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In introductory mechanics, the momentum of a particle is its mass times its velocity. In electrodynamics, the momentum of a field is proportional to the cross-product of the electric field with the magnetic field. In special relativity, momentum is generalized to four-momentum.

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Given the S.E $$ \Psi(x,0) = \begin{cases} A,&0 \leq x \leq \frac{a}{2}, \\ 0,& \rm elsewhere \end{cases} $$ I'm supposed to find the uncertainty product at $t=0$. However since the equation is const …
asked Apr 5 '18 by bullbo