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 _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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
Quoting Dan Asimov <dasimov@earthlink.net>:
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?
Yes! The point is, there isn't any; but that needn't deter someone who doesn't know that. Think of Dirac's delta function (unit matrix), or Heaviside's operator calculus. The formula in question defines a shift operator, which is nicely enough related to Taylor's series. When and where do shift operators apply? Schroedinger was impressed with the fact that p + iq (d/dx + ix) and its conjugate ran you through the harmonic oscillator spectrum, and conjectured that other Hamiltonians had cimilar operators. They almost do, but the whole subject is fairly complicated. Mike's ideas are valid enough for finite matrices (over the reals (or complex)); it is getting a limit which would satisfy a "real" mathematician which is the problem, and he could be reassured to know that multiple valuedness of logarithms isn't the source of the problem. Wintner's paper, as I recall, used the trace argument and the ladder operators had infinite traces, so there was a convergence anomaly; in 1950 this did seem to strike at the roots of quantum mechanics. -hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
From: "mcintosh@servidor.unam.mx" <mcintosh@servidor.unam.mx> To: Dan Asimov <dasimov@earthlink.net>; math-fun <math-fun@mailman.xmission.com>; Dan Asimov <dasimov@earthlink.net> Cc: mcintosh@servidor.unam.mx Sent: Mon, December 28, 2009 1:24:53 PM Subject: Re: [math-fun] discrete version of exp(d/dz) .... Wintner's paper, as I recall, used the trace argument and the ladder operators had infinite traces, so there was a convergence anomaly; in 1950 this did seem to strike at the roots of quantum mechanics. -hvm ________________________________ Can you explain the "strike at the roots of quantum mechanics"? -- Gene
Quoting Eugene Salamin <gene_salamin@yahoo.com>:
Can you explain the "strike at the roots of quantum mechanics"?
As Dirac noted, that commutation relation distinguished classical hamiltonian mechanics from the new quantum mechanics; he saw it as the harbinger of a whole new kind of science (oops! Ive quoted the wrong authority), with q-numbers and c-numbers. That was his reaction to Heisenberg's preprint (or rather, galley proofs). -hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
From: "mcintosh@servidor.unam.mx" <mcintosh@servidor.unam.mx> To: math-fun <math-fun@mailman.xmission.com> Cc: mcintosh@servidor.unam.mx Sent: Mon, December 28, 2009 3:53:34 PM Subject: Re: [math-fun] discrete version of exp(d/dz) Quoting Eugene Salamin <gene_salamin@yahoo.com>:
Can you explain the "strike at the roots of quantum mechanics"?
As Dirac noted, that commutation relation distinguished classical hamiltonian mechanics from the new quantum mechanics; he saw it as the harbinger of a whole new kind of science (oops! Ive quoted the wrong authority), with q-numbers and c-numbers. That was his reaction to Heisenberg's preprint (or rather, galley proofs). -hvm ________________________________ To quote you more fully: "Wintner's paper, as I recall, used the trace argument and the ladder operators had infinite traces, so there was a convergenceanomaly; in 1950 this did seem to strike at the roots of quantum mechanics." But what was the "strike at the roots of quantum mechanics" in 1950? By that date, quantum mechanics was generally well accepted. And what were the trace arguments that led to this strike? That the Heisenberg algebra has no finite dimensional representation should have no bearing on everyday quantum theory, though it perhaps excludes a [q,p]= i hbar type mechanics over finite spaces. -- Gene
Quoting Eugene Salamin <gene_salamin@yahoo.com>:
But what was the "strike at the roots of quantum mechanics" in 1950? By that date, quantum mechanics was generally well accepted. And what were the trace arguments that led to this strike? That the Heisenberg > algebra has no finite dimensional representation should have no bearing on everyday quantum theory, though it perhaps excludes a [q,p]= i hbar type mechanics over finite spaces.
There are all kinds of acceptance, which is why mentioning Heaviside may have been worthwhile. His methods worked, excellently for electrical engineers, while thoroughly offending mathematicians. Or at least he seems to have thought so. For a famous example, consider Einstein's objections to quantum mechanics, which were studied in depth by Bohm and later Bell. That didn't keep anyone from working with quantum mechanics, although it certainly did lead eventually to quite interesting experimental results. Seventy years later. (1935 paper = 40 yr) There is the dictum "shut up and compute" which we heard often when "what does quantum mechanics really mean? was asked. But if all was well in 1950, why was so much attention given to Laurent Schwarz's distribution theory? Dyson certainly took it seriously enough to decide that that it didn't cure quantum electrodynamic's divergences. For some background to Wintner's paper, recall that he did some of the early work defining Hilbert spaces, which somehow got deemphasized when von Neumann wrote "Mathematische Grundlagen ..." specifically to counter Dirac's presentation and the "fiction that all operators are self-adjoint." Also, who coined the term "Heisenberg Algebra, and at what date? Coming to grips with quantum mechanics certainly took time, and that people bothered must be some indication that all was not crystal clear. Another Diracism: "All of Chemistry and most of Physics is now in principle understood." -hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
participants (3)
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Dan Asimov -
Eugene Salamin -
mcintosh＠servidor.unam.mx