Matrix Elements of One Dimensional Explicitly Correlated Gaussian Basis Functions

Abstract

Kinetic, potential and overlap matrix elements of one dimensional correlated Gaussians multiplied by polynomial factors are presented. These matrix elements can be used to calculate energies of one dimensional cold atom systems, or to construct a tensorial product to calculate energies in 2 or 3 dimensional systems with a nonspherical potential.

Appendices

Appendix A

In this appendix we show an example how to generate integers satisfying Eq. (17). First we rewrite Eq. (17) as

$$\begin{aligned} k_{1,p} + k_{2,p}+ \cdots + 2k_{p,p} + \cdots + k_{p,2N}= m_{p} \end{aligned}$$

(56)

This equation can be written in form of a linear equation by reindexing $k_{1,p},{\ldots },k_{p,2N}$ to form a vector, q, and defining an S matrix for the the coefficients of q:

$$\begin{aligned} \sum _{i=1}^{2N} q_{i}S_{ip} + \sum _{k=2N+1}^{2N + \frac{2N(2N-1)}{2}} q_{k}S_{kp}=m_{p} \end{aligned}$$

(57)

To illustrate how this can be done we give a $N=2$ example showing the equivalence between Eqs. (56) and (57).

Starting with Eq. (56) for $N=2$,

$$\begin{aligned}&2k_{1,1} + k_{1,2} + k_{1,3} + k_{1,4}=m_{1}\nonumber \\&2k_{2,2} + k_{1,2} + k_{2,3} + k_{2,4}=m_{2}\nonumber \\&2k_{3,3} + k_{1,3} + k_{2,3} + k_{3,4}=m_{3}\nonumber \\&2k_{4,4} + k_{1,4} + k_{2,4} + k_{3,4}=m_{4}\nonumber \\ \end{aligned}$$

(58)

Writing out the summations of Eq. (57),

$$\begin{aligned}&q_{1}S_{1,1} + q_{2}S_{2,1} + q_{3}S_{3,1} + q_{4}S_{4,1} \nonumber \\&\quad + q_{5}S_{5,1}+ q_{6}S_{6,1} + q_{7}S_{7,1} + q_{8}S_{8,1} + q_{9}S_{9,1} + q_{10}S_{10,1} = m_{1} \end{aligned}$$

(59)

$$\begin{aligned}&q_{1}S_{1,2} + q_{2}S_{2,2} + q_{3}S_{3,2} + q_{4}S_{4,2} \nonumber \\&\quad + q_{5}S_{5,2}+ q_{6}S_{6,2} + q_{7}S_{7,2} + q_{8}S_{8,2} + q_{9}S_{9,2} + q_{10}S_{10,2} = m_{2} \end{aligned}$$

(60)

$$\begin{aligned}&q_{1}S_{1,3} + q_{2}S_{2,3} + q_{3}S_{3,3} + q_{4}S_{4,3} \nonumber \\&\quad + q_{5}S_{5,3}+ q_{6}S_{6,3} + q_{7}S_{7,3} + q_{8}S_{8,3} + q_{9}S_{9,3} + q_{10}S_{10,3} = m_{3} \end{aligned}$$

(61)

$$\begin{aligned}&q_{1}S_{1,4} + q_{2}S_{2,4} + q_{3}S_{3,4} + q_{4}S_{4,4} \nonumber \\&\quad + q_{5}S_{5,4}+ q_{6}S_{6,4} + q_{7}S_{7,4} + q_{8}S_{8,4} + q_{9}S_{9,4} + q_{10}S_{10,4} = m_{4} \end{aligned}$$

(62)

To define a mapping between q and k, let $q_i = k_{i,i}$ for $i=1,2N$. This is the diagonal of the matrix k. For $i>2N$, we assign the remaining $q_{i}$ to the elements for which $i<j$, left to right and top to down (with x denoting redundant elements due to symmetry):

$$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} q_1 &{} q_5 &{} q_6 &{} q_{7}\\ x &{} q_2 &{} q_8 &{} q_9\\ x &{} x &{} q_3 &{} q_{10}\\ x &{} x &{} x &{} q_4\\ \end{array} \right) \rightarrow \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} k_{11} &{} k_{12} &{} k_{13} &{} k_{14}\\ x &{} k_{22} &{} k_{23} &{} k_{24}\\ x &{} x &{} k_{33} &{} k_{34}\\ x &{} x &{} x &{} k_{44}\\ \end{array} \right) \end{aligned}$$

(63)

Note that q is not a matrix, it is only illustrated this way for clarity.

The selection matrix S contains integer elements of 0, 1, or 2. This will account for (by setting the value of S equal to 0) the extra terms in Eqs. (59)–(62) that are not in Eq. (58), and account for the coefficient of 2 in front of some terms.

Rewriting Eq. (58) with this mapping,

$$\begin{aligned}&2q_1 + q_5 + q_6 + q_7 = m_1 \nonumber \\&2q_2 + q_5 + q_8 + q_9 = m_2 \nonumber \\&2q_3 + q_6 + q_8 + q_{10} = m_3 \nonumber \\&2q_4 + q_7 + q_9 + q_{10} = m_4 \end{aligned}$$

(64)

The matrix S must then be

$$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 2 &{} 0 &{} 0 &{} 0\\ 0 &{} 2 &{} 0 &{} 0\\ 0 &{} 0 &{} 2 &{} 0\\ 0 &{} 0 &{} 0 &{} 2\\ 1 &{} 1 &{} 0 &{} 0\\ 1 &{} 0 &{} 1 &{} 0\\ 1 &{} 0 &{} 0 &{} 1\\ 0 &{} 1 &{} 1 &{} 0\\ 0 &{} 1 &{} 0 &{} 1\\ 0 &{} 0 &{} 1 &{} 1\\ \end{array} \right) \end{aligned}$$

(65)

Once the S matrix is defined, Eq. (57) can be solved for any given set of $m_p$ by finding the integers satisfying the equation.

Appendix B

In this appendix we show how to calculate a Gaussian integral used in the derivation of the matrix elements. Equations similar to this have been published before (see e.g. in Ref. [41]), but here we prove it for a general complex symmetric matrix.

Let $A \in \mathbb {C}^{N\times N}$ be a complex-symmetric matrix with positive-definite real part. Then for any $Q \in \mathbb {C}^{N \times N}$ and $s \in \mathbb {C}^N$

$$\begin{aligned}&\int _{\mathbb {R}^N} \frac{1}{2} \mathbf x ^T Q \mathbf x \exp \left( -\frac{1}{2} \mathbf x ^T A \mathbf x + \mathbf s ^T \mathbf x \right) d\mathbf x \nonumber \\&\quad =\sqrt{\frac{2\pi }{\det B}}\exp \left( \frac{1}{2} \mathbf s ^T A^{-1} \mathbf s \right) \left[ \frac{1}{2} \hbox {Tr}(Q A^{-1} + \frac{1}{2}(A^{-1} \mathbf s )^T Q (A^{-1} \mathbf s )\right] \end{aligned}$$

(66)

Proof

As in the preceding proof, pick an invertible matrix $V \in \mathbb {R}^{N \times N}$ such that $D \equiv {V^T AV}$ is diagonal, and let $R \equiv {V^T QV}$ and $\mathbf u \equiv {V^T} \mathbf s$. Then by performing a change of variable $\mathbf y = {V}^T \mathbf x$, we can write

$$\begin{aligned}&I \equiv \int _{\mathbb {R}^N} \frac{1}{2} \mathbf x ^T Q \mathbf x \exp \left( -\frac{1}{2} \mathbf x ^T A \mathbf x + \mathbf s ^T \mathbf x \right) d\mathbf x \nonumber \\&\quad = \int _{\mathbb {R}^N} \frac{1}{2} \mathbf x ^T Q \mathbf x \exp \left( -\frac{1}{2} \mathbf x ^T V^{-T} D V^{-1} \mathbf x + \mathbf s ^T \mathbf x \right) d\mathbf x \nonumber \\&\quad = \int _{\mathbb {R}^N} \frac{1}{2} \mathbf y ^T V^T Q V \mathbf y \exp \left( -\frac{1}{2} \mathbf y ^T D \mathbf y + \mathbf s ^T V\mathbf y \right) \det V d\mathbf y \nonumber \\&\quad = \frac{\det V}{2} \int _{\mathbb {R}^N} \mathbf y ^T R \mathbf y \exp \left( -\frac{1}{2} \mathbf y ^T R \mathbf y + \mathbf u ^T \mathbf y \right) d\mathbf y \nonumber \\&\quad = \frac{\det V}{2} \int _{\mathbb {R}^N} \sum _{j,k=1}^N R_{jk} y_j y_k \exp \left( \frac{1}{2} \sum _{l=1}^N d_l y_{l}^2 + \sum _{l=1}^N u_l y_l \right) d\mathbf y \nonumber \\&\quad = \frac{\det V}{2} \sum _{j,k=1}^N R_{jk} \int _{\mathbb {R}^N} y_j y_k \prod _{l=1}^N \exp \left( -\frac{1}{2}d_l y_{l}^2 + u_l y_l \right) d\mathbf y \end{aligned}$$

(67)

At this point, we consider separately the $j=k$ and $j \ne k$ terms. When $j=k$, we want to evaluate the integral

$$\begin{aligned} I_j \equiv \int _{\mathbb {R}^N} y_{j}^2 \prod _{l=1}^N \exp \left( -\frac{1}{2} d_l y_{l}^2 + u_l y_l \right) d\mathbf y \end{aligned}$$

(68)

Observe that each term with $l \ne j$ contributes a factor of

$$\begin{aligned} \int _{-\infty }^\infty \exp \left( -\frac{1}{2} d_l y_{l}^2 + u_l y_l\right) dy_l = \sqrt{\frac{2 \pi }{d_l}} \exp \left( \frac{u_{l}^2}{2d_l} \right) \end{aligned}$$

(69)

while the $l=j$ term contributes a factor of

$$\begin{aligned} \int _{-\infty }^\infty \exp \left( -\frac{1}{2} d_j y_{j}^2 + u_j y_j\right) dy_j = \left( \frac{d_j + u_{j}^2}{d_{j}^2}\right) \sqrt{\frac{2 \pi }{d_j}} \exp \left( \frac{u_{j}^2}{2d_j} \right) \end{aligned}$$

(70)

multiplying these factors gives the result

$$\begin{aligned} I_j = \left( \frac{d_j + u_{j}^2}{d_{j}^2} \right) \prod _{l=1}^N \sqrt{\frac{2 \pi }{d_l}} \exp \left( \frac{u_{j}^2}{2d_j} \right) = \left( \frac{d_j + u_{j}^2}{d_{j}^2}\right) \sqrt{\frac{(2 \pi )^N}{\det D}} \exp \left( \frac{1}{2} \mathbf s ^T A^{-1} \mathbf s \right) \end{aligned}$$

(71)

When $j \ne k$ we want to evaluate the integral

$$\begin{aligned} I_{jk} \equiv \int _{{\mathbb {R}^N}} y_j y_k \prod _{l=1}^N \exp \left( -\frac{1}{2} d_l y_{l}^2 + u_l y_l \right) d\mathbf y \end{aligned}$$

(72)

Again, each term with $l \ne j,k$, contributes a factor of $\sqrt{\frac{2 \pi }{d_l}}\exp (\frac{u_{l}^2}{2d_l})$, while the $l=j,k$ terms contribute a factor of

$$\begin{aligned} \int _{-\infty }^\infty y_j \exp \left( -\frac{1}{2} d_j y_{j}^2 + u_j y_j \right) dy_j = (\frac{u_j}{d_j})\sqrt{\frac{2 \pi }{d_j}} \exp \left( \frac{u_{j}^2}{2d_j} \right) \end{aligned}$$

(73)

Multiplying these factors gives the result

$$\begin{aligned} I_{jk} = \left( \frac{u_j u_k}{d_j d_k}\right) \prod _{l=1}^N \sqrt{\frac{2 \pi }{2d_l}} \exp \left( \frac{u_{j}^2}{2d_j} \right) = \left( \frac{u_j u_k}{d_j d_k}\right) \sqrt{\frac{(2 \pi )^N}{\det D}} \exp \left( \frac{1}{2} \mathbf s ^T A^{-1} \mathbf s \right) \end{aligned}$$

(74)

Returning to the integral, we have

$$\begin{aligned} I= & {} \frac{\det V}{2} \Big [\sum _{j=1}^N R_{jj} I_j + \sum _{j \ne k}^N R_{jk} I_{jk} \Big ] \nonumber \\= & {} \frac{\det V}{2} \sqrt{\frac{(2 \pi )^N}{\det D}} \exp \left( \frac{1}{2} \mathbf s ^T A^{-1} \mathbf s \right) \Big [\sum _{j=1}^N R_{jj} \frac{d_j + u_{j}^2}{d_{j}^2} + \sum _{j \ne k}^N R_{jk} \frac{u_j u_k}{d_j d_k} \Big ] \nonumber \\= & {} \frac{1}{2} \sqrt{\frac{(2 \pi )^N}{\det (V^{-T} D V^{-1})}} \exp \left( \frac{1}{2} \mathbf s ^T A^{-1} \mathbf s \right) \Big [ \sum _{j=1}^N \frac{R_{jj}}{d_j} + \sum _{j,k=1}^N R_{jk} \frac{u_j u_k}{d_j d_k} \nonumber \\= & {} \sqrt{\frac{(2 \pi )^N}{\det D}} \exp \left( \frac{1}{2} \mathbf s ^T A^{-1} \mathbf s \right) \Big [ \frac{1}{2} \hbox {Tr}(R D^{-1}) + \frac{1}{2} \mathbf u ^T D^{-T} T D^{-1} \mathbf u \Big ] \nonumber \\= & {} \sqrt{\frac{(2 \pi )^N}{\det D}} \exp \left( \frac{1}{2} \mathbf s ^T A^{-1} \mathbf s \right) \Big [\frac{1}{2} \hbox {Tr}((V^T Q V)(V^T A V)^{-1}) \nonumber \\&+ \frac{1}{2} \mathbf u ^T (V^T A V)^{-T}(V^T Q V)(V^T A V)^{-1} \mathbf u \Big ] \nonumber \\= & {} \sqrt{\frac{(2 \pi )^N}{\det D}} \exp \left( \frac{1}{2} \mathbf s ^T A^{-1} \mathbf s \right) \Big [ \frac{1}{2} \hbox {Tr}(Q A^{-1}) + \frac{1}{2}(A^{-1}{} \mathbf s )^{T} Q (A^{-1} \mathbf s ) \Big ], \end{aligned}$$

(75)

which is the desired result. $\square$