采采流水

An Essay Concerning Human Understanding

LaTex $E=mc^{2}$

2016-09-07


Problem


质能公式

$$ E=mc^{2} $$

$$ \frac{\partial u}{\partial t} = h^2 \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right) $$

Proof

$$ e^{i\sigma_{z}\phi}\sigma_ye^{-i\sigma_z\phi}=e^{2i\sigma_z\phi}\sigma_y $$

Solution

$$ \begin{align} e^{i\sigma_{z}\phi}\sigma_ye^{-i\sigmaz\phi} &=e^{i\sigma{z}\phi}(e^{-i\sigma_z\phi}\sigma_y+[\sigma_y,e^{-i\sigma_z\phi}])
&=\sigmay+e^{i\sigma{z}\phi}[\sigma_y,cos(\phi)-i sin(\sigma_z\phi)]
&=\sigmay+e^{i\sigma{z}\phi}[\sigma_y,-i\sigma_zsin(\phi)]
&=\sigmay-i sin(\phi) e^{i\sigma{z}\phi}[\sigma_y,\sigma_z]
&=\sigma_y+2isin(\phi) e^{i\sigma_z\phi}\sigma_z\sigma_y
&=e^{i\sigma_z\phi}(e^{-i\sigma_z\phi}\sigma_y+2i\sigma_zsin(\phi)\sigma_z\sigma_y)
&=e^{i\sigma_z\phi}(cos(\sigma_z\phi)+i\sigma_zsin(\phi))\sigma_y
&=e^{2i\sigma_z\phi}\sigma_y \end{align
} $$

$$ \begin{align} x &= \sqrt{4^2-1^2}
&= \sqrt{15}. \end{align
}$$ $$

$$ \begin{align} \dot{x} & = & \sigma(y-x) \
\dot{y} & = &\rho x - y - xz \
\dot{z} & = &-\beta z + xy \end{align} $$ $$ \begin{align} \nabla\cdot\vec{E} &=& \frac{\rho}{\epsilon_0} \
\nabla\cdot\vec{B} &=& 0 \
\nabla\times\vec{E} &=& -\frac{\partial B}{\partial t} \
\nabla\times\vec{B} &=& \mu_0\left(\vec{J}+\epsilon_0\frac{\partial E}{\partial t} \right) \end{align} $$

Useful formulas $$ \begin{align} \hat{A}\hat{B}\hat{C}&=\hat{A}(\hat{C}\hat{B}+[\hat{B},\hat{C}])\
\hat{I}&=e^{i\sigma_z\phi}e^{-i\sigma_z\phi}\
e^{i\sigma_z\phi}&=cos(\sigma_z\phi)+isin(\sigma_z\phi)\
&=cos(\phi)+i\sigma_zsin(\phi)\
[\sigma_i,\sigma_j]&=i\varepsilon_{ijk}\sigma_k\
\sigma_i\sigmaj&=i\varepsilon{ijk}\sigma_k \end{align} $$

插图测试