Advanced Chemistry

June 9, 2026

Magnetic properties of a solid

The magnetic properties of a solid arise primarily from the spin magnetic moments of electrons and the quantum-mechanical interactions that govern their ordering within the material.

In classical electrodynamics, the relation between the magnetic dipole moment $\boldsymbol{\mathit{\mu}}$ and the angular momentum $\boldsymbol{\mathit{L}}$ is given by (see this article for derivation):

$\boldsymbol{\mathit{\mu}}=\gamma\boldsymbol{\mathit{L}}$

where $\gamma=\frac{q}{2m}$ is the classical gyromagnetic ratio.

When transitioning from classical electrodynamics to quantum mechanics, these physical observables are represented by operators: $\hat{\mu}=\gamma\hat{L}$ . It follows that the magnitude of an atom’s magnetic moment depends on $\gamma\propto \frac{1}{m}$ . Consequently, even though protons and neutrons possess spin angular momentum, their contributions to the magnetic moment are much smaller than those of electrons because $m_p=m_n\approx 1836m_e$ .

Question

Does the orbital angular momentum of an electron contribute to the magnetic moment of an atom?

Answer

Yes, it does. However, the contribution of orbital angular momentum to the magnetic moment is about half that of spin angular momentum. For an electron, the gyromagnetic ratio can be written as $\gamma=-g \frac{e}{2m_e}$ , where $g$ is the $g$ -factor. Since orbital angular momentum arises from the electron’s spatial degrees of freedom, its associated gyromagnetic ratio has the classical form, corresponding to $g=1$ . In contrast, spin angular momentum is an intrinsic quantum property of the electron, with $g$ experimentally determined to be approximately 2.

According to the Pauli Exclusion Principle, two electrons occupying the same atomic orbital have opposite spins, and therefore opposite spin magnetic moments. This is why isolated atoms with one or more unpaired electrons, such as Ag and Fe, have a net magnetic moment and are deflected in an external magnetic field.

In a solid, the situation is more complicated because atoms are no longer isolated. Instead, their outer electrons are influenced by neighbouring atoms, and the atomic orbitals broaden into energy bands, with electrons becoming delocalised throughout the crystal. Even when individual atoms carry magnetic moments, these moments do not automatically align throughout the material because the formation of magnetic order is governed by a competition between different energy contributions, in particular exchange interactions and magnetostatic energy.

The exchange interaction (also known as exchange force) is a purely quantum-mechanical effect arising from the combination of the Pauli exclusion principle and electron–electron Coulomb repulsion. As shown in a previous article, the total energy of a two-electron state depends on the symmetry of the spatial wavefunction. Due to exchange effects arising from the Pauli exclusion principle, states with parallel spins (antisymmetric spatial wavefunction) can have lower energy than those with antiparallel spins (symmetric spatial wavefunction), making the parallel configuration more stable in cases where the exchange interaction is positive. However, if a large region of the material were uniformly magnetised, it would produce a strong external stray magnetic field. This field is energetically costly because it contributes magnetostatic (demagnetising) energy.

To minimise the total energy, the material breaks up into microscopic regions called magnetic domains during formation (see diagram above). Within each domain, exchange interactions favour a roughly uniform spin alignment. Across the material, however, different domains may point in different directions, reducing the net external field and thereby lowering the magnetostatic energy. Furthermore, spins are not perfectly rigidly aligned within a domain, and small deviations can occur due to thermal fluctuations. When an external magnetic field is applied, domain walls can move so that domains aligned with the field grow at the expense of others, producing a net magnetisation of the material.

The measure of how strongly a material becomes magnetised in response to an external magnetic field strength $H$ is called the magnetic susceptibility $\chi$ of the material. Mathematically,

$\chi=\frac{M}{H}$

where the magnetisation $M$ is the magnetic moment per unit volume of the material.

In general, the magnetic behaviour of solids depends on how how electron spins respond to exchange interactions and external fields:

- Ferromagnetism: In materials such as Fe, Co and Ni, exchange interactions favour parallel alignment. Domain walls can move or domains can rotate under an external field, leading to strong magnetisation. In many ferromagnets, the magnetisation is hysteretic, meaning it depends on the history of the applied field: even after the external field is removed, a residual magnetisation (remanence) can remain (see diagram below). Ferromagnets also lose their permanent magnetic properties above certain temperatures (Curie temperature or Curie point) and become paramagnetic because the higher thermal energy causes greater atomic vibrations, disrupting spin alignments. For example, Fe has a Curie temperature of approximately 770°C. Below this temperature, Fe is ferromagnetic and can be permanently magnetised. Above it, Fe becomes paramagnetic and only exhibits weak magnetism in the presence of an external magnetic field.

- Antiferromagnetism: In materials such as MnO and NiO, the atoms are arranged with the oxide ion positioned between two metal ions, forming an M²⁺–O²⁻–M²⁺ linkage. The oxygen 2p orbital oriented along the M²⁺–O²⁻–M²⁺ bond axis contains a pair of electrons that are antiparallel, as required by the Pauli exclusion principle. One electron in this bridging orbital interacts with the metal ion on the left, while the other interacts with the metal ion on the right. Because the electron pair occupying this specific oxygen orbital is initially antiparallel, the exchange interactions mediated through the oxygen ion favour an antiparallel alignment of the neighbouring metal-ion spins, with zero net magnetisation in the absence of an external magnetic field. However, the system can still respond weakly to external fields due to thermal fluctuations.

- Ferrimagnetism: In materials such as Fe₃O₄ (magnetite), the spins of neighbouring Fe ions are aligned antiparallel, as in antiferromagnets, but the opposing magnetic moments are unequal. This is because Fe₃O₄ can be represented as Fe³⁺[Fe²⁺Fe³⁺]O₄, with Fe³⁺ occupying tetrahedral holes and both Fe³⁺ and Fe²⁺ occupying octahedral holes (see diagram above). Because Fe³⁺ and Fe²⁺ ions are present in a 2:1 ratio, the magnetic moments of the Fe³⁺ ions largely cancel one another, leaving an uncompensated magnetic moment arising from the Fe²⁺ ions. As a result, the material possesses a non-zero net magnetisation and exhibits domain behaviour and hysteresis similar to those of ferromagnets. In fact, the strongest permanent magnets, such as neodymium magnets, exhibit ferrimagnetism.

- Paramagnetism: In materials such as Al and Pt (including O₂ gas), atomic magnetic moments are largely independent in the absence of strong exchange interactions. They align weakly and only temporarily with an applied magnetic field due to thermal agitation, and the magnetisation disappears when the field is removed.

- Diamagnetism: In materials such as Cu, Ag and Au (including H₂O), there are no permanent magnetic moments from unpaired electrons. An applied magnetic field induces small circulating currents that oppose the applied field, producing a weak, negative magnetisation that is present only while the field is applied. In fact, all materials exhibit diamagnetism to some extent due to Lenz’ law. In paramagnetic and ferromagnetic materials, however, this effect is usually overwhelmed by stronger magnetic mechanisms. Because diamagnetic materials are repelled by magnetic fields, they can be levitated in sufficiently strong magnetic field gradients. A notable example is the levitation of superconductors, which behave as perfect diamagnets due to the Meissner effect.

Question

Explain the hysteresis loop in detail.

Answer

Consider a piece of iron with randomly oriented magnetic domains such that $M=0$ . As $H$ increases from zero, the domains begin to align with the applied field. This initial magnetisation is represented by the dotted curve, which rises rapidly and reaches saturation magnetisation $+M_s$ when all domains are aligned. A common misconception is that $M$ returns to zero when $H$ is subsequently reduced. Instead, the material retains a magnetisation known as the remanent magnetisation $+M_r$ , because the aligned domains remain energetically stable, and only some domains reverse as $H$ decreases to zero. If a negative magnetic field is then applied, more domains begin to reverse and $M$ decreases to zero at $-H_c$ (the coercive field). Beyond this point, a sufficiently strong negative field aligns all domains in the opposite direction, producing the negative saturation magnetisation $-M_s$ . When the negative field is subsequently reduced back to zero, a remanent magnetisation of $-M_r$ remains. Next, increasing $H$ in the positive direction causes the domains to reverse once again. The magnetisation returns to zero at $+H_c$ , and further increases in $H$ eventually bring the material back to the positive saturation magnetisation $+M_s$ , thereby completing the hysteresis loop.

The area enclosed by the hysteresis loop represents the energy dissipated as heat during one complete magnetisation cycle. Consequently, transformer cores are typically made from materials with narrow hysteresis loops to minimise energy losses, whereas permanent magnets are designed to have wide hysteresis loops and large coercivities so that they retain their magnetisation more effectively.

Ultimately, a solid’s magnetic identity is not a static property, but a dynamic equilibrium dictated by this delicate balance between quantum-scale spin ordering and large-scale thermodynamic and electrostatic forces.

Question

What properties of Fe, Co and Ni, compared to Al and Pt, make them ferromagnetic?

Answer

Fe, Co and Ni have partially filled spatially compact 3d orbitals that form smaller overlaps with each other, leading to narrower bands. These narrower bands consist of a large number of available electronic states concentrated near the Fermi level, giving a high density of states (see diagram below). In contrast, the 5d orbitals in Pt are more spatially extended and overlap more strongly, producing broader bands with a lower density of states at the Fermi level. Similarly, the valence s–p bands in Al are relatively broad and have a low density of states at the Fermi level. Consequently, the energy gained from exchange interactions in Fe, Co and Ni through parallel spin alignment exceeds the energy cost associated with the redistribution of electrons between states (the kinetic energy cost of spin polarisation), leading to spontaneous ferromagnetic ordering. An applied magnetic field can then readily enlarge favourably-oriented domains and rotate domain magnetisations, producing a strong net magnetisation.

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June 9, 2026

Neodymium magnets

Neodymium magnets are permanent magnets made primarily from an alloy of neodymium, iron and boron (NdFeB), and are the strongest commercially available type of permanent magnet.

Pure iron is strongly ferromagnetic because of exchange interactions among its 3d electrons. However, despite its large magnetisation, it is not an ideal permanent magnet because its magnetic domains can reorient relatively easily in response to external magnetic fields, reflecting its weak magnetocrystalline anisotropy.

When iron is alloyed with rare-earth elements such as neodymium, additional exchange interactions arise between the rare-earth and iron atoms. Although the 4f orbitals of rare-earth elements are partially filled and carry substantial magnetic moments, they are spatially compact and lie deep within the atom. Consequently, the 4f electrons are highly localised and do not interact directly with the 3d electrons of iron. Instead, the coupling occurs through an indirect exchange mechanism involving the rare-earth 5d electrons.

This mechanism is a two-step coupling process. While the 5d shell of an isolated rare-earth atom is empty in its ground state, the formation of a metallic crystal lowers the energy of the 5d-derived states through bonding interactions, allowing some valence electrons to occupy them. In the first step, the localised 4f electrons interact through strong intra-atomic exchange with these more spatially extended 5d electrons, causing the spins of the 4f and 5d electrons to align (see diagram below).

In the second step, the diffuse rare-earth 5d orbitals overlap with the 3d orbitals of neighbouring iron atoms, lowering the total energy of the crystal. This resulting exchange interaction is governed by the Pauli exclusion principle, which restricts how electrons of the same spin can occupy overlapping states. The lowest-energy configuration generally corresponds to antiparallel alignment between the rare-earth 5d spin and the iron 3d spin. Since the rare-earth 4f and 5d spins are already aligned from the first step, it follows that the rare-earth 4f spin and the iron 3d spin tend to align antiparallel. In this way, the localised 4f moments of the rare-earth ions become indirectly coupled to the iron sublattice.

However, because the total magnetic moment of a rare-earth ion is contingent on both spin and orbital angular momentum, the resulting alignment of the rare-earth magnetic moment relative to iron depends on Hund’s rules.

Consider the following rare-earth electron configurations, noting that the 4f subshell can accommodate up to 14 electrons:

Nd: [Xe]4f⁴6s² (less than half-filled 4f subshell)

Tb: [Xe]4f⁹6s² (more than half-filled 4f subshell)

Dy: [Xe]4f¹⁰6s² (more than half-filled 4f subshell)

According to Hund’s third rule, the ground state of Nd is characterised by J = L − S, whereas those of Tb and Dy are characterised by J = L + S. Consequently, the total Nd magnetic moment is oriented opposite to its spin angular momentum, while the Tb and Dy total magnetic moments are aligned with their spins. Since the Nd 4f spin is coupled antiparallel to the Fe 3d spin through indirect exchange mediated by the rare-earth 5d states, the total magnetic moment of Nd aligns parallel to that of iron, resulting in ferrimagnetic alignment between the Nd and Fe sublattices. In contrast, the total magnetic moments of Tb and Dy align antiparallel to the magnetic moment of iron, producing ferrimagnetic ordering in Tb–Fe and Dy–Fe systems with a reduced net magnetic moment.

Nevertheless, Nd and Fe alone do not readily form a stable crystal structure. This issue is resolved by adding boron to produce the inter-metallic compound Nd₂Fe₁₄B. In this phase, boron plays a crucial structural role in stabilising a complex tetragonal lattice that accommodates both a high density of Fe atoms and well-ordered Nd sites, resulting in a crystal with strong exchange-driven magnetisation and large magnetocrystalline anisotropy.

In practice, neodymium magnets are engineered so that the Nd₂Fe₁₄B phase constitutes the majority of the material. The remaining volume (10–15%) consists primarily of Nd-rich grain-boundary phases that separate neighbouring magnetic grains (see diagram above). If neighbouring Nd₂Fe₁₄B grains were strongly exchange-coupled throughout the entire magnet, reversal of the magnetic moment in one grain could more easily propagate into adjacent grains, reducing coercivity (i.e. the resistance to demagnetisation). By partially isolating neighbouring grains, the Nd-rich boundary phase inhibits the propagation of reversed magnetic domains and significantly improves coercivity.

Neodymium magnets are typically manufactured using a powder-metallurgy process. An alloy containing neodymium, iron and boron is first melted and cast, then crushed into a fine powder. The powder particles are aligned in a strong magnetic field so that their crystallographic-easy axes point in the same direction, maximising the magnetic performance of the final magnet. The aligned powder is then compacted and sintered at high temperature to form a dense solid. Subsequent heat-treatment steps optimise the microstructure by promoting the formation of Nd-rich grain-boundary phases that enhance coercivity. In addition, small amounts of Dy and Tb are often incorporated into the alloy to further increase coercivity and improve magnetic stability at elevated operating temperatures. Finally, the magnet is machined to the required shape, coated to improve corrosion resistance, and magnetised using a strong external magnetic field.

Applications that require strong magnetic fields in a compact volume often rely on neodymium magnets. Examples include electric vehicle traction motors, wind turbine generators, computer hard disk drives, magnetic resonance imaging (MRI) systems, industrial actuators, and consumer electronics such as headphones, loudspeakers and smartphones.

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Parity of spatial wavefunctions

Parity of spatial wavefunctions is a property describing how a wavefunction behaves under inversion, being classified as even (g) if it remains unchanged or odd (u) if it changes sign.

Consider the hydrogenic wavefunction $\psi=R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)$ , where $R_{n,l}(r)$ and $Y_l^{\vert m_l\vert}(\theta,\phi)=P_l^{\vert m_l\vert}(cos\theta)e^{im_l\phi}$ are the radial and angular wavefunctions respectively. In spherical coordinates, an inversion ( $\hat{i}=\hat{\sigma}\hat{C_2}$ ) through the origin results in:

- $r\rightarrow r$ (the radius is always positive)
- $\theta\rightarrow \pi-\theta$ (polar angle reflects across the $xy$ -plane)
- $\phi\rightarrow\phi+\pi$ (azimuthal angle rotates by half a circle)

Therefore,

$\hat{i}R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)=\hat{\sigma}R_{n,l}(r)P_l^{\vert m_l\vert}(cos\theta)e^{im_l(\phi+\pi)}=\hat{\sigma}R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)e^{im_l\pi}$

Since $e^{im_l\pi}=(e^{i\pi})^{m_l}=(-1)^{m_l}$ ,

$\hat{i}R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)=(-1)^{m_l}R_{n,l}(r)\hat{\sigma}Y_l^{\vert m_l\vert}(\theta,\phi)$

$P_l^{\vert m_l\vert}(cos\theta)$ are the associated Legendre polynomials given by $P_l^{\vert m_l\vert}(cos\theta)=\frac{1}{2^ll!}(1-cos^2\theta)^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{dcos\theta^{l+\vert m_l\vert}}(cos^2\theta-1)^l$ , or equivalently, $P_l^{\vert m_l\vert}(x)=\frac{1}{2^ll!}(1-x^2)^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{dx^{l+\vert m_l\vert}}(x^2-1)^l$ , where $x=cos\theta$ . So,

$\hat{\sigma}P_l^{\vert m_l\vert}(x)=P_l^{\vert m_l\vert}(-x)=\frac{1}{2^ll!}[1-(-x)^2]^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{d(-x)^{l+\vert m_l\vert}}[(-x)^2-1]^l$

Question

Prove by induction the expression $\frac{d^n}{d(-x)^n}=(-1)^n\frac{d^n}{dx^n}$ .

Answer

For $n=1$ , let $u=-x$ . Applying the chain rule, $\frac{d}{du}=\frac{dx}{du}\frac{d}{dx}=(-1)\frac{d}{dx}$ . So, $\frac{d}{d(-x)}=\frac{d}{du}=(-1)\frac{d}{dx}$ and the expression is valid. Assume the expression holds for some $n$ , i.e. $\frac{d^n}{d(-x)^n}=(-1)^n\frac{d^n}{dx^n}$ . Then for $n+1$ ,

$\frac{d^{n+1}}{d(-x)^{n+1}}=\frac{d}{d(-x)}\frac{d^n}{d(-x)^n}=(-1)^n\frac{d}{d(-x)}\frac{d^n}{dx^n}=(-1)^{n+1}\frac{d^{n+1}}{dx^{n+1}}$

By mathematical induction, $\frac{d^{n}}{d(-x)^{n}}=(-1)^{n}\frac{d^{n}}{dx^{n}}$ .

It follows that

$\hat{\sigma}P_l^{\vert m_l\vert}(x)=(-1)^{l+\vert m_l\vert}\frac{1}{2^ll!}[1-x^2]^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{dx^{l+\vert m_l\vert}}(x^2-1)^l=(-1)^{l+\vert m_l\vert}P_l^{\vert m_l\vert}(x)$

and

$\hat{i}\psi=(-1)^{m_l}(-1)^{l+\vert m_l\vert}\psi=(-1)^{l+2\vert m_l\vert}\psi=(-1)^l\psi$

Hence, the parity of hydrogenic wavefunctions is given by $(-1)^l$ , where $\psi$ is even if $l$ is even and odd if $l$ is odd. For an $N$ -electron atom with a wavefunction expressed as a product of hydrogenic orbitals $\Psi(\boldsymbol{\mathit{r}}_1,\cdots,\boldsymbol{\mathit{r}}_N)=\psi_1(\boldsymbol{\mathit{r}}_1)\cdots\psi_N(\boldsymbol{\mathit{r}}_N)$ , its parity is

$\prod^N_i(-1)^{l_i}=(-1)^{\sum^N_il_i}$

because $\hat{i}\Psi(\boldsymbol{\mathit{r}}_1,\cdots,\boldsymbol{\mathit{r}}_N)=\Psi(-\boldsymbol{\mathit{r}}_1,\cdots,-\boldsymbol{\mathit{r}}_N)=\psi_1(-\boldsymbol{\mathit{r}}_1)\cdots\psi_N(-\boldsymbol{\mathit{r}}_N)=\hat{i}\psi_1(\boldsymbol{\mathit{r}}_1)\cdots\hat{i}\psi_N(\boldsymbol{\mathit{r}}_N)$ .

So, the wavefunction is even if $\sum^N_il_i$ is even and odd if $\sum^N_il_i$ is odd.

If the multielectron wavefunction is properly antisymmetrised, such as in a Slater determinant, each term in the determinant is a permutation of the same set of one-electon orbitals. Since parity depends only on the set of occupied orbitals and not their ordering, every permutation acquires the same overall factor $(-1)^{\sum^N_il_i}$ . Therefore, the parity of the Slater determinant remains $(-1)^{\sum^N_il_i}$ .

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May 16, 2026

Continuous point groups

Continuous point groups are symmetry groups described by continuous transformations (rather than discrete ones), where operations like rotations can vary smoothly and form a continuum of symmetries about a fixed point.

Examples include:

- SO(3): the group of all possible rotations about a point in three-dimensional space, with no reflections or inversions. It represents full rotational symmetry of a sphere and is a fundamental example of a continuous symmetry group in chemistry, physics and mathematics. This symmetry is a good approximation for spherically symmetric systems such as isolated atoms or idealised spherical tops.

- $C_{\infty v}$ : contains all possible rotations about an axis (infinite-fold rotational symmetry) plus an infinite number of vertical mirror planes. Heteronuclear diatomic molecules belong to this group.

- $D_{\infty h}$ : includes all the symmetry elements of $C_{\infty v}$ , along with a horizontal mirror plane, inversion symmetry, and an infinite set of two-fold rotation axes perpendicular to the main axis. Homonuclear diatomic molecules belong to this group.

As the SO(3) group has been covered extensively in a previous article, we shall focus on the $C_{\infty v}$ and $D_{\infty h}$ point groups.

${\color{red}C_{\infty v}}$ point group

To understand how the $C_{\infty v}$ character table is derived, we must first define the basis functions. In general, the spherical harmonics $Y_l^{m_l}$ are used because they form a complete basis set for any point group. $Y_1^0=zf(r)$ is equivalent to the linear function $z$ because $f(r)$ is invariant under all symmetry operations of a point group. Therefore, $Y_1^0$ is invariant under all symmetry operations of the point group. Mathematically, this means that it has an eigenvalue of +1 under every operation, for example $\hat{C}_{\infty}Y_1^0=Y_1^0$ . Hence, $Y_1^0$ transforms according to the totally symmetric irreducible representation $A_1$ . The rotation vector (or axial vector) $\boldsymbol{\mathit{R}}_z$ , however, transforms according to $A_2$ because its curved arrow around the $z$ -axis reverses under $\infty\hat{\sigma}$ and returns an eigenvalue of -1.

The remaining irreducible representations are doubly-degenerate and are generated from linear combinations of the basis functions $Y_l^{m_l}$ and $Y_l^{-m_l}$ . Under a rotation about the $z$ -axis by an angle $\alpha$ , each component transforms as $\hat{C}_{\alpha}Y_l^{m_l}=P(\theta)e^{im_l(\phi-\alpha)}=e^{-im_l\alpha}Y_l^{m_l}$ :

$\hat{C}_{\alpha}\begin{pmatrix}Y_l^{m_l}\\Y_l^{-m_l}\end{pmatrix}=\begin{\pmatrix}e^{-im_l\alpha}&0\\0&e^{im_l\alpha}\end{pmatrix}\begin{pmatrix}Y_l^{m_l}\\Y_l^{-m_l}\end{pmatrix}$

The character $\chi$ corresponding to $\hat{C}_{\alpha}$ is the trace of this 2×2 matrix. Using Euler’s formula $e^{\pm ix}=cosx\pm isinx$ , we obtain

$\chi(\hat{C}_{\alpha})=2cos(m_l\alpha)$

Although, the quantum numbers $l$ and $m_l$ in $Y_l^{m_l}$ are associated with a single electron, they can be replaced by $L$ and $M_L$ for many-electron systems. In such cases, the basis functions are constructed as linear combinations of products of $Y_l^{m_l}(\theta,\phi)$ . A properly coupled state with definite $L$ and projection $M_L$ is an eigenfunction of $\hat{L}_z$ , with $M_L=\sum_im_{l_i}$ . Since each component of the linear combination has an azimuthal dependence proportional to $e^{im_{l_1}}\phi e^{im_{l_2}\phi}\cdots=e^{iM_L\phi}$ , the overall function transforms as $\Psi_L^{M_L}\rightarrow e^{-iM_L\alpha}\Psi_L^{M_L}$ , where $M_L=-L,-L+1,\cdots,L$ . Therefore,

$\chi(\hat{C}_{\phi})=2cos(M_L\phi)$

where we have swapped the arbitrary symbol $\alpha$ with $\phi$ to be consistent with the above character table.

Since $+M_L$ and $-M_L$ correspond to the doubly-degenerate states $\Psi_L^{M_L}$ and $\Psi_L^{-M_L}$ , we can rewrite:

$\chi(\hat{C}_{\phi})=2cos(\vert M_L\vert\phi)=2cos(\Lambda\phi)\;\;\;\;\;\;\;\;130$

where $\Lambda=\vert M_L\vert$ is the projection of $\boldsymbol{\mathit{L}}$ in a many-electron system.

This is why $\Lambda$ takes only non-negative integer values (including 0) and is denoted by special symbols known as molecular term symbols:

It follows that the irreducible representations $A_1,A_2,E_1,E_2,\cdots...$ are also labelled as $\Sigma^+,\Sigma^-,\Pi,\Delta,\cdots$ respectively. Setting $\phi=0$ in eq130 corresponds to the identity operation $\hat{E}$ , which gives $\chi(\hat{E})=2$ for all doubly-degenerate irreducible representations. Furthermore, a reflection of the basis functions $\Psi_L^{M_L}$ and $\Psi_L^{-M_L}$ in a plane containing the $z$ -axis yields:

$\hat{\sigma}_v\Psi_L^{M_L}=\hat{\sigma}(\Psi_{polar}e^{iM_L\phi})=\Psi_{polar}e^{-iM_L\phi}=\Psi_L^{-M_L}$

Thus,

$\hat{\sigma}_{v}\begin{pmatrix}\Psi_L^{M_L}\\\Psi_L^{-M_L}\end{pmatrix}=\begin{\pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}\Psi_L^{M_L}\\\Psi_L^{-M_L}\end{pmatrix}$

with the character of $\hat{\sigma}$ and hence $\infty\hat{\sigma}_v$ being 0 for all doubly-degenerate irreducible representations.

${\color{red}D_{\infty h}}$ point group

The $D_{\infty h}$ point group, unlike the $C_{\infty v}$ point group, includes the inversion symmetry element. Although molecules that belong to this group, such as homonuclear diatomic molecules, are symmetric under inversion, the basis functions (orbitals or rotations) may or may not be. Therefore, every irreducible representation of $D_{\infty h}$ must be either symmetric or antisymmetric with respect to inversion. These are labelled with the subscripts $g$ (gerade, German for “even”) or $u$ (ungerade, German for “odd”).

The characters associated with all $g$ and $u$ irreducible representations for $\hat{E}$ , $\hat{C}_{\infty}$ and $\hat{\sigma}_v$ are the same as those in the $C_{\infty v}$ point group. Since performing the inversion operation twice returns every point to its original position, we have $\hat{i}^2=I$ , where $I$ is the identity matrix. For a one-dimensional representation, this restricts the possible characters of $\hat{i}$ to +1 (for a gerade irreducible representation) or -1 (for an ungerade irreducible representation). The matrix representation of $\hat{i}$ in a two-dimensional representation must be:

$\hat{i}=\begin{pmatrix}\pm 1&0\\0&\pm 1\end{pmatrix}$

with $\hat{i}=\begin{pmatrix}+1&0\\0&+ 1\end{pmatrix}$ , $\chi(\hat{i})=2$ for gerade representations, and $\hat{i}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ , $\chi(\hat{i})=-2$ for ungerade representations.

Since $\hat{S}_{\infty}=\hat{\sigma}_h\hat{C}_{\infty}$ and the basis function is invariant under the rotation for one-dimensional irreducible representations, $\hat{S}_{\infty}=\hat{i}$ . Therefore, the characters $\chi(\hat{S}_{\infty})$ for $g$ and $u$ representations are +1 and -1 respectively. Noting that $\hat{i}=\hat{\sigma}_h\hat{C}_{2}$ , multiplication on the right by $\hat{C}_{2}$ gives $\hat{i}\hat{C}_{2}=\hat{\sigma}_h$ . A rotation by 180° about the $z$ -axis gives the eigenvalue of -1 for odd $\Lambda$ states (e.g. $d_{xz},d_{yz}$ ) and +1 for even $\Lambda$ states (e.g. $d_{x^2-y^2},d_{xy}$ ). Thus, $\chi(\hat{C}_{2})$ is given by $(-1)^{\Lambda}$ for $\Lambda>0$ . It follows that for two-dimensional representations:

$\chi(\hat{S}_{\infty})=\chi(\hat{\sigma}_h)\chi(\hat{C}_{\infty})=\chi(\hat{i})\chi(\hat{C}_2)2cos(\Lambda\phi)=\chi(\hat{i})(-1)^{\Lambda}2cos(\Lambda\phi)$

where $\chi(\hat{i})=+1$ for gerade irreducible representations and -1 for ungerade irreducible representations.

Therefore, $\chi(\hat{S}_{\infty})$ for two-dimensional $g$ and $u$ irreducible representations are $(-1)^{\Lambda}2cos(\Lambda\phi)$ and $(-1)^{\Lambda+1}2cos(\Lambda\phi)$ respectively.

Finally, $C'_2$ denotes a rotation about an axis perpendicular to the $z$ -axis. It can be expressed as $\hat{C}'_2=\hat{i}\hat{\sigma}_v$ because $\hat{\sigma}_v$ transforms $(x,y,z)\rightarrow(x,-y,z)$ and $\hat{i}$ then transforms $(x,-y,z)\rightarrow(-x,y,-z)$ , which is equivalent to a net rotation of 180° about the $y$ -axis (see diagram above). Therefore, $\chi(\hat{C}'_2)=\chi(\hat{i})\chi(\hat{\sigma}_v)$ , resulting in the corresponding characters in the $D_{\infty h}$ character table.

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Electronic selection rules for molecules

Electronic selection rules for molecules are quantum-mechanical conditions (based on changes in quantum numbers and symmetry) that determine whether transition between energy levels in the molecules are allowed or forbidden.

According to the time-dependent perturbation theory, the transition probability between the initial and final states, $\Psi$ and $\Psi'$ , of a molecule is proportional to the square of the matrix element $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ , where $\hat{\boldsymbol{\mathit{\mu}}}=-e\sum_{j=1}^N\boldsymbol{\mathit{r}}_j$ is the operator for the molecule’s electric dipole moment.

Consider a homonuclear diatomic molecule. Although it has no permanent dipole moment in a stationary state, it can acquire a transient, time-dependent dipole moment (just as an atom does) when interacting with oscillating incident radiation. Like multi-electron atoms, $\Psi$ and $\Psi'$ must satisfy the Pauli exclusion principle and can be represented by Slater determinants built from orthogonal spin-orbitals $\chi(\boldsymbol{\mathit{x}})=\psi(\boldsymbol{\mathit{r}})\sigma(\omega)$ , with $\psi(\boldsymbol{\mathit{r}})$ and $\sigma(\omega)$ being the spatial and spin wavefunctions respectively, and $\sigma=\alpha\;or\;\beta$ . Since the dipole moment operator does not act on the spin wavefunctions, which are identical for the initial and final states, the total spin quantum number does not change, i.e. $\Delta S=0$ .

The selections rules involving angular momentum are determined using the two vanishing integral rules in group theory:

Rule 1

If $\boldsymbol{\mathit{f_j}}$ and $\boldsymbol{\mathit{g_l}}$ transform according to two different non-equivalent irreducible representations $\boldsymbol{\mathit{\Gamma_f}}$ and $\boldsymbol{\mathit{\Gamma_g}}$ respectively, the integral of their product over all space is necessarily zero

Rule 2

If a function transforms according to an irreducible representation that is not the totally symmetric representation of a group, its integral over all space is necessarily zero

For linear molecules belonging to the $C_{\infty v}$ point group (see this link for details), the electric dipole moment operator can be resolved into parallel and perpendicular components and transforms according to either the basis function $z$ or the pair $(x,y)$ , the latter forming a linear combination of $x$ and $y$ . If the operator transforms as $z$ , then the product $\hat{\boldsymbol{\mathit{\mu}}}\Psi$ in $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ transforms according to a direct product representation. In this case, the product transforms as the same irreducible representation as $\Psi$ , since $z$ transforms as the totally symmetric irreducible representation $\Sigma^+$ . Therefore, Rule 1 requires that the initial and final states transform according to the same irreducible representation ( $\Sigma^+\leftrightarrow \Sigma^+$ and $\Sigma^-\leftrightarrow \Sigma^-$ ) for the integral to be non-zero. This also implies that the transition $\Sigma^+\leftrightarrow \Sigma^-$ is forbidden. Since states transforming as $\Sigma^{\pm}$ correspond to $\Lambda=0$ , it follows that

$\Delta\Lambda=0$

If the dipole moment operator transforms as $(x,y)$ , we have the following cases:

Case 1: $\Psi_{\Lambda=0}\leftrightarrow\Psi'_{\Lambda=1}$ with $\Delta\Lambda=\pm 1$

The direct product of $\Psi'\hat{\boldsymbol{\mathit{\mu}}}\Psi$ is $\Gamma_a=E_1\otimes E_1\otimes A_1$ or $\Gamma_b=E_1\otimes E_1\otimes A_2$ , both corresponding to the reducible representation

where we have used the identity $2cos^2x=1+cos2x$ .

Decomposing the reducible representation gives: $A_1\oplus A_2\oplus E_2$ . Since $\int\psi_{\Gamma_1}\oplus \psi_{\Gamma_2}\oplus\cdots d\tau=\biggr$\int\psi_{\Gamma_1}d\tau\biggr$\oplus\biggr$\int\psi_{\Gamma_2}d\tau\biggr$\oplus\cdots$ (see this link for proof), the matrix element $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ is non-zero according to Rule 2 because it includes the term $\int\psi_{A_1}d\tau$ .

Case 2: $\Psi_{\Lambda=0}\leftrightarrow\Psi'_{\Lambda=2}$ with $\Delta\Lambda=\pm 2$

The direct product of $\Psi'\hat{\boldsymbol{\mathit{\mu}}}\Psi$ is $\Gamma_c=E_2\otimes E_1\otimes A_1$ or $\Gamma_d=E_2\otimes E_1\otimes A_2$ , both corresponding to the reducible representation

where we have used the identity $cosAcosB=\frac{1}{2}[cos(A+B)+cos(A-B)]$ .

Decomposing the reducible representation gives: $E_1\oplus E_3$ . Since $\int\psi_{\Gamma_1}\oplus \psi_{\Gamma_2}\oplus\cdots d\tau=\biggr$\int\psi_{\Gamma_1}d\tau\biggr$\oplus\biggr$\int\psi_{\Gamma_2}d\tau\biggr$\oplus\cdots$ , the matrix element $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ is zero according to Rule 2. In fact, if $\Delta\Lambda>1$ or $\Delta\Lambda<-1$ , the character for the rotation operator is always $\chi(C_{\infty})=4cos\phi cos(\Lambda\phi)=2[cos(\Lambda+1)\phi+cos(\Lambda-1)\phi)$ . Hence, $\chi(C_{\infty})$ , for $\Delta\Lambda>1$ , will never include the $cos(0)$ term, which is necessary for the $A_1$ component to exist.

Therefore, the selection rules involving $\Lambda$ are:

$\Delta\Lambda=0,\pm 1\;\;\;(with\;\Sigma^+\leftrightarrow\Sigma^-\;forbidden)$

For linear molecules, $\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}$ becomes $\Omega=\vert\Lambda+\Sigma\vert$ (see this link for explanation), where

$\Omega$ is the projection of the total electronic angular momentum onto the molecular axis.
$\Lambda$ is the projection of the total electronic orbital angular momentum ( $\boldsymbol{\mathit{L}}$ ) onto the molecular axis.
$\Sigma$ is the projection of the total electronic spin angular momentum ( $\boldsymbol{\mathit{S}}$ ) onto the molecular axis.

Since the values of $\Sigma$ range from $-S$ to $+S$ , and $\Delta S=0$ , we have $\Delta\Sigma=0$ . This implies that $\Delta\Omega=\Delta\Lambda$ or equivalently,

$\Delta\Omega=0,\pm 1\;\;\;(with\;\Sigma^+\leftrightarrow\Sigma^-\;forbidden)$

Repeating the same analysis for linear molecules belonging to the $D_{\infty h}$ point group (see this link for details) yields the same angular momentum-based selection rules as for molecules belonging to the $C_{\infty v}$ point group. However, because $D_{\infty h}$ possesses inversion symmetry, additional selection rules apply:

$g\leftrightarrow u\;\;\;(allowed)$

$g\leftrightarrow g,\;\;u\leftrightarrow u\;\;\;(forbidden)$

These arise from the Laporte selection rule, which states that the electric dipole transition matrix element is non-zero only if the overall integrand is symmetric (even) under spatial inversion over all space. Since the dipole moment operator transforms as $u$ , the direct product of the initial and final states symmetries must also be $u$ . Consequently, the initial and final states must have opposite parity for the integrand to be overall $g$ , allowing the transition. In summary,

For non-linear polyatomic molecules, $\Delta S=0$ remains applicable to all systems. However, the Laporte selection rule applies only to molecules possessing an inversion centre (e.g. those belonging to $D_{2h}$ or $O_h$ ). The determination of whether the transition moment integral $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ is nonzero follows the same logic as described above: one first identifies the point group of the molecule, then determines the symmetries of the initial and final states, and finally applies the vanishing integral rules.

Question

Does the selection rule $\Delta n=0$ apply to molecules?

Answer

No. $n$ is not a good quantum number for molecular states. Instead, molecular states are described using term symbols and symmetry labels, so no selection rule involving $\Delta n$ applies.

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Electronic spectroscopy

Electronic spectroscopy is the study of how atoms or molecules absorb or emit light due to transitions of electrons between different energy levels.

When atoms or molecules absorb energy, typically in the ultraviolet or visible regions of the spectrum, electrons can be promoted from lower-energy (ground) states to higher-energy (excited) states. Conversely, when these excited electrons return to lower energy levels, energy is released, often in the form of light. These absorption and emission processes form the basis of electronic spectra.

The resulting spectra are highly characteristic of the substance being studied. Because each element or compound has a unique arrangement of electrons and energy levels, the wavelengths of light they absorb or emit act like a fingerprint. This makes electronic spectroscopy a powerful analytical tool in chemistry, physics, and materials science for identifying substances and studying their structure.

An important branch of electronic spectroscopy is atomic spectroscopy, which focuses specifically on isolated atoms, usually in the gas phase. Unlike molecules, atoms have simpler electronic structures, so their spectra consist of sharp, well-defined lines rather than broad bands. Techniques such as atomic absorption spectroscopy (AAS) and atomic emission spectroscopy (AES) measure the specific wavelengths of light absorbed or emitted by atoms as their electrons transition between discrete energy levels. Because each element produces a unique line spectrum, atomic spectroscopy is widely used for precise elemental analysis in fields such as environmental testing, metallurgy and forensic science.

There are several other types of electronic spectroscopy, with ultraviolet-visible (UV-Vis) spectroscopy being one of the most common. In UV-Vis spectroscopy, a sample absorbs light in the ultraviolet or visible region, causing electronic transitions, particularly in molecules with conjugated systems or transition metal complexes. The intensity and position of absorption bands can provide information about concentration (via Beer–Lambert law), molecular structure and the nature of chemical bonds.

Another important form is fluorescence spectroscopy, where a substance absorbs light at one wavelength and then emits light at a longer wavelength. This technique is especially useful in biological and environmental studies due to its high sensitivity.

Overall, electronic spectroscopy serves as a bridge between light and matter, revealing the electronic structure and properties of substances through their interaction with radiation.

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Atomic absorption and emission spectroscopy

Atomic Absorption Spectroscopy (AAS) and Atomic Emission Spectroscopy (AES) are analytical techniques used to determine the concentration of elements by measuring the interaction of free atoms in the gaseous state with electromagnetic radiation.

A typical atomic absorption spectrometer consists of a light source, such as a hollow cathode lamp (HCL), which emits light characteristic of the element being analysed (see diagram above). The light is split into two beams by mirrors, with one beam passing through an atomiser, where the sample atoms are introduced using a nebuliser and then vaporised by an air-acetylene flame. The other beam is directed through a vacuum reference cell.

Question

Explain how the HCL works.

Answer

A HCL emits narrow wavelengths of light because its operation is based on the specific electronic energy levels of a single element. Inside the lamp, the cathode is made of the element to be analysed (for example, copper or sodium), and the tube is filled with an inert gas such as neon or argon. When a high voltage is applied, the inert gas becomes ionised, producing positively charged gas ions. These ions accelerate towards the cathode and strike its surface, knocking out atoms of the cathode material in a process called sputtering. Some of these sputtered atoms are excited by collisions within the lamp. When their electrons return from higher energy levels to lower ones, they emit light. Because the energy differences between electronic levels in atoms are fixed and discrete, the emitted photons have very specific energies (wavelengths). As a result, the lamp produces a spectrum consisting of sharp emission lines rather than a broad range of wavelengths. Since the cathode is made of only one element, the emitted lines correspond almost exclusively to that element, giving the hollow cathode lamp its narrow, characteristic output. For example, if the cathode is made of sodium, the emitted light involves the 3p → 3s energy levels, corresponding to:

²P_3/2 → ²S_1/2 ( ≈ 589.0 nm: D2 line)

²P_1/2 → ²S_1/2 ( ≈ 589.6 nm: D1 line)

N.B: Click this link to understand how the above atomic term symbols are derived.

As the light passes through the cloud of gaseous atoms in the atomiser, it is absorbed at specific wavelengths corresponding to electronic energy level transitions. The transmitted light then passes into a monochromator, which isolates the specific wavelength absorbed by the element of interest, removing any unwanted radiation.

The two beams, emerging from the monochromator and the reference cell respectively, are subsequently directed to separate detectors, typically photomultiplier tubes. Here, the intensities of the transmitted and reference light are measured. The signals are then compared by a signal processor, which compensates for fluctuations in the light source and instrumental noise. Finally, the processed signal is converted into an electrical output and displayed, typically as absorbance, allowing the concentration of the element in the sample to be determined.

To analyse the concentrations of multiple elements in a sample (see above diagram), the light source may adopt one of the following designs:

- Lamp turret: Several HCLs, each with a cathode made of a different element, are automatically rotated to emit different wavelengths of light. Measurements are then taken sequentially.
- Multi-element lamps: A single HCL, in which the cathode is made from an alloy or a mixture of metal powders (e.g. a combination of cobalt, chromium, copper, iron, manganeses and nickel), is used. Sensitivity is usually lower than that of single-element lamps.
- Continuum source: More modern instruments use a broadband light source, such as a xenon lamp. This provides a broad spectrum of light, allowing the detector to measure multiple elements at different wavelengths without changing lamps.

Atomic Emission Spectroscopy (AES) differs from AAS in that it does not use an external light source (see diagram above); instead, the sample itself is energetically excited using a high-energy source such as a flame, electric arc, spark or inductively coupled plasma (ICP). As the excited atoms spontaneously return to lower energy levels, they emit light at characteristic wavelengths, and this emitted radiation is measured directly. In AES, there is no need for a reference beam or a hollow cathode lamp, and the signal is based on emission intensity rather than absorption.

Modern AES instruments use monochromators with high resolving power, which can separate wavelengths that are extremely close together, enabling multi-element analysis. The computer records a composite spectrum but presents the results as individual element concentrations by filtering the data using its wavelength database. The emission spectra of individual elements can also be derived from the composite spectrum by the software (see diagram below).

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Electronic selection rules for atoms

Electronic selection rules for atoms are quantum-mechanical conditions (based on changes in quantum numbers and symmetry) that determine whether transitions between energy levels in the atoms are allowed or forbidden.

According to the time-dependent perturbation theory, the transition probability $\vert\mu_{fi}\vert^2$ between the states $\psi(r,\theta,\phi)_{n,l,m_l}$ and $\psi(r,\theta,\phi)_{n^{'},l^{'},m_l^{'}}$ of an atom is proportional to $\vert\langle\psi(r,\theta,\phi)_{n^{'},l^{'},m_l^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\psi(r,\theta,\phi)_{n,l,m_l}\rangle\vert^2$ , where $\hat{\boldsymbol{\mathit{\mu}}}$ is the operator for the atom’s electric dipole moment. Although an atom has no permanent dipole moment in a stationary state, it can acquire a transient, time-dependent dipole moment when interacting with an oscillating electromagnetic field. During this interaction, the atomic state is described as a time-dependent superposition of the initial and final states: $\Psi(t)=c'(t)\psi_{n^{'},l^{'},m_l^{'}}+c(t)\psi_{n,l,m_l}$ . When the incident frequency is far from resonance, the coefficient $c^{'}(t)$ remains very small. Near resonance, however, the mixing between the states becomes significant ( $c'\approx c$ ). This coherent superposition (for example between $\psi_{n^{'},l^{'},m_l^{'}}=\psi_p$ and $\psi_{n,l,m_l}=\psi_s$ ) leads to interference between the wavefunctions of the two states, creating an asymmetric, time-dependent electron distribution. As a result, the centre of negative charge oscillates relative to the nucleus, generating a transient dipole moment that oscillates at (or near) the driving frequency of the incident radiation.

Hydrogenic atoms

Consider a hydrogenic atom, for which $\psi(r,\theta,\phi)_{n,l,m_l}=R_{n,l}(r)Y_l^{m_l}(\theta,\phi)$ and $\mu_{fi}\propto I=\langle\psi(r,\theta,\phi)_{n^{'},l^{'},m_l^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\psi(r,\theta,\phi)_{n,l,m_l}\rangle$ . For a plane-polarised electromagnetic wave with its electric field oscillating along the $z$ -direction, only the $z$ -component of the dipole operator contributes. Thus, $\hat{\boldsymbol{\mathit{\mu}}}=-ez=-ercos\theta$ and

$I=-e\int_0^{\infty}R_{n^{'},l^{'}}(r)rR_{n,l}(r)r^2dr\int_0^{\pi}\int_0^{2\pi}Y_{l^{'}}^{m^{'}_l*}(\theta,\phi)cos\theta Y_l^{m_l}(\theta,\phi)sin\theta d\theta d\phi$

The angular integral involving the spherical harmonics is non-zero only when (see this link for derivation):

$\Delta l=\pm 1\;\;\;\;\Delta m_l=0,\pm 1$

This leaves the radial integral:

$I'=\int_0^{\infty}R_{n^{'},l^{'}}(r)rR_{n,l}(r)r^2dr$

where $l'=l\pm 1$ .

Question

Show that $rR_{n,l}(r)$ is square-integrable.

Answer

For $rR_{n,l}(r)$ to be square-integrable,

$\int_0^{\infty}\vert rR_{n,l}(r)\vert^2r^2dr=\int_0^{\infty}\vert R_{n,l}(r)\vert^2r^4dr<\infty$

When $r\rightarrow 0$ , we have $R_{n,l}(r)=e^{-\frac{r}{a_0n}}\frac{r^l}{(na_0)^{l+1}}L^{2l+1}_{n-l-1}\biggr$\frac{2r}{na_0}\biggr$\propto r^l$ . So, $\vert R_{n,l}(r)\vert^2r^4\propto r^{2l+4}$ . Consider the integral $\int^b_0r^adr$ . If $a\neq -1$ , then $\int^b_0r^adr=\frac{b^{a+1}}{a+1}$ . If $a=-1$ , then $\int^b_0\frac{1}{r}dr=lnb$ . As $b\rightarrow 0$ ,

Thus, $\int_0^{\infty}\vert R_{n,l}(r)\vert^2r^4dr\propto\int_0^{\infty}r^{2l+4}dr$ converges as $r\rightarrow 0$ because $2l+4>-1$ . Furthermore, $R_{n,l}(r)=e^{-\frac{r}{a_0n}}\frac{r^l}{(na_0)^{l+1}}L^{2l+1}_{n-l-1}\biggr$\frac{2r}{na_0}\biggr$=e^{-\frac{r}{a_0n}}f(r)$ , so that $\vert R_{n,l}(r)\vert^2r^4=g(r)^2e^{-\frac{2r}{a_0n}}$ , where $f(r)$ and $g(r)$ are polynomial functions of $r$ . As $r\rightarrow \infty$ , the exponential term decays to zero faster than $g(r)^2$ , ensuring that $\vert R_{n,l}(r)\vert^2r^4$ converges. Therefore, $\int_0^{\infty}\vert rR_{n,l}(r)\vert^2r^2dr<\infty$ .

Since $rR_{n,l}(r)$ is square-integrable with respect to $r^2dr$ , it belongs to the radial Hilbert space. Additionally, $\{R_{n',l'}(r)\}$ forms a complete set for the radial Hilbert space for a fixed $l'$ . We can therefore expand $rR_{n,l}(r)$ as a linear combination of $\{R_{n',l'}(r)\}$ for any fixed $l'$ :

$rR_{n,l}(r)=\sum_{n'}c_{n'}R_{n',l'}(r)$

with

$c_{n'}=\langle R_{n',l'}(r)\vert r\vert R_{n,l}(r)\rangle$

If there were a restriction such as $n'=n\pm 1$ , the expansion $\sum_{n'}c_{n'}R_{n',l'}(r)$ would contain only a finite number of terms. However, multiplying $R_{n,l}(r)$ by $r$ generally produces a function with a different shape that requires infinitely many basis functions to represent. This corresponds to an infinite number of nonzero coefficients $c_{n'}$ , and hence infinitely many possible values of $n'$ . In other words, $c_{n'}=\langle R_{n',l'}(r)\vert r\vert R_{n,l}(r)\rangle=I'$ is nonzero for infinitely many values of $n'$ , resulting in no selection rule for $\Delta n$ . Therefore, the combined selection rules for electronic transitions in hydrogenic atoms are:

$\Delta n\;(no\;restriction)\;\;\;\;\Delta l=\pm 1\;\;\;\;\Delta m_l=0,\pm 1$

Multi-electron atoms

For multi-electron atoms, the initial and final states, $\Psi$ and $\Psi'$ , must satisfy the Pauli exclusion principle. This can be accomplished by representing them using Slater determinants built from orthogonal spin-orbitals $\chi(\boldsymbol{\mathit{x}})$ :

$\Psi_a=\frac{1}{\sqrt{N!}}\begin{vmatrix} \chi_1(1)&\chi_2(1)&\cdots &\chi_N(1)\\ \chi_1(2)&\chi_2(2)& \cdots &\chi_N(2)\\ \vdots&\vdots &\ddots &\vdots \\ \chi_1(N)&\chi_2(N)&\cdots &\chi_N(N) \end{vmatrix}$

where

$\chi(\boldsymbol{\mathit{x}})=\psi(\boldsymbol{\mathit{r}})\sigma(\omega)$ , with $\psi(\boldsymbol{\mathit{r}})$ and $\sigma(\omega)$ being the spatial and spin wavefunctions respectively, and $\sigma=\alpha\;or\;\beta$ .
$\Psi_a$ denotes the antisymmetric form of $\Psi$ and $\Psi'$ , i.e. $\Psi_a$ or $\Psi'_a$ .
$\Psi_s$ denotes the symmetric form of $\Psi$ and $\Psi'$ , i.e. $\Psi_s$ or $\Psi'_s$ .

The electric dipole operator is a sum of one-electron operators:

$\hat{\boldsymbol{\mathit{\mu}}}=-e\sum_{j=1}^N\boldsymbol{\mathit{r}}_j$

Determining the selection rules requires evaluating $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ , in which the initial and final states must differ. Let us assume that $\Psi$ and $\Psi'$ differ by one spin orbital, i.e. $\Psi_s=\psi_1(x_1)\sigma_1(\omega_1)\cdots\psi_j(x_j)\sigma_j(\omega_j)\cdots\psi_N(x_N)\sigma_N(\omega_N)$ and $\Psi'_s=\psi_1(x_1)\sigma_1(\omega_1)\cdots\psi_k(x_j)\sigma_k(\omega_j)\cdots\psi_N(x_N)\sigma_N(\omega_N)$ . Applying the Slater-Condon rule for one-electron operators, where $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle=\langle\Psi'_s\vert\hat{\boldsymbol{\mathit{\mu}}}\hat{A}^2\vert\Psi_s\rangle$ , in which $\hat{A}=\frac{1}{\sqrt{N!}}\sum_{r=1}^{N!}(-1)^rP_r$ is the antisymmetriser, gives:

$\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle=-e\sum_{j=1}^N\langle\Psi'_s\vert\boldsymbol{\mathit{r}}_j\sum_{t=1}^{N!}(-1)^t\hat{P}_t\Psi_s\rangle$

Thus,

$\begin{align}\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =&-e\sum_{j=1}^N\int\cdots\int[\psi_1(x_1)\cdots\psi_k(x_j)\cdots\psi_N(x_N)]^*\boldsymbol{\mathit{r}}_j\times\\&[\psi_1(x_1)\psi_2(x_2)\cdots\psi_j(x_j)\cdots\psi_N(x_N)-\\&\psi_2(x_1)\psi_1(x_2)\cdots\psi_j(x_j)\cdots\psi_N(x_N)+\cdots]dx_1\cdots dx_N\times\\&\int\cdots\int\sigma_1^*(\omega_1)\sigma_1(\omega_1)\cdots\sigma^*_k(\omega_j)\sigma_j(\omega_j)\cdots\sigma^*_N(\omega_N)\sigma_N(\omega_N)d\omega_1\cdots d\omega_N\end{align}$

Due to spin orthogonality, $\int\sigma_i^*(\omega)\sigma_j(\omega)d\omega=\delta_{ij}$ . So, $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle\neq 0$ only if $\sigma_k(\omega_j)=\sigma_j(\omega_j)$ . This implies $\Delta m_s=0$ and therefore $\Delta M_S=0$ because the electron spins in a multi-electron atom are coupled such that $M_S=\sum_im_{s,i}$ , where $M_S=-S,-S+1,\cdots,S-1,S$ . Since the spin wavefunctions of the initial and final states are identical, this further requires that the total spin quantum number does not change, i.e. $\Delta S=0$ . It follows that:

$\begin{align}\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =&-e\sum_{j=1}^N\biggr\{\int\cdots\int[\psi_1(x_1)\cdots\psi_k(x_j)\cdots\psi_N(x_N)]^*\boldsymbol{\mathit{r}}_j\times\\&\psi_1(x_1)\psi_2(x_2)\cdots\psi_j(x_j)\cdots\psi_N(x_N)-\\&\int\cdots\int[\psi_1(x_1)\psi_2(x_2)\cdots\psi_k(x_j)\cdots\psi_N(x_N)]^*\boldsymbol{\mathit{r}}_j\times\\&[\psi_2(x_1)\psi_1(x_2)\cdots\psi_j(x_j)\cdots\psi_N(x_N)+\cdots]\biggr\}dx_1\cdots dx_N\end{align}$

Due to the orthogonality of the spatial wavefunctions, only the first term on the RHS survives, giving:

$\begin{align}\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =&-e\sum_{j=1}^N\int\psi_1^*(x_1)\psi_1(x_1)dx_1\cdots\int\psi_k^*(x_j)\boldsymbol{\mathit{r}}_j\psi_j(x_j)dx_j\\&\cdots\int\psi_N^*(x_N)\psi_N(x_N)dx_N\end{align}$

All terms vanish except for $j=k$ , yielding

$\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =-e\int\psi_k^*(x_k)\boldsymbol{\mathit{r}}_k\psi_k(x_k)dx_k$

It is evident from the integrals in the derivation above that if $\Psi$ and $\Psi'$ differ by more than one spin orbital, $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle =0$ . This is why double or multiple electon excitations are forbidden in electric dipole transitions. Therefore, $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle$ reduces to a single one-electron integral, whose selection rules are those of the hydrogenic atom:

$\Delta n\;(no\;restriction)\;\;\;\;\Delta l=\pm 1\;\;\;\;\Delta m_l=0,\pm 1$

$\Delta n$ remains unrestricted for multi-electron atoms. The dipole operator acts on a single electron and enforces the hydrogenic selection rule $\Delta l=\pm 1$ . In a multi-electron atom, the total orbital angular momentum is obtained by vector coupling $\boldsymbol{\mathit{L}}\sum_i\boldsymbol{\mathit{l}}_i$ . A change in one electron’s orbital angular momentum by one unit therefore changes the coupled total orbital angular momentum by at most one unit, leading to the selection rule $\Delta L=0,\pm 1$ (with $L=0\leftrightarrow L'=0$ forbidden). Since $\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}$ and $\Delta S=0$ , any change in the total angular momentum between the initial and final states ( $\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}$ and $\boldsymbol{\mathit{J}}'=\boldsymbol{\mathit{L}}'+\boldsymbol{\mathit{S}}'$ ) must arise entirely from the change in $\boldsymbol{\mathit{L}}$ , which is governed by $\Delta L=0,\pm 1$ . Therefore, $\Delta J=0,\pm 1$ , with $J=0\leftrightarrow J'=0$ forbidden. Furthermore, since $J_z=L_z+S_z$ , it follows that $M_J=M_L+M_S$ and hence $\Delta M_J=\Delta M_L+\Delta M_S$ . Because $\Delta M_S=0$ and $\Delta M_L=\sum_i\Delta m_{l,i}$ in the Slater-Condon reduction, we obtain $\Delta M_J=0,\pm 1$ .

The final selection rule for multi-electron atoms is based on the Laporte selection rule, which states that the electric dipole transition matrix element $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle$ is non-zero only if the integrand is even under spatial inversion over all space. This follows from the fact that if the integrand is odd under the transformation $(\boldsymbol{\mathit{r}}_1,\cdots,\boldsymbol{\mathit{r}}_N)\rightarrow(-\boldsymbol{\mathit{r}}_1,\cdots,-\boldsymbol{\mathit{r}}_N)$ , then each configuration has a corresponding inverted configuration contributing equal magnitude with opposite sign, leading to cancellation and a vanishing integral.

Since $\hat{\boldsymbol{\mathit{\mu}}}=-e\sum_{j=1}^N\boldsymbol{\mathit{r}}_j$ is odd under inversion ( $\boldsymbol{\mathit{r}}_j\rightarrow -\boldsymbol{\mathit{r}}_j$ ), the matrix element $\langle\Psi'_a\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi_a\rangle$ is non-zero only if the initial and final states have opposite parity. For a multi-electron atom, the parity of a Slater determinant is given by $(-1)^{\sum_i^Nl_i}$ . So, the wavefunction is even if $\sum_i^Nl_i$ is even and odd if $\sum_i^Nl_i$ is odd.

The combined selection rules for multi-electron atoms are:

$\Delta n\;(no\;restriction)$
$\Delta S=0$
$\Delta L=0,\pm 1\;(with\;L=0\leftrightarrow L'=0\;forbidden)$
$\Delta J=0,\pm 1\;(with\;J=0\leftrightarrow J'=0\;forbidden)$
$\Delta M_J=0,\pm 1$
$Parity\;of\;states\;must\;change\;(g\leftrightarrow u)$

Question

Why is $L=0\leftrightarrow L'=0$ forbidden? Evaluate whether the transition $^3P_1\rightarrow ^3P_2$ is allowed (see this link for how atomic term symbols are derived).

Answer

Electric dipole transitions in centrosymmetric atoms involve the generation of a transient dipole moment, which is directional, leading to the redistribution of charge density. Such a transition cannot occur if both the initial and final states are spherically symmetric. Since an $L=0$ state is spherically symmetric and has no directional dependence, transitions with $L=0\leftrightarrow L'=0$ are forbidden. The same reasoning explains why $J=0\leftrightarrow J'=0$ transitions are forbidden.

For the transition $^3P_1\rightarrow ^3P_2$ ,

1. $\Delta n=0$ : allowed, since there is no restriction on $\Delta n$ .
2. $\Delta S=1-1=0$ : satisfies the selection rule.
3. $\Delta L=1-1=0$ : allowed, since $L\neq 0$ .
4. $\Delta J=2-1=+1$ : satisfies the selection rule.
5. For $J=1$ and $J'=2$ , the magnetic quantum numbers are $M_J=-1,0,+1$ and $M_J=-2,-1,0,+1,+2$ respectively. Transitions are allowed only for $\Delta M_J=0,\pm 1$ .
6. The parity of both the initial and final states is $(-1)^{\sum_i^Nl_i}=(-1)^{1+1}=+1$ , which violates the parity rule.

Therefore, the transition $^3P_1\rightarrow ^3P_2$ is forbidden.

The selection rules for hydrogenic and multi-electron atoms are best illustrated by Grotrian diagrams (see above Grotrian diagram for hydrogen).

In summary, electronic transitions in hydrogenic and multi-electron atoms produce discrete spectral lines (see diagram above), even when fine structures is taken into account. However, in molecules, each electronic transition is accompanied by many possible simultaneous vibrational and rotational transitions. These thousands of closely-spaced lines overlap, creating the appearance of a continuous band.

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Vibronic and ro-vibronic selection rules for molecules

Vibronic and ro-vibronic selection rules specify which combined electronic, vibrational, and rotational transitions in molecules are allowed based on quantum number and symmetry constraints.

An electronic transition in a molecule is often accompanied by vibration-rotation transitions. To determine the selection rules for these transitions, we refer to the Born-Oppenheimer approximation, which states that the total wavefunction $\Psi$ of the molecule can be expressed as:

$\Psi=\psi_e(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})\psi_v(Q)\psi_r(\tau_r)$

where

$\psi_e$ is the electronic wavefunction that is a function of nuclear coordinates $\boldsymbol{\mathit{R}}$ and electronic coordinates $\boldsymbol{\mathit{r}}$ .
$\psi_v$ is the vibrational wavefunction that is a function of normal coordinates $Q$ .
$\psi_r$ is the rotational wavefunction that is a function of rotational coordinates $\tau_r$ , typically the Euler angles.

According to the time-dependent perturbation theory, the transition probability between the initial and final states, $\Psi$ and $\Psi'$ is proportional to the square of $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ , where the electric dipole moment operator $\hat{\boldsymbol{\mathit{\mu}}}$ is the sum of all charge–position contributions in the molecule:

$\hat{\boldsymbol{\mathit{\mu}}}=\hat{\boldsymbol{\mathit{\mu}}}_N+\hat{\boldsymbol{\mathit{\mu}}}_{el}$

where

$\hat{\boldsymbol{\mathit{\mu}}}_N=\sum_AZ_Ae\boldsymbol{\mathit{R}}_A$
$\hat{\boldsymbol{\mathit{\mu}}}_{el}=-\sum_ie\boldsymbol{\mathit{r}}_i$
$Z_Ae$ and $\boldsymbol{\mathit{R}}_A$ are the charge and position of nucleus $A$ .
$e$ and $\boldsymbol{\mathit{r}}_i$ are the charge and position of electron $i$ .

The total transition matrix $\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle$ is:

$\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle=\langle\psi_{e^{'}}\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_N\vert\psi_e\psi_v\psi_r\rangle+\langle\psi_{e^{'}}\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_{el}\vert\psi_e\psi_v\psi_r\rangle$

Although $\psi_e$ depends parametrically on the nuclear coordinates, the nuclear dipole operator $\hat{\boldsymbol{\mathit{\mu}}}_N$ does not act on the electronic coordinates. Therefore, we can write:

$\langle\psi_{e^{'}}\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_N\vert\psi_e\psi_v\psi_r\rangle=\langle\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_{N}\langle\psi_{e^{'}}\vert\psi_e\rangle\vert\psi_v\psi_r\rangle$

Since electronic spectroscopy involves transitions between different electronic states, and the eigenfunctions in the set $\{\psi_e\}$ are orthonormal, $\langle\psi_{e^{'}}\vert\psi_e\rangle=0$ and thus this term vanishes. In contrast, for microwave or infrared spectroscopy, where no change in electronic state occurs, $\langle\psi_{e^{'}}\vert\psi_e\rangle=1$ , and so:

$\langle\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_N\langle\psi_{e^{'}}\vert\psi_e\rangle\vert\psi_v\psi_r\rangle=\langle\psi_{v^{'}}\psi_{r^{'}}\vert\hat{\boldsymbol{\mathit{\mu}}}_{N}\vert\psi_v\psi_r\rangle$

which corresponds to vibration-rotation transitions.

Therefore, the total transition matrix reduces to:

$\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle=\int_Q\int_{\tau_r}\psi_{v^{'}}^{*}(Q)\psi_{r^{'}}^{*}(\tau_r)\biggr\[\int_r\psi^{*}_{e^{'}}(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})\hat{\boldsymbol{\mathit{\mu}}}_{el}\psi_e(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})d\boldsymbol{\mathit{r}}\biggr\]\psi_v(Q)\psi_r(\tau_r)dQd\tau_r$

Because the integral in square brackets is performed over all electronic coordinates for each set of nuclear coordinates $\boldsymbol{\mathit{R}}$ , we can define it as a function of $\boldsymbol{\mathit{R}}$ : $\hat{\mu}_{el}(\boldsymbol{\mathit{R}})=\int_r\psi^{*}_{e^{'}}(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})\hat{\boldsymbol{\mathit{\mu}}}_{el}\psi_e(\boldsymbol{\mathit{r}};\boldsymbol{\mathit{R}})d\boldsymbol{\mathit{r}}$ . To further simplify the total transition matrix, we apply the Condon approximation, which assumes that electronic transitions occur so rapidly (on the order of 10^-15 s) that the nuclei can be considered stationary during the transition. Evaluating at the equilibrium nuclear geometry $\boldsymbol{\mathit{R}}_{eq}$ (i.e. an average internuclear separation), $\hat{\mu}_{el}(\boldsymbol{\mathit{R}}_{eq})$ can be treated as a constant. Thus,

$\langle\Psi'\vert\hat{\boldsymbol{\mathit{\mu}}}\vert\Psi\rangle=\hat{\mu}_{el}(\boldsymbol{\mathit{R}}_{eq})\int_Q\psi_{v^{'}}^{*}(Q)\psi_v(Q)dQ\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)\psi_r(\tau_r)d\tau_r$

Question

Is $\hat{\boldsymbol{\mathit{\mu}}}_{el}=-\sum_ie\boldsymbol{\mathit{r}}_i$ defined in the molecular frame or the laboratory frame?

Answer

It is defined in the molecular frame. In the Born-Oppenheimer approximation, the electronic coordinates $\boldsymbol{\mathit{r}}_i$ are defined relative to the nuclear centre of mass.

Although $\hat{\mu}_{el}(\boldsymbol{\mathit{R}}_{eq})$ is defined in the molecular frame $(a,b,c)$ , the rotational wavefunctions are defined with respect to the laboratory frame $(x,y,z)$ . For the integral to be physically meaningful, the dipole operator and the wavefunctions must be expressed in the same coordinate system (usually the laboratory frame) as that is where the interaction with the external electromagnetic field occurs. To simplify the integral, we shall consider the projection of $\hat{\boldsymbol{\mathit{\mu}}}_{el}$ onto the lab $z$ -axis (see this article for derivation):

$\hat{\mu}_{z}=\mu_a\Phi_a+\mu_b\Phi_b+\mu_c\Phi_c$

where $\Phi_a=sin\theta cos\chi$ , $\Phi_b=-sin\theta sin\chi$ , $\Phi_c=cos\theta$ .

Since $\mu_a$ , $\mu_b$ , $\mu_c$ are constants (Condon approximation),

$\langle\Psi'\vert\hat{\mu}_z\vert\Psi\rangle=\sum_{g=a,b,c}\int_{\tau_{el}}\psi^*_{e^{'}}\hat{\mu}_g\psi_ed\tau_{el}\int_Q\psi_{v^{'}}^{*}(Q)\psi_v(Q)dQ\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)\Phi_g\psi_r(\tau_r)d\tau_r$

with $d\tau_r=sin\theta d\theta d\phi d\chi$ (see this article for derivation).

$\int_{\tau_{el}}\psi^*_{e^{'}}\hat{\mu}_g\psi_ed\tau_{el}$ defines the electronic selection rules. As explained in the previous article, these rules are governed by

1. Symmetry: One first identifies the point group of the molecule, then determines the symmetries of the initial and final states, and finally applies the vanishing integral rules.
2. Parity: The Laporte selection rule applies only to molecules possessing an inversion centre (e.g. those belonging to $D_{2h}$ or $O_h$ ).
3. Spin: $\Delta S=0$ , assuming spin-orbit coupling is weak.

For example (see previous article for derivation),

Unlike pure molecular vibrational motion in infrared spectroscopy, where the initial and final states are part of a complete set of eigenfunctions of the same Hamiltonian (resulting in the pure vibrational selection rule $\Delta v=\pm 1$ ), the initial state in $\int_Q\psi_{v^{'}}^{*}(Q)\psi_v(Q)dQ$ (known as the Frank-Condon overlap integral) belongs to a complete set of eigenfunctions of a Hamiltonian that is different from that of the final state. This is because the two states correspond to different electronic states, each with a different potential-energy function. Therefore, the final vibrational state may not be orthogonal to the initial vibrational state. However, both complete sets of eigenfunctions span the same Hilbert space, which implies that we can expand $\psi_{v^{'}}^{*}(Q)$ as a linear combination of $\psi_v(Q)$ :

$\psi_{v^{'}}^{*}(Q)=\sum_vc_v\psi_v(Q)$

with

$c_v=\langle\psi_v(Q)\vert\psi_{v^{'}}^{*}(Q)\rangle$

If there were a restriction such as $v=v'\pm 1$ , the expansion $\sum_vc_v\psi_v(Q)$ would contain only a finite number of terms. However, the excited state generally requires infinitely many initial state basis functions to represent. This corresponds to an infinite number of nonzero coefficients $c_v$ and hence infinitely many possible values of $v$ , resulting in no selection rule for $\Delta v$ . In other words,

$\Delta v=0,\pm 1,\pm 2,\cdots$

For linear molecules, $\mu_a=0$ and $\mu_b=0$ , while $\mu_c\neq 0$ , even for homonuclear diatomic molecules, because of the generation of a transient dipole moment in electronic spectroscopy. Therefore, $\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)\Phi_g\psi_r(\tau_r)d\tau_r$ in $\langle\Psi'\vert\hat{\mu}_z\vert\Psi\rangle$ simplifies to $\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)cos\theta\psi_r(\tau_r)d\tau_r$ , where the spherical harmonics are the wavefunctions, and the volume element reduces to $d\tau_r=sin\theta d\theta d\phi$ because the spherical harmonics do not depend on $\chi$ . As shown in another article, this integral yields the selection rules: $\Delta J=0,\pm 1$ . The inclusion of $\Delta J=0$ reflects the fact that conservation of total angular momentum is satisfied during a photon-mediated electronic transition in linear molecules when $\Delta \Omega=\pm 1$ , even though $\Delta J=0$ is forbidden for pure rotational transitions in linear molecules. More specifically, the ro-vibronic transition selection rules for linear molecules are:

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=\pm 1\;\;\;\;for\;\Delta\Omega=0\;transitions$

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=0,\pm 1\;\;\;\;for\;\Delta\Omega=\pm 1\;transitions$

This is why homonuclear diatomic molecules are vibrationally and rotationally inactive in standard infrared and microwave spectroscopy, but vibrationally and rotationally active in electronic spectroscopy.

For symmetric rotors, the rotational wavefunctions are represented by the Wigner D-functions $D^J_{M_JK}(\phi,\theta,\chi)=e^{-i\chi K}\langle J,M_J\vert e^{-i\phi\hat{J}_z}e^{-i\theta\hat{J}_y}\vert J,K\rangle$ . As shown in another article, $\int_{\tau_r}\psi_{r^{'}}^{*}(\tau_r)\Phi_g\psi_r(\tau_r)d\tau_r$ results in $\Delta J=0,\pm 1,\Delta K=0$ for parallel transitions, and $\Delta J=0,\pm 1,\Delta K=\pm 1$ for perpendicular transitions. Therefore, for allowed electronic transitions,

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=0,\pm 1,\;\;\;\Delta K=0\;\;\;\;for\;parallel\;transitions$

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=0,\pm 1,\;\;\;\Delta K=\pm 1\;\;\;\;for\;perpendicular\;transitions$

Although a spherical top is rotationally inactive in pure rotational spectroscopy, it is rotationally active in ro-vibronic transitions due to the generation of a transcient dipole moment. Since a spherical top is a special case of a symmetric top for which the energy does not depend on $K$ , the selection rules for allowed electronic transitions are simply:

$\Delta v=0,\pm 1,\pm 2,\cdots,\;\;\;\Delta J=0,\pm 1$

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May 16, 2026

Vibronic and ro-vibronic spectra

A vibronic spectrum records simultaneous changes in the electronic and vibrational energies of molecules, whereas a ro-vibronic spectrum additionally includes resolved rotational energy changes.

Unlike an atomic absorption spectrometer, which typically uses element-specific line sources such as hollow-cathode lamps, the spectrometers used to measure vibronic and ro-vibronic absorption spectra generally employ broadband continuous radiation sources (e.g. deuterium or tungsten lamps) together with monochromators to isolate the desired wavelength range (see diagram below).

Vibronic and ro-vibronic spectra can also be observed as emission spectra. In such cases, the sample molecules can be excited using:

1. Electric discharges: Passing a high-voltage current through a low-pressure gaseous sample.
2. Inductively coupled plasma arcs: Heating the sample to extremely high temperatures.
3. Laser excitation: Using a laser tuned to a specific transition. The molecules absorb the laser light and then emit it as fluorescence, allowing for very “clean” ro-vibronic spectra.
4. Electron beams: Bombarding the gas with high-energy electrons.

Electronic spectroscopic analyses of very large or heavy molecules, such a protein, usually produce vibronic spectra. In such cases, the rotational constants are so small that the rotational energy levels are closer together than the natural linewidth imposed by the Heisenberg uncertainty principle. Studies of molecules in the liquid or solid phase also result in broad vibronic bands because collisions and intermolecular interactions shorten the lifetimes of the excited states and broaden the spectral lines, preventing the individual rotational transitions from being resolved. Conversely, small molecules in the gaseous phase, like H₂ or CO, have larger rotational constants and more widely spaced rotational energy levels, making ro-vibronic spectrum observable.

Consider the homonuclear diatomic molecule N₂. Its first few electronic states are: $X^1\Sigma_g^+$ (ground state), $A^3\Sigma_u^+$ (first excited state), $B^3\Pi_g$ (next higher excited state), $C^3\Pi_u$ (another higher excited state), and so on (click this link for derivation of term symbols). The transition $X^1\Sigma_g^+\leftrightarrow A^3\Sigma_u^+$ is forbidden due to the selection rule $\Delta S=0$ . Therefore, the first two observed emission spectroscopy transitions are $B^3\Pi_g\rightarrow A^3\Sigma_u^+$ and $C^3\Pi_u\rightarrow B^3\Pi_g$ , which, along with their associated vibrational and rotational fine structures, are known as the first positive system and the second positive system respectively (see diagram above). For illustration purposes, three vibrational transitions ( $v=0\leftrightarrow v=0$ , $v=0\leftrightarrow v=1$ and $v=0\leftrightarrow v=2$ ), along with their associated rotational fine structures, are shown for each system. These emissions lines typically group into bands, with the wavelength corresponding to each vibrational transition located approximately at the band head (the point of highest intensity within a band). Notably, the parity rule $g\leftrightarrow u$ is satisfied for both systems, but the Q-branches are absent in the second positive system because $\Delta J=0$ is forbidden for $\Delta\Omega=0$ transitions (see this link for details). In practice, these bands may overlap, forming a “forest of peaks” rather than distinct, well-separated bands.

The intensities of the bands of any ro-vibronic spectrum vary according to the Frank-Condon principle, which states that because atomic nuclei are much more massive than electrons, an electronic transition occurs so rapidly that the nuclear configuration of the molecule remains practically unchanged during the process. In general, a molecule’s nuclear configuration is described by vibrational wavefunctions, each expressed as a function of the internuclear separation $R$ corresponding to a given electronic state, where the equilibrium separation $R_e$ satisfies the relationship: $R_e(ground)\neq R_e(1st\;excited)\neq R_e(2nd\;excited)\neq\cdots$ . For example, the $\Delta v=+1$ band for the $\vert\Omega=0,v=0\rangle\leftrightarrow\vert\Omega'=1,v'=1\rangle$ transition in the diagram below is expected to have a higher intensity than the $\Delta v=0$ band.

This is because, classically, the amplitude of the wavefunction of an oscillator in an excited vibrational state is greatest near the turning points, where the nuclei move most slowly. In contrast, the ground-state wavefunction has its largest amplitude near $R_e$ . Since the vibrational transition intensity is proportional to the square of the overlap integral $\int_Q\psi^*_{v^{'}}(Q)\psi_v(Q)dQ$ , it is determined, according to the Frank-Condon principle, by the extent of overlap between the initial-state wavefunction and the final-state wavefunction lying vertically above it. Since $\int_Q\psi^*_{v^{'}}(Q)\psi_v(Q)dQ$ is constant for every rotational line within a given vibrational band, it acts as a global scaling factor. Consequently, the intensities of rotational fine structures within a higher-intensity vibrational band are generally higher than those within a lower-intensity vibrational band for a given electronic transition.