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Next: Results for D + H Up: Flux and real wave Previous: Real wave packet propagation

Flux analysis of real wave packets

If one has just $q_\psi ({\bf x},t)$ = Re[ $\psi ({\bf x},t)$], propagated using the time-dependent Schrödinger equation, one may use $q_\psi ({\bf x},t)= (1/2) \left[\psi({\bf x},t)
+ \psi({\bf x},t)^\ast\right]$, and expand as with Eq. (1),

$\displaystyle q_\psi ({\bf x},t)$ = $\displaystyle \frac{1}{2}\left[\frac{1}{2\pi\hbar} \int e^{-iEt/\hbar}
\psi^+({\bf x},E) a_\psi (E) dE\right.$  
    $\displaystyle +\left. \frac{1}{2\pi\hbar} \int e^{iEt/\hbar}
{\psi^+}^\ast({\bf x},E) a_\psi^\ast(E) dE \right].$ (15)

From this it follows that

\begin{displaymath}\int e^{iEt/\hbar} q_\psi ({\bf x},t) dt = \half\left[
..._\psi (E) + {\psi^+}^\ast ({\bf x},-E) a_\psi^\ast(-E)\right].
\end{displaymath} (16)

We require that the initial wave packet be such that $a_\psi(-E)$, for the Evalues of interest, be zero, i.e., that the initial wave packet has no appreciable amplitude for energies -E. [This can be achieved if the energy scale is chosen such that only E > 0 corresponds to scattering states. Alternatively, as with the energies f(E) associated with the modified Schrödinger equation, no physical states correspond to -f(E).] We obtain

\psi^+ ({\bf x},E) = \frac{2}{a_\psi (E)}\int e^{iEt/\hbar}
q_\psi ({\bf x},t) dt,
\end{displaymath} (17)

which differs from Eq. (1), the complex wave packet result, by simply a factor of 2. We want to point out here that this is a general result. Therefore, we can use it to adapt any method mentioned in Sec. 2. For reaction probabilities using the flux method from Refs. [5,6], we get for real wave packets that Pr(E) is simply four times the right hand site of Eq. (13).

Consider now the modified Schrödinger equation and its corresponding real part $q_\chi({\bf x},t)$, which also satisfies Eq. (14). One simply repeats the entire analysis above, but with $\hat H$, E, $\psi$, and $q_\psi$ replaced by $f(\hat{H})$, f(E) = $(-\hbar / \tau )$arccos (asE+bs), $\chi$, and $q_\chi$, respectively. To get from Pr(f) back to Pr(E), one has to realize that ``f'' normalized scattering functions are related to E normalized ones by $\chi^+
= \sqrt{ \vert dE/df \vert} \psi^+$. Thus, if the initial condition for the modified Schrödinger equation is the same as for the ordinary one, $\chi (t=0)=
\psi(t=0)$, then $a_\chi (f) = \left\langle \chi^+ \left\vert\chi (0)
\right.\right\rangle = \sqrt{ \vert dE/df\vert } a_\psi (E)$. Moreover, since we have the real part $q_\chi({\bf x},t)$ at discrete times (or iterations) $k\tau$, we can employ a Fourier series to evaluate the integrals in Eq. (11). Using this and the fact that Pr(E) = |df/dE| Pr(f) we find the useful relation

Pr(E) = $\displaystyle \frac{4\hbar^3}{\mu_s}\frac{a_s^2}{\left(1-E_s^2\right)}
\sum_k^\infty e^{i \theta (E) k} q_\chi ({\bf x},k\tau)
    $\displaystyle \times \left\vert
\sum_{k'}^\infty e^{i \theta (E) k'}
...s_0}) \frac{\partial}{\partial s} q_\chi ({\bf x},k'\tau)\right]
\right\rangle,$ (18)

where $\theta (E) = -\arccos(a_sE+b_s)$ and df/dE has been explicitly evaluated. (Note that $\tau$ has canceled out of the final expression.)

For the T-matrix formalism, the same analysis yields

T^{c,a}_{FI}(E)=\frac{\hbar a_s}{\sqrt{1-E_s^2}} \frac{2}{a_...
q_\chi({\bf x},k\tau)\right]\right.\right.\right\rangle.
\end{displaymath} (19)

next up previous
Next: Results for D + H Up: Flux and real wave Previous: Real wave packet propagation
Anthony J. H. M. Meijer