# Velocity of measurement

As per to Heisenberg uncertainty we will not be able to calculate the position and momentum at same instant because by the time we calculate the next of the one, it changes (i.e.) the changes are very fast. Suppose my speed of calculation is near to the speed of light or equal to the speed of light. Can I at that instant calculate the accurate position and momentum of a particle

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Hi. Welcome to Physics.SE. What do you men by "my speed of calculation is near $c$"..? I'm quite confused of that phrase ;-) – Waffle's Crazy Peanut Mar 27 '13 at 13:50
Calculation is irrelevant. Nature was obeying the uncertainty principle long before any physicists were around to calculate things. – Michael Brown Mar 27 '13 at 13:53

Heisenberg's uncertainty principle isn't based on any limited "speed of calculation". It simply says that the particle has an uncertain location and an uncertain velocity and they obey the inequality $$\Delta x \cdot m\Delta v \geq \frac\hbar 2$$ These errors are "inevitable errors of the measurement" of the position or the velocity. They really mean that the particle has an uncertain location and an uncertain velocity so if we repeat the same experiment with the very same initial conditions many times, the measured values of $x$ will be spread with the width of the distribution $\Delta x$ and similarly for the velocity.
There isn't any slow or fast calculation involved here. The calculation of the predicted $x$ may be immediate or superfast but it's still true that the right prediction will say that $x$ can't be a sharp number, it may only be specified with the error margin $\Delta x$. Again, this is not due to some technical limitations or slow calculation or anything of the sort. The reason is that the particle simply does not possess a sharp value of the position if it has a pretty accurate velocity, and vice versa.