エーレンフェストの定理/メモ

(1165d) 更新

解答:エーレンフェストの定理

(1)

  \frac{d}{dt}\langle x\rangle &=\frac{1}{i\hbar}\iiint \psi^*\Big[x\hat H-\hat Hx\Big]\psi\,d\bm r\\ &=\frac{1}{i\hbar}\iiint \psi^*\Big[x\Big(\frac{1}{2m}\hat p^2+\cancel{V}\Big) -\Big(\frac{1}{2m}\hat p^2+\cancel{V}\Big)x\Big]\psi\,d\bm r\\ &=\frac{1}{i\hbar}\iiint\frac{1}{2m}\psi^*\Big(x\hat p^2-\hat p^2x\Big)\psi\,d\bm r

(2)

  \hat p_x^2x&=\hat p_x(x\hat p_x-i\hbar)\\ &=\hat p_xx\hat p_x-i\hbar\hat p_x\\ &=(x\hat p_x-i\hbar)\hat p_x-i\hbar\hat p_x\\ &=x\hat p_x^2-2i\hbar\hat p_x

(3)

  x\hat p_y^2-\hat p_y^2x&=x\hat p_y\hat p_y-\hat p_y\hat p_yx\\ &=\hat p_yx\hat p_y-\hat p_yx\hat p_y\\ &= 0

同様に、 x\hat p_z^2-\hat p_z^2x=0 より、

  x\hat p^2-\hat p^2x&=x(\hat p_x^2+\hat p_y^2+\hat p_z^2)-(\hat p_x^2+\hat p_y^2+\hat p_z^2)x\\ &=(x\hat p_x^2-\hat p_x^2x)+(x\hat p_y^2-\hat p_y^2x)+(x\hat p_z^2-\hat p_z^2x)\\ &=2i\hbar\hat p_x

(4)

  \frac{d}{dt}\langle x\rangle &=\frac{1}{\cancel{i\hbar}}\iiint\frac{1}{\cancel 2m}\psi^*\Big(\cancel 2\cancel{i\hbar}\hat p_x\Big)\psi\,d\bm r\\ &=\frac{1}{m}\iiint\psi^*\hat p_x\psi\,d\bm r\\ &=\frac{\left\langle p_x\right\rangle}{m}

(5)

  \hat p_x\hat p^2 &=\frac{\hbar^3}{i^3}\frac{\PD}{\PD x}\left(\frac{\PD^2}{\PD x^2}+\frac{\PD^2}{\PD y^2}+\frac{\PD^2}{\PD z^2}\right)\\ &=\frac{\hbar^3}{i^3}\left(\frac{\PD^2}{\PD x^2}+\frac{\PD^2}{\PD y^2}+\frac{\PD^2}{\PD z^2}\right)\frac{\PD}{\PD x}\\ &=\hat p^2\hat p_x

(6)

  \hat p_xV\psi&=\frac{\hbar}{i}\frac{\PD}{\PD x}(V\psi)\\ &=\frac{\hbar}{i}\frac{\PD V}{\PD x}\psi+V\frac{\hbar}{i}\frac{\PD}{\PD x}\psi\\ &=\Big(\frac{\hbar}{i}\frac{\PD V}{\PD x}+V\hat p_x\Big)\psi\\

(7)

  \frac{d}{dt}\langle p_x\rangle &=\frac{1}{i\hbar}\iiint\psi^*\left[\hat p_x\left(\frac{1}{2m}\hat p^2+V\right) -\left(\frac{1}{2m}\hat p^2+V\right)\hat p_x\right]\psi\,d\bm r\\ &=\frac{1}{i\hbar}\iiint\psi^*\Big(\frac{1}{2m}\cancel{(\hat p_x\hat p^2-\hat p^2\hat p_x)}+(\hat p_xV-V\hat p_x)\Big)\psi\,d\bm r\\ &=\frac{1}{i\cancel\hbar}\iiint\psi^*\Big(\frac{\cancel\hbar}{i}\frac{\PD V}{\PD x}\Big)\psi\,d\bm r\\ &=-\left\langle\frac{\PD V}{\PD x}\right\rangle


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