Basis Set Spectra
We’ll assume we’ve already done a basis-set calculation to get a set of energies ${E_n}$ and wavefunctions ${\psi_n}$.
The key relationship we want to model is
\[I_{n,m} \propto \nu_{n,m} {\left\lvert \left\langle \psi_n | \mu | \psi_m \right\rangle \right\rvert}^{2}\]From what we already have, $\nu_{n,m} = E_m - E_n$ is easy to calculate, so we’ll give most of our attention to to the transition moment.
Recalling our discussion on evaluating properties with a basis set approach, we know that to do this we first need to build a representation for the dipole moment in our original basis, ${\phi_i}$. This is the hard part, and we’ll look at the special case of a harmonic oscillator basis in a bit.
Assuming, however, that we have the representation
\[\textbf{M}_{i,j} = \left\langle \phi_i \lvert \mu \rvert \phi_j \right\rangle\]we then know that we can get
\[\left\langle \psi_n | \mu | \psi_m \right\rangle = C^{n} \textbf{M} C^{m}\]The Dipole Moment Vector
There’s one subtlety to this, which we’ve been sweeping under the rug until now, which is that the dipole moment is a three-dimensional vector. I.e. for every set of molecular coordinates $\textbf{r}$ we have a component of the dipole moment that points along the x-axis, one that points along the y-axis, and one that points along the z-axis.1
That means when we evaluate
\[\textbf{M}_{i,j} = \left\langle \phi_i \lvert \mu \rvert \phi_j \right\rangle\]we really need to calculate, e.g.
\[\begin{align} \textbf{M}^{x}_{i,j} &= \left\langle \phi_i \lvert \mu^{x} \rvert \phi_j \right\rangle \\ \textbf{M}^{y}_{i,j} &= \left\langle \phi_i \lvert \mu^{y} \rvert \phi_j \right\rangle \\ \textbf{M}^{z}_{i,j} &= \left\langle \phi_i \lvert \mu^{z} \rvert \phi_j \right\rangle \end{align}\]and then our squared-transition moment is actually the norm of the transition moment squared, i.e.
\[{\left\lvert \left\langle \psi_n | \mu | \psi_m \right\rangle \right\rvert}^{2} = {\left\lvert \left\langle \psi_n | \mu^{x} | \psi_m \right\rangle \right\rvert}^{2} + {\left\lvert \left\langle \psi_n | \mu^{y} | \psi_m \right\rangle \right\rvert}^{2} + {\left\lvert \left\langle \psi_n | \mu^{z} | \psi_m \right\rangle \right\rvert}^{2}\]The 1D Harmonic Oscillator
To help build intuition, we’ll return to the simplest basis for studying vibrations, the 1D harmonic oscillator eigenfunctions. We won’t go over their properties again, since we’ve already done that. Instead we’ll just remember one thing. For the coordinate operator, $\hat{r}$, we have
\[\langle i \lvert \hat{r} \rvert j \rangle = \begin{cases} \sqrt{\frac{i+1}{2}} & j=i+1 \\ \sqrt{\frac{i}{2}} & j=i-1 \\ 0 & \text{else} \end{cases}\]Why is this relevant? Well we can write out a Taylor expansion of our dipole function about our equilibrium structure like
\[\mu(r) = \sum_k \frac{1}{k!} \frac{d^k \mu}{d r^k}(r_e) \Delta r^k\]and in general it’s not a bad approximation to truncate this at first order. We often call this the linear dipole approximation.
Again, you may be thinking, “okay…why do I care?”, well we’ll first say that $r_e = 0$, since the origin is arbitrary in the first place, and then our truncated expression becomes
\[\mu(r)^{(1)} = \mu_0 + \mu_0^{\prime} r\]now when we build our representation of the dipole like
\[\textbf{M}_{i,j} = \left\langle \phi_i \lvert \mu(r)^{(1)} \rvert \phi_j \right\rangle\]we get
\[\langle i \lvert \mu(r)^{(1)} \rvert j \rangle = \begin{cases} \mu_0 & j=i \\ \sqrt{\frac{i+1}{2}} \mu_0^{\prime} & j=i+1 \\ \sqrt{\frac{i}{2}} \mu_0^{\prime} & j=i-1 \\ 0 & \text{else} \end{cases}\]and so for a harmonic oscillator under the linear dipole approximation, transitions are allowed only for states that differ by one quantum, i.e. only when $j = i+1$ or $j=i-1$ is the transition moment non-zero. For many systems, this is actually a very good approximation. For what we tend to be interested in, it is not.
One thing we often want to account for is the non-linearity in the dipole moment. That means truncating our Taylor expansion at some higher order, say at second order
\[\mu(r)^{(2)} = \mu_0 + \mu_0^{\prime} r + \frac{\mu_0^{\prime\prime}}{2} r^2\]then when we evaluate matrix elements, writing $r^2 = rr$, we get (try showing this for yourself)
\[\langle i \lvert \mu(r)^{(2)} \rvert j \rangle = \begin{cases} \mu_0 + \frac{(2i + 1)}{2} \frac{\mu_0^{\prime\prime}}{2} & j=i \\ \sqrt{\frac{i+1}{2}} \mu_0^{\prime} & j=i+1 \\ \sqrt{\frac{i}{2}} \mu_0^{\prime} & j=i-1 \\ \sqrt{\frac{i+1}{2}}\sqrt{\frac{i+2}{2}} \frac{\mu_0^{\prime\prime}}{2} & j=i+2 \\ \sqrt{\frac{i-1}{2}}\sqrt{\frac{i}{2}} \frac{\mu_0^{\prime\prime}}{2} & j=i-2 \\ 0 & \text{else} \end{cases}\]What is noteworthy about this is that transitions are allowed for states that differ by one or two quanta. In general, we find that if we want to use harmonic oscillators to see transition with up to $n$ quanta of excitation, we need to expand our dipole moment out the the $n^{th}$ order.
Obviously, real systems are not perfectly harmonic and real dipoles aren’t perfectly linear. To investigate real world systems like that, we’ll often move to something like a discrete variable representation approach.
From Transition Moment to Intensity
It is importnant to note that these methods outline how to get a dipole moment vector and not a proper intesity. Not to worry though, the dipole moment vector is the tricky part.
First, recall the intensity formula
we have half of this, but to get a proper intensity we need to multiply by $\nu_{n,m}$ and of course, make sure you are tracking your units. (Wavenumbers is probably a good place to start.) If you choose to stick with this, you will probably want to convert it to a relative intensity where your transistions are normalized so that either the sum is 1 or one of the fundamental transitions is 1.
Second, if don’t want a relative intensity, you can implement one of the unit conversions described here.
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