Question

Câu 7.

Hãy chứng minh các hệ thức giao hoán sau đây rồi rút ra các kết luận cần thiết:

a) \({[\hat{M_x},\hat{M_y}] = i.\hbar.\hat{M_z}}\)

b) \({[\hat{S^2},\hat{S_x}]}= 0\)

Answer 1

a. CM:         [ M^{^}_x , M^{^}_y ]  =    ih.M^{^}_z  

ta có :

 M^{^}_x M^{^}_y   =    ( - i.h )^2.\(\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\)

               =    (  i.h )^2.\(\left(y\frac{\partial}{\partial x}-xy\frac{\partial^2}{\partial z^2}\right)\)

M^{^}_y.M^{^}_x    =    ( - i.h )².\(\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\)

suy ra :

[ M^{^}_x , M^{^}_y ] = M^{^}_x M^{^}_y  - M^{^}_y.M^{^}_x  

                 = ( i.h )^2.\(\left(y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}\right)\)

                 = ih.( - i.h)\(\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)\)

                 =  ih.M^{^}_{z               (dpcm)}

b.CM:    [S^{^}_x, S^{^}_y] = 0

 ta có :

S^{^2} =   S^{^2}_{x  +}  S^{^2}_{y  +}   S^{^2}_z

        ₌ ( h⁴/4) \(\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\)  + ( h⁴/4) \(\left(\begin{matrix}0&-i\\1&0\end{matrix}\right)\left(\begin{matrix}0&-i\\1&0\end{matrix}\right)\)  +  ( h⁴/4)\(\left(\begin{matrix}1&0\\0&-1\end{matrix}\right)\left(\begin{matrix}1&0\\0&-1\end{matrix}\right)\)

         ⁼   (3h/4).\(\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\)

mặt khác :

S^{^2}.S^{^}_x  =   (3h²/4)\(\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\)(h/2).\(\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\)

            =    (3h³/8)\(\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\) S^{^}_x.S^{^2}  = (h/2).\(\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\) (3h²/4)\(\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\)            =(3h³/8).\(\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\)suy ra : [S^{^}_x, S^{^}_y] =  S^{^2}.S^{^}_x    -  S^{^}_x.S^{^2}   = 0