In GAMER ELBDM GaussianWavePacket

• Gau_v0 = $v_0$
• Gau_Width= $w_0$ = $\sqrt{2}\times(\sigma_{x,0}=\mathrm{standard~ deviation~of~the~initial~density~distribution~}\rho(x,0)=|\psi(x,0)|^2)$
• Gau_Center = $x_0$
• $C_1=1+\Big(\frac{t}{\frac{m}{\hbar}w_0^2}\Big)^2$
• $\theta_1=-\frac{1}{2}\mathrm{cos^{-1}}(C_1^{-\frac{1}{2}})=-\frac{1}{2}\mathrm{cos^{-1}}\Big( \frac{1}{\sqrt{1+(\frac{t}{\frac{m}{\hbar}w_0^2})^2}}\Big)$
• $C_2=(w_0^2\pi C_1) ^{-1/4}\exp[-\frac{1}{2}(\frac{(x-v_0t-x_0)^2}{w_0^2C_1})]=\frac{1}{\{\pi w_0^2[1+(\frac{t}{\frac{m}{\hbar}w_0^2})^2]\}^{1/4}}\exp\Big[{-\frac{1}{2}\frac{(x-v_0t-x_0)^2}{w_0^2[1+(\frac{t}{\frac{m}{\hbar}w_0^2})^2]}}\Big]$
• $\theta_2=\frac{1}{2}(x-v_0t-x_0)^2\frac{\frac{m}{\hbar}t}{(\frac{m}{\hbar}w_0^2)^2+t^2}+v_0\frac{m}{\hbar}(x-\frac{1}{2}v_0t-x_0)=\frac{1}{2}(x-v_0t-x_0)^2\frac{(\frac{t}{\frac{m}{\hbar}w_0^2})}{w_0^2[1+(\frac{t}{\frac{m}{\hbar}w_0^2})^2]}+v_0\frac{m}{\hbar}(x-\frac{1}{2}v_0t-x_0)$

→

$\begin{aligned} \psi&=C_2e^{i(\theta_1+\theta_2)}\\& =\frac{1}{\{\pi w_0^2[1+(\frac{t}{\frac{m}{\hbar}w_0^2})^2]\}^{1/4}}\exp\Big[{-\frac{1}{2}\frac{(x-v_0t-x_0)^2}{w_0^2[1+(\frac{t}{\frac{m}{\hbar}w_0^2})^2]}}\Big]\exp\Big[i\Big(-\frac{1}{2}\mathrm{cos^{-1}}\big( \frac{1}{\sqrt{1+(\frac{t}{\frac{m}{\hbar}w_0^2})^2}}\big)+\frac{1}{2}(x-v_0t-x_0)^2\frac{(\frac{t}{\frac{m}{\hbar}w_0^2})}{w_0^2[1+(\frac{t}{\frac{m}{\hbar}w_0^2})^2]}+v_0\frac{m}{\hbar}(x-\frac{1}{2}v_0t-x_0)\Big)\Big] \\& =\frac{1}{\{2\pi\sigma_{x,0}^2[1+(\frac{t}{2\frac{m}{\hbar}\sigma_{x,0}^2})^2]\}^{1/4}}\exp\Big[{-\frac{1}{2}\frac{(x-v_0t-x_0)^2}{2\sigma_{x,0}^2[1+(\frac{t}{2\frac{m}{\hbar}\sigma_{x,0}^2})^2]}}\Big]\exp\Big[i\Big(-\frac{1}{2}\mathrm{cos^{-1}}\big( \frac{1}{\sqrt{1+(\frac{t}{2\frac{m}{\hbar}\sigma_{x,0}^2})^2}}\big)+\frac{1}{2}(x-v_0t-x_0)^2\frac{(\frac{t}{2\frac{m}{\hbar}\sigma_{x,0}^2})}{2\sigma_{x,0}^2[1+(\frac{t}{2\frac{m}{\hbar}\sigma_{x,0}^2})^2]}+v_0\frac{m}{\hbar}(x-\frac{1}{2}v_0t-x_0)\Big)\Big] \end{aligned}$

$\rho(x,t)=|\psi(x,t)|^2=\frac{1}{\{2\pi\sigma_{x,0}^2[1+(\frac{t}{2\frac{m}{\hbar}\sigma_{x,0}^2})^2]\}^{1/2}}\exp\Big[{-\frac{(x-v_0t-x_0)^2}{2\sigma_{x,0}^2[1+(\frac{t}{2\frac{m}{\hbar}\sigma_{x,0}^2})^2]}}\Big]=\frac{1}{\sigma_{x,t}\sqrt{2\pi}}\exp\Big[{-\frac{(x-v_0t-x_0)^2}{2\sigma_{x,t}^2}}\Big]$

$\int_{-\infty}^{\infty} \rho(x,t) dx = 1$

$\psi(x,t)=\sqrt{\frac{\alpha}{\alpha+i\frac{\hbar}{m}t}}\exp\Big[-\frac{(x+x_0-ik\alpha)^2}{2(\alpha+i\frac{\hbar}{m}t)}\Big]\exp\Big[-\frac{\alpha k^2}{2}\Big]$

• $\begin{aligned} \sqrt{\frac{\alpha}{\alpha+i\frac{\hbar}{m}t}}&=\sqrt{\frac{1}{1+i\frac{\hbar t}{m\alpha}}}\\ &=\sqrt{\frac{(1-i\frac{\hbar t}{m \alpha})}{(1+i\frac{\hbar t}{m\alpha})(1-i\frac{\hbar t}{m \alpha})}}\\&=\sqrt{\frac{1}{1+\frac{\hbar^2t^2}{m^2\alpha^2}}}\sqrt{1-i\frac{\hbar t}{m \alpha}}\\&=\sqrt{\frac{1}{1+\frac{\hbar^2t^2}{m^2\alpha^2}}}\Big[\sqrt{(1+\frac{\hbar^2 t^2}{m^2\alpha^2})}\exp\Big(-i{\cos^{-1}\big(\frac{1}{\sqrt{1+\frac{\hbar^2 t^2}{m^2\alpha^2}}}\big)}\Big)\Big]^{1/2}\\&=\Big(\frac{1}{1+\frac{\hbar^2t^2}{m^2\alpha^2}}\Big)^{1/4}\exp\Big[-i\frac{1}{2}{\cos^{-1}\big(\frac{1}{\sqrt{1+\frac{\hbar^2 t^2}{m^2\alpha^2}}}\big)}\Big] \end{aligned}$
• $\begin{aligned} \exp\Big[-\frac{(x+x_0-ik\alpha)^2}{2(\alpha+i\frac{\hbar}{m}t)}\Big]&=\exp\Big[-\frac{(x+x_0)^2-k^2\alpha^2-2i(x+x_0)k\alpha}{2\alpha(1+i\frac{\hbar t}{m \alpha})}\Big] \\&=\exp\Big[-\frac{\big((x+x_0)^2-k^2\alpha^2-2i(x+x_0)k\alpha\big)\big(1-i\frac{\hbar t}{m \alpha}\big)}{2\alpha(1+i\frac{\hbar t}{m \alpha})(1-i\frac{\hbar t}{m \alpha})}\Big] \\&=\exp\Big[-\frac{\big((x+x_0)^2-k^2\alpha^2-2(x+x_0)k\alpha\frac{\hbar t}{m\alpha}\big)-i\big((x+x_0)^2\frac{\hbar t}{m\alpha}-k^2\alpha^2\frac{\hbar t}{m\alpha}+2(x+x_0)k\alpha\big)}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big] \\&=\exp\Big[-\frac{(x+x_0)^2-k^2\alpha^2-2(x+x_0)k\alpha\frac{\hbar t}{m\alpha}}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big]\times\exp\Big[i\frac{(x+x_0)^2\frac{\hbar t}{m\alpha}-k^2\alpha^2\frac{\hbar t}{m\alpha}+2(x+x_0)k\alpha}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big] \end{aligned}$
  • $\begin{aligned} \exp\Big[-\frac{(x+x_0)^2-k^2\alpha^2-2(x+x_0)k\alpha\frac{\hbar t}{m\alpha}}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big]&= \exp\Big[-\frac{(x+x_0-k\frac{\hbar}{m}t)^2-k^2\frac{\hbar^2}{m^2}t^2-k^2\alpha^2}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big] \end{aligned}$
  • $\begin{aligned} \exp\Big[i\frac{(x+x_0)^2\frac{\hbar t}{m\alpha}-k^2\alpha^2\frac{\hbar t}{m\alpha}+2(x+x_0)k\alpha}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big]&= \exp\Big[i\frac{\big((x+x_0)^2\frac{\hbar t}{m\alpha}-(2(x+x_0)k\alpha\frac{\hbar t}{m\alpha})\frac{\hbar t}{m\alpha}+(k^2\alpha^2\frac{\hbar^2 t^2}{m^2\alpha^2})\frac{\hbar t}{m\alpha}\big)+(2(x+x_0)k\alpha\frac{\hbar t}{m\alpha})\frac{\hbar t}{m\alpha}-(k^2\alpha^2\frac{\hbar^2 t^2}{m^2\alpha^2})\frac{\hbar t}{m\alpha}-k^2\alpha^2\frac{\hbar t}{m\alpha}+2(x+x_0)k\alpha}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big]\\&= \exp\Big[i\frac{\big((x+x_0-k\alpha\frac{\hbar t}{m\alpha})^2\frac{\hbar t}{m\alpha}\big)-k^2\alpha^2\frac{\hbar t}{m\alpha}\big(1+\frac{\hbar^2 t^2}{m^2 \alpha^2}\big)+2(x+x_0)k\alpha\big(1+\frac{\hbar^2 t^2}{m^2 \alpha^2}\big)}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big]\\&= \exp\Big[i\frac{(x+x_0-k\alpha\frac{\hbar t}{m\alpha})^2\frac{\hbar t}{m\alpha}}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big]\times\exp\Big[i\Big(-\frac{1}{2}k^2\frac{\hbar}{m}t+k(x+x_0)\Big)\Big]\\&= \exp\Big[i\frac{(x+x_0-k\frac{\hbar t}{m})^2\frac{\hbar t}{m\alpha}}{2\alpha(1+\frac{\hbar^2 t^2}{m^2 \alpha^2})}\Big]\times\exp\Big[i\Big(k(x+x_0-\frac{1}{2}k\frac{\hbar}{m}t)\Big)\Big] \end{aligned}$