群速度と波束の崩壊/メモ のバックアップソース(No.3)

更新 
* 最小波束の時間発展 [#b94b4a9f]

　&math(
\psi(x,t)&=\int_{-\infty}^{\infty}\varphi(k)\frac{1}{\sqrt{2\pi}}e^{ikx}e^{-i\omega_kt}dk\\
&=\sqrt{\frac{\sigma_{x0}}{\pi\sqrt{2\pi}}}\int_{-\infty}^{\infty}e^{-4\sigma_{x0}^2(k-k_0)^2/4+ikx-i\hbar k^2t/2m}dk\\
);

指数部を整理すると、

　&math(
&-\sigma_{x0}^2(k-k_0)^2+ikx-i\hbar k^2t/2m\\
&=-\sigma_{x0}^2(1+i\underbrace{\hbar t/2m\sigma_{x0}^2}_{\xi t})k^2+\sigma_{x0}^2\{ix/\sigma_{x0}^2+2k_0\}k-\sigma_{x0}^2k_0^2\\
&=-\sigma_{x0}^2\left[\sqrt{1+i\xi t}\ k-\frac{ix/\sigma_{x0}^2+2k_0}{2\sqrt{1+i\xi t}}\right]^2+
\frac{\sigma_{x0}^2\{ix/\sigma_{x0}^2+2k_0\}^2}{4(1+i\xi t)}-\sigma_{x0}^2k_0^2\\
);

２項目以降は、

　&math(
&\frac{\sigma_{x0}^2\{ix/2\sigma_{x0}^2+k_0\}^2}{1+i\xi t}-\sigma_{x0}^2k_0^2\\
&=\frac{-x^2/4\sigma_{x0}^2+ik_0x+\sigma_{x0}^2k_0^2}{1+i\xi t}-\sigma_{x0}^2k_0^2\\
&=\frac{-x^2/4\sigma_{x0}^2+ik_0x-i\overbrace{\sigma_{x0}^2k_0^2\xi}^{\omega_{k0}} t}{1+i\xi t}\\
&=\frac{-x^2/4\sigma_{x0}^2+i(k_0x-\omega_{k0} t)}{1+i\xi t}\\
);

であるから、

　&math(
\psi(x,t)&=\sqrt{\frac{\sigma_{x0}}{\pi\sqrt{2\pi}}}
\frac{\exp\left[\frac{-x^2/4\sigma_{x0}^2+i(k_0x-\omega_{k0} t)}{1+i\xi t}\right]}{\sigma_{x0}\sqrt{1+i\xi t}}
\underbrace{\int_{-\infty}^{\infty}e^{-\sigma_{x0}^2\left[\sqrt{1+i\xi t}\ k-\frac{ix/\sigma_{x0}^2+2k_0}{2\sqrt{1+i\xi t}}\right]^2}\left(\sigma_{x0}\sqrt{1+i\xi t}\ dk\right)}_{\sqrt{\pi}}\\
&=\sqrt{\frac{1}{\sqrt{2\pi}\sigma_{x0}(1+i\xi t)}}\exp\left[\frac{-x^2/4\sigma_{x0}^2+i(k_0x-\omega_{k0} t)}{1+i\xi t}\right]\\
);

ただし、&math(\xi=\frac{\hbar}{2m\sigma_{x0}^2});、&math(\omega_0=\frac{\hbar k_0^2}{2m});

このとき、

　&math(
|\psi(x,t)|^2&=\frac{1}{\sqrt{2\pi}\sigma_{x0}\sqrt{(1-i\xi t)(1+i\xi t)}}
\exp\left[\frac{-x^2/4\sigma_{x0}^2-i(k_0x-\omega_{k0} t)}{1-i\xi t}\right]
\exp\left[\frac{-x^2/4\sigma_{x0}^2+i(k_0x-\omega_{k0} t)}{1+i\xi t}\right]\\
&=\frac{1}{\sqrt{2\pi}\sigma_{x0}\sqrt{1+\xi^2 t^2}}
\exp\left[\frac{-x^2/2\sigma_{x0}^2+2(k_0x-\omega_{k0}t)\xi t}{1+\xi^2t^2}\right]\\
&=\frac{1}{\sqrt{2\pi}\sigma_{x0}\sqrt{1+\xi^2 t^2}}
\exp\left[\frac{-x^2/2\sigma_{x0}^2+2(k_0x-\hbar k_0^2 t/2m)\hbar t/2m\sigma_{x0}^2}{1+\xi^2t^2}\right]\\
&=\frac{1}{\sqrt{2\pi}\sigma_{x0}\sqrt{1+\xi^2 t^2}}
\exp\left[\frac{-x^2+(2x-\hbar k_0 t/m)\hbar k_0 t/m}{2\sigma_{x0}^2(1+\xi^2t^2)}\right]\\
&=\frac{1}{\sqrt{2\pi}\sigma_{x0}\sqrt{1+\xi^2 t^2}}
\exp\left[\frac{-\{x-(\hbar k_0/m) t\}^2}{2\sigma_{x0}^2(1+\xi^2t^2)}\right]\\
);

* 群速度と位相速度 [#pccbf754]

 LANG:mathematica
 F[x_, t_, k_] := 
   Sqrt[Sqrt[2 Pi] (1 + I t)]^(-1) Exp[(-x^2/4 + I (k x - k^2 t))/(1 + I t)]
 
 Animate[
   Plot[ 
     Re[F[x, t, 10]], {x, -5, 20}, 
     PlotRange -> {-0.7, 0.7}, PlotPoints -> 200
   ], 
   {t, 0, 1}
 ]
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