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- Thread starter Silviu
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DrDu

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I know this. But based on this definition of Klein-Gordon operator, you know the position of particle with 0 error, which means that you should know the momentum with infinite error, which is, the particle can have any momentum from 0 to infinity. But this is not possible, as the particle can't travel faster than light and you know the mass of the particle as you know what particle you create, and the particle is not virtual, so its mass is known (of course with a certain uncertainty, but again a finite one). Is this logic wrong?

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DrDu

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Namely ## p=\frac{mv}{\sqrt{1-v^2/c^2}}##.

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