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Sep 14, 2013 at 13:17 vote accept CommunityBot
Sep 11, 2013 at 14:42 answer added orange_soda timeline score: 8
Sep 11, 2013 at 12:43 answer added Qmechanic timeline score: 3
Sep 11, 2013 at 12:35 answer added xuanji timeline score: 4
Sep 11, 2013 at 11:41 comment added Trimok Note that the position vector $x^j$ is a "contravariant" vector, with upper indices, while momentum vector $p_i$ is fundamentally a "covariant" vector, with lower indices. Transformation laws for contravariant and covariant vectors are different. Of course, in special relativity, the momentum is sometimes written with upper indices like $p^i = m\frac{dx^i}{d \tau}$, but it hides the true nature of the momentum. This said, your expressions may be written in the compact form $\Delta p_i \Delta x^j \geq \delta_i^j \large \frac{\hbar}{2}$
Sep 11, 2013 at 11:23 history edited Qmechanic
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Sep 11, 2013 at 11:14 history asked user20250 CC BY-SA 3.0