A Paradox?

  • Thread starter eljose
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  • #1
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Let,s suppose we have a particle moving under a potential V:
By Heisenberg picture we know that:

[tex]\frac{dx}{dt}=p/m [/tex]

so if we knew x(t) and x(t+h) we could calculate the expresion:

[tex]x(t+\epsilon)-x(t)=p\epsilon/m [/tex]

so knowing x(t) and x(t+e) we could calculate the momentum of the particle:
 
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Answers and Replies

  • #2
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eljose said:
Let,s suppose we have a particle moving under a potential V:
By Heisenberg picture we know that:

[tex]\frac{dx}{dt}=p/m [/tex]

so if we knew x(t) and x(t+h) we could calculate the expresion:

[tex]x(t+\epsilon)-x(t)=p\epsilon/m [/tex]

so knowing x(t) and x(t+e) we could calculate the momentum of the particle:
Please do not mix the differential equation between operators and eigenvalues. They are completely different.


Seratend.
 

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