Symmetry of diatomic molecules

A group is a mathematical structure (usually denoted in the form (G,*)) consisting of a set G and a binary operation ${\displaystyle '*'}$ (sometimes loosely called 'multiplication'), satisfying the following properties:

1. closure: For every pair of elements ${\displaystyle x,y\in G}$, the product ${\displaystyle x*y\in G}$.
2. associativity: For every x and y and z in G, both (x*y)*z and x*(y*z) result with the same element in G  (in symbols, ${\displaystyle (x*y)*z=x*(y*z)\forall x,y,z\in G}$).
3. existence of identity: There must be an element (say e ) in G such that product any element of G with e make no change to the element (in symbols, ${\displaystyle x*e=e*x=x;\forall x\in G}$ ).
4. existence of inverse: For each element ( x ) in G, there must be an element y in G such that product of x and y is the identity element e  (in symbols, for each ${\displaystyle x\in G}$${\displaystyle {\text{ }}\exists {\text{ }}y\in G}$ such that ${\displaystyle x*y=y*x=e}$).

• In addition to the above four, if it so happens that ${\displaystyle \forall x,y\in G}$,${\displaystyle x*y=y*x}$, i.e., the operation in commutative, then the group is called an abelian group. Otherwise it is called a non-abelian group.

The set of all symmetry transformations of a Hamiltonian has the structure of a group, with group multiplication equivalent to applying the transformations one after the other. The group elements can be represented as matrices, so that the group operation becomes the ordinary matrix multiplication. In quantum mechanics, the evolution of an arbitrary superposition of states are given by unitary operators, so each of the elements of the symmetry groups are unitary operators. Now any unitary operator can be expressed as the exponential of some Hermitian operator. So, the corresponding Hermitian operators are the 'generators' of the symmetry group. These unitary transformations act on the Hamiltonian operator in some Hilbert space in a way that the Hamiltonian remains invariant under the transformations. In other words, the symmetry operators commute with the Hamiltonian. If ${\displaystyle U}$ represents the unitary symmetry operator and acts on the Hamiltonian ${\displaystyle H}$, then;

${\displaystyle [H,U]=0}$
${\displaystyle {\begin{aligned}&{H}'={{U}^{\dagger }}HU=H\\&\Rightarrow HU=UH\\&\Rightarrow [H,U]=0;\forall U\in G\\\end{aligned}}}$

These operators have the above-mentioned properties of a group:

• The symmetry operations are closed under multiplication.
• Application of symmetry transformations are associative.
• There is always a trivial transformation, where nothing is done to the original co-ordinates. This is the identity element of the group.
• And as long as an inverse transformation exists, it is a symmetry transformation, i.e. it leaves the Hamiltonian invariant. Thus the inverse is part of this set.

So, by the symmetry of a system, we mean a set of operators, each of which commutes with the Hamiltonian, and they form a symmetry group. This group may be abelian or non-abelian. Depending upon which one it is, the properties of the system changes (for example, if the group is abelian, there would be no degeneracy). Corresponding to every different kind of symmetry in a system, we can find a symmetry group associated with it.

It follows that the generator ${\displaystyle T}$ of the symmetry group also commutes with the Hamiltonian. Now, it follows that:

Some specific examples can be systems having rotational, translational invariance etc. For a rotationally invariant system, the symmetry group of the Hamiltonian is the general rotation group. Now, if (say) the system is invariant about any rotation about Z-axis (i.e., the system has axial symmetry), then the symmetry group of the Hamiltonian is the group of rotation about the symmetry axis. Now, this group is generated by the Z-component of the orbital angular momentum, ${\displaystyle {L}_{z}}$ (general group element ${\displaystyle R(\alpha )={{e}^{\frac {-i\alpha {{L}_{z}}}{\hbar }}}}$). Thus, ${\displaystyle {L}_{z}}$ commutes with ${\displaystyle H}$ for this system and Z-component of the angular momentum is conserved. Similarly, translation symmetry gives rise to conservation of linear momentum, inversion symmetry gives rise to parity conservation and so on.

A molecule at equilibrium in a certain electronic state usually has some geometrical symmetry. This symmetry is described by a certain point group which consists of operations (called symmetry operations) that produce a spatial orientation of the molecule that is indistinguishable from the starting configuration. There are five types of point group symmetry operation: identity, rotation, reflection, inversion and improper rotation or rotation-reflection. Common to all symmetry operations is that the geometrical center-point of the molecule does not change its position; hence the name point group. One can determine the elements of the point group for a particular molecule by considering the geometrical symmetry of its molecular model. However, when one uses a point group, the elements are not to be interpreted in the same way. Instead the elements rotate and/or reflect the vibronic (vibration-electronic) coordinates and these elements commute with the vibronic Hamiltonian. The point group is used to classify by symmetry the vibronic eigenstates. The symmetry classification of the rotational levels, the eigenstates of the full (rovibronic nuclear spin) Hamiltonian, requires the use of the appropriate permutation-inversion group as introduced by Longuet-Higgins.[1] See the Section Inversion symmetry and nuclear permutation symmetry below. The elements of permutation-inversion groups commute with the full molecular Hamiltonian. In addition to point groups, there exists another kind of group important in crystallography, where translation in 3-D also needs to be taken care of. They are known as space groups.

The five basic symmetry operations mentioned above are:[2]

1. Identity Operation E (from the German 'Einheit' meaning unity): The identity operation leaves the molecule unchanged. It forms the identity element in the symmetry group. Though its inclusion seems to be trivial, it is important also because even for the most asymmetric molecule, this symmetry is present. The corresponding symmetry element is the entire molecule itself.
2. Inversion, i : This operation inverts the molecule about its center of inversion (if it has any). The center of inversion is the symmetry element in this case. There may or may not be an atom at this center. A molecule may or may not have a center of inversion. For example: the benzene molecule, a cube, and spheres do have a center of inversion, whereas a tetrahedron does not.
3. Reflection σ: The reflection operation produces a mirror image geometry of the molecule about a certain plane. The mirror plane bisects the molecule and must include its center of geometry. The plane of symmetry is the symmetry element in this case. A symmetry plane parallel with the principal axis (defined below) is dubbed vertical (σ_v) and one perpendicular to it horizontal (σ_h). A third type of symmetry plane exists: If a vertical symmetry plane additionally bisects the angle between two 2-fold rotation axes perpendicular to the principal axis, the plane is dubbed dihedral (σ_d).
4. n-Fold Rotation ${\displaystyle {{c}_{n}}}$: The n-fold rotation operation about a n-fold axis of symmetry  produces molecular orientations indistinguishable from the initial for each rotation of  ${\displaystyle {\frac {{360}^{0}}{n}}}$ (clockwise and counter-clockwise).It is denoted by ${\displaystyle {{c}_{n}}}$. The axis of symmetry is the symmetry element in this case. A molecule can have more than one symmetry axis; the one with the highest n is called the principal axis, and by convention is assigned the z-axis in a Cartesian coordinate system.
5. n-Fold Rotation-Reflection or improper rotation S_n : The n-fold improper rotation operation about an n-fold axis of improper rotation is composed of two successive geometry transformations: first, a rotation through ${\displaystyle {\frac {{360}^{0}}{n}}}$ about the axis of that rotation, and second, reflection through a plane perpendicular (and through the molecular center of geometry) to that axis. This axis is the symmetry element in this case. It is abbreviated S_n.

All other symmetry present in a specific molecule are a combination of these 5 operations.

The Schoenflies (or Schönflies) notation, named after the German mathematician Arthur Moritz Schoenflies, is one of two conventions commonly used to describe point groups. This notation is used in spectroscopy and is used here to specify a molecular point group.

Here, the system is not spherically symmetric. So, ${\displaystyle [H,{{L}^{2}}]\neq 0}$, and the state cannot be depicted in terms of ${\displaystyle l}$ as an eigenstate of the Hamiltonian is not an eigenstate of ${\displaystyle {{L}^{2}}}$ anymore (in contrast to the atomic term symbol, where the states were written as ${\displaystyle ^{2S+1}{{L}_{J}}}$). But, as ${\displaystyle [H,{{L}_{z}}]=0}$, the eigenvalues corresponding to ${\displaystyle {{L}_{z}}}$ can still be used. If,

${\displaystyle {\begin{aligned}&{{L}_{z}}\left|\Psi \right\rangle ={{M}_{L}}\hbar \left|\Psi \right\rangle$ ;{{M}_{L}}=0,\pm 1,\pm 2,..........\\&\Rightarrow {{L}_{z}}\left|\Psi \right\rangle =\pm \Lambda \hbar \left|\Psi \right\rangle ;\Lambda =0,1,2,...........\\\end{aligned}}}

where ${\displaystyle \Lambda =\left|{{M}_{L}}\right|}$ is the absolute value (in a.u.) of the projection of the total electronic angular momentum on the internuclear axis; ${\displaystyle \Lambda }$ can be used as a term symbol. By analogy with the spectroscopic notation S, P, D, F, ... used for atoms, it is customary to associate code letters with the values of ${\displaystyle \Lambda }$ according to the correspondence:

For the individual electrons, the notation and the correspondence used are:

${\displaystyle \lambda =\left|{{m}_{l}}\right|}$ and

Again, ${\displaystyle [{{A}_{y}},H]=0}$, and in addition: ${\displaystyle {{A}_{y}}{{L}_{z}}=-{{L}_{z}}{{A}_{y}}}$ [as ${\displaystyle {{L}_{z}}=-i\hbar (x{\frac {\partial }{\partial y}}-y{\frac {\partial }{\partial x}})}$]. It follows immediately that if ${\displaystyle \Lambda \neq 0}$ the action of the operator ${\displaystyle {{A}_{y}}}$ on an eigenstate corresponding to the eigenvalue ${\displaystyle \Lambda \hbar }$ of ${\displaystyle {L}_{z}}$ converts this state into another one corresponding to the eigenvalue ${\displaystyle -\Lambda \hbar }$, and that both eigenstates have the same energy. The electronic terms such that ${\displaystyle \Lambda \neq 0}$ (that is, the terms ${\displaystyle \Pi ,\Delta ,\Phi ,................}$) are thus doubly degenerate, each value of the energy corresponding to two states which differ by the direction of the projection of the orbital angular momentum along the molecular axis. This twofold degeneracy is actually only approximate and it is possible to show that the interaction between the electronic and rotational motions leads to a splitting of the terms with ${\displaystyle \Lambda \neq 0}$ into two nearby levels, which is called ${\displaystyle \Lambda }$-doubling.[3]

${\displaystyle \Lambda =0}$ corresponds to the ${\displaystyle \Sigma }$ states. These states are non-degenerate, so that the states of a ${\displaystyle \Sigma }$ term can only be multiplied by a constant in a reflection through a plane containing the molecular axis. When ${\displaystyle \Lambda =0}$, simultaneous eigenfunctions of ${\displaystyle H}$,${\displaystyle {L}_{z}}$ and ${\displaystyle {A}_{y}}$ can be constructed. Since ${\displaystyle A_{y}^{2}=1}$, the eigenfunctions of ${\displaystyle {A}_{y}}$ have eigenvalues ${\displaystyle \pm 1}$. So to completely specify ${\displaystyle \Sigma }$ states of diatomic molecules, ${\displaystyle {{\Sigma }^{+}}}$ states, which is left unchanged upon reflection in a plane containing the nuclei, needs to be distinguished from ${\displaystyle {{\Sigma }^{-}}}$ states, for which it changes sign in performing that operation.

Homonuclear diatomic molecules have a center of symmetry at their midpoint. Choosing this point (which is the nuclear center of mass) as the origin of the coordinates, the electronic Hamiltonian is invariant under the point group operation i of inversion of the coordinates of all electrons at that origin. This operation is not the parity operation P (or E*); the parity operation involves the inversion of nuclear and electronic spatial coordinates at the molecular center of mass. Electronic states either remain unchanged by the operation i, or they are changed in sign by i. The former are denoted by the subscript g and are called gerade, while the latter are denoted by the subscript u and are called ungerade. The subscripts g or u are therefore added to the term symbol, so that for homonuclear diatomic molecules electronic states can have the symmetries ${\displaystyle \Sigma _{g}^{+},\Sigma _{g}^{-},\Sigma _{u}^{+},\Sigma _{u}^{-},{{\Pi }_{g}},{{\Pi }_{u}}}$,......according to the irreducible representations of the ${\displaystyle {{D}_{\infty h}}}$ point group.

The complete Hamiltonian of a diatomic molecule (as for all molecules) commutes with the parity operation P or E* and rovibronic (rotation-vibration-electronic) energy levels (often called rotational levels) can be given the parity symmetry label + or -. The complete Hamiltonian of a homonuclear diatomic molecule also commutes with the operation of permuting (or exchanging) the coordinates of the two (identical) nuclei and rotational levels gain the additional label s or a depending on whether the total wavefunction is unchanged (symmetric) or changed in sign (antisymmetric) by the permutation operation. Thus, the rotational levels of heteronuclear diatomic molecules are labelled + or -, whereas those of homonuclear diatomic molecules are labelled +s, +a, -s or -a. The rovibronic nuclear spin states are classified using the appropriate permutation-inversion group.[4]

The complete Hamiltonian of a homonuclear diatomic molecule (as for all centro-symmetric molecules) does not commute with the point group inversion operation i because of the effect of the nuclear hyperfine Hamiltonian. The nuclear hyperfine Hamiltonian can mix the rotational levels of g and u vibronic states (called ortho-para mixing) and give rise to ortho-para transitions[5][6]

If S denotes the resultant of the individual electron spins, ${\displaystyle s(s+1){{\hbar }^{2}}}$ are the eigenvalues of S and as in the case of atoms, each electronic term of the molecule is also characterised by the value of S. If spin-orbit coupling is neglected, there is a degeneracy of order ${\displaystyle 2s+1}$ associated with each ${\displaystyle s}$ for a given ${\displaystyle \Lambda }$. Just as for atoms, the quantity ${\displaystyle 2s+1}$ is called the multiplicity of the term and.is written as a (left) superscript, so that the term symbol is written as ${\displaystyle {}^{2s+1}\Lambda }$. For example, the symbol ${\displaystyle {}^{3}\Pi }$denotes a term such that ${\displaystyle \Lambda =1}$ and ${\displaystyle s=1}$. It is worth noting that the ground state (often labelled by the symbol ${\displaystyle X}$) of most diatomic molecules is such that ${\displaystyle s=0}$ and exhibits maximum symmetry. Thus, in most cases it is a ${\displaystyle {}^{1}{{\Sigma }^{+}}}$ state (written as ${\displaystyle X{}^{1}{{\Sigma }^{+}}}$, excited states are written with ${\displaystyle A,B,C,...}$ in front) for a heteronuclear molecule and a ${\displaystyle {}^{1}\Sigma _{g}^{+}}$ state (written as ${\displaystyle X{}^{1}\Sigma _{g}^{+}}$) for a homonuclear molecule.

Spin–orbit coupling lifts the degeneracy of the electronic states. This is because the z-component of spin interacts with the z-component of the orbital angular momentum, generating a total electronic angular momentum along the molecule axis J_z. This is characterized by the quantum number ${\displaystyle {{M}_{J}}}$, where ${\displaystyle {{M}_{J}}={{M}_{S}}+{{M}_{L}}}$. Again, positive and negative values of ${\displaystyle {{M}_{J}}}$ are degenerate, so the pairs (M_L, M_S) and (−M_L, −M_S) are degenerate. These pairs are grouped together with the quantum number ${\displaystyle \Omega }$, which is defined as the sum of the pair of values (M_L, M_S) for which M_L is positive: ${\displaystyle \Omega =\Lambda +{{M}_{S}}}$

So, the overall molecular term symbol for the most general diatomic molecule is given by:

${\displaystyle {}^{2S+1}\!\Lambda _{\Omega ,(g/u)}^{(+/-)}}$

where

• S is the total spin quantum number
• ${\displaystyle \Lambda }$ is the projection of the orbital angular momentum along the internuclear axis
• ${\displaystyle \Omega }$ is the projection of the total angular momentum along the internuclear axis
• u/g is the effect of the point group operation i
• +/− is the reflection symmetry along an arbitrary plane containing the internuclear axis

The electronic terms or potential curves ${\displaystyle {{E}_{S}}(R)}$ of a diatomic molecule depend only on the internuclear distance ${\displaystyle R}$, and it is important to investigate the behaviour of these potential curves as R varies. It is of considerable interest to examine the intersection of the curves representing the different terms.

The non-crossing rule of von Neumann and Wigner. Two potential curves ${\displaystyle {{E}_{1}}(R)}$ and ${\displaystyle {{E}_{2}}(R)}$ cannot cross if the states 1 and 2 have the same point group symmetry

Let ${\displaystyle {{E}_{1}}(R)}$ and ${\displaystyle {{E}_{2}}(R)}$ two different electronic potential curves. If they intersect at some point, then the functions ${\displaystyle {{E}_{1}}(R)}$ and ${\displaystyle {{E}_{2}}(R)}$ will have neighbouring values near this point. To decide whether such an intersection can occur, it is convenient to put the problem as follows. Suppose at some internuclear distance ${\displaystyle {R}_{c}}$ the values ${\displaystyle {{E}_{1}}({{R}_{C}})}$ and ${\displaystyle {{E}_{2}}({{R}_{C}})}$ are close, but distinct (as shown in the figure). Then it is to be examined whether or ${\displaystyle {{E}_{1}}(R)}$ and ${\displaystyle {{E}_{2}}(R)}$ can be made to intersect by the modification ${\displaystyle {{R}_{C}}\to {{R}_{C}}+\Delta R}$. The energies ${\displaystyle E_{1}^{(0)}={{E}_{1}}({{R}_{C}})}$ and ${\displaystyle E_{2}^{(0)}={{E}_{2}}({{R}_{C}})}$ are eigenvalues of the Hamiltonian ${\displaystyle {{H}_{0}}=H({{R}_{C}})}$. The corresponding orthonormal electronic eigenstates will be denoted by ${\displaystyle \left|\Phi _{1}^{(0)}\right\rangle }$ and ${\displaystyle \left|\Phi _{2}^{(0)}\right\rangle }$ and are assumed to be real. The Hamiltonian now becomes ${\displaystyle H\equiv H({{R}_{C}}+\Delta R)={{H}_{0}}+H'}$, where ${\displaystyle H'={\frac {\partial {{H}_{0}}}{\partial {{R}_{C}}}}\Delta R}$ is the small perturbation operator (though it is a degenerate case, so ordinary method of perturbation won't work). setting ${\displaystyle H_{ij}^{'}=\left\langle \Phi _{i}^{(0)}|H'|\Phi _{j}^{(0)}\right\rangle ;i,j=1,2}$, it can be deduced that in order for ${\displaystyle {{E}_{1}}(R)}$ and ${\displaystyle {{E}_{2}}(R)}$ to be equal at the point ${\displaystyle {{R}_{C}}+\Delta R}$ the following two conditions are required to be fulfilled:

${\displaystyle E_{1}^{(0)}-E_{2}^{(0)}+H_{11}^{'}-H_{22}^{'}=0}$ and ${\displaystyle H_{12}^{'}=0}$

As an initial zero-order approximation, instead of ${\displaystyle \left|\Phi _{1}^{(0)}\right\rangle }$ and ${\displaystyle \left|\Phi _{2}^{(0)}\right\rangle }$ themselves, linear combinations of them of the form ${\displaystyle \left|{{\Phi }^{(0)}}\right\rangle ={{c}_{1}}\left|\Phi _{1}^{(0)}\right\rangle +{{c}_{2}}\left|\Phi _{2}^{(0)}\right\rangle }$, can be taken as the eigenstate of the Hamiltonian ${\displaystyle H}$ (where ${\displaystyle {{c}_{1}}}$ and ${\displaystyle {{c}_{2}}}$ are, in general, complex). Substituting this expression in the perturbed Schrödinger  equation: ${\displaystyle ({{H}_{0}}+H')\left|{{\Phi }^{(0)}}\right\rangle =E\left|{{\Phi }^{(0)}}\right\rangle }$ Expanding: ${\displaystyle {{c}_{1}}(E_{1}^{(0)}+H'-E)\left|\Phi _{1}^{(0)}\right\rangle +{{c}_{2}}(E_{2}^{(0)}+H'-E)\left|\Phi _{2}^{(0)}\right\rangle =0}$ Taking inner product with the respective bra's: ${\displaystyle {{c}_{1}}(E_{1}^{(0)}+H'-E)\left\langle \Phi _{1}^{(0)}|\Phi _{1}^{(0)}\right\rangle +{{c}_{2}}(E_{2}^{(0)}+H'-E)\left\langle \Phi _{1}^{(0)}|\Phi _{2}^{(0)}\right\rangle =0}$; and ${\displaystyle {{c}_{1}}(E_{1}^{(0)}+H'-E)\left\langle \Phi _{2}^{(0)}|\Phi _{1}^{(0)}\right\rangle +{{c}_{2}}(E_{2}^{(0)}+H'-E)\left\langle \Phi _{2}^{(0)}|\Phi _{2}^{(0)}\right\rangle =0}$ Now, ${\displaystyle \left|\Phi _{1}^{(0)}\right\rangle }$ and ${\displaystyle \left|\Phi _{2}^{(0)}\right\rangle }$ are eigenstates of the Hamiltonian ${\displaystyle {{H}_{0}}}$ corresponding to different eigenvalues and as ${\displaystyle {{H}_{0}}}$ is itself Hermitian, they are orthonormal: ${\displaystyle \left\langle \Phi _{i}^{(0)}|\Phi _{j}^{(0)}\right\rangle ={{\delta }_{ij}}}$ Thus: ${\displaystyle {{c}_{1}}(E_{1}^{(0)}+H_{11}^{'}-E)+{{c}_{2}}H_{12}^{'}=0}$; and ${\displaystyle {{c}_{1}}H_{21}^{'}+{{c}_{2}}(E_{2}^{(0)}+H_{22}^{'}-E)=0}$ Since the operator ${\displaystyle H'}$ is Hermitian, the matrix elements ${\displaystyle H_{11}^{'}}$ and ${\displaystyle H_{22}^{'}}$ are real, while ${\displaystyle H_{12}^{'}=H_{21}^{'*}}$. The compatibility condition for these equations is (such that both ${\displaystyle {{c}_{1}}}$ and ${\displaystyle {{c}_{2}}}$ are not simultaneously zero): ${\displaystyle \left|{\begin{matrix}E_{1}^{(0)}+H_{11}^{'}-E&H_{12}^{'}\\H_{21}^{'}&E_{2}^{(0)}+H_{22}^{'}-E\\\end{matrix}}\right|=0}$ This gives: ${\displaystyle E={\frac {1}{2}}(E_{1}^{(0)}+E_{2}^{(0)}+H_{11}^{'}+H_{22}^{'})\pm {\sqrt {{\frac {1}{4}}{{(E_{1}^{(0)}-E_{2}^{(0)}+H_{11}^{'}-H_{22}^{'})}^{2}}+{{\left|H_{12}^{'}\right|}^{2}}}}}$ This formula gives the required eigenvalues of the energy in the first approximation. If the energy values of the two terms become equal at the point ${\displaystyle ({{R}_{C}}+\Delta R)}$ (i.e. the terms intersect), this means that the two values of ${\displaystyle E}$ given by formula, are the same. For this to happen, the expression under the radical must vanish. Since it is the sum of two squares, both are simultaneously zero. So, it gives the conditions: ${\displaystyle E_{1}^{(0)}-E_{2}^{(0)}+H_{11}^{'}-H_{22}^{'}=0}$ and ${\displaystyle H_{12}^{'}=0}$

However, we have at our disposal only one arbitrary parameter ${\displaystyle \Delta R}$ giving the perturbation ${\displaystyle H'}$. Hence the

two conditions involving more than one parameter cannot in general be simultaneously satisfied (the initial assumption that ${\displaystyle \left|\Phi _{1}^{(0)}\right\rangle }$ and ${\displaystyle \left|\Phi _{2}^{(0)}\right\rangle }$ real, implies that ${\displaystyle H_{12}^{'}}$ is also real). So, two case can arise: 

1. The matrix element ${\displaystyle H_{12}^{'}}$ vanishes identically. It is then possible to satisfy the first condition independently. Therefore, it is possible for the crossing to occur if, for a certain value of ${\displaystyle \Delta R}$ (i.e., for a certain value of ${\displaystyle R}$) the first equation is satisfied. As the perturbation operator ${\displaystyle H'}$ (or ${\displaystyle H}$) commutes with the symmetry operators of the molecule, this case will happen if the two electronic states ${\displaystyle \left|\Phi _{1}^{(0)}\right\rangle }$ and ${\displaystyle \left|\Phi _{2}^{(0)}\right\rangle }$ have different point group symmetries (for example if they correspond to two electronic terms having different values of ${\displaystyle \Lambda }$, different electronic parities g and u, different multiplicities, or for example are the two terms ${\displaystyle {{\Sigma }^{+}}}$ and ${\displaystyle {{\Sigma }^{-}}}$) as it can be shown that, for a scalar quantity whose operator commutes with the angular momentum and inversion operators, only the matrix elements for transitions between states of the same angular momentum and parity are non-zero and the proof remains valid, in essentially the same form, for the general case of an arbitrary symmetry operator.  
2. If the electronic states ${\displaystyle \left|\Phi _{1}^{(0)}\right\rangle }$ and ${\displaystyle \left|\Phi _{2}^{(0)}\right\rangle }$ have the same point group symmetry, then ${\displaystyle H_{12}^{'}}$ can be, and will in general be, non-zero. Except for accidental crossing which would occur if, by coincidence, the two equations were satisfied at the same value of ${\displaystyle R}$, it is in general impossible to find a single value of ${\displaystyle \Delta R}$ (i.e., a single value of ${\displaystyle R}$) for which the two conditions are satisfied simultaneously.  

Thus, in a diatomic molecule, only terms of different symmetry can intersect, while the intersection of terms of like symmetry is forbidden. This is, in general, true for any case in quantum mechanics where the Hamiltonian contains some parameter and its eigenvalues are consequently functions of that parameter. This general rule is known as von Neumann - Wigner non-crossing rule. [notes 1]

This general symmetry principle has important consequences is molecular spectra. In fact, in the applications of valence bond method in case of diatomic molecules, three main correspondence between the atomic and the molecular orbitals are taken care of:

1. Molecular orbitals having a given value of ${\displaystyle \lambda }$ (the component of the orbital angular momentum along the internuclear axis) must connect with atomic orbitals having the same value of ${\displaystyle \lambda }$ (i.e. the same value of ${\displaystyle \left|m\right|}$).
2. The electronic parity of the wave function (g or u) must be preserved as ${\displaystyle R}$ varies from ${\displaystyle 0}$ to ${\displaystyle \infty }$.
3. The von Neumann-Wigner non-crossing rule must be obeyed, so that energy curves corresponding to orbitals having the same symmetry do not cross as ${\displaystyle R}$ varies from ${\displaystyle 0}$ to ${\displaystyle \infty }$.

Thus, von Neumann-Wigner non-crossing rule also acts as a starting point for valence bond theory.

In the electric dipole approximation the transition amplitude for emission or absorption of radiation can be shown to be proportional to the vibronic matrix element of the component of the electric dipole operator ${\displaystyle D}$ along the molecular axis. This is the permanent electric dipole moment. In homonuclear diatomic molecules, the permanent electric dipole moment vanishes and there is no pure rotation spectrum (but see N.B. below). Heteronuclear diatomic molecules possess a permanent electric dipole moment and exhibit spectra corresponding to rotational transitions, without change in the vibronic state. For ${\displaystyle \Lambda =0}$, the selection rules for a rotational transition are: ${\displaystyle {\begin{aligned}&\Delta \Im =\pm 1\\&\Delta {{M}_{\Im }}=0,\pm 1\\\end{aligned}}}$. For ${\displaystyle \Lambda \neq 0}$, the selection rules become: ${\displaystyle {\begin{aligned}&\Delta \Im =0,\pm 1\\&\Delta {{M}_{\Im }}=0,\pm 1\\\end{aligned}}}$.This is due to the fact that although the photon absorbed or emitted carries one unit of angular momentum, the nuclear rotation can change, with no change in ${\displaystyle \Im }$, if the electronic angular momentum makes an equal and opposite change. Symmetry considerations require that the electric dipole moment of a diatomic molecule is directed along the internuclear line, and this leads to the additional selection rule ${\displaystyle \Delta \Lambda =0}$.The pure rotational spectrum of a diatomic molecule consists of lines in the far infra-red or the microwave region, the frequencies of these lines given by:

${\displaystyle \hbar {{\omega }_{\Im +1,\Im }}={{E}_{r}}(\Im +1)-{{E}_{r}}(\Im )=2B(\Im +1)}$; where ${\displaystyle B={\frac {{\hbar }^{2}}{2\mu R_{0}^{2}}}}$, and ${\displaystyle \Im \geq \Lambda }$

• N.B. In exceptional circumstances the hyperfine Hamiltonian can mix the rotational levels of g and u vibronic states of homonuclear diatomic molecules giving rise to pure rotational (ortho - para) transitions in a homonuclear diatomic molecule.[6]

The transition matrix elements for pure vibrational transition are ${\displaystyle {{\mu }_{v,v'}}=\left\langle v'|\mu |v\right\rangle }$, where ${\displaystyle \mu }$ is the dipole moment of the diatomic molecule in the electronic state ${\displaystyle \alpha }$. Because the dipole moment depends on the bond length ${\displaystyle R}$, its variation with displacement of the nuclei from equilibrium can be expressed as: ${\displaystyle \mu ={{\mu }_{0}}+{{({\frac {d\mu }{dx}})}_{0}}x+{\frac {1}{2}}{{({\frac {{{d}^{2}}\mu }{d{{x}^{2}}}})}_{0}}{{x}^{2}}+.......}$; where ${\displaystyle {{\mu }_{0}}}$ is the dipole moment when the displacement is zero. The transition matrix elements are, therefore: ${\displaystyle \left\langle v'|\mu |v\right\rangle ={{\mu }_{0}}\left\langle v'|v\right\rangle +{{({\frac {d\mu }{dx}})}_{0}}\left\langle v'|x|v\right\rangle +{\frac {1}{2}}{{({\frac {{{d}^{2}}\mu }{d{{x}^{2}}}})}_{0}}\left\langle v'|{{x}^{2}}|v\right\rangle +.......={{({\frac {d\mu }{dx}})}_{0}}\left\langle v'|x|v\right\rangle +{\frac {1}{2}}{{({\frac {{{d}^{2}}\mu }{d{{x}^{2}}}})}_{0}}\left\langle v'|{{x}^{2}}|v\right\rangle +.......}$using orthogonality of the states. So, the transition matrix is non-zero only if the molecular dipole moment varies with displacement, for otherwise the derivatives of ${\displaystyle \mu }$ would be zero. The gross selection rule for the vibrational transitions of diatomic molecules is then: To show a vibrational spectrum, a diatomic molecule must have a dipole moment that varies with extension. So, homonuclear diatomic molecules do not undergo electric-dipole vibrational transitions. So, a homonuclear diatomic molecule doesn't show purely vibrational spectra.

For small displacements, the electric dipole moment of a molecule can be expected to vary linearly with the extension of the bond. This would be the case for a heteronuclear molecule in which the partial charges on the two atoms were independent of the internuclear distance. In such cases (known as harmonic approximation), the quadratic and higher terms in the expansion can be ignored and ${\displaystyle {{\mu }_{v,v'}}=\left\langle v'|\mu |v\right\rangle ={{({\frac {d\mu }{dx}})}_{0}}\left\langle v'|x|v\right\rangle }$. Now, the matrix elements can be expressed in position basis in terms of the harmonic oscillator wavefunctions: Hermite polynomials. Using the property of Hermite polynomials: ${\displaystyle 2(\alpha x){{H}_{v}}(\alpha x)=2v{{H}_{v-1}}(\alpha x)+{{H}_{v+1}}(\alpha x)}$, it is evident that ${\displaystyle x\left|v\right\rangle }$ which is proportional to ${\displaystyle x{{H}_{v}}(\alpha x)}$, produces two terms, one proportional to ${\displaystyle \left|v+1\right\rangle }$ and the other to ${\displaystyle \left|v-1\right\rangle }$. So, the only non-zero contributions to ${\displaystyle {{\mu }_{v,v'}}}$ comes from ${\displaystyle v'=v\pm 1}$. So, the selection rule for heteronuclear diatomic molecules is: ${\displaystyle \Delta v=\pm 1}$

• Conclusion: Homonuclear diatomic molecules show no pure vibrational spectral lines, and the vibrational spectral lines of heteronuclear diatomic molecules are governed by the above-mentioned selection rule.

Homonuclear diatomic molecules show neither pure vibrational nor pure rotational spectra. However, as the absorption of a photon requires the molecule to take up one unit of angular momentum, vibrational transitions are accompanied by a change in rotational state, which is subject to the same selection rules as for the pure rotational spectrum. For a molecule in a ${\displaystyle \Sigma }$ state, the transitions between two vibration-rotation (or rovibrational) levels ${\displaystyle (v,\Im )}$ and ${\displaystyle (v',\Im ')}$, with vibrational quantum numbers ${\displaystyle v}$ and ${\displaystyle v'=v+1}$, fall into two sets according to whether ${\displaystyle \Delta \Im =+1}$ or ${\displaystyle \Delta \Im =-1}$. The set corresponding to ${\displaystyle \Delta \Im =+1}$ is called the R branch. The corresponding frequencies are given by: ${\displaystyle \hbar {{\omega }^{R}}=E(v+1,\Im +1)-E(v,\Im )=2B(\Im +1)+\hbar {{\omega }_{0}};{\text{ }}\Im =0,1,2,......}$

The set corresponding to ${\displaystyle \Delta \Im =-1}$ is called the P branch. The corresponding frequencies are given by: ${\displaystyle \hbar {{\omega }^{P}}=E(v+1,\Im -1)-E(v,\Im )=-2B\Im +\hbar {{\omega }_{0}};{\text{ }}\Im =1,2,3,......}$

Both branches make up what is called a rotational-vibrational band or a rovibrational band. These bands are in the infra-red part of the spectrum.

If the molecule is not in a ${\displaystyle \Sigma }$ state, so that ${\displaystyle \Lambda \neq 0}$, transitions with ${\displaystyle \Delta \Im =0}$ are allowed. This gives rise to a further branch of the vibrational-rotational spectrum, called the Q branch. The frequencies ${\displaystyle {{\omega }^{Q}}}$ corresponding to the lines in this branch are given by a quadratic function of ${\displaystyle \Im }$ if ${\displaystyle {{B}_{v}}}$ and ${\displaystyle {{B}_{v+1}}}$ are unequal, and reduce to the single frequency: ${\displaystyle \hbar {{\omega }^{Q}}=E(v+1,\Im )-E(v,\Im )=\hbar {{\omega }_{0}}}$ if ${\displaystyle {{B}_{v+1}}={{B}_{v}}}$.

For a heteronuclear diatomic molecule, this selection rule has two consequences:

1. Both the vibrational and rotational quantum numbers must change. The Q-branch is therefore forbidden.
2. The energy change of rotation can be either subtracted from or added to the energy change of vibration, giving the P- and R- branches of the spectrum, respectively.

Homonuclear diatomic molecules also show this kind of spectra. The selection rules, however, are a bit different.

• Conclusion: Both homo- and hetero-nuclear diatomic molecules show rovibrational spectra. A Q-branch is absent in the spectra of heteronuclear diatomic molecules.