In Feynman's path integral formulation, in order to calculate the probability amplitude, we sum up all the possible trajectories of the particle between the points $A$ and $B$.

Since we know precisely that the particle will be at $A$ and $B$, does it mean that the uncertainty of the momentum is infinite?

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    $\begingroup$ Er...this is about five questions. Please ask one at a time. That way other users can judge the correctness and value of each answer. $\endgroup$ – dmckee Jun 22 '13 at 1:20
  • $\begingroup$ @dmckee Right. I just wasn't sure if they would be too short by separate. $\endgroup$ – jinawee Jun 22 '13 at 1:26

Position/momentum Uncertainty, and path integral formulation are exactly the same thing.

Suppose you cut the time interval in time $t_0= t_A, t_1,....,t_n=t_B$.

At time $t_0$, the particle is at the position $x_0=x_A$. Because we know the position, uncertainty about momentum is infinite, but this simply means, that, at time $t_1$, the particle could be in any position $x_1$.

Now, if the particle is, at time $t_1$ at position $x_1$, we can repeat the same argument as above, and say, that, at time $t_2$, the particle could be in any position $x_2$.

So, we see, that all the intermediary positions $x_1, x_2, ....x_{n-1}$ at times $t_1, t_2, ....t_{n-1}$ could take all values.

(The only constraints are the initial and final values of position $x_A,x_B$.)

That means that all paths, beginning at $t_A,x_A$ and ending at $t_B,x_B$ have to be taken in account.

And this is precisely the definition of the path integral formulation.


If you are using non-relatavistic quantum mechanics then yes the momentum uncertainty is infinite. If you want to include Lorentz invariance you need to use quantum field theory in which case you describe the evolution of a field with the path integral formalism and interpret particles as disturbances in the field.


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