Eliezer_Yudkowsky comments on Which Basis Is More Fundamental? - Less Wrong

14 Post author: Eliezer_Yudkowsky 24 April 2008 04:17AM

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Comment author: Eliezer_Yudkowsky 12 June 2008 12:00:20AM 1 point [-]

The derivative is still pretty local because it only depends on the immediate neighborhood in the continuous space. Does either the kinetic portion of the Hamiltonian in the position basis, or the potential portion of the Hamiltonian in the momentum basis, require looking at distant portions of configuration space? My understanding is that this is true for the latter but not the former; please correct me if this is not so.

Comment author: wnoise 27 March 2010 12:18:46AM *  6 points [-]

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.

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