This is not so, or it is equally so, depending on how exactly you interpret things.
To get expectation values of either x or p, we need to multiply by x (or p), and integrate over the entire configuration space. In that sense, they are both non-local.
In order to apply the Schroedinger equation:
-i h d/dt (psi(x)) = -h^2 (d/dx)^2(psi(x))/2m + V(x)(psi(x))
we can act locally in the position basis: we only need to examine the area around x to update psi(x) in the next timestep.
Or we can look at it in the momentum basis:
-i h d/dt (psi(p)) = p^2(psi(p))/2m + V(i h d/dp)(psi(p))
This is exactly as local in the momentum basis as the position basis. We only need to look at the area around p to update psi(p) for the next timestep.
They really are on equal footing.
EDIT: There is one slight complication—an infinite number of derivatives truly can become non-local: one nice example is exp(- i a p/h) psi(x) = exp(- a d/dx) psi(x) = psi(x-a). This is a reflection of momentum being the generator of displacement.
This is not so, or it is equally so, depending on how exactly you interpret things.
To get expectation values of either x or p, we need to multiply by x (or p), and integrate over the entire configuration space. In that sense, they are both non-local.
In order to apply the Schroedinger equation:
we can act locally in the position basis: we only need to examine the area around x to update psi(x) in the next timestep.
Or we can look at it in the momentum basis:
This is exactly as local in the momentum basis as the position basis. We only need to look at the area around p to update psi(p) for the next timestep.
They really are on equal footing.
EDIT: There is one slight complication—an infinite number of derivatives truly can become non-local: one nice example is exp(- i a p/h) psi(x) = exp(- a d/dx) psi(x) = psi(x-a). This is a reflection of momentum being the generator of displacement.