Is it true that one can calculate the vibrational spectrum of a bond by the Fourier transform of the dipole moment vector autocorrelation function $C_{\mu \mu}(t)$?

For example, suppose that I have a diatomic molecule $\text{A-B}$, containing atoms $\text{A}$ and $\text{B}$.

$\text{A}$ and $\text{B}$ have partial charges $q_A$ and $q_B$ that sum to zero: $q_A + q_B = 0$.

I can find the dipole moment vector $\vec{\mu}$ of this diatomic molecule:

$$\vec{\mu}(t) = \sum_i q_i \vec{r}_i^{\prime} = q_A \vec{r}_A^{\prime} + q_B \vec{r}_B^{\prime}$$

where $\vec{r}_A^{\prime}(t)$ and $\vec{r}_B^{\prime}(t)$ are the position vectors of the charges (atoms). These position vectors depend on time $t$ because the charges (atoms) move -- the bond vibrates. Thus the dipole moment vector $\vec{\mu}(t)$ also depends on time $t$.

I think that the autocorrelation function $C_{\mu\mu}(t)$ can be written

$$C_{\mu\mu}(t) = \langle \vec{\mu}(t) \cdot \vec{\mu}(0) \rangle$$

where $\langle ... \rangle$ denotes an ensemble average.

Now, $C_{\mu\mu}(t)$ is in the time domain. It has dimensions of $\text{(dipole moment)}^2$ (typically, $\text{Debye}^2$ in chemistry). But I would like to compute the vibrational spectrum (a plot of intensity, arbitrary units, etc. versus frequency). Is it true that the vibrational spectrum is the Fourier transform of $C_{\mu\mu}(t)$?

  • 1
    $\begingroup$ It is the imaginary part of the Fourier transform. $\endgroup$
    – Fabian
    Dec 15, 2012 at 20:01
  • 1
    $\begingroup$ This isn't right. $\endgroup$
    – KDN
    Dec 16, 2012 at 4:20

2 Answers 2


The power-spectral density is the Fourier transform of the autocorrelation function. It is not simply the imaginary part; it will depend on your data. Your data is real, so the spectrum is hermitian, i.e., $A(z)=A(-z)*$. If your data is an even function (symmetric about 0, like cosine), then the transform will be purely real. If it is an odd function, (like sine) then the transform will be purely imaginary. If it is neither even nor odd, then the transform will have both real and imaginary parts.

When you want to know what the power spectrum for a periodic signal is, you generally mean to include the power at both the frequency and it's negation, i.e., colloquially, the "power at frequency $f$" is actually $P(f)+P(-f)$. It is this condition that keeps you from having imaginary power, since your real-valued data has a hermitian Fourier transform.

You have to be careful what you mean by "Fourier transform", since this applies to periodic/infinite signals, and you are almost certainly dealing with finite time, and (at least slightly) aperiodic signals. You can handle this by performing a normalized (short-time) Fourier transform. This is the usual method, and it is frequently referred to as a Fourier transform, but mathematically, it is somewhat different.


What you are considering is the permanent dipole moment \begin{equation} \mu = \langle\psi(r)|\hat{\mu}|\psi(r)\rangle \tag{1} \end{equation} and its fluctuation and correlation in time. However, the vibrational spectrum is related to the transition dipole moment \begin{equation} \mu_{fi} = \langle\psi_{f}(r)|\hat{\mu}|\psi_{i}(r)\rangle \tag{2} \end{equation} where subscripts $i$ and $f$ stand for initial and final vibrational states. To see what kind of dipole moment is relevant, let me derive below why the correlation function of dipole moment gives a vibrational (or any other type of) spectrum.

A macroscopic polarization $P$ associated with $N$ atoms (molecules) in a volume $V$ is given by the expectation value of the dipole operator $\hat{\mu}$ as \begin{equation} P(t) = \frac{N}{V}\langle\hat{\mu}(t)\rangle = \frac{N}{V}\mathrm{Tr}\left[\hat{\mu}(t)\rho(t)\right] \tag{3} \end{equation} Schrödinger equation for density matrix in the interaction picture reads as \begin{equation} i\hbar\frac{d}{dt}\rho(t) = \left[\hat{V}(t),\rho(t)\right] \tag{4} \end{equation} where $\hat{V}(t) = \hat{\mu}(t)\cdot \hat{E}(t)$ is the light-matter interaction term. From the perturbative expansion, the density matrix at time $t$ is \begin{equation} \rho(t) = \rho(0) - \frac{i}{\hbar}\int_{0}^{t}dt'\left[\hat{V}(t),\rho(t)\right] \tag{5} \end{equation} By plugging Eq. (5) into Eq. (3), the macroscopic polarization turns out to be proportional to the correlation function of the transition dipole moment \begin{equation} P(t) \propto \langle\hat{\mu}(t)\hat{\mu}(0)\rangle \equiv C_{\mu \mu}(t) \tag{6} \end{equation} and it is proportional to the transmitted electric field, that is, vibrational spectrum. Here, the dipole moments are the transition dipole moment defined in Eq. (2) associated with two vibrational states. In practice, the interaction between the incident light field and multiple vibrational states need to be considered.

Note that power spectral density is a Fourier transform of electric-field correlation function (Wiener-Khintchine theorem) \begin{equation} F(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}d\tau\langle E^{*}(t)E(t+\tau)\rangle e^{i\omega\tau} \tag{7} \end{equation} where $E(t)$ is the transmitted light field that has interacted with ensemble of diatomic molecules, which contains the information of $C_{\mu \mu}(t)$. Therefore, the Fourier transform of the correlation function of transition dipole moment is directly proportional to the vibrational spectra.


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