21
$\begingroup$

In the time-dependent Kohn-Sham formalism the effective potential on electrons is given by $$ v_s[\rho(\mathbf{r},t)]=v(\mathbf{r},t)+v_H(\mathbf{r},t)+v_{xc}[\rho(\mathbf{r},t)] $$ where, $v(\mathbf{r},t)$ is the external TD potential, $v_H(\mathbf{r},t)$ is the TD Hartree potential and $v_{xc}[\rho(\mathbf{r},t)]$ is the TD exchange-correlation potential. To get accurate results, we need to find a good approximation to $v_{xc}$ and a common starting point is the adiabatic approximation: $$ v_{xc}[\rho(\mathbf{r},t)]= \left.v_{xc}[\rho_0(\mathbf{r})]\right|_{\rho_0(\mathbf{r})=\rho(\mathbf{r},t)} $$ where $v_{xc}[\rho_0(\mathbf{r})]$ is the static XC potential. As said in Ullrich's TDDFT book,

this approximation means that $v_{xc}[\rho(\mathbf{r},t)]$ becomes exact in the limit where the adiabatic theorem of quantum mechanics applies, i.e., a physical system remains in its instantaneous eigenstate if a perturbation that is acting on it is slow enough.

What is the time scale on which the adiabatic approximation works? In other words, when is this approximation valid?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
8
$\begingroup$

The time scale is related to time derivative of the kinetic energy of electrons defined as:

$$T(t) = \sum_{i} \int |\nabla \phi_{i}(\mathbf{r},t)|^{2} d^{3} \mathbf{r}$$

You have this for time-derivative of the kinetic energy:

$$\frac{d T(t)}{d t} \simeq \frac{T(t=0)}{\tau}$$

Where $\tau$ is the relaxation time for kinetic energy in the order of period associated with the lowest excitation energy of the system.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ I don't think the second equation is justified, because $T(t)$ does not decrease monotonically, nor does it become anywhere close to zero; instead, the fluctuation of $T(t)$ is usually much smaller than the magnitude of $T(t)$ itself, so the l.h.s. of the second equation should be much smaller in magnitude than the r.h.s. $\endgroup$
    – wzkchem5
    Jun 15, 2022 at 9:24
  • $\begingroup$ @wzkchem5 Well, $T(t)$ is the summation of integrals of gradient of orbital functionals, so it might fluctuate quite a bit in a very short time scale, but eventually it should asymptotically decrease when the system is going towards equilibrium (Note that from Newtonian mechanics, we know that $\frac{dT(t)}{dt} = F v$, where $F$ is the net force and $v$ is the velocity. In equilibrium, when $t = t_{\mathrm{equilibrium}}$, $\frac{dT}{dt} = 0$). Also, the second expression is not an equation, just a rough estimate of how kinetic energy would behave with respect to time. $\endgroup$
    – Mithridates the Great
    Jun 15, 2022 at 16:25
  • $\begingroup$ I agree that at equilibrium $\frac{dT(t)}{dt}=0$. But unless the fluctuation of $T(t)$ is on the order of $T(t)$ itself, you cannot estimate $\frac{dT(t)}{dt}$ by dividing the $T(t=0)$ by a timescale. In fact, when the external time-dependent potential is arbitrarily small in magnitude and there is no resonant absorption, the magnitude of the fluctuation of $T(t)$ can also be arbitrarily small. This makes the second expression unreliable even as an order-of-magnitude estimate, when the magnitude of the time-dependent external potential is unknown. $\endgroup$
    – wzkchem5
    Jun 15, 2022 at 17:21
4
$\begingroup$

A probably more useful heuristic is that the highest frequency mode of the perturbation is smaller than the excitation energy of the first excited state that is dominated by double (or higher) excitations. In the linear response regime, this rationale can be justified by the following argument: the adiabatic approximation recovers all single excitations but no double and higher excitations, while the latter are described exactly by the full non-adiabatic XC kernel. Therefore, when the external potential varies at a higher and higher rate, the first time that the adiabatic approximation fails qualitatively is when the external potential oscillates at the frequency consistent with the first state that is missed by the adiabatic approximation, where the exact response diverges, but the approximate response does not have any divergences at that frequency, not even at nearby frequencies.

Of course the heuristic is not completely rigorous, for the following reasons: (1) the true response is not necessarily in the linear-response regime; (2) even though the heuristic rules out qualitative errors, there may still be large quantitative errors. Ultimately, to conclusively say when an approximation is valid, one must also provide the desired precision of the simulation; an approximation that is valid for a crude calculation is not necessarily valid for an accurate one.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .