From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Mon, December 28, 2009 12:42:08 PM Subject: Re: [math-fun] discrete version of exp(d/dz) Harold wrote: << Quoting Mike Stay <metaweta@gmail.com>: << Is there a general formula for "d/dz" in dimension D? You are in good company with Paul Dirac and Oliver Heaviside. As long as D is finite, there is no problem, even with a multiply valued complex logarithm. It is the limit that hurts, and for which distribution theory was sort of invented. I recall that in the late forties, Aurel Wintner proved that there are no matrices A and B such that AB - BA = I, the unit matrix. .. . . .. . .
Can someone please elaborate? I'm not getting the connection between Mike's question and being in good company with anyone, or with a complex logarithm, or with the equation AB - BA = I. Since Mike first mentioned that exp(d/dz)(f)(z) = f(z+1) (presumably on R or C), I thought he was looking for a D-dimensional analogue of this formula. No? If Harold's comment addresses this, apologies but I'm missing it. On the other hand, I haven't received Harold's post directly but only because Gene quoted it. Maybe I've missed some other intervening posts as well? --Dan _____________________________________________________________________ For the corresponding result in D-dimensional space, let z be a point in R^D or C^D, and let a be a displacement vector in the same space. Then [exp(a . grad)(f)](z) = f(z + a). The connection to commutators and quantum theory is that if q and p are respectively the position and momentum operators, then [q,p] = i h-bar (where h-bar is Planck's constant h divided by 2 pi). One way to represent q and p is by their action on the coordinate wave function psi(q). The action of operator q is psi(q) --> q psi(q), i.e. multiplication by q. The action of p is psi(q) --> (-i h-bar)(d/dq) psi(q). Now, the momentum is the infinitesimal generator of displacements: exp(-i p a / h-bar) psi(q) = psi(q - a), which is the same as displacing the wavefunction by a. Another representation is by the action on the momentum wavefunction phi(p). The action of p is multiplication by p. The action of q is (+i h-bar) (d/dp). As one might guess, psi(q) and phi(p) are a Fourier transform pair. -- Gene