Section3Application to Quantum Chemistry

Subsection3.1Introduction of the Problem

In the field of quantum chemistry, linear algebra is heavily utilized and higher order tensors can arise in some specialized problems. In my personal experience, I have encountered 3rd-order tensors while working with the spectroscopic properties of molecules. In practice the issues arise when trying to manipulate 3rd-order tensors, and specifically when trying to rotate a 3rd-order tensor analogous to how a matrix would be rotated for a change in coordinates. For these problems the 3rd-order tensors can arise as an outer product between a matrix and a vector that represent the polarizability and dipole moment, respectively, of a molecule, the outer product of which is the hyperpolarizability tensor. In this case the method used to rotate the hyperpolarizability tensor is to rotate the vector and matrix independently and then recalculating the outer product after each rotation. This method maybe more computationally expensive and could result in numerical inaccuracies after repeated iterations, especially when dealing with small values. This solution is the only one that could be found by my colleagues and me. The true issue arises when the hyperpolarizability tensor is calculated independently by the computational program unless an intuitive user explicitly specifies the separate polarizability and dipole moments, but even then the outer product of the polarizability and dipole moment is only approximately equal to the hyperpolarizability tensor. For this reason it is desirable to have a way to rotate the hyperpolarizability tensor independently.

Subsection3.2Rotation Matrices

In the context of linear algebra and quantum chemistry, a rotation matrix is a matrix that is used to rotate a set of points in space. In the specific context of spectroscopy, a rotation matrix is used to rotate a molecules properties, such as hyperpolarizability represented by a 3rd-order tensor, about an axis or axes. This process is done to orientationally average molecules over axes that are known to allow free rotation. The rotation matrix commonly used to do this is below and the corresponding axes of rotation are shown in Figure 1 [4.2], \[R = \begin{bmatrix} \cos (\phi ) \cos (\psi )-\cos (\theta ) \sin (\phi ) \sin (\psi ) & -\cos (\theta ) \cos (\psi ) \sin (\phi )-\cos (\phi ) \sin (\psi ) & \sin (\theta) \sin (\phi ) \\ \cos (\psi ) \sin (\phi )+\cos (\theta ) \cos (\phi ) \sin (\psi ) & \cos (\theta ) \cos (\phi ) \cos (\psi )-\sin (\phi ) \sin (\psi ) & -\cos (\phi ) \sin (\theta ) \\ \sin (\theta ) \sin (\psi ) & \cos (\psi ) \sin (\theta ) & \cos (\theta ) \\ \end{bmatrix}\] It is immediately evident that the rotation matrix gets complex quickly when rotating an object in 3 dimensions around 3 rotational axes. Despite the complexity, this matrix still maintains a very nice property that all rotational matrices have in common, they are unitary. This is significant because it gives us an easy-to-find inverse for an otherwise complex looking matrix and determinant \(1\) guarantees the matrix operated on by the rotation matrix will not be stretched or compressed.

Figure3.1Axes of rotation as described in Wang et al. (2005)[4.2]

Subsection3.3Rotation by CP Decomposition

The most natural solution to this problem after reviewing the decompositions of higher order tensors is to use a CP decomposition. Above it is shown that the CP decomposition produces a sum of vectors, so to rotate a 3rd order tensor we can simply rotate each of the vectors individually by the same angle and then perform the necessary outer products and summations to form the rotated 3rd order tensor. This is mathematically illustrated for the case of \(\mathcal{X} \in \mathbb{R}^{3 \times 3 \times 3}\) where there exists a CP decomposition. In this particular problem we also know that using a value of \(r=3\) will provide an exact decomposition because the tensor should be the outer product of a \( 3 \times 3\) matrix and vector in \(\mathbb{R}^3\) so the resulting tensor should have the same rank as the matrix. Since the matrix is a calculated numerical matrix, its rank is almost certainly 3, thus the tensor rank is 3. Fortunately in this situation, unlike most, the tensor rank is unambiguous so we can write, \[\mathcal{X} = \sum_{j=1}^{3}a_j \circ b_j \circ c_j.\] Then, \begin{align*} \mathcal{X}_{Rot} & = \sum_{j=1}^{3} (R a_j) \circ (R b_j) \circ (R c_j)\\ & = [R a_1|R a_2|R a_3] \odot [R b_1|R b_2|R b_3] \odot [R c_1|R c_2|R c_3]\\ & = R[a_1|a_2|a_3] \odot R[b_1|b_2|b_3] \odot R[c_1|c_2|c_3]\\ & = R\textbf{A} \odot R\textbf{B} \odot R\textbf{C}\\ & = [[R\textbf{A},R\textbf{B},R\textbf{C}]] \end{align*} This results in only an additional 3 matrix multiplications, which are computationally cheap, but the CP decomposition is computationally expensive. In this case where our 3rd-order tensor is relatively small, the more accurate but more computationally intense PMF3 algorithm can be utilized over the ALS method. This is very beneficial because often in these computations every decimal point is significant and sometimes the number are very small making it more difficult to distinguish between a numerical zero and a significant number. Another thing to keep in mind is that this procedure only applies to a \(3 \times 3 \times 3\) system or a \(2 \times 2 \times 2\) (with a \(2\times 2\) rotational matrix) because rotational matrices have no larger analogs, so the tensors will always be relatively small. However, the decomposition is still nontrivial and is not guaranteed to converge in every situation.

Subsection3.4Proposal of a 3rd-Order Rotational Tensor

Disclaimer: This portion is not necessarily mathematically accurate and will be rather conversational.

After spending hours searching, I have been unable to come across a higher order tensor analog to a rotational matrix. In the literature they repeatedly refer to rotating a 3rd-order tensor as, \[\mathcal{X}^{\prime}_{ijk}=R_{li}R_{mj}R_{nk}\mathcal{X}_{lmn}\] but have never been able to find the matrices that behave in this manner plus I would expect that the 3rd-order tensor would have to be rotated by another 3rd-order tensor. For this reason I have been trying to figure out a way to construct a rotational tensor. While working with the CP decomposition and rotation by multiplying each of the vectors with a rotational matrix, I thought what if I did a similar routine multiplying different combinations of the standard unit vectors with a rotational matrix to form the entries of a 3rd-order rotational tensor. \[\mathcal{R}=\sum R e_i \circ R e_j \circ R e_k\] It is often said it is good enough to see what a matrix/linear transformation does on a basis, so if the right basis was used this could be used to form a 3rd-order rotational tensor. In addition, after such a tensor is created, it would be difficult to define it as a rotational tensor because it is hard to translate the concepts like unitary, inverse, and determinant into terms of a higher order tensor.