15
$\begingroup$

I am interested in calculating the transition dipole moment (TDM) from the information from two wavefunctions of different states. This is somewhat similar to calculating the molecular dipole moment which was previously answered here:

How to calculate molecular dipole moment from a known wavefunction?

The information I have is the molecular orbital coefficients of alpha and beta from states 1 and 2 and the multipole matrix. In considering just one Cartesian direction X, calculating the dipole moment along X would be something like

$$\mathbf{P_a} = \mathbf{C_a} \cdot \mathbf{C_a^T}$$ $$\mathbf{P_b} = \mathbf{C_b} \cdot \mathbf{C_b^T}$$ $$\mathbf{P} = \mathbf{P_a} + \mathbf{P_b}$$ $$\mathbf{\mu_{matrix}} = \mathbf{P} \cdot \mathbf{\text{multipoleMatrixX}}$$ $$\mu_{electronic} = trace(\mathbf{\mu_{matrix}})$$

$$\mu_{nuclear} = \sum_{\text{all atoms}}(Z_{nucleus} * \overrightarrow{r_x})$$

$$\mu_{X} = - \mu_{electronic} + \mu_{nuclear}$$

where $\mathbf{C_a}$ and $\mathbf{C_b}$ are the occupied alpha and beta molecular orbital coefficients, $\mathbf{P_a}$ and $\mathbf{P_b}$ are the alpha and beta density, $\mathbf{P}$ is the total density, and $\mathbf{\text{multipoleMatrixX}}$ is the multipole matrix for the X direction.

I might think that to calculate the transition dipole moment I might just change the densities to the transition densities doing something like I have show below. The transition denisty is based on page 10 of the article here:

$$\mathbf{S_a} = \mathbf{C_{a1}} \cdot \mathbf{C_{a2}^T}$$ $$\mathbf{S_b} = \mathbf{C_{b1}} \cdot \mathbf{C_{b2}^T}$$ $$\mathbf{P_a} = \mathbf{C_{a1}} \cdot \mathbf{C_{a2}^T} \cdot \mathbf{S_a^{-1}}$$ $$\mathbf{P_b} = \mathbf{C_{b1}} \cdot \mathbf{C_{b2}^T} \cdot \mathbf{S_b^{-1}}$$ $$\mathbf{P} = \mathbf{P_a} + \mathbf{P_b}$$ $$\mathbf{\mu_{matrix}} = \mathbf{P} \cdot \mathbf{\text{multipoleMatrixX}}$$ $$\mu_{electronic} = trace(\mathbf{\mu_{matrix}})$$ $$\mu_{nuclear} = \sum_{\text{all atoms}}(Z_{nucleus} * \overrightarrow{r_x})$$

$$\mu_{X,TDM} = - \mu_{electronic} + \mu_{nuclear}$$

where $\mathbf{C_{a1}}$ and $\mathbf{C_{a2}}$ are the occupied alpha molecular orbital coefficients from state 1 and 2, $\mathbf{C_{b1}}$ and $\mathbf{C_{b2}}$ are the occupied beta molecular orbital coefficients from state 1 and 2, $\mathbf{S_a}$ and $\mathbf{S_b}$ are the alpha and beta overlap matricies, $\mathbf{S_a^{-1}}$ and $\mathbf{S_b^{-1}}$ are the inverse of the overlap matrices.

I am unsure how to calculating the TDM since it is in disagreement with known calculated TDM values; some of these transitions should have a TDM of zero due to symmetry considerations. I have been unable to find any discussion on calculating the TDM from two wavefunctions, and most discussions I have seen are in the context of some perturbation on the ground state (i. e. TD or CI). I would be grateful if anyone could provide some suggestions or references to look at.

$\endgroup$
  • 1
    $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. You might want to take a shot of expressing the equations you use in these terms. It is a bit hard to follow as of now. $\endgroup$ – Martin - マーチン Mar 7 '16 at 7:39
  • 1
    $\begingroup$ Can I assume that $\mathbf{C}_{a}$ is a vector and hence $P_a$ should be a scalar, i.e. $P_a = \mathbf{C}_a\cdot\mathbf{C}_a^{\mathbf{T}}$. Or are $\mathbf{C}_{a/b}$ matrices, hence $\mathbf{P}_{a/b}$ are those, too. I'd assume the latter. However I don't know what you mean with $\mathbf{multipoleMatrixX}$. And with $\text{X coordinate}$ do you mean $\mathbf{r}_x$, $\vec{r}_x$ if you prefer arrows? $\endgroup$ – Martin - マーチン Mar 7 '16 at 8:56
  • 1
    $\begingroup$ $\mathbf{P}$ and $\mathbf{C}$ are both matrices. I tried to represent matrices as bold and have scalars be not bold. $\mathbf{MultipoleMatrixX}$ is the multipole matrix along the X direction. The $\overrightarrow{r_x}$ is meant to be the X coordinate $\endgroup$ – brose Mar 7 '16 at 10:49
  • 1
    $\begingroup$ Right now, your use of MathJax makes it even harder to understand, that's why I was asking. It looks like computer pseudo code and not like mathematics. So $$\mathbf{P_a} = dot[\mathbf{C_a},transpose(\mathbf{C_a})]$$ is $$\mathbf{P}_a = \mathbf{C}_a\cdot\mathbf{C}_a^{\mathbf{T}}?$$ And I have no idea what a multipole matrix along the X direction is, can you please include the definition of such a thing. $\endgroup$ – Martin - マーチン Mar 7 '16 at 11:06
  • 1
    $\begingroup$ Unfortunately, it is not that simple. The of diagonal elements are needed. I believe this is to account for the phase, i. e. if the relevant orbitals are perfectly out of phase the coupling is zero, regardless of what the state dipoles are. $\endgroup$ – brose Dec 10 '16 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.