群速度と波束の崩壊/メモ のバックアップ(No.1)

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最小波束の時間発展

 &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、 \xi=\frac{\hbar}{2m\sigma_{x0}^2} \omega_0=\frac{\hbar k_0^2}{2m}


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