My question stems from a derivation given in chapter 12 of R. McWeeny's Methods of Molecular Quantum Mechanics for solving linear response equations via variational perturbation theory. (Linear response correlation functions for a little more context).

My issue stems from a strange choice the author makes in defining the perturbation part of the Hamiltonian. They write the perturbation as $\mathbf{H}'(t)=F(t)\mathbf{A}$ where they describe $F(t)$ as a time dependent strength function and $A$ as some sort of shape function for the perturbation.

He then expresses the Hamiltonian using a Fourier Transform as $$\mathbf{H}'(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(\omega)\frac{1}{2}(\mathbf{A}_\omega e^{-i\omega t}+\mathbf{A}_{-\omega} e^{i\omega t}) d\omega$$ with $\mathbf{A}=\frac{1}{2}(\mathbf{A}_\omega+\mathbf{A}_{-\omega})$, assuming $\mathbf{A}_{-\omega}=\mathbf{A}^\dagger_\omega$, and $f(\omega)=f(-\omega)$ with $f(\omega)$ the Fourier transform of $F(t)$.

Now here is where I get confused. He goes on to define $\mathbf{H}'(\omega)=\frac{1}{2}(\mathbf{A}_\omega e^{-i\omega t}+\mathbf{A}_{-\omega}e^{i\omega t})$. To me the notation is already misleading enough, as it seems to imply that this is the Fourier Transform of $\mathbf{H}'$, but he also directly replaces instances of $\mathbf{H}'(t)$ with $\mathbf{H}'(\omega)$ to solve them. For instance, he gives an initial expression for the first order coefficients of the perturbed wavefunction as: $$c_n(t)=\frac{1}{i\hbar}\int_{-\infty}^t \left<n|\mathbf{A}|0\right> F(t')\exp(i\omega_{0n}t')dt'$$ Then on the next page, he gives another expression by substituting $\mathbf{H}'(\omega)$ (with a convergence factor $\epsilon$ also introduced): $$c_n(t)=-\frac{1}{2\hbar}\big[\frac{\left<n|\mathbf{A}|0\right>}{\omega_{0n}-\omega-i\epsilon}\exp(i(\omega_{0n}-\omega-i\epsilon))+(\omega\to -\omega)\big]$$

Which would be the expression obtained if $\mathbf{H}'(\omega)$ directly replaced $\mathbf{H}'(t)$. I had initially glossed over this section because it didn't have much bearing on the later sections, but this use of $\mathbf{H}'(\omega)$ has cropped up again and I'm still uncertain how/why it is used.

Is there an explanation for what the author is doing with this factor or whether there is some error in the formulas? My guess was that maybe the second expression for the coefficients should be a function of $\omega$ instead of time, but I can't confirm for certain if that makes sense.

In searching online, I found another use of this format, though it seems to possibly be derived from my source. Has anyone else encountered this before?


1 Answer 1


I think the author is simply considering a single Fourier component of the perturbation at a time. The mathematics is exactly the same for each Fourier component, and so at the end you would simply need to sum (or integrate) over all the Fourier components of $F(t)$ to get the total transition amplitude.

More explicitly, the author is looking specifically at the case where $f(\omega)=\delta(\omega-\omega_0)$, which implies that $F(t)=e^{i\omega_0 t}$. Once you have solved this case, you can use superposition to solve the case where $f(\omega)$ is a sum of delta-functions, or even some continuous function of $\omega$. All you would need to do is properly integrate the expression you have for $c_n(t)$ over the frequency components.

However, the key point with linear response is that you just look at the response at a single frequency at a time. One could, in principle, look at the response to a distribution of frequencies $F(t)$, but that is not very useful in general. It is easy to go from the response at individual frequencies to the response to a sum of frequencies, but not vice-versa.

Practically, we want to know the response of the system at every $\omega$ individually, and not some integral over $\omega$. If you imagine the response of a molecule to light, then you would want to know the absorption of the molecule as a function of the incident light frequency $\omega$. This gives a lot more information that the response of the molecule to some arbitrary distribution of light (say from a thermal lamp).

  • $\begingroup$ So are the $c_n(t)$ not the same between the two expressions? Because they suggest that to get the time dependent response, I would multiply the expression by $f(\omega)$ and then integrate over $\omega$, but that would seem to give me something different than $c_n(t)$ on the left hand side. $\endgroup$
    – Tyberius
    May 31, 2018 at 19:29
  • 1
    $\begingroup$ I think the author is being a bit sloppy with notation overall. In the equations you give, think of $c_n(t)$ as one piece of the true $\mathbf{c}_{n}(t)$ that corresponds to just a single frequency response. $\endgroup$
    – KF Gauss
    May 31, 2018 at 19:49
  • $\begingroup$ That makes more sense. I've recognized that he is sloppy and has a lot of typos, so the crux of the question was really getting at what he was actually trying to say. $\endgroup$
    – Tyberius
    May 31, 2018 at 19:54
  • $\begingroup$ @Tyberius which aspect of your question do you need more detail on? $\endgroup$
    – KF Gauss
    Jun 1, 2018 at 20:38
  • $\begingroup$ you might have answered it, I just need the free time to make sure I'm understanding it right. $\endgroup$
    – Tyberius
    Jun 1, 2018 at 20:46

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