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# Maths is working!

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This fact is also true in all inertial frames:

$\begin{bmatrix} c\,t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma&-\beta_x\,\gamma&-\beta_y\,\gamma&-\beta_z\,\gamma\\ -\beta_x\,\gamma&1+(\gamma-1)\frac{\beta_{x}^{2}}{\beta^{2}}&(\gamma-1)\frac{\beta_{x}\beta_{y}}{\beta^{2}}&(\gamma-1)\frac{\beta_{x}\beta_{z}}{\beta^{2}}\\ -\beta_y\,\gamma&(\gamma-1)\frac{\beta_{y}\beta_{x}}{\beta^{2}}&1+(\gamma-1)\frac{\beta_{y}^{2}}{\beta^{2}}&(\gamma-1)\frac{\beta_{y}\beta_{z}}{\beta^{2}}\\ -\beta_z\,\gamma&(\gamma-1)\frac{\beta_{z}\beta_{x}}{\beta^{2}}&(\gamma-1)\frac{\beta_{z}\beta_{y}}{\beta^{2}}&1+(\gamma-1)\frac{\beta_{z}^{2}}{\beta^{2}}\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix}\$

And with minimum uncertainty:

$i\hbar\frac{\partial}{\partial t} \Psi(\mathbf{r},\,t) = \hat H \Psi(\mathbf{r},t)$