哈密顿算子重要性质

• 重要公式
  $\triangledown \cdot(\vec A \times \vec B)= \vec B \cdot \triangledown \times\vec A-\vec A\cdot\triangledown\times\vec B$
• 证明
  $\triangledown=\frac{\partial}{\partial x}\vec e_x+\frac{\partial }{\partial y}\vec e_y+\frac{\partial }{\partial z}\vec e_z$
  $\vec A \times \vec B=\left | \begin{matrix} \vec e_x &\vec e_y &\vec e_z\\ A_x & A_y & A_z\\ B_x & B_y & B_z \end{matrix} \right |\\ =(A_yB_z-A_zB_y)\vec e_x+(A_zB_x-A_xB_z)\vec e_y+(A_xB_y-A_yB_x)\vec e_z$
  $\triangledown \cdot(\vec A \times \vec B)=(\frac{\partial}{\partial x}\vec e_x+\frac{\partial }{\partial y}\vec e_y+\frac{\partial }{\partial z}\vec e_z) \cdot\bigg((A_yB_z-A_zB_y)\vec e_x+(A_zB_x-A_xB_z)\vec e_y+(A_xB_y-A_yB_x)\vec e_z\bigg)\\ =\frac{\partial{(A_yB_z-A_zB_y)}}{\partial x}+\frac{\partial{(A_zB_x-A_xB_z)}}{\partial y}+\frac{\partial{(A_xB_y-A_yB_x)}}{\partial z}\\ =\left | \begin{matrix} \frac{\partial}{\partial x}&\frac{\partial }{\partial y}&\frac{\partial }{\partial z}\\ A_x & A_y & A_z\\ B_x & B_y & B_z \end{matrix} \right |\\$

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$\vec B \cdot \triangledown \times\vec A-\vec A\cdot\triangledown\times\vec B=\vec B \cdot \left | \begin{matrix} \vec e_x &\vec e_y &\vec e_z\\ \frac{\partial}{\partial x}&\frac{\partial }{\partial y}&\frac{\partial }{\partial z}\\ A_x & A_y & A_z \end{matrix} \right | - \vec A \cdot \left | \begin{matrix} \vec e_x &\vec e_y &\vec e_z\\ \frac{\partial}{\partial x}&\frac{\partial }{\partial y}&\frac{\partial }{\partial z}\\ B_x & B_y & B_z \end{matrix}\right | \\ = (B_x\vec e_x+B_y\vec e_y+B_z\vec e_z)\cdot\bigg( (\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z})\vec e_x+(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x})\vec e_y+(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})\vec e_z \bigg)\\ -(A_x\vec e_x+A_y\vec e_y+A_z\vec e_z)\cdot\bigg( (\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z})\vec e_x+(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x})\vec e_y+(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y})\vec e_z \bigg)\\ =B_x( \frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z})+ B_y(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x})+ B_z(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})\\ -\bigg(A_x( \frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z})+ A_y(\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x})+ A_z(\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y})\bigg)\\ \because B_z\frac{\partial A_y}{\partial x}-B_y\frac{\partial A_z}{\partial x}+ A_y\frac{\partial B_z}{\partial x}- A_z\frac{\partial B_y}{\partial x}= \bigg(B_z\frac{\partial A_y}{\partial x}+ A_y\frac{\partial B_z}{\partial x}\bigg)-\bigg(B_y\frac{\partial A_z}{\partial x}+A_z\frac{\partial B_y}{\partial x} \bigg)\\ =\frac{\partial( A_yB_z)}{\partial x}-\frac{\partial( A_zB_y)}{\partial x}=\frac{\partial( A_yB_z-A_zB_y)}{\partial x}$

• 同理其他三项可得，证毕