Eckart conditions

Separation of external and internal coordinates

The N position vectors ${\displaystyle {\overrightarrow{R}}_A}$ of the nuclei constitute a 3N dimensional linear space R^3N: the configuration space. The Eckart conditions give an orthogonal direct sum decomposition of this space

${\displaystyle {\mathbf{R}}^{3N}={\mathbf{R}}_{\text{ext}}\oplus {\mathbf{R}}_{\text{int}}.}$

The elements of the 3N-6 dimensional subspace R_int are referred to as internal coordinates, because they are invariant under overall translation and rotation of the molecule and, thus, depend only on the internal (vibrational) motions. The elements of the 6-dimensional subspace R_ext are referred to as external coordinates, because they are associated with the overall translation and rotation of the molecule.

To clarify this nomenclature we define first a basis for R_ext. To that end we introduce the following 6 vectors (i=1,2,3):

${\displaystyle \begin{array}{cc}{\overrightarrow{s}}_i^A& \equiv {\overrightarrow{f}}_i\\ {\overrightarrow{s}}_{i+3}^A& \equiv {\overrightarrow{f}}_i\times {\overrightarrow{R}}_A^0.\end{array}}$

An orthogonal, unnormalized, basis for R_ext is,

${\displaystyle {\overrightarrow{S}}_t\equiv \mathrm{row}(\sqrt{M_1}{\overrightarrow{s}}_t^1,\dots ,\sqrt{M_N}{\overrightarrow{s}}_t^N)\mathrm{f}\mathrm{o}\mathrm{r}t=1,\dots ,6.}$

A mass-weighted displacement vector can be written as

${\displaystyle \overrightarrow{D}\equiv \mathrm{col}(\sqrt{M_1}{\overrightarrow{d}}^1,\dots ,\sqrt{M_N}{\overrightarrow{d}}^N)\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}{\overrightarrow{d}}^A\equiv \overrightarrow{\mathbf{F}}\cdot {\mathbf{d}}_A.}$

For i=1,2,3,

${\displaystyle {\overrightarrow{S}}_i\cdot \overrightarrow{D}=\sum _{A=1}^N{M}_A{\overrightarrow{s}}_i^A\cdot {\overrightarrow{d}}^A=\sum _{A=1}^N{M}_A{d}_{Ai}=0,}$

where the zero follows because of the translational Eckart conditions. For i=4,5,6

${\displaystyle {\overrightarrow{S}}_i\cdot \overrightarrow{D}=\sum _{A=1}^N{M}_A({\overrightarrow{f}}_i\times {\overrightarrow{R}}_A^0)\cdot {\overrightarrow{d}}^A={\overrightarrow{f}}_i\cdot \sum _{A=1}^N{M}_A{\overrightarrow{R}}_A^0\times {\overrightarrow{d}}^A=\sum _{A=1}^N{M}_A({\mathbf{R}}_A^0\times {\mathbf{d}}_A{)}_i=0,}$

where the zero follows because of the rotational Eckart conditions. We conclude that the displacement vector ${\displaystyle \overrightarrow{D}}$ belongs to the orthogonal complement of R_ext, so that it is an internal vector.

We obtain a basis for the internal space by defining 3N-6 linearly independent vectors

${\displaystyle {\overrightarrow{Q}}_r\equiv \mathrm{row}(\frac{1}{\sqrt{M_1}}{\overrightarrow{q}}_r^1,\dots ,\frac{1}{\sqrt{M_N}}{\overrightarrow{q}}_r^N),\mathrm{f}\mathrm{o}\mathrm{r}r=1,\dots ,3N-6.}$

The vectors ${\displaystyle {\overrightarrow{q}}_r^A}$ could be Wilson's s-vectors or could be obtained in the harmonic approximation by diagonalizing the Hessian of V. We next introduce internal (vibrational) modes,

${\displaystyle q_r\equiv {\overrightarrow{Q}}_r\cdot \overrightarrow{D}=\sum _{A=1}^N{\overrightarrow{q}}_r^A\cdot {\overrightarrow{d}}^A\mathrm{f}\mathrm{o}\mathrm{r}r=1,\dots ,3N-6.}$

The physical meaning of q_r depends on the vectors ${\displaystyle {\overrightarrow{q}}_r^A}$. For instance, q_r could be a symmetric stretching mode, in which two C—H bonds are simultaneously stretched and contracted.

We already saw that the corresponding external modes are zero because of the Eckart conditions,

${\displaystyle s_t\equiv {\overrightarrow{S}}_t\cdot \overrightarrow{D}=\sum _{A=1}^N{M}_A{\overrightarrow{s}}_t^A\cdot {\overrightarrow{d}}^A=0\mathrm{f}\mathrm{o}\mathrm{r}t=1,\dots ,6.}$

Overall translation and rotation

The vibrational (internal) modes are invariant under translation and infinitesimal rotation of the equilibrium (reference) molecule if and only if the Eckart conditions apply. This will be shown in this subsection.

An overall translation of the reference molecule is given by

${\displaystyle {\overrightarrow{R}}_A^0\mapsto {\overrightarrow{R}}_A^0+\overrightarrow{t}}$'

for any arbitrary 3-vector ${\displaystyle \overrightarrow{t}}$. An infinitesimal rotation of the molecule is given by

${\displaystyle {\overrightarrow{R}}_A^0\mapsto {\overrightarrow{R}}_A^0+\mathrm{\Delta }\phi (\overrightarrow{n}\times {\overrightarrow{R}}_A^0)}$

where Δφ is an infinitesimal angle, Δφ >> (Δφ)², and ${\displaystyle \overrightarrow{n}}$ is an arbitrary unit vector. From the orthogonality of ${\displaystyle {\overrightarrow{Q}}_r}$ to the external space follows that the ${\displaystyle {\overrightarrow{q}}_r^A}$ satisfy

${\displaystyle \sum _{A=1}^N{\overrightarrow{q}}_r^A=\overrightarrow{0}\mathrm{a}\mathrm{n}\mathrm{d}\sum _{A=1}^N{\overrightarrow{R}}_A^0\times {\overrightarrow{q}}_r^A=\overrightarrow{0}.}$

Now, under translation

${\displaystyle q_r\mapsto \sum _A{\overrightarrow{q}}_r^A\cdot ({\overrightarrow{d}}^A-\overrightarrow{t})=q_r-\overrightarrow{t}\cdot \sum _A{\overrightarrow{q}}_r^A=q_r.}$

Clearly, ${\displaystyle {\overrightarrow{q}}_r^A}$ is invariant under translation if and only if

${\displaystyle \sum _A{\overrightarrow{q}}_r^A=0,}$

because the vector ${\displaystyle \overrightarrow{t}}$ is arbitrary. So, the translational Eckart conditions imply the translational invariance of the vectors belonging to internal space and conversely. Under rotation we have,

${\displaystyle q_r\mapsto \sum _A{\overrightarrow{q}}_r^A\cdot ({\overrightarrow{d}}^A-\mathrm{\Delta }\phi (\overrightarrow{n}\times {\overrightarrow{R}}_A^0))=q_r-\mathrm{\Delta }\phi \overrightarrow{n}\cdot \sum _A{\overrightarrow{R}}_A^0\times {\overrightarrow{q}}_r^A=q_r.}$

Rotational invariance follows if and only if

${\displaystyle \sum _A{\overrightarrow{R}}_A^0\times {\overrightarrow{q}}_r^A=\overrightarrow{0}.}$

The external modes, on the other hand, are not invariant and it is not difficult to show that they change under translation as follows:

${\displaystyle \begin{array}{cc}{s}_i& \mapsto s_i+M{\overrightarrow{f}}_i\cdot \overrightarrow{t}\mathrm{f}\mathrm{o}\mathrm{r}i=1,2,3\\ s_i& \mapsto s_i\mathrm{f}\mathrm{o}\mathrm{r}i=4,5,6,\end{array}}$

where M is the total mass of the molecule. They change under infinitesimal rotation as follows

${\displaystyle \begin{array}{cc}{s}_i& \mapsto s_i\mathrm{f}\mathrm{o}\mathrm{r}i=1,2,3\\ s_i& \mapsto s_i+\mathrm{\Delta }\varphi {\overrightarrow{f}}_i\cdot {\mathbf{I}}^0\cdot \overrightarrow{n}\mathrm{f}\mathrm{o}\mathrm{r}i=4,5,6,\end{array}}$

where I⁰ is the inertia tensor of the equilibrium molecule. This behavior shows that the first three external modes describe the overall translation of the molecule, while the modes 4, 5, and, 6 describe the overall rotation.

Vibrational energy

The vibrational energy of the molecule can be written in terms of coordinates with respect to the Eckart frame as

${\displaystyle 2T_{\mathrm{v}\mathrm{i}\mathrm{b}}=\sum _{A=1}^N{M}_A{\dot{\mathbf{R}}}_A\cdot {\dot{\mathbf{R}}}_A=\sum _{A=1}^N{M}_A{\dot{\mathbf{d}}}_A\cdot {\dot{\mathbf{d}}}_A.}$

Because the Eckart frame is non-inertial, the total kinetic energy comprises also centrifugal and Coriolis energies. These stay out of the present discussion. The vibrational energy is written in terms of the displacement coordinates, which are linearly dependent because they are contaminated by the 6 external modes, which are zero, i.e., the d_A's satisfy 6 linear relations. It is possible to write the vibrational energy solely in terms of the internal modes q_r (r =1, ..., 3N-6) as we will now show. We write the different modes in terms of the displacements

${\displaystyle \begin{array}{cc}{q}_r=\sum _{Aj}{d}_{Aj}& (q_{rj}^A)\\ s_i=\sum _{Aj}{d}_{Aj}& (M_A{\delta }_{ij})=0\\ s_{i+3}=\sum _{Aj}{d}_{Aj}& (M_A\sum _k{ϵ}_{ikj}{R}_{Ak}^0)=0\end{array}}$

The parenthesized expressions define a matrix B relating the internal and external modes to the displacements. The matrix B may be partitioned in an internal (3N-6 x 3N) and an external (6 x 3N) part,

${\displaystyle \mathbf{v}\equiv \left(\begin{array}{c}{q}_1\\ ⋮\\ ⋮\\ q_{3N-6}\\ 0\\ ⋮\\ 0\end{array}\right)=\left(\begin{array}{c}{\mathbf{B}}^{\mathrm{i}\mathrm{n}\mathrm{t}}\\ \cdots \\ {\mathbf{B}}^{\mathrm{e}\mathrm{x}\mathrm{t}}\end{array}\right)\mathbf{d}\equiv \mathbf{B}\mathbf{d}.}$

We define the matrix M by

${\displaystyle \mathbf{M}\equiv \mathrm{diag}({\mathbf{M}}_1,{\mathbf{M}}_2,\dots ,{\mathbf{M}}_N)\text{and}{\mathbf{M}}_A\equiv \mathrm{diag}(M_A,M_A,M_A)}$

and from the relations given in the previous sections follow the matrix relations

${\displaystyle {\mathbf{B}}^{\mathrm{e}\mathrm{x}\mathrm{t}}{\mathbf{M}}^{-1}({\mathbf{B}}^{\mathrm{e}\mathrm{x}\mathrm{t}}{)}^{\mathrm{T}}=\mathrm{diag}(N_1,\dots ,N_6)\equiv \mathbf{N},}$

and

${\displaystyle {\mathbf{B}}^{\mathrm{i}\mathrm{n}\mathrm{t}}{\mathbf{M}}^{-1}({\mathbf{B}}^{\mathrm{e}\mathrm{x}\mathrm{t}}{)}^{\mathrm{T}}=\mathbf{0}.}$

We define

${\displaystyle \mathbf{G}\equiv {\mathbf{B}}^{\mathrm{i}\mathrm{n}\mathrm{t}}{\mathbf{M}}^{-1}({\mathbf{B}}^{\mathrm{i}\mathrm{n}\mathrm{t}}{)}^{\mathrm{T}}.}$

By using the rules for block matrix multiplication we can show that

${\displaystyle ({\mathbf{B}}^{\mathrm{T}}{)}^{-1}\mathbf{M}{\mathbf{B}}^{-1}=\left(\begin{array}{ccc}{\mathbf{G}}^{-1}& & \mathbf{0}\\ \mathbf{0}& & {\mathbf{N}}^{-1}\end{array}\right),}$

where G⁻¹ is of dimension (3N-6 x 3N-6) and N⁻¹ is (6 x 6). The kinetic energy becomes

${\displaystyle 2T_{\mathrm{v}\mathrm{i}\mathrm{b}}={\dot{\mathbf{d}}}^{\mathrm{T}}\mathbf{M}\dot{\mathbf{d}}={\dot{\mathbf{v}}}^{\mathrm{T}}({\mathbf{B}}^{\mathrm{T}}{)}^{-1}\mathbf{M}{\mathbf{B}}^{-1}\dot{\mathbf{v}}=\sum _{r,r^{\prime }=1}^{3N-6}(G^{-1}{)}_{r{r}^{\prime }}{\dot{q}}_r{\dot{q}}_{r^{\prime }}}$

where we used that the last 6 components of v are zero. This form of the kinetic energy of vibration enters Wilson's GF method. It is of some interest to point out that the potential energy in the harmonic approximation can be written as follows

${\displaystyle 2V_{\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{m}}={\mathbf{d}}^{\mathrm{T}}\mathbf{H}\mathbf{d}={\mathbf{v}}^{\mathrm{T}}({\mathbf{B}}^{\mathrm{T}}{)}^{-1}\mathbf{H}{\mathbf{B}}^{-1}\mathbf{v}=\sum _{r,r^{\prime }=1}^{3N-6}{F}_{r{r}^{\prime }}{q}_r{q}_{r^{\prime }},}$

where H is the Hessian of the potential in the minimum and F, defined by this equation, is the F matrix of the GF method.

Relation to the harmonic approximation

In the harmonic approximation to the nuclear vibrational problem, expressed in displacement coordinates, one must solve the generalized eigenvalue problem

${\displaystyle \mathbf{H}\mathbf{C}=\mathbf{M}\mathbf{C}\mathbf{\Phi },}$

where H is a 3N × 3N symmetric matrix of second derivatives of the potential ${\displaystyle V({\mathbf{R}}_1,{\mathbf{R}}_2,\dots ,{\mathbf{R}}_N)}$. H is the Hessian matrix of V in the equilibrium ${\displaystyle {\mathbf{R}}_1^0,\dots ,{\mathbf{R}}_N^0}$. The diagonal matrix M contains the masses on the diagonal. The diagonal matrix ${\displaystyle \mathbf{\Phi }}$ contains the eigenvalues, while the columns of C contain the eigenvectors.

It can be shown that the invariance of V under simultaneous translation over t of all nuclei implies that vectors T = (t, ..., t) are in the kernel of H. From the invariance of V under an infinitesimal rotation of all nuclei around s, it can be shown that also the vectors S = (s x R₁⁰, ..., s x R_N⁰) are in the kernel of H :

${\displaystyle \mathbf{H}\left(\begin{array}{c}\mathbf{t}\\ ⋮\\ \mathbf{t}\end{array}\right)=\left(\begin{array}{c}\mathbf{0}\\ ⋮\\ \mathbf{0}\end{array}\right)\mathrm{a}\mathrm{n}\mathrm{d}\mathbf{H}\left(\begin{array}{c}\mathbf{s}\times {\mathbf{R}}_1^0\\ ⋮\\ \mathbf{s}\times {\mathbf{R}}_N^0\end{array}\right)=\left(\begin{array}{c}\mathbf{0}\\ ⋮\\ \mathbf{0}\end{array}\right)}$

Thus, six columns of C corresponding to eigenvalue zero are determined algebraically. (If the generalized eigenvalue problem is solved numerically, one will find in general six linearly independent linear combinations of S and T). The eigenspace corresponding to eigenvalue zero is at least of dimension 6 (often it is exactly of dimension 6, since the other eigenvalues, which are force constants, are never zero for molecules in their ground state). Thus, T and S correspond to the overall (external) motions: translation and rotation, respectively. They are zero-energy modes because space is homogeneous (force-free) and isotropic (torque-free).

By the definition in this article, the non-zero frequency modes are internal modes, since they are within the orthogonal complement of R_ext. The generalized orthogonalities: ${\displaystyle {\mathbf{C}}^{\mathrm{T}}\mathbf{M}\mathbf{C}=\mathbf{I}}$ applied to the "internal" (non-zero eigenvalue) and "external" (zero-eigenvalue) columns of C are equivalent to the Eckart conditions.