Vibrational selection rules for molecules

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

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

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

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

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

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

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

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

With reference to eq32a,

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

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

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

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

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

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

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

Question

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

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

Answer

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

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

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

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

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

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

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

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

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

Expanding eq103,

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

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

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

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

Consequently, we end up with the same two selection rules. For example, the normal mode of the symmetric stretch of $CO_2$ is IR inactive because the net electric dipole moment is always zero and hence ${\biggr$\frac{\partial\boldsymbol{\mathit{\mu}}}{\partial Q_i}\biggr$_e=0}$ , while the normal mode of the molecule undergoing an antisymmetric stretch is IR active because it has a permanent electric dipole moment that varies with the displacement of the molecule.

Vibrational selection rules for molecules

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