エーレンフェストの定理/メモ のバックアップソース(No.2)

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* 解答:エーレンフェストの定理 [#wbe3d8ad]

(1)

 &math(
\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)

 &math(
\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)

 &math(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、&math(x\hat p_z^2-\hat p_z^2x=0); より、

 &math(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)

 &math(
\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)

 &math(
\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)

 &math(
\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)

 &math(
\frac{d}{dt}\langle p_x\rangle
\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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