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Alfredo A. Correa edited figures/energy_dis7/caption.tex
over 8 years ago
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\label{fig:fit_graph} (color online). The total energy increase as a function of proton position at velocities $v = 0.06 ~\mathrm{a.u.}$ (green
dashed line) along [100] channeling trajectory and the adiabatic energy (magenta
line). dot-dashed line) (that would correspond to a proton moving infinitely slowly).
The initial (ground
state) energy state energy) is subtracted.
The lower dashed lines are the TDDFT
($v= 0.06~\mathrm{a.u.}$) and the adiabatic (ground state) energies as a function of proton
position moving at $v=0.06~\mathrm{a.u.}$. position. The red line shows the energy difference (subtraction of adiabatic energy from the $v = 0.06 ~\mathrm{a.u.}$ curves). The blue and black lines shows the slope of linear and oscillatory fits of the red line from $x = 5.0 ~\mathrm{a_0}$ to a given maximum position $x$ as a function of this maximum position respectively. For visualization purposes the
linear and oscillatory fit curves black line have been shifted vertically for clarity.
%%$v %
%$v = 0 ~\mathrm{a.u.}$ from $v = 0.06 ~\mathrm{a.u.}$ curves)%
%%should be added a constant value of $-16856.62389 ~\mathrm{a.u.}$.%
The oscillations in the curves reflect the periodicity of the $\mathrm{Cu}$ lattice.
To obtain the slope for $v = 0.06 ~\mathrm{a.u.}$ which represents the $S_\text{e}$ at this velocity, the adiabatic results ($v = 0 ~\mathrm{a.u.}$) are subtracted from those of $v = 0.06 ~\mathrm{a.u.}$ to obtain only the non-adiabatic contributions; it is shown as a solid line (red) with little oscillations. This subtraction is done to remove the oscillations that result from the periodic lattice. A linear fit , $y = a + bx$ (blue line) yields a slope of $6.989 \times 10^{-3}~E_\text{h}/a_0$ with an error of $\pm 6.936 \times 10^{-5}~E_\text{h}/a_0$. We then proceed to do a linear fit in addition to an oscillatory function $y = a + bx + A\cos(k x + \phi)$ (black line) to capture any remnant oscillation. This oscillation fit generates a slope of $7.435 \times 10^{-3}~E_\text{h}/a_0$ with an error of $\pm 8.52 \times 10^{-7}~E_\text{h}/a_0$.