Molecular Degrees of Freedom - Quantum Mechanics for Chemistry

Energies of molecules¶

Molecules consisting of N nuclei and n electrons are described by wave functions that depend on 3(n+N) variables in 3D space. These coordinates, or degrees of freedom (DOF), are usefully broken down into different kinds of motion classified as translational, rotational, vibrational and electronic. Because molecules are microscopic objects, we expect all of these energies to be quantized. The relative spacings, however, differ significantly because of the boundary conditions that restrict the motions of these DOFs. This is why different kinds of spectroscopy exist for probing the specific degrees of freedom in molecules.

E= \epsilon_{trans}+ \epsilon_{rot}+ \epsilon_{vib}+\epsilon_{elec}

(1)

mol deg freed — Figure 1:Energy levels associated with the different degrees of freedom in molecules.

3N Nuclear degrees of freedom¶

The Born-Oppenheimer approximation allows us to separate the nuclear and electronic degrees of freedom. The nuclear Hamiltonian for $N$ nuclei can now be written in such a way that the electronic part appears as a potential term:

{\hat{H} = \sum\limits_{i=1}^{N} -\frac{\hbar^2}{2m_i}\nabla_{R_i}^2 + E(R_1, R_2, ..., R_N)}

(2)

In the absence of external electric or magnetic fields, the potential term $E$ depends only on the relative positions of the nuclei, as shown above, and not on the overall position of the molecule or its orientation in space. The above Hamiltonian $H$ can often be approximately written as a sum of the following terms:

{\hat{H} = \hat{H}_{tr} + \hat{H}_{rot} + \hat{H}_{vib}}

(3)

where $H_{tr}$ is the translational, $H_{rot}$ the rotational, and $H_{vib}$ the vibrational Hamiltonian. The translational and rotational terms have no potential part, but the vibrational part contains the potential $E$ , which depends on the distances between the nuclei.

In some cases the different degrees of freedom become coupled and one cannot use the following separation technique. Separation of $H$ means that we can write the wavefunction as a product:

{\psi = \psi_{tr}\psi_{rot}\psi_{vib}}

(4)

Separation of degrees of freedom¶

The resulting three Schrödinger equations are then:

{\hat{H}_{tr}\psi_{tr} = E_{tr}\psi_{tr}}

(5)

{\hat{H}_{rot}\psi_{rot} = E_{rot}\psi_{rot}}

(6)

{\hat{H}_{vib}\psi_{vib} = E_{vib}\psi_{vib}}

(7)

The translational part is not interesting, since there is no external potential or boundary condition that could lead to quantization (i.e., it produces a continuous spectrum).
On the other hand, the rotational part is subject to a cyclic boundary condition and the vibrational part to the potential $E$ , so we expect these to produce quantization, which can be probed by spectroscopic methods.
The original number of variables in the Hamiltonian is $3\times N$ (i.e. the $x,y,z$ coordinates for each nucleus). We can neglect the translational motion, and we are left with $3N - 3$ coordinates.
To account for molecular rotation, three variables are required, or only two variables for a linear molecule. Therefore the vibrational part must have either $3N - 6$ variables for a non-linear molecule or $3N - 5$ variables for a linear molecule. These are referred to as vibrational degrees of freedom or internal coordinates.