_l\phi}$ by an angle $\alpha$ around the $z$ -axis, only the azimuthal angle $\phi$ is affected and each basis is transformed into $P(\theta)e^{im_l$\phi-\alpha$}$ . The transformation for all basis functions is summarised as:

$\hat{C}_{\alpha}\begin{pmatrix}Y^l_l\\Y^{l-1}_l\\\vdots\\Y^{-l}_l\end{pmatrix}=\begin{bmatrix}e^{-il\alpha} &0&\cdots &0\\0&e^{-i(l-1)\alpha}&\cdots &0\\\vdots &\vdots &\ddots &\vdots\\0&0&\cdots &e^{il\alpha}\end{bmatrix}\begin{pmatrix}Y^l_l\\Y^{l-1}_l\\\vdots\\Y^{-l}_l\end{pmatrix}\;\;\;\;\;\;\;\;120$

Question

Prove that if $\hat{A}\vert a\rangle=\lambda\vert a\rangle$ , then $e^{\hat{A}}\vert a\rangle=e^{\lambda}\vert a\rangle$ .

Answer

Consider the Taylor series of the exponential function $f$\hat{A}$=e^{\hat{A}}=\sum_{n=0}^{\infty}\frac{\hat{A}^n}{n!}$ . Since

$\hat{A}^n\vert a\rangle=\hat{A}^{n-1}\hat{A}\vert a\rangle=\lambda\hat{A}^{n-1}\vert a\rangle=\cdots=\lambda^n\vert a\rangle$

We have

$e^{\hat{A}}\vert a\rangle=\sum_{n=0}^{\infty}\frac{\hat{A}^n}{n!}\vert a\rangle=\sum_{n=0}^{\infty}\frac{\lambda^n}{n!}\vert a\rangle=e^{\lambda}\vert a\rangle$

In other words, $\hat{C}_{\alpha}Y^{m_l}_l(\theta,\phi)=Y^{m_l}_l(\theta,\phi-\alpha)=e^{-im_l\alpha}Y^{m_l}_l(\theta,\phi)$ , and according to the above Q&A,

$\hat{C}_{\alpha}=e^{-i\alpha\hat{l}_z}\;\;\;\;\;\;\;\;121$

where $\hat{l}_z$ is the $z$ -component of the angular momentum operator, expressed in $\hbar$ units.

Since an irreducible representation of a group is expressed by a set of matrices (or matrix-valued functions) that represent the group elements as linear operators on a vector space, an irreducible representation of the SO(3) group is $(2l+1)$ –dimensional (see eq120), with the set $\{Y^{m_l}_l(\theta,\phi)\}^l_{m_l=-l}$ forming a basis for the irreducible representation for a fixed $l$ .

Question

Why does the matrix representation of $\hat{C}_{\alpha}$ about the $z$ -axis belong to an irreducible representation of SO(3) when it is diagonal?

Answer

An irreducible representation is a group representation whose matrices cannot be simultaneously transformed, via the same invertible similarity transformation, into block diagonal form. For SO(3), an irreducible representation consists of matrices representing rotation operators about different axes. Even though the matrix of $\hat{C}_{\alpha}$ about the $z$ -axis is diagonal, which implies reducibility on its own, other matrices in the same representation, such as $\hat{C}_{\alpha}$ about the $y$ -axis, contain non-zero off-diagonal elements. These non-diagonal matrices cannot all be simultaneously diagonalised by the same similarity transformation. Therefore, the matrix representation of $\hat{C}_{\alpha}$ about the $z$ -axis belongs to an irreducible representation of SO(3).

The character of the rotation matrix is $\chi(C_{\alpha})=e^{-il\alpha}+e^{-i(l-1)\alpha}+\cdots+e^{il\alpha}$ , which is a geometric series $a+ar+ar^2+\cdots +ar^n=\frac{a(r^{n+1}-1)}{r-1}$ , where $a=e^{-il\alpha}$ , $r=e^{i\alpha}$ and $n=2l$ . This implies that

$\chi (C_{\alpha})=\frac{e^{i(l+1/2)\alpha}-e^{-i(l+1/2)\alpha}}{e^{i\alpha/2}-e^{-i\alpha/2}}$

Using Euler’s formula of $e^{ix}=cosx+isinx$ ,

$\chi (C_{\alpha})=\frac{sin(l+\frac{1}{2})\alpha}{sin\frac{\alpha}{2}}\;\;\;\;\;\;\;\;122$

The SO(3) group does not have a standard character table in the same way that finite point groups do. Its irreducible representations are labelled by a non-negative integer $l=0,1,2,\cdots$ , which form a discrete set. However, the group elements and their associated characters are continuous functions that depend on the rotation angle.

Nevertheless, for illustrative purposes, we can express the relationships between group elements, their irreducible representations and corresponding characters in the following way:

where each $\hat{R}$\alpha,\boldsymbol{\mathit{k}}_n$$ can be further expanded as follows:

Question

Why is $\chi$C_{\alpha=0}$=3$ for $l=1$ ?

Answer

From eq122, $\chi$C_{\alpha}$=\frac{sin(3\alpha/2)}{sin(\alpha/2)}$ for $l=1$ . At $\alpha=0$ , this expression is indeterminate, but we can evaluate the limit. Substituting $x=\alpha/2$ into the trigonometric identity $sin3x=3sinx-4sin^3x$ gives $sin(3\alpha/2)=3sin(\alpha/2)-4sin^3(\alpha/2)$ . So,

$\chi$C_{\alpha}$=\frac{3sin(\alpha/2)-4sin^3(\alpha/2)}{sin(\alpha/2)}=3-4sin^2(\alpha/2)$

Substituting the half-angle formula $sin^2(\alpha/2)=\frac{1-cos\alpha}{2}$ into the above equation yields $\chi(C_{\alpha})=1+2cos\alpha$ . Therefore, $\chi(C_{\alpha=0})=3$ . Since $\hat{R}(\alpha=0)$ is represented by the identity matrix, the trace $\chi(C_{\alpha=0})=3$ equals the dimension of the representation of $2l+1$ . Using the same logic, $\chi(C_{\alpha=0})=5$ for $l=2$ .

In general, a rotation operator of the SO(3) group is given by (see this article for derivation):

$\hat{D}(\phi,\theta,\chi)=e^{-i\phi\hat{J}_z}e^{-i\theta\hat{J}_y}e^{-i\chi\hat{J}_z}\;\;\;\;\;\;\;\;123$

where $0\leq\phi\leq 2\pi$ , $0\leq\theta\leq \pi$ and $0\leq\chi\leq 2\pi$ are the Euler angles.

The question, then, is whether the table relating the general rotation group elements, their irreducible representations and corresponding characters will be the same as that of a single-axis, single-angle rotation operator such as $\hat{C}_{\alpha}$ ?

The character of an irreducible representation of SO(3) depends only on the total rotation angle, not on the specific rotation axis or the individual Euler angles. Since SO(3) is the infinite group of all possible 3D rotations, there exists a one-to-one correspondence in symmetry between three consecutive rotations described by $\hat{D}(\phi,\theta,\chi)$ and a single rotation described by $\hat{C}_{\alpha}$ . In other words, even though $\hat{D}(\phi,\theta,\chi)$ is expressed in terms of Euler angles, it represents the same group element as some $\hat{C}_{\alpha}=e^{-i\alpha\hat{l}_k}$ , because we can always find a $\hat{C}_{\alpha}$ symmetry operation about an appropriate axis $\boldsymbol{\mathit{k}}$ that produces the same $\hat{D}(\phi,\theta,\chi)$ group element and therefore the same character value. Hence,

$\chi\[\hat{D}(\phi,\theta,\chi)\]=\chi(C_{\alpha})$

Finally, the rotation of $D^J_{M_JK}(\phi,\theta,\chi)=d^J_{M_JK}(\theta)e^{-i\phi M_J}e^{-i\chi K}$ , the Wigner D-matrix elements, for fixed $J$ and fixed $K$ is given by:

$\hat{C}_{\alpha}\begin{pmatrix}D^J_{J,K}\\D^J_{J-1,K}\\\vdots\\D^J_{-J,K}\end{pmatrix}=\begin{bmatrix}e^{iJ\alpha} &0&\cdots &0\\0&e^{i(J-1)\alpha}&\cdots &0\\\vdots &\vdots &\ddots &\vdots\\0&0&\cdots &e^{-iJ\alpha}\end{bmatrix}\begin{pmatrix}D^J_{J,K}\\D^J_{J-1,K}\\\vdots\\D^J_{-J,K}\end{pmatrix}$

This shows that, for fixed $K$ , the set $\{D^J_{M_JK}(\phi,\theta,\chi)\}^l_{m_l=-l}$ forms a basis of the same $(2J+1)$ -dimensional irreducible representation of SO(3) as the set $\{Y^{m_l}_l(\theta,\phi)\}^l_{m_l=-l}$ . Since $Y^{m_l}_l(\theta,\phi)$ are eigenfunctions of the angular momentum operators ( $\hat{L}^2$ and $\hat{l}_z$ ), it follows that $D^J_{M_JK}(\phi,\theta,\chi)$ are also eigenfunctions of the corresponding operators $\hat{J}^2$ and $\hat{J}_z$ . In other words, $D^J_{M_JK}(\phi,\theta,\chi)$ are also rotational wavefunctions.

Question

Elaborate on why $D^J_{M_JK}(\phi,\theta,\chi)$ are also eigenfunctions of $\hat{J}^2$ and $\hat{J}_z$ if $Y^{m_l}_l(\theta,\phi)$ are eigenfunctions of $\hat{L}^2$ and $\hat{l}_z$ .

Answer

As mentioned earlier, $\hat{l}_z$ (or equivalently $\hat{J}_z$ ) is a generator of an irreducible representation of SO(3). Its matrix representation is

$\hat{l}_z =\hbar\begin{pmatrix}l &0&\cdots &0\\0&l-1&\cdots &0\\\vdots &\vdots &\ddots &\vdots\\0&0&\cdots &-l\end{pmatrix}$

However, the matrix itself, is not an element of an irreducible representation of SO(3). $\hat{l}_z$ , or any component $\hat{l}_k$ of the angular momentum operator, is related to the rotation operator $\hat{D}(R)$ , by $\hat{D}(R)=e^{f(\hat{l}_k)}$ (see eq121 for an example). Using the Taylor series definition of the exponential function yields:

$\hat{L}^2,\hat{D}(R)\]=\[\hat{L}^2,e^{f(\hat{l}_k)}\]=\biggr\[\hat{L}^2,\sum_{n=0}^{\infty}\frac{\hat{l}_k^n}{n!}\biggr$

Applying the commutation relation identities $\[\hat{A},\hat{B}+\hat{C}+\cdots\]=\[\hat{A},\hat{B}\]+\[\hat{A},\hat{C}\]+\cdots$ and $\hat{A},\hat{B}\hat{C}\]=\[\hat{A},\hat{B}\]\hat{C}+\hat{B}\[\hat{A},\hat{C}$ gives:

$\begin{align}\[\hat{L}^2,\hat{D}(R)\]&=\[\hat{L}^2,\frac{\hat{l}_k^0}{0!}\]+\[\hat{L}^2,\frac{\hat{l}_k^1}{1!}\]+\[\hat{L}^2,\frac{\hat{l}_k^2}{2!}\]+\cdots\\&=\[\hat{L}^2,1\]+\[\hat{L}^2,\hat{l}_k\]+\[\hat{L}^2,\frac{\hat{l}_k}{2}\]\hat{l}_k+\frac{\hat{l}_k}{2}\[\hat{L}^2,\hat{l}_k\]+\cdots\\&=\[\hat{L}^2,1\]+\[\hat{L}^2,\hat{l}_k\]+\[\hat{L}^2,\hat{l}_k\]\frac{\hat{l}_k}{2}+\[\hat{L}^2,\frac{1}{2}\]\hat{l}_k^2+\frac{\hat{l}_k}{2}\[\hat{L}^2,\hat{l}_k\]+\cdots\end{align}$

Since $\[\hat{L}^2,c\]=0$ , where $c$ is a constant, and $\hat{L}^2$ commutes with $\hat{l}_k$ , i.e. $\[\hat{L}^2,\hat{l}_k\]=0$ , we have:

$\[\hat{L}^2,\hat{D}(R)\]=0$

According to Schur’s first lemma, any non-zero matrix that commutes with all matrices of an irreducible representation of a group is a multiple of the identity matrix. Therefore, $\hat{L}^2$ (or $\hat{J}^2$ ) must be a multiple of the identity operator within the $(2J+1)$ -dimensional subspace:

$\hat{L}^2=\lambda\hat{I}$

When $\hat{L}^2$ (or $\hat{J}^2$ ) acts on any basis function $\vert\psi\rangle$ within this irreducible subspace, we obtain:

$\hat{L}^2\vert\psi\rangle=\lambda\hat{I}\vert\psi\rangle=\lambda\vert\psi\rangle$

This shows that the basis functions $Y^{m_l}_l(\theta,\phi)$ are eigenfunctions of $\hat{L}^2$ and $\hat{l}_z$ . Similarly, $D^J_{M_JK}(\phi,\theta,\chi)$ , are eigenfunctions of $\hat{J}^2$ and $\hat{J}_z$ .

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October 16, 2025October 29, 2025

Coriolis effect and Coriolis coupling

The Coriolis effect is the apparent deflection of a moving object when observed from a rotating reference frame.

Consider a person holding a ball and standing on the edge of a large circular platform that is rotating anti-clockwise as viewed from above. At the twelve o’clock position, the person throws the ball directly towards the centre point O of the platform.

From the perspective of a stationary observer hovering above the platform (i.e. in an inertia frame of reference), the ball follows a straight-line path towards point X, slightly to the left of O. This trajectory occurs because, at the moment of release, the ball has two components of velocity:

- A radial component directed towards O.
- A tangential velocity due to the rotation of the platform at the point of release (in this case, towards the nine o’clock direction).

The resulting trajectory is the vector sum of these two components, which points towards X.

However, to the thrower, who is in the rotating frame of reference, the motion of the ball appears quite different. In this rotating frame, the ball seems to curve clockwise, as if being deflected to the right of its intended path. This illusion arises because, after release, the ball retains its original tangential speed, while the platform between the thrower and point O are rotating more slowly (closer to the center, where tangential speed is lower). In other words, the ball appears to be moving ahead of the rotating platform beneath it.

It follows that an apparent force, acting perpendicular to the direction of the ball’s motion, is influencing its path. This fictitious force, introduced to account for the apparent deflection experienced in rotating frames, is known as the Coriolis force.

This phenomenon isn’t just a curiosity of rotating platforms. On the molecular level, the Coriolis force introduces a crucial vibration-rotation interaction (also known as Coriolis coupling) in the dynamics of a rotating molecule — an interaction that would otherwise be neglected in a first-order approximation, where vibration and rotation are treated as independent motions.

When a molecule rotates, its internal motion (vibration) occurs simultaneously within the rotating molecular frame. To an observer in this rotating frame, the atoms undergoing vibration appear to be deflected perpendicular to their vibrational velocity. The resulting Coriolis force, associated with this apparent deflection, acts to couple the molecule’s vibrational angular momentum with its overall rotational angular momentum.

In quantum mechanics, this coupling appears as an additional term $\hat{H}_{cor}$ in the Hamiltonian that couples vibrational and rotational states:

$\hat{H}=\hat{H}_{vib}+\hat{H}_{rot}+\hat{H}_{cor}$

where $\hat{H}_{cor}$ is proportional to the product of vibrational and rotational angular momentum operators.

Explicitly,

$\hat{H}_{cor}\propto\sum_k\hat{J}_k\hat{p}_k$

where

$k$ denotes the molecular principal axes ( $a,b,c$ ).
$\hat{J}_{k}$ is the operator for the $k$ -component of the total angular momentum of the molecule (rotation).
$\hat{p}_{k}$ is the operator for the $k$ -component of the vibrational angular momentum.

This leads to shifts in rotational energy levels depending on the vibrational state, as well as the splitting or mixing of rotational levels associated with degenerate vibrations. In the infrared spectrum, this manifests as an anomalous splitting of the P, Q and R branches.

On a much larger scale, the Coriolis effect plays a crucial role in shaping natural systems on Earth. Because Earth is a rotating sphere, this effect influences the motion of air masses, ocean currents and weather patterns. For instance, a cyclone’s spin results from two forces acting on the atmosphere simultaneously:

- The pressure gradient force moves air from a high-pressure region to a low-pressure region.
- The Coriolis effect deflects that moving air

A low-pressure system is essentially a partial vacuum that draws in air from all directions. Consider a low-pressure region in the Northern Hemisphere (the upper yellow region in the diagram below). To an observer in a rotating frame of reference (e.g. a satellite moving at the same angular velocity as Earth from west to east), winds moving from the equator towards the low-pressure region appear to be deflected to the right, while winds flowing towards the region from the north appear to be deflected to the left. The combined movements result in an anti-clockwise sprial. The same principle causes winds to sprial clockwise in the Southern Hemisphere.

In conclusion, the Coriolis effect demonstrates the profound influence of rotational motion across vastly different scales — from affecting molecular behaviour to shaping global weather systems. At the molecular level, the principle manifests as Coriolis coupling, where rotational motion interacts with vibrational motion, subtly altering energy levels and spectral properties. On Earth, it governs the deflection of winds and ocean currents, giving rise to the rotation of cyclones and other large-scale atmospheric patterns. Together, these phenomena highlight the unifying power of rotational dynamics in both macroscopic and microscopic systems.

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October 16, 2025

Vibration-rotation spectra of polyatomic molecules

Vibration-rotation spectra arise from the simultaneous vibrational and rotational transitions of molecules, typically observed in the infrared (IR) region of the electromagnetic spectrum. These spectra provide detailed information about molecular structure, bond strength, and moment of inertia.

For a molecule to be IR active, it must have a permanent dipole or a dipole moment that changes over time. Since polyatomic molecules have $3N-6$ vibrational modes, some which do not result in a change in the molecules’ dipole moments, only certain vibration-rotation transitions are IR active. The transition selection rules are given by eq7 or eq8b, depending on whether the molecule is linear or non-linear.

Linear molecules

An example of a polyatomic linear molecule is CO₂, which has the same expressions for $\tilde{v}_P$ , $\tilde{v}_Q$ and $\tilde{v}_R$ as those for diatomic molecules. The symmetric stretch of CO₂ is IR inactive because the net electric dipole moment is always zero (see diagram above).

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 $\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{R}}+\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{G}}+\boldsymbol{\mathit{S}}$ , where

$\boldsymbol{\mathit{R}}$ is the nuclear rotation angular momentum (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.

In contrast, the antisymmetric stretch is IR active because it produces a dipole moment that varies with displacement. Since CO₂is a closed-shell molecule, $\boldsymbol{\mathit{L}}=\boldsymbol{\mathit{S}}=\boldsymbol{\mathit{0}}$ . Furthermore, this vibrational mode does not produce vibrational angular momentum ( $\boldsymbol{\mathit{G}}=\boldsymbol{\mathit{0}}$ ). Therefore, the absorption of the photon’s angular momentum must be accounted for by $\Delta\boldsymbol{\mathit{R}}=\pm\boldsymbol{\mathit{1}}$ , resulting in $J=0\rightarrow J=\pm 1$ (P-branch or R-branch).

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. In other words, $\Delta\boldsymbol{\mathit{G}}=\pm\boldsymbol{\mathit{1}}$ after photon absorption, but the vector addition of $\boldsymbol{\mathit{R}}$ and $\boldsymbol{\mathit{G}}$ can result in the magnitude of $\boldsymbol{\mathit{J}}$ being unchanged (possible when $\boldsymbol{\mathit{R}}\neq\boldsymbol{\mathit{0}}$ ), enabling $\vert n=0,J\rangle\rightarrow\vert n=1,J\rangle$ transitions to appear in the IR spectrum as the Q-branch.

Symmetric tops

For a symmetric top, its molecular rotational energy is given by eq51. Therefore, eq9 becomes:

$E_{v,J,K}=\biggr$v+\frac{1}{2}\biggr$\tilde\nu+\tilde{B}J(J+1)+K^2(\tilde{A}-\tilde{B})\;\;\;\;\;\;\;\;17$

Any allowed vibrational transition of a symmetric rotor like CH₃I can involve a change in the dipole moment $\mu_{c,e}$ along the molecule’s symmetry axis $c$ (known as a parallel transition), as well as changes in $\mu_{a,e}$ and $\mu_{b,e}$ , which are perpendicular to $c$ (perpendicular transitions). The general selection rules governing parallel transitions for symmetric rotors are given by eq8:

$\Delta J=0,\pm 1,\Delta K=0,\Delta v=\pm 1$

while those for perpendicular transitions are given by eq8a:

$\Delta J=0,\pm 1,\Delta K=\pm 1,\Delta v=\pm 1$

It follows that the parallel transition expressions for $\tilde{v}_P$ , $\tilde{v}_Q$ and $\tilde{v}_R$ are the same as those for linear molecules. For perpendicular transitions, each branch has two sub-branches depending on $\Delta K=\pm 1$ :

Therefore, the IR spectrum for CH₃I (see diagram above) is more complicated than that for CO₂. Furthermore, there are $3N-6=9$ vibrational modes, which are categorised into six types:

The first three modes are symmetric vibrations (see diagram below), with a change in dipole moment parallel to the C₃ axis. The remaining three modes are doubly degenerate asymmetric vibrations, with a change in dipole moment perpendicular to the C₃ axis.

Spherical tops

Although a spherical top is non-polar overall, some of its vibrational modes are IR active due to temporary dipole moments during vibration. The first-order approximation of the vibration-rotation energy levels of a spherical top is also given by eq9, with the selection rules governing IR-active transitions between these levels being the same as those for linear molecules. An example is CH₄, which has $3N-6=9$ vibrational modes (see table and diagram below): one symmetric stretch, one doubly degenerate bend, and two triply degenerate modes (one stretch and one bend).

Mode	Symmetry	Description	Wavenumber /cm^-1
$v_1$	$A_1$	Symmetric C-H stretch	IR inactive
$v_2$	$E$	Symmetric bend (scissoring)	IR inactive
$v_3$	$T_2$	Asymmetric C-H stretch	3020
$v_4$	$T_2$	Asymmetric bend (umbrella)	1300

However, only the $T_2$ modes are IR active due to their symmetry and ability to cause a dynamic dipole moment (see diagram below).

In conclusion, the derivation of the three branches using the first-order vibration-rotation energy levels (eq9 and eq17) of an IR-active polyatomic molecule allows us to analyse most IR spectra with ease. However, these energy levels are more complex due to changes in $\tilde{B}$ at higher $v$ , centrifugal distortion, anharmonicity and vibration-rotation interactions such as Coriolis coupling.

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October 16, 2025

Vibration-rotation spectra of diatomic molecules

For a molecule to be IR active, it must have a permanent dipole or a dipole moment that changes over time. Homonuclear diatomic molecules, such as O₂, lack these properties and are therefore IR inactive. In contrast, when a heteronuclear diatomic molecule absorbs IR radiation, it can undergo a change in its vibrational energy level along with a change in its rotational state, resulting in a distinctive pattern of spectral lines. These energy levels (expressed in wavenumbers), characterised by the rigid rotor–harmonic oscillator approximation, are the sum of its allowed rotational and vibrational energies:

$E_{v,J}=\biggr$v+\frac{1}{2}\biggr$\tilde\nu+BJ(J+1)\;\;\;\;\;\;\;\;9$

The selection rules governing transitions are given by eq7:

$\Delta J=0,\pm 1,\Delta v=\pm 1$

However, the condition $\Delta J=0$ applies only to molecules with a nonzero projection of the electronic orbital angular momentum onto the internuclear axis (i.e. $\Lambda\neq 0$ ), such as NO. To elaborate, the rotational quantum number $J$ represents the magnitude of the coupled total angular momentum vector, given by

$\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{R}}+\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{G}}+\boldsymbol{\mathit{S}}\;\;\;\;\;\;\;\;10$

where:

Note: Nuclear spin angular momentum is excluded because it couples only weakly to the other components.

Question

What is the relationship between the component vectors of $\boldsymbol{\mathit{J}}$ and their corresponding quantum numbers?

Answer

Note: The spin quantum number $\Sigma$ should not be confused with the electronic state $\Sigma$ .

The selection rules for a pure vibrational transition are $\Delta v=\pm 1$ , while those for pure rotational transitions (ignoring the $M_J$ -related fine structure transitions, which are often unresolved in most experimental spectra) are $\Delta J=\pm 1$ . Additionally, the total angular momentum $\boldsymbol{\mathit{J}}$ of the system must be conserved during a photon-mediated transition, as a photon carries one unit of angular momentum.

For a closed shell heteronuclear diatomic molecule like HCl ( $^1\Sigma$ ), $\boldsymbol{\mathit{L}}=\boldsymbol{\mathit{S}}=\boldsymbol{\mathit{G}}=\boldsymbol{\mathit{0}}$ , with $\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{R}}$ . Therefore, the absorption of the IR photon’s angular momentum during $v=0\rightarrow v=1$ must be accounted for by $\Delta\boldsymbol{\mathit{R}}=\pm 1$ , resulting in $\vert v=0,J\rangle\rightarrow \vert v=1,J\pm 1\rangle$ .

Conversely, the ground state of NO ( $^2\Pi$ ) has an unpaired electron in a degenerate $\pi^*$ molecular orbital, giving $\boldsymbol{\mathit{L}}\neq\boldsymbol{\mathit{0}}$ , $\boldsymbol{\mathit{G}}=\boldsymbol{\mathit{0}}$ , $\boldsymbol{\mathit{S}}\neq\boldsymbol{\mathit{0}}$ . This allows the coupling (vector addition) of $\boldsymbol{\mathit{L}}$ and $\boldsymbol{\mathit{S}}$ to reorient internally in such a way that it can exactly compensate for the photon’s angular momentum (see diagram above). In other words, $\boldsymbol{\mathit{J}}$ can be preserved without changing $\boldsymbol{\mathit{R}}$ , making the transitions $\vert v=0,J\rangle\rightarrow \vert v=1,J\rangle$ , where $\Delta J=0$ , no longer forbidden.

In general, the absorption lines in a vibration-rotation spectrum can be grouped into three types, called branches (see diagram below). The R branch consists of all $\vert v,J\rangle\rightarrow \vert v+1,J+1\rangle$ transitions, with energies given by:

$\tilde{v}_R=E_{v+1,J+1}-E_{v,J}=\tilde{v}+2B(J+1)\;\;\;\;\;\;\;\;11$

where $J=0,1,2,\cdots$ .

$\vert v,J\rangle\rightarrow \vert v+1,J-1\rangle$ transitions form the P branch, with energies:

$\tilde{v}_P=E_{v+1,J-1}-E_{v,J}=\tilde{v}-2BJ\;\;\;\;\;\;\;\;12$

where $J=1,2,3,\cdots$ .

The energy separation between adjacent lines in the P and R branches corresponds to $2B$ . Therefore, the vibration-rotation spectrum, like the pure rotation spectrum, allows the moment of inertia and bond length of molecules to be calculated. The relative intensities of the lines in the P and R branches depend on the product of $e^{-\varepsilon_J/kT}$ and $(2J+1)$ , the same mechanism that governs the intensities in a pure rotational spectrum.

Question

According to the above diagram, the rotational constant $B$ is smaller for NO than for HCl. Why?

Answer

The rotational constant is given by $B=\frac{h}{8\pi^2cI$ , where $I=\mu r^2$ . Since the reduced mass $\mu$ of HCl is smaller than that of NO, $B_{NO}< B_{HCl}$ .

Finally, the Q branch, if allowed by selection rules, consists of $\vert v,J\rangle\rightarrow \vert v+1,J\rangle$ transitions, all of which have the same energy:

$\tilde{v}_Q=E_{v+1,J}-E_{v,J}=\tilde{v}\;\;\;\;\;\;\;\;13$

This means that transitions such as $\vert v,J=0\rangle\rightarrow \vert v+1,J=0\rangle$ and $\vert v,J=1\rangle\rightarrow \vert v+1,J=1\rangle$ appear as a single line in the spectrum under the rigid rotor-harmonic oscillator approximation. In reality, the Q branch appears as a series of closely spaced lines rather than a single line. This is because $B$ changes with the vibrational quantum number $v$ , where the bond length $r$ increases slightly in higher vibrational states, causing the moment of inertia $I$ to increase and $B$ to decrease. From eq9,

$\begin{align}\tilde{v}_Q&=E_{v+1,J}-E_{v,J}\\&=\biggr$v+\frac{3}{2}\biggr$\tilde{v}+B_1J(J+1)-\biggr$v+\frac{1}{2}\biggr$\tilde{v}-B_0J(J+1)\\&=\tilde{v}+(B_1-B_0)J(J+1)\;\;\;\;\;\;\;\;14\end{align}$

Eq14 shows that transitions such as $\vert v,J=0\rangle\rightarrow \vert v+1,J=0\rangle$ and $\vert v,J=1\rangle\rightarrow \vert v+1,J=1\rangle$ now appear as separate lines. Since $B_1< B_0$ , the spacing between these lines decreases with $J$ . At lower resolution, these closely spaced lines give the Q-branch the characteristic appearance of a single, strong, broad peak.

Using the same logic,

$\begin{align}\tilde{v}_R&=E_{v+1,J+1}-E_{v,J}\\&=\biggr$v+\frac{3}{2}\biggr$\tilde{v}+B_1(J+1)(J+2)-\biggr$v+\frac{1}{2}\biggr$\tilde{v}-B_0J(J+1)\end{align}$

which rearranges to:

$\tilde{v}_R=\tilde{v}+(B_1+B_0)(J+1)+(B_1-B_0)(J+1)^2\;\;\;\;\;\;\;\;15$

As $B_1< B_0$ the spacing between the lines of the R-branch decreases slightly as $J$ increases.

Question

Is the change in $B$ at higher $v$ the same as centrifugal distortion?

Answer

No, it is different from centrifugal distortion. When a molecule vibrates at higher $v$ , the average bond length increases slightly. This is purely a vibrational effect. In contrast, centrifugal distortion is a rotational effect, which also causes the bond to stretch slightly as the rotational speed increases at higher $J$ . In other words, the former phenomenon depends on $v$ , while the latter depends on $J$ . If centrifugal distortion in considered, eq9 becomes $E_{v,J}=\biggr$v+\frac{1}{2}\biggr$\tilde\nu+BJ(J+1)-DJ^2(J+1)^2$ and

$\tilde{v}_R=\tilde{v}+B_1(J+1)(J+2)-B_0J(J+1)-D_1(J+1)^2(J+2)^2+D_0J^2(J+1)^2$

For the P-branch,

$\begin{align}\tilde{v}_P&=E_{v+1,J-1}-E_{v,J}\\&=\biggr$v+\frac{3}{2}\biggr$\tilde{v}+B_1J(J-1)-\biggr$v+\frac{1}{2}\biggr$\tilde{v}-B_0J(J+1)\\&=\tilde{v}-(B_1+B_0)J+(B_1-B_0)J^2\;\;\;\;\;\;\;\;16\end{align}$

Since the third term on RHS of eq16 is negative, the lines of the P-branch diverge slightly as $J$ increases. Note that eq14, eq15 and eq16 reduce to eq13, eq11 and eq12 respectively if $B_1=B_0$ .

Question

Why do many IR spectra have broad, continuous peaks rather than discrete lines?

Answer

Many common IR spectra are taken in the liquid phase (or solid), which leads to broadened, continuous-looking peaks. In the liquid or solid phase, molecules interact strongly with each other via van der Waals forces and hydrogen bonding. These interactions distort the the dipole moments of molecules and broaden energy levels. On the other hand, intermolecular forces are minimal in dilute samples of gaseous molecules. When these gas-phase samples are analysed with high-resolution instruments, discrete peaks — especially rotational fine structure — can be observed.

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October 16, 2025October 27, 2025

Selection rules for vibration-rotation transitions

The selection rules for vibration-rotation transitions govern which simultaneous changes in vibrational and rotational energy levels are allowed when a molecule absorbs or emits infrared radiation.

These rules arise from quantum mechanical principles and the requirement that transitions involve a change in the molecule’s electric dipole moment $\boldsymbol{\mathit{\mu}}$ , and they combine the selection rules for pure vibrational and pure rotational transitions.

In determining the selections rules for pure vibrational and pure rotational transitions, we evaluated conditions under which $\langle\psi_v'\vert\boldsymbol{\mathit{\mu}}\vert\psi_v\rangle\neq 0$ and $\langle\psi_r'\vert\boldsymbol{\mathit{\mu}}\vert\psi_r\rangle\neq 0$ separately. However, the selection rules for vibration-rotation transitions require that $\langle\psi_v'\psi_r'\vert\boldsymbol{\mathit{\mu}}\vert\psi_v\psi_r\rangle\neq 0$ . To show that this ultimately reduces to the selection rules for pure vibrational and pure rotational transitions, we begin with

$\boldsymbol{\mathit{\mu}}=\mu_x\boldsymbol{\mathit{e}}_x+\mu_y\boldsymbol{\mathit{e}}_y+\mu_z\boldsymbol{\mathit{e}}_z=\mu_a\boldsymbol{\mathit{e}}_a+\mu_b\boldsymbol{\mathit{e}}_b+\mu_c\boldsymbol{\mathit{e}}_c\;\;\;\;\;\;\;\;1$

where $x$ , $y$ and $z$ are the orthogonal lab-frame axes; $a$ , $b$ and $c$ are the orthogonal principal axes of inertia of the molecule (molecular frame); and the $\boldsymbol{\mathit{e}}$ ’s are unit vectors along the axes.

The relationship between the two frames is shown in the diagram above, with the origin $o$ of both frames commonly placed at the centre of mass of the molecule. The lab-frame is stationary in space and does not rotate with the molecule, whereas the molecular frame moves with the molecule and rotates as it does. $N$ , known as the line of nodes, denotes the line passing through the intersections of the $xy$ and $ab$ planes. The orientation of the molecular frame with respect to the lab frame is described by the Euler angles $0\leq\theta\leq\pi$ , $0\leq\phi\leq 2\pi$ , and $0\leq\chi\leq 2\pi$ .

From eq1,

$\mu_z=\boldsymbol{\mathit{e}}_z\cdot\boldsymbol{\mathit{\mu}}=\mu_a cos(\angle zoa)+\mu_b cos(\angle zob)+\mu_c cos(\angle zoc)\;\;\;\;\;\;\;\;2$

where $cos(\angle zoa)=\boldsymbol{\mathit{e}}_z\cdot\boldsymbol{\mathit{e}}_a$ , $cos(\angle zob)=\boldsymbol{\mathit{e}}_z\cdot\boldsymbol{\mathit{e}}_b$ and $cos(\angle zoc)=\boldsymbol{\mathit{e}}_z\cdot\boldsymbol{\mathit{e}}_c$ .

Similarly, $\mu_y=\mu_a cos(\angle yoa)+\mu_b cos(\angle yob)+\mu_c cos(\angle yoc)$ and $\mu_x=\mu_a cos(\angle xoa)+\mu_b cos(\angle xob)+\mu_c cos(\angle xoc)$ . However, we can simplify the analysis to $\mu_z$ , which can always be arbitrarily chosen as the polarisation direction of the incident oscillating radiation.

To express the cosines in terms of the Euler angles, we need to derive the full rotation matrix according to the convention of a specific sequence of three consecutive elemental rotations to describe any 3D orientation.

The first rotation is about the lab-frame $z$ -axis by $\phi$ :

$R_z\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}cos\phi&-sin\phi&0\\sin\phi&cos\phi&0\\0&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x'\\y'\\z'\end{pmatrix}$

The second rotation is about the $y'$ axis, also known as the line of nodes $N$ , by $\theta$ . It lies in the initial $xy$ -plane but is fixed in its direction after the first rotation and is perpendicular to both the initial $z$ -axis and the final $c$ -axis:

$R_N\begin{pmatrix}x'\\y'\\z'\end{pmatrix}=\begin{pmatrix}cos\theta&0&-sin\theta\\0&1&0\\sin\theta&0&cos\theta\end{pmatrix}\begin{pmatrix}x'\\y'\\z'\end{pmatrix}=\begin{pmatrix}x''\\y''\\z''\end{pmatrix}$

The last rotation is about the $z''$ axis, which is also the molecular $c$ -axis, by $\chi$ :

$R_c\begin{pmatrix}x''\\y''\\z''\end{pmatrix}=\begin{pmatrix}cos\chi&-sin\chi&0\\sin\chi&cos\chi&0\\0&0&1\end{pmatrix}\begin{pmatrix}x''\\y''\\z''\end{pmatrix}=\begin{pmatrix}a\\b\\c\end{pmatrix}$

Therefore, the full rotation matrix is:

$R=R_zR_NR_c=\begin{pmatrix}cos\phi cos\theta cos\chi-sin\phi sin\chi&-cos\phi cos\theta sin\chi-sin\phi cos\chi&-cos\phi sin\theta\\sin\phi cos\theta cos\chi+cos\phi sin\chi&-sin\phi cos\theta sin\chi+cos\phi cos\chi&-sin\phi sin\theta\\sin\theta cos\chi&-sin\theta sin\chi&cos\theta\end{pmatrix}$

A rotation matrix is the transpose of the change of basis matrix $R^T$ , which in this case is

$R^T=\begin{pmatrix}\boldsymbol{\mathit{e}}_a\cdot\boldsymbol{\mathit{e}}_x&\boldsymbol{\mathit{e}}_a\cdot\boldsymbol{\mathit{e}}_y&\boldsymbol{\mathit{e}}_a\cdot\boldsymbol{\mathit{e}}_z\\\boldsymbol{\mathit{e}}_b\cdot\boldsymbol{\mathit{e}}_x&\boldsymbol{\mathit{e}}_b\cdot\boldsymbol{\mathit{e}}_y&\boldsymbol{\mathit{e}}_b\cdot\boldsymbol{\mathit{e}}_z\\\boldsymbol{\mathit{e}}_c\cdot\boldsymbol{\mathit{e}}_x&\boldsymbol{\mathit{e}}_c\cdot\boldsymbol{\mathit{e}}_y&\boldsymbol{\mathit{e}}_c\cdot\boldsymbol{\mathit{e}}_z\end{pmatrix}$

Therefore,

$R^T=\begin{pmatrix}\boldsymbol{\mathit{e}}_x\cdot\boldsymbol{\mathit{e}}_a&\boldsymbol{\mathit{e}}_x\cdot\boldsymbol{\mathit{e}}_b&\boldsymbol{\mathit{e}}_x\cdot\boldsymbol{\mathit{e}}_c\\\boldsymbol{\mathit{e}}_y\cdot\boldsymbol{\mathit{e}}_a&\boldsymbol{\mathit{e}}_y\cdot\boldsymbol{\mathit{e}}_b&\boldsymbol{\mathit{e}}_y\cdot\boldsymbol{\mathit{e}}_c\\\boldsymbol{\mathit{e}}_z\cdot\boldsymbol{\mathit{e}}_a&\boldsymbol{\mathit{e}}_z\cdot\boldsymbol{\mathit{e}}_b&\boldsymbol{\mathit{e}}_z\cdot\boldsymbol{\mathit{e}}_c\end{pmatrix}$

and eq2 becomes

$\mu_z=\mu_asin\theta cos\chi-\mu_bsin\theta sin\chi+\mu_ccos\theta\;\;\;\;\;\;\;\;3$

We can also determine $\mu_x$ and $\mu_y$ in terms of the Euler angles by repeating the steps above.

Now, certain vibrational motions change a molecule’s electric dipole moment, such that $\frac{d\boldsymbol{\mathit{\mu}}}{dQ}\neq 0$ , where $Q$ represents the normal coordinates typically used to describe vibrational motions. Similarly, rotational motions also affect the dipole moment, with $\frac{d\boldsymbol{\mathit{\mu}}}{d\Omega}\neq 0$ , where $\Omega$ collectively denotes the Euler angles. This implies that the dipole moment is a function of both $Q$ and $\Omega$ , i.e. $\boldsymbol{\mathit{\mu}}(Q,\Omega)$ . Since the components $\mu_a$ , $\mu_b$ and $\mu_c$ in eq3 are projections of $\boldsymbol{\mathit{\mu}}$ onto the molecular axes, which rotate with the molecule, they are independent of the Euler angles, and must therefore be functions of $Q$ alone.

If we assume that the atoms in the molecule vibrate with a small amplitude about their equilibrium positions, we can expand $\mu_a$ , $\mu_b$ and $\mu_c$ in a Taylor series, e.g.

$\mu_a=\mu_{a,e}+\sum_{k=1}^f\biggr$\frac{\partial \mu_a}{\partial Q_k}\biggr$_eQ_k+\cdots\;\;\;\;\;\;\;\;4$

where $f=3N-6$ for non-linear polyatomic molecules and $f=3N-5$ for linear molecules.

Substituting eq4 (by ignoring the higher terms) and the corresponding expansions for $\mu_b$ and $\mu_c$ into $\langle\psi_v'\psi_r'\vert\mu_z\vert\psi_v\psi_r\rangle$ gives:

$\begin{align}\langle\psi_v'\psi_r'\vert\mu_z\vert\psi_v\psi_r\rangle &=\biggr$\int\psi_v^{*'}\psi_vdQ\biggr$\[\mu_{a,e}I_a-\mu_{b,e}I_b+\mu_{c,e}I_c\]\\&+\sum_{k=1}^f\biggr$\int\psi_v^{*'}Q_k\psi_vdQ\biggr$\biggr\[\biggr$\frac{\partial\mu_a}{\partial Q_k}\biggr$_eI_a-\biggr$\frac{\partial\mu_b}{\partial Q_k}\biggr$_eI_b+\biggr$\frac{\partial\mu_c}{\partial Q_k}\biggr$_eI_c\biggr\]\;\;\;\;\;\;\;\;5\end{align}$

where

$\int\psi_v^{*'}\psi_vdQ=\prod_{k=1}^f\int\psi_k^{*'}\psi_kdQ_k$ .
$I_a=\int\psi_r^{*'}\psi_rsin\theta cos\chi d\tau_r$ .
$I_b=\int\psi_r^{*'}\psi_rsin\theta sin\chi d\tau_r$ .
$I_c=\int\psi_r^{*'}\psi_rcos\theta d\tau_r$ .
$d\tau_r=sin\theta d\theta d\phi d\chi$ .

Question

Why is $d\tau_r=sin\theta d\theta d\phi d\chi$ ?

Answer

Consider an arrow in a 3D space. To fully describe its rotation, we must determine its complete orientation. This is achieved by pointing the arrow in a certain direction and spinning it around its shaft. In other words, we must first specify the direction of a chosen axis of a 3D object (like the arrow’s shaft), and then the rotation about that axis to describe its rotation. In the Euler angle framework, the first two angles $\theta$ and $\phi$ define a direction on the unit sphere (like latitude and longitude on Earth). Small changes to this direction are expressed as $sin\theta d\theta d\phi$ (see diagram below, where $r=1$ ). The third angle $\chi$ determines the extent of rotation around that direction. Therefore, the combined infinitesimal change in the object’s orientation in rotation space is $sin\theta d\theta d\phi d\chi$ .

Since the vibrational wavefunctions are orthonormal, the first integral on RHS of eq5 is non-zero only if $\psi_k'=\psi_k$ for all $k$ , i.e. when there is no vibrational transition. Assuming $I_a\leq I_b\leq I_c$ , the selection rules associated with this term pertain to pure rotational transitions. The second term is non-zero only if both $\int\psi_v^{*'}Q_k\psi_vdQ$ and the terms in the square bracket are non-zero. This implies that the selection rules associated with the 2^nd term correspond to those of both pure vibrational and pure rotational transitions.

For a polar linear molecule with $c$ as the molecular axis, $\mu_{a,e}=\mu_{b,e}=0$ while $\mu_{c,e}\neq 0$ , and the first term on RHS of eq5 reduces to $\mu_{c,e}I_c$ if $\psi_k'=\psi_k$ for all $k$ . Furthermore, $\biggr$\frac{\partial\mu_a}{\partial Q_k}\biggr$_e=\biggr$\frac{\partial\mu_b}{\partial Q_k}\biggr$_e=0$ for non-degenerate normal modes like stretching, but may not be zero for degenerate modes like bending. Despite that, $I_a\leq I_b\leq I_c$ and therefore, the general selection rules for vibration-rotation transitions of polar linear molecules, which require $\langle\psi_v'\psi_r'\vert\mu_z\vert\psi_v\psi_r\rangle\neq 0$ for linearly polarised light, are:

$\Delta J=\pm 1,\Delta M_J=0,\pm 1,\Delta v=\pm 1\;\;\;\;\;\;\;\;6$

In practice, eq6 is written more simply as:

$\Delta J=0,\pm 1,\Delta v=\pm 1\;\;\;\;\;\;\;\;7$

The condition $\Delta M_J=0,\pm 1$ is omitted because the $M_J$ -related fine structure is often unresolved in most experimental spectra. Additionally, the inclusion of $\Delta J=0$ highlights that total angular momentum conservation must be satisfied during a photon-mediated vibrational transition, even though $\Delta J=0$ is forbidden for most linear molecules in electric dipole transitions.

Eq7 also corresponds to be the vibration-rotation selection rules for spherical tops.

For a symmetric rotor, any allowed vibrational transition can involve a change in $\mu_{c,e}$ along the symmetry axis $c$ (known as parallel transitions), as well as changes in $\mu_{a,e}$ and $\mu_{b,e}$ , which are perpendicular to $c$ (perpendicular transitions). Since the pure rotational selection rules for a symmetric rotor (neglecting $M_J$ -related fine structure), where the permanent dipole lies along the symmetry axis, are $\Delta J=\pm 1,\Delta K=0$ , the selection rules for parallel transitions are:

$\Delta J=0,\pm 1,\Delta K=0,\Delta v=\pm 1\;\;\;\;\;\;\;\;8$

In the case of a perpendicular transition, $\mu_{a,e}\neq 0$ and $\mu_{b,e}\neq 0$ , while $\mu_{c,e}=0$ . This results in the following selection rules:

$\Delta J=0,\pm 1,\Delta K=\pm 1,\Delta v=\pm 1\;\;\;\;\;\;\;\;8a$

Question

Explain why $\Delta K=\pm 1$ for a perpendicular transition.

Answer

The rotational wavefunction can be approximated as the Wigner D-functions $\psi_r=D^J_{M_JK}(\theta,\psi,\chi)=d^J_{M_JK}(\theta)e^{-M_J\phi}e^{-iK\chi}$ . Since $\mu_{a,e}\neq 0$ ,

$\mu_{a,e}I_a=\mu_{a,e}\int\psi_r^{*'}\psi_rsin\theta cos\chi d\tau_r\propto\frac{1}{2}\int^{2\pi}_0e^{iK'\chi}e^{-iK\chi}(e^{i\chi}+e^{-i\chi})d\chi$

The integral $\int_0^{2\pi}\[e^{i(K'-K+1)\chi}+e^{i(K'-K-1)\chi}\]d\chi$ is nonzero only if either exponent is zero, which happens when $K'-K+1=0$ or $K'-K-1=0$ , i.e. $\Delta K=\pm 1$ . The same logic when applied to $\mu_{b,e}\neq 0$ also results in $\Delta K=\pm 1$ .

Combining eq8 and eq8a, the general vibration-rotation selection rules for a symmetric rotor are:

$\Delta J=0,\pm 1,\Delta K=0,\pm 1,\Delta v=\pm 1\;\;\;\;\;\;\;\;8b$

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Mean molecular energies of non-interacting molecules

The mean molecular energy of a system of non-interacting molecules is defined as the average internal energy per molecule, calculated over all possible quantum states accessible to the molecule at a given temperature.

This average is weighted by the Boltzmann probability of each state and reflects the contributions from translational, rotational, vibrational and electronic motions, depending on the system’s complexity.

Mathematically, the mean molecular energy $\langle \varepsilon\rangle$ of a molecule in thermal equilibrium is given by:

$\langle \varepsilon\rangle=\sum_i\varepsilon_ip_i\;\;\;\;\;\;\;\;300$

where $\varepsilon_i$ is energy the $i$ -th state measured relative to the ground state energy of the molecule, and $p_i$ is the probability of the molecule being in that state.

Substituting eq251 into eq300 gives:

$\langle \varepsilon\rangle=\frac{1}{q}\sum_i\varepsilon_ie^{-\beta\varepsilon_i}\;\;\;\;\;\;\;\;301$

where

$q=\sum_ie^{-\beta\varepsilon_i}$ is the molecular partition function,
$k$ is the Boltzmann constant,
$T$ is the absolute temperature,
$\beta=\frac{1}{kT}$ .

Since $\frac{\partial q}{\partial\beta}=\sum_i\frac{\partial}{\partial\beta}e^{-\beta\varepsilon_i}=-\sum_i\varepsilon_ie^{-\beta\varepsilon_i}$ and $\frac{\partial lnq}{\partial q}=\frac{1}{q}$ , eq301 can be expressed as:

$\langle \varepsilon\rangle=-\frac{1}{q}\frac{\partial q}{\partial\beta}\;\;\;\;\;\;\;\;302$

$\langle \varepsilon\rangle=-\frac{\partial lnq}{\partial\beta}\;\;\;\;\;\;\;\;303$

Question

Why does eq303 (or eq302) involve a partial derivative?

Answer

A partial derivative is used because $q$ may be dependent on a few variables, such as $T$ and $V$ (see eq266).

Substituting eq257 into eq303 yields:

$\langle \varepsilon\rangle=\langle \varepsilon^T\rangle+\langle \varepsilon^R\rangle +\langle \varepsilon^V\rangle+\langle \varepsilon^E\rangle\;\;\;\;\;\;\;\;304$

where $\langle \varepsilon^T\rangle=-\frac{\partial lnq^T}{\partial\beta}$ , $\langle \varepsilon^R\rangle=-\frac{\partial lnq^R}{\partial\beta}$ , $\langle \varepsilon^V\rangle=-\frac{\partial lnq^V}{\partial\beta}$ and $\langle \varepsilon^E\rangle=-\frac{\partial lnq^E}{\partial\beta}$ .

Mean translational energy

The mean translational energy $\langle \varepsilon^T\rangle$ of a non-interacting molecule is derived by substituting eq266 into $\langle \varepsilon^T\rangle=-\frac{\partial lnq^T}{\partial\beta}$ to give:

$\langle \varepsilon^T\rangle=-\frac{\partial ln\biggr\[V\biggr$\frac{2\pi m}{h^2\beta}\biggr$^{\frac{3}{2}}\biggr\]}{\partial\beta}=-\frac{\partial ln\biggr$\frac{1}{\beta}\biggr$}{\partial \beta}^{\frac{3}{2}}=\frac{3}{2}kT\;\;\;\;\;\;\;\;305$

Question

Show that each of the three components of $\langle \varepsilon^T\rangle$ is equal to $\frac{1}{2}kT$ .

Answer

Substituting eq261 into $\langle \varepsilon^T\rangle=-\frac{\partial lnq^T}{\partial\beta}$ results in:

$\langle \varepsilon^T\rangle=-\frac{\partial ln q^T_x}{\partial \beta}-\frac{\partial ln q^T_y}{\partial \beta}-\frac{\partial ln q^T_z}{\partial \beta}\;\;\;\;\;\;\;\;306$

Let’s focus on the first derivative in eq306. Substituting eq265 into it yields:

$-\frac{\partial ln q^T_x}{\partial \beta}=-\frac{\partial ln\biggr\[a\biggr$\frac{2\pi m}{h^2\beta}\biggr$^{\frac{1}{2}}\biggr\]}{\partial\beta}=-\frac{\partial ln\sqrt{\frac{1}{\beta}}}{\partial\beta}=\frac{1}{2}kT$

Similarly, each of the other two components is equal to $\frac{1}{2}kT$ because $q^T_y=\sqrt{\frac{2\pi m}{h^2\beta}}b$ and $q^T_z=\sqrt{\frac{2\pi m}{h^2\beta}}c$ .

Thus, each translational degree of freedom contributes $\frac{1}{2}kT$ , consistent with the equipartition theorem.

Mean rotational energy

The mean rotational energy $\langle \varepsilon^R\rangle$ of a non-interacting heteronuclear linear molecule at low temperatures is derived by substituting eq270 into $\langle \varepsilon^R\rangle=-\frac{1}{q^R}\frac{\partial q^R}{\partial\beta}$ to give:

$\langle \varepsilon^R\rangle=-\frac{1}{\sum_J(2J+1)e^{-hcBJ(J+1)\beta}}\frac{\partial\sum_J(2J+1)e^{-hcBJ(J+1)\beta}}{\partial\beta}\;\;\;\;\;\;\;\;307$

At very low temperatures, almost all molecules occupy the ground state with $J=0$ , so eq307 reduces to $\langle \varepsilon^R\rangle\approx 0$ . As the temperature increases, $\langle \varepsilon^R\rangle$ is approximately given by expanding the summations in eq307 and differentiating to yield:

$\langle \varepsilon^R\rangle=-\frac{hcB(6e^{-2hcB\beta}+30e^{-6hcB\beta}+\cdots)}{1+3e^{-2hcB\beta}+5e^{-6hcB\beta}+\cdots}\;\;\;\;\;\;\;\;308$

At higher temperatures, we substitute eq275, where $\theta_R=\frac{hcB}{k}$ and $\beta=\frac{1}{kT}$ , into $\langle \varepsilon^R\rangle=-\frac{1}{q^R}\frac{\partial q^R}{\partial\beta}$ to give:

$\langle \varepsilon^R\rangle=-\sigma\beta hcB\frac{\partial}{\partial\beta}\frac{1}{\sigma hcB\beta}=kT\;\;\;\;\;\;\;\;309$

which is consistent with the equipartition theorem.

Since the symmetry number $\sigma$ cancels out, the mean rotational energy of a non-interacting symmetrical linear molecule at high temperatures is also $\langle \varepsilon^R\rangle=kT$ . Unlike linear rotors, which have two independent rotational degrees of freedom, spherical rotors, symmetric rotors and asymmetric rotors have three. Substituting eq280, eq286 and eq287 separately into $\langle \varepsilon^R\rangle=-\frac{1}{q^R}\frac{\partial q^R}{\partial\beta}$ and differentiating results in $\langle \varepsilon^R\rangle=\frac{3}{2}kT$ for each of the three types of rotors, which is again consistent with the equipartition theorem.

Mean vibrational energy

The mean vibrational energy $\langle \varepsilon^V\rangle$ of a non-interacting diatomic molecule oscillating harmonically at low temperatures is derived by substituting eq291 into $\langle \varepsilon^V\rangle=-\frac{1}{q^V}\frac{\partial q^V}{\partial\beta}$ , where $\theta_V=\frac{hc\tilde\nu}{k}$ , to give:

$\langle \varepsilon^V\rangle=-\frac{1-e^{-hc\tilde\nu\beta}}{e^{-\frac{1}{2}hc\tilde\nu\beta}}\frac{\partial}{\partial\beta}\frac{e^{-\frac{1}{2}hc\tilde\nu\beta}}{1-e^{-hc\tilde\nu\beta}}\;\;\;\;\;\;\;\;310$

If we compute the derivative in eq310 and multiply the result by $\frac{e^{hc\tilde\nu\beta}}{e^{hc\tilde\nu\beta}}$ , we get:

$\langle \varepsilon^V\rangle=-\frac{1}{2}hc\tilde\nu\frac{e^{hc\tilde\nu\beta}+1}{e^{hc\tilde\nu\beta}-1}\;\;\;\;\;\;\;\;311$

At high temperatures, $hc\tilde\nu\beta\ll 1$ , which allows us to expand $e^{hc\tilde\nu\beta}$ as a Taylor series ( $e^x=1+x+\cdots$ ) to give:

$\langle \varepsilon^V\rangle=\frac{1}{2}hc\tilde\nu\frac{(1+hc\tilde\nu\beta+\cdots)+1}{(1+hc\tilde\nu\beta+\cdots)-1}=kT+\frac{1}{2}hc\tilde\nu\;\;\;\;\;\;\;\;312$

As mentioned in an earlier article, eq291 evaluates absolute energy levels, including the zero-point energy. If we derive $\langle \varepsilon^V\rangle$ using eq293 instead of eq291, we obtain:

$\langle \varepsilon^V\rangle=kT\;\;\;\;\;\;\;\;313$

Although the equipartition theorem states that the mean vibrational energy of a classical oscillator at high temperatures is equal to $kT$ , both eq312 and eq313 are consistent with this result. This is because the theorem accounts only for the thermal contribution to vibrational energy; the zero-point energy term $\frac{1}{2}hc\tilde\nu$ is a quantum mechanical artifact that is independent of temperature.

For polyatomic molecules, each normal mode of vibration behaves approximately like a separate harmonic oscillator, with the total vibrational partition function given by eq294:

$q^V_{total}=\prod^f_{i=1}q^V_i$

where $f=3N-6$ for non-linear molecules, $f=3N-5$ for linear molecules, and $N$ is the number of atoms.

So,

$lnq^V_{total}=ln\prod^f_{i=1}q^V_i=\sum_{i=1}^flnq^V_i$

Since $\langle \varepsilon^V_i\rangle=-\frac{1}{q^V_i}\frac{\partial q^V_i}{\partial\beta}=-\frac{\partial lnq^V_i}{\partial\beta}$ ,

$\langle\varepsilon^V_{total}\rangle=-\frac{\partial lnq^V_{total}}{\partial\beta}=-\frac{\partial \sum_{i=1}^f lnq^V_i}{\partial\beta}=\sum_{i=1}^f\langle\varepsilon^V_i\rangle\;\;\;\;\;\;\;\;314$

In other words, the total mean vibrational energy of a non-interacting polyatomic molecule oscillating harmonically is the sum of the mean energy of each normal mode. This implies that

$\langle\varepsilon^V_{total}\rangle=(3N-6)kT\;\;\;\;\;non-linear\;molecule$

$\langle\varepsilon^V_{total}\rangle=(3N-5)kT\;\;\;\;\;linear\;molecule$

which are consistent with the equipartition theorem at high temperatures.

Mean electronic energy

The mean electronic energy $\langle \varepsilon^E\rangle$ of a non-interacting molecule is derived by substituting eq296, where $\beta=\frac{1}{kT}$ , into $\langle \varepsilon^E\rangle=-\frac{1}{q^E}\frac{\partial q^E_i}{\partial\beta}$ to give:

$\langle \varepsilon^E\rangle=\frac{g_1\varepsilon_1e^{-\varepsilon_1\beta}+g_2\varepsilon_2e^{-\varepsilon_2\beta}+\cdots}{g_0+g_1e^{-\varepsilon_1\beta}+g_2e^{-\varepsilon_2\beta}+\cdots}\;\;\;\;\;\;\;\;315$

At low temperatures, $\beta\rightarrow\infty$ . Therefore, all the numerator terms in eq315 approach zero, giving:

$\langle \varepsilon^E\rangle\approx 0\;\;\;\;\;\;\;\;316$

This can be explained qualitatively: the energy gap between the ground electronic state and the first excited electronic state of a typical molecule is very large, which results in the molecule occupying only the electronic ground state, which has zero energy.

At higher temperatures, the mean electronic energy increases as excited electronic states become thermally accessible. However, this regime is often not reached before the molecule dissociates

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The canonical ensemble

A canonical ensemble is a collection of identical, non-interacting copies of a physical system, each with a fixed number of particles $N$ , fixed volume $V$ , and constant temperature $T$ (maintained by thermal equilibrium with a heat bath), in which particles within each system may interact.

The word “canonical” means “according to a rule”, which, in this case, refers to fixed $N$ , $V$ and $T$ . As mentioned in the above definition, a canonical ensemble clearly states that even though the individual systems in the ensemble are identical from a thermodynamic perspective, they may not be identical at the molecular level. Such a theoretical construct serves to connect the microscopic behaviour of systems (governed by quantum mechanics or classical mechanics) to the macroscopic thermodynamic properties.

Question

Why do we only consider three parameters for each system?

Answer

While a physical system, such as a salt solution, may involve many microscopic variables (like individual particle positions and velocities), its macroscopic behaviour can often be fully characterised by a small set of thermodynamic parameters. In the canonical ensemble, fixing $N$ , $V$ and $T$ is sufficient to statistically describe the equilibrium properties of the system.

All thermodynamic properties are, in essence, measured as averages. Consider each system in the ensemble as a rigid container holding a gas. The pressure of the gas in each system is the time average of countless rapid collisions of gas particles with the walls of the container. The force on the wall fluctuates constantly, but a pressure gauge cannot register each individual fluctuation. Instead, the gauge averages the force over a short time to provide a stable reading. Similarly, while the kinetic energy of individual molecules varies moment to moment, the temperature reflects the average kinetic energy of all molecules at thermal equilibrium.

Due to fluctuations in properties such as pressure, kinetic energy or even quantum-mechanical variables, each system can occupy many possible energy states (microstates) at any given moment. In quantum mechanical systems, the energy corresponding to each microstate is determined by solving the Schrodinger equation $\hat{H}\psi_i=E_i\psi_i$ , where $\psi_i$ is the wavefunction that describes the quantum state resulting from the positions and momenta of all $N$ particles in the system, and $E_i$ is the total energy of system for the $i$ -th microstate.

If the number of systems $\tilde N$ in the ensemble is taken to be infinitely large, we postulate that:

Postulate 1

The macroscopic properties of a system are given by statistical averages over all possible microstates in the ensemble at any given moment (ensemble average).

For example, the thermodynamic internal energy $U$ is equal to the statistical average energy of the systems $\langle E\rangle$ in the ensemble:

$U=\langle E\rangle=\sum_ip_iE_i\;\;\;\;\;\;\;\;120$

where $p_i$ is the probability that a system in the ensemble is in the $i$ -th microstate with energy $E_i$ .

To determine $p_i$ , we further postulate that:

Postulate 2

All microstates with equal energy have equal probability of occurring for any system of fixed volume, composition and temperature.

Therefore, if we consider any such system in thermal equilibrium with a constant temperature heat bath, and identify two of its microstates, 1 and 2, with energies $E_1$ and $E_2$ respectively, the corresponding probabilities are given by $p_1=\frac{\tilde N_1}{\tilde N}=f(E_1)$ and $p_2=\frac{\tilde N_2}{\tilde N}=f(E_2)$ , where $\tilde N_i$ is the number of systems with energy $E_i$ . It follows that the relative probability is:

$\frac{\tilde N_2}{\tilde N_1}=f(E_1,E_2)\;\;\;\;\;\;\;\;121$

In thermodynamics and quantum mechanics, energy is always referenced to a zero of energy $E_0$ , which can be defined in different ways. Since the function $f(E_1,E_2)\equiv f(E_1-E_0,E_2-E_0)$ represents a physical quantity, it must yield the same result regardless of how $E_0$ is defined. This is only possible if the function $f(E_1,E_2)$ depends on the difference between $E_1$ and $E_2$ , i.e. $f(E_1,E_2)= f(E_1-E_2)$ .

Question

Show that the function $f(E_1-E_2)$ is invariant to $E_0$ .

Answer

Let $E_3=E_1-E_0$ and $E_4=E_2-E_0$ . We have $E_3-E_4=E_1-E_2$ , which does not depend on $E_0$ .

Multiplying $\frac{\tilde N_2}{\tilde N_1}=f(E_1-E_2)$ by $\frac{\tilde N_3}{\tilde N_2}=f(E_2-E_3)$ and using $\frac{\tilde N_3}{\tilde N_1}=f(E_1-E_3)$ gives:

$f(E_1-E_3)=f(E_1-E_2)f(E_2-E_3)\;\;\;\;\;\;\;\;122$

Letting $E_5=E_1-E_2$ and $E_6=E_2-E_3$ , we can rewrite eq122 as:

$f(E_5+E_6)=f(E_5)f(E_6)\;\;\;\;\;\;\;\;123$

Taking the natural logarithm on both sides of eq123 and then differentiating with respect to $E_5$ yields:

$\biggr\[\frac{\partial lnf(E_5+E_6)}{\partial E_5}\biggr\]_{E_6}=\frac{d lnf(E_5)}{d E_5}$

Using the chain rule,

$\frac{dlnf(E_5+E_6)}{d(E_5+E_6)}\biggr\[\frac{\partial (E_5+E_6)}{\partial E_5}\biggr\]_{E_6}=\frac{d lnf(E_5)}{d E_5}$

$\frac{dlnf(E_5+E_6)}{d(E_5+E_6)}=\frac{d lnf(E_5)}{d E_5}\;\;\;\;\;\;\;\;124$

Similarly, taking the natural logarithm on both sides of eq123 and then differentiating with respect to $E_6$ yields:

$\frac{dlnf(E_5+E_6)}{d(E_5+E_6)}=\frac{d lnf(E_6)}{d E_6}\;\;\;\;\;\;\;\;125$

Equating eq124 and eq125 results in:

$\frac{d lnf(E_5)}{d E_5}=\frac{d lnf(E_6)}{d E_6}\;\;\;\;\;\;\;\;126$

Each side of eq126 is a function of a different independent variable. The only way eq126 can be valid for all values of $E_5$ and $E_6$ is if both functions are equal to the same constant $\beta$ :

$\frac{d lnf(E_5)}{d E_5}=\frac{d lnf(E_6)}{d E_6}=\beta\;\;\;\;\;\;\;\;127$

Integrating $\frac{d lnf(E_5)}{d E_5}=\beta$ , i.e. $\int d lnf(E_5)=\int\beta dE_5$ , gives $lnf(E_5)=\beta E_5+c$ or equivalently,

$f(E_5)=ae^{\beta E_5}\;\;\;\;\;\;\;\;128$

where $a=e^c$ .

Accordingly, we have:

$f(E_h-E_i)=ae^{\beta (E_h-E_i)}\;\;\;\;\;\;\;\;129$

Substituting eq129 into $\frac{\tilde N_i}{\tilde N_h}=f(E_h-E_i)$ yields $\frac{\tilde N_i}{\tilde N_h}=ae^{\beta (E_h-E_i)}$ or:

$\tilde N_i=Ce^{-\beta E_i}\;\;\;\;\;\;\;\;130$

where $C=\tilde N_hae^{\beta E_h}$ .

$C$ can be evaluated by summing both sides of eq130 over $i$ :

$\sum_i\tilde N_i=C\sum_ie^{-\beta E_i}$

Since $\sum_i\tilde N_i=\tilde N$ ,

$C=\frac{\tilde N}{\sum_ie^{-\beta E_i}}\;\;\;\;\;\;\;\;131$

Substituting eq131 back into eq130 results in:

$p_i=\frac{e^{-\beta E_i}}{\sum_ie^{-\beta E_i}}\;\;\;\;\;\;\;\;133$

where $p_i=\frac{\tilde N_i}{\tilde N}$ .

Eq133 represents the general probability of finding a system, which is in thermal equilibrium with a constant temperature heat bath, in a specific microstate $i$ with energy $E_i$ . Its derivation relies on fundamental principles of probability and the nature of equilibrium, not on the specific nature of the particles themselves (e.g., whether they’re distinguishable or indistinguishable). In other words, eq133 is applicable to both classical and quantum mechanical systems. Therefore, eq120 becomes:

$U=\langle E\rangle=\frac{\sum_iE_ie^{-\beta E_i}}{Q}\;\;\;\;\;\;\;\;134$

where

$Q=\sum_je^{-\beta E_j}\;\;\;\;\;\;\;\;135$

is known as the canonical partition function.

The fact that eq133 yields the same expression as the Boltzmann distribution is not a coincidence. It demonstrates that the Boltzmann distribution naturally emerges from the statistical treatment of systems in thermal equilibrium, and illustrates the deep consistency between thermodynamic principles and statistical reasoning.

Question

Can two different microstates have the same energy?

Answer

Yes. Consider a system with two particles. In one microstate, particle A has an energy of 1 unit, and Particle B has an energy of 2 units. In another microstate, particle A has an energy of 2 units, while Particle B has an energy of 1 unit. These two distinct microstates (or configurations) clearly have the same total energy of 3 units. Microstates that share the same energy are called degenerate states.

Since different microstates can be degenerate, the probability that a system has energy $E_i$ is:

$p(E_i)=\frac{g_ie^{-\beta E_i}}{\sum_je^{-\beta E_j}}\;\;\;\;\;\;\;\;136$

where $g_i$ is the degeneracy of $E_i$ .

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Canonical partition function for a system of non-interacting particles

The canonical partition function is a central quantity in statistical mechanics, encapsulating the thermodynamic behaviour of a system in equilibrium with a heat reservoir at fixed temperature. It allows us to derive key thermodynamic quantities of a system, such as pressure (see eq146) and internal energy (see eq148). These relationships, e.g. $U=-\biggr$\frac{\partial Q}{\partial \beta}\biggr$_{V,N}$ , applies to both systems with interacting and non-interacting particles.

For a system with interacting particles, the Schrödinger equation is incredibly complex due to the interaction terms between the $N$ particles, making it difficult to calculate the exact energy $E_j$ of a given microstate . On the other hand, in a system of non-interacting particles, intermolecular forces are absent. This simplifies the evaluation of $Q$ significantly, since the Hamiltonian operator for a system of $N$ non-interacting particles can be expressed as $\hat{H}=\hat{H}_1+\hat{H}_2+\cdots+\hat{H}_N$ . The total energy of the system then becomes the sum of the individual particle energies,

$E_j=\varepsilon^j_{1,m}+\varepsilon^j_{2,n}+\cdots+\varepsilon^j_{N,y}\;\;\;\;\;\;\;\;160$

Here $=\varepsilon^j_{1,m}$ refers to the energy of particle 1 in quantum state $=m$ when the system is in microstate $j$ . The energy of each particle is determined by solving the corresponding one-particle Schrödinger equation, e.g. $\hat{H}_1\psi^j_{1,m}=\varepsilon_{1,m}\psi^j_{1,m}$ .

If the non-interacting particles are distinguishable (e.g. atoms or molecules with fixed positions in a solid), then the microstate of the system is defined by the quantum state of each individual particle. For example, microstate 1 might have the configuration $\langle m,n,\cdots,y\rangle$ , while microstate 2 is $\langle n,m,\cdots,y\rangle$ . The sum over all microstates $j$ of the system, by substituting eq160 into eq155, is given by:

$Q=e^{-\frac{E_1}{kT}}+e^{-\frac{E_2}{kT}}+\cdots=e^{-\frac{\varepsilon^1_{1,m}+\varepsilon^1_{2,n}+\cdots}{kT}}+e^{-\frac{\varepsilon^2_{1,n}+\varepsilon^2_{2,m}+\cdots}{kT}}+\cdots$

which is equivalent to:

$Q=e^{-\frac{\varepsilon^1_{1,1}}{kT}}e^{-\frac{\varepsilon^1_{2,2}}{kT}}\cdots+e^{-\frac{\varepsilon^2_{1,2}}{kT}}e^{-\frac{\varepsilon^2_{2,1}}{kT}}\cdots+\cdots\;\;\;\;\;\;\;\;161$

where we have set $m=1$ and $n=2$ for illustrative purposes.

Each term on the RHS of eq161 corresponds to a distinct microstate, represented by a specific permutation of $N$ exponential factors. Eq161 can also be written as:

$Q=\sum_me^{-\frac{\varepsilon^m_{1,m}}{kT}}\sum_ne^{-\frac{\varepsilon^n_{2,n}}{kT}}\cdots\sum_ye^{-\frac{\varepsilon^y_{N,y}}{kT}}\;\;\;\;\;\;\;\;162$

We can show that eq162 is equivalent to eq161 by expanding it:

$Q=\biggr$e^{-\frac{\varepsilon^1_{1,1}}{kT}}+e^{-\frac{\varepsilon^2_{1,2}}{kT}}+\cdots\biggr$\biggr$e^{-\frac{\varepsilon^1_{2,1}}{kT}}+e^{-\frac{\varepsilon^2_{2,2}}{kT}}+\cdots\biggr$\cdots\;\;\;\;\;\;\;\;163$

If we multiply out all the terms in eq163, each resulting term corresponds to a unique microstate, as expressed in eq161. Therefore,

$Q=\sum_je^{-\frac{E_j}{kT}}=q_1q_2\cdots q_N\;\;\;\;\;\;\;\;164$

where $q_1=\sum_me^{-\frac{\varepsilon^m_{1,m}}{kT}}$ , $q_2=\sum_ne^{-\frac{\varepsilon^n_{2,n}}{kT}}$ ,…, $q_N=\sum_ye^{-\frac{\varepsilon^y_{N,y}}{kT}}$ are known as the molecular partition functions.

Since each $q_N$ represents the total statistical weight of a single particle, the superscript $y$ in $\varepsilon^y_{N,y}$ becomes redundant. Thus, we can express a molecular partition function more simply as $q_N=\sum_ye^{-\frac{\varepsilon_{N,y}}{kT}}$ . If $N_1$ molecules are of the same species, where $N_1 < N$ , then

$Q=(q_1)^{N_1}q_2\cdots q_{N-N_1}\;\;\;\;\;\;\;\;165$

Question

Using the example of a system of 3 distinguishable, identical, non-interacting particles, show that the sum over all microstates of the system then becomes the product of the separate sums of the quantum states of each particle.

Answer

Suppose each particle can be in one of two quantum states with energy $\varepsilon_{0}=0$ and $\varepsilon_{1}=\varepsilon$ . The total possible microstates is $2^3=8$ , comprising of:

Microstate	Total energy
$\langle 0,0,0\rangle$	$0$
$\langle \varepsilon,0,0\rangle$	$\varepsilon$
$\langle 0,\varepsilon,0\rangle$	$\varepsilon$
$\langle 0,0,\varepsilon\rangle$	$\varepsilon$
$\langle \varepsilon,\varepsilon,0\rangle$	$2\varepsilon$
$\langle \varepsilon,0,\varepsilon\rangle$	$2\varepsilon$
$\langle 0,\varepsilon,\varepsilon\rangle$	$2\varepsilon$
$\langle \varepsilon,\varepsilon,\varepsilon\rangle$	$3\varepsilon$

So,

$Q=1+3e^{-\frac{\varepsilon}{kT}}+3e^{-\frac{2\varepsilon}{kT}}+e^{-\frac{3\varepsilon}{kT}}$

Similarly,

$Q=q^3=\biggr$1+e^{-\frac{\varepsilon}{kT}}\biggr$^3=1+3e^{-\frac{\varepsilon}{kT}}+3e^{-\frac{2\varepsilon}{kT}}+e^{-\frac{3\varepsilon}{kT}}$

It follows that for $N$ distinguishable, identical, non-interacting particles,

$Q=q^N\;\;\;\;\;\;\;\;166$

If the $N$ identical, non-interacting particles are indistinguishable, such as those of an ideal gas, and if each particle occupies a distinct quantum state, then the total number of microstates is no longer given by the number of permutations of the quantum states, but by the number of combinations. One way to approximate $Q$ in this case is to divide eq166 by a factor $N!$ :

$Q=\frac{q^N}{N!}\;\;\;\;\;\;\;\;167$

This gives $Q_{indist}=\frac{Q_{dist}}{N!}$ at high temperatures, where each particle is assumed to be in a distinct quantum state. It is important to note that the relation $Q_{indist}=\frac{Q_{dist}}{N!}$ is not a strict mathematical identity, because $Q_{dist}$ is not a literal count of microstates but a weighted sum of exponential terms, each corresponding to a microstate. Nevertheless, this approximation works well in the classical limit and successfully resolves the Gibbs paradox.

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Thermodynamic functions involving the canonical partition function

Thermodynamic functions involving the canonical partition function describe thermodynamic properties of a canonical ensemble.

In statistical mechanics, the canonical ensemble provides a powerful framework for connecting microscopic states to macroscopic thermodynamic properties. Central to this approach is the canonical partition function, $Q$ , which encapsulates the statistical behaviour of a system in thermal equilibrium at fixed temperature, volume, and particle number. From $Q$ , one can derive key thermodynamic functions such as pressure, internal energy and entropy through straightforward mathematical relations.

In the previous article, we derived the expressions for pressure and internal energy:

$P=\frac{1}{\beta}\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}\;\;\;\;\;\;\;\;146$

$U=-\biggr$\frac{\partial lnQ}{\partial \beta}\biggr$_{V,N}\;\;\;\;\;\;\;\;148$

where $\beta=\frac{1}{kT}$ and $Q=\sum_je^{-\frac{E_j}{kT}}$ .

Question

How does $P=\frac{1}{\beta}\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}$ connect quantum mechanics to thermodynamics?

Answer

$P$ , the pressure of a system, is a macroscopic thermodynamic property, while $E_i$ is the energy of a quantum mechanical microstate, which is an eigenvalue of the Hamiltonian. Therefore, the equation provides a bridge between the microscopic quantum description and macroscopic thermodynamic observables.

The corresponding expression for entropy $S$ can be derived by first dividing the fundamental equation of thermodynamics by $T$ to give:

$dS=d\biggr$\frac{U}{T}\biggr$+\frac{U}{T^2}dT+\frac{P}{T}dV\;\;\;\;\;\;\;\;156$

Applying the chain rule to eq148 gives:

$U=-\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}\biggr$\frac{d T}{d \beta}\biggr$=kT^2\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}\;\;\;\;\;\;\;\;157$

Substituting eq146 and eq157 into eq156 yields:

$\begin{align}dS&=d\biggr$\frac{U}{T}\biggr$+k\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}dT+k\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}dV\\&=d\biggr$\frac{U}{T}\biggr$+kdlnQ\;\;\;\;\;\;\;\;158\end{align}$

Integrating eq158 results in $S=\frac{U}{T}+klnQ+C$ , where $C$ is a constant. As $T\rightarrow 0$ , only the ground state $E_0$ is populated. Assuming the ground state is non-degenerate, the partition function becomes $Q=e^{-\frac{E_0}{kT}}$ , with $S=\frac{E_0}{T}-\frac{E_0}{T}+C=C$ . According to the third law of thermodynamics, entropy approaches zero as the temperature approaches absolute zero. Therefore, $C$ must be zero to satisfy this requirement, resulting in:

$dS=\frac{U}{T}+klnQ=kT\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}+klnQ\;\;\;\;\;\;\;\;159$

It follows that the Helmholtz energy expression is:

$A=U-TS=-kTlnQ\;\;\;\;\;\;\;\;159a$

Using the pressure and internal energy equations, the expression for enthalpy is:

$H=U+PV=KT^2\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}+VkT\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}\;\;\;\;\;\;\;\;159b$

Question

Does eq159b contradict the application of enthalpy, which describes a system at constant pressure?

Answer

The definition of enthalpy, $H=U+PV$ , holds for any thermodynamic state, regardless of whether the pressure is constant. The condition of constant pressure becomes relevant when evaluating the change in enthalpy, $\Delta H$ , which equals the heat exchanged in a constant-pressure process $q_p$ . Eq159b expresses the enthalpy of a system in terms of $N$ , $V$ and $T$ . It gives the enthalpy for a particular equilibrium state, not just under constant pressure. Therefore, we can calculate the enthalpy of a system at any given state using eq159b. By evaluating it at two different states (e.g. the initial and final states of a process), we obtain $\Delta H$ . If the process connecting those states occur at constant pressure, then $\Delta H=q_p$ . So, there is no contradiction.

The expression for constant-volume heat capacity is:

$\begin{align}C_V&=\biggr$\frac{\partial U}{\partial T}\biggr$_{V,N}\\&=k\biggr\[\frac{\partial T^2\biggr$\frac{\partial lnQ}{\partial T}\biggr$}{\partial T}\biggr\]_{V,N}\\&=KT^2\biggr$\frac{\partial^2 lnQ}{\partial^2 T}\biggr$_{V,N}+2kT\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}\;\;\;\;\;\;\;\;159c\end{align}$

For Gibbs energy, we have $G=H-TS=U+PV-TS=A+PV$ . Hence,

$G=-kTlnQ+KTV\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}\;\;\;\;\;\;\;\;159d$

If we define chemical potential as $\mu=\biggr$\frac{\partial A}{\partial N}\biggr$_{T,V}$ , and $N$ is an integer-valued discrete variable, then the derivative becomes a finite difference, so:

$\mu\approx A_N-A_{N-1}=-kTln\biggr$\frac{Q_N}{Q_{N-1}}\biggr$\;\;\;\;\;\;\;\;159e$

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