When Canonical Quantization Fails, Here is How to Fix It by John Rider Klauder

John Klauder

Essay Abstract

Following Dirac [1], the rules of canonical quantization include classical and quantum contact transformations of classical and quan- tum phase space variables. While arbitrary classical canonical coor- dinate transformations exist that is not the case for some analogous quantum canonical coordinate transformations. This failure is due to the rigid connection of quantum variables arising by promoting the corresponding classical variable from a c-number to a q-number. A different relationship of c-numbers and q-numbers in the procedures of Enhanced Quantization [2] shows the compatibility of all quantum operators with all classical canonical coordinate transformations.

Author Bio

Employed as Member of Technical Staff, AT&T Bell Laboratories, for 35 years, and Professor, University of Florida, for 22 years. Head, Theoretical Physics Research Department, 1966-1967 and 1969-1971. Head, Solid State Spectroscopy Research Department, 1971-1976. National Science Foundation Physics Advisory Panel, 1972-1975. Consultant, Los Alamos National Laboratory, Theoretical Division, 1978-1989. Editor, Journal of Mathematical Physics, 1979-1985. Associate Secretary-General, Executive Council of International Union of Pure and Applied Physics, 1985-1990. Member Executive Committee of the International Association of Mathematical Physics, 1985-1991. President, International Association of Mathematical Physics, 1988-1991.

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Scott Gordon

Hi John, Your paper is an exercise in pure math and to be very straight forward, I am wondering what is being presented as "fundamental" here? Can you help us out?

John Klauder

The rules of quantization are fundamental. The usual rule of the relation of a classical/quantum connection fails sometimes. It's replacement with a new rule performs much better. That is a fundamental change.

Scott Gordon

In this fundamental change in mathematical representation only or has the physical model changed or become more specific as to how the math comes out of a physical model? Or has the physical model remained unchanged with only the mathematical representation and notation changed? If only the mathematical notation changed, do you think that this math will bring about the answers bewildering physics such as why matter over anti-matter or dark energy, etc...

Anthony Garrett

John

Does this generalised canonical quantization procedure still guarantee the unitary property? That was the main strength of canonical quantization over the more manifestly covarient path-integral formulation; is this preserved?

Anton

John Klauder

Anton,

Then new rule preserves all unitary properties of the old way. What is different,is the choice of the Hamiltonian operator in relation to the classical Hamiltonian, when needed. This may effect properties of the path integral formulation. In short, that basic principles are preserved, just the choice of Hamiltonian operators may change.

John

John Klauder

Scott

The change I propose is to the physical rule relating a classical and quantum system and not simply a change of the mathematical formalism. This change preserves the classical/quantum connection that works for some systems but provides a different result when the usual classical/quantum fails, e.g., when the classical system has a nonlinear interaction but it's normal quantization has no nonlinear action. Such failures of the rule of quantization are unwelcome and the new rule provides an acceptable solution.

I do not expect that the change I propose will directly help in understanding the inequality you mention, but it could be part of the answer.

John

John Cox

Professor Klauder,

you have succeeded in getting me to follow that! I think I see your point. From the classical perspective, we might have an analytical observed momentum that falls one side or the other of a Planck value position on the real line, which might normalize (be made orthogonal) within a probable h-bar limit in real time. But in some cases, the convergence of two classical continuous functions will result in a real time window which would exceed that h-bar limit, and would nor renormalize. This of course is an analytical result, but must still be made true to the Planck value being the averaged least observable Action. And thus the metric for the Principle of Least Action being h-bar/2 in one dimension.

Did I get that correctly? Thank-you jrc

John Klauder

John

What you say has some merit but it is not connected with my essay. Let me say in simple terms the basic point I wish to make. A simple classical system involves two variables a position called q and a momentum p (normally a mass m times a velocity v, so p=m v), These variables are finite real numbers with no limitations. The Hamiltonian function H represents the energy of a system and is given as a function of p and q, e.g., H= p^2 +q^4, where p^2 = p squared, etc. One can change variables by a coordinate transformation, e,g, in 2 dimensions changing Cartesian coordinates x and y into polar variables r and u (u=the Greek letter theta) by letting x=r cos(u) and y= r sin(u), etc. A similar change can be made for p and q, e.g. b=p/q^2 and c=q^3/3, That leads to H having a different form, e.g., for our example H=b^2q^4+(3c)^{4/3}. Both of these forms are correct even though they are different. By choosing the right values for b and c the Hamiltonian H has the same numerical value as it did when we used p and q. Such equalities do NOT hold for quantum operators

such as P, Q, B, and C related to p, q, b, and c. One needs a rule to choose the right classical coordinates to promote to quantum operators. The present rule is what I want to change, i.e., what is the right quantum Hamiltonian operator for a given classical Hamiltonian. The current rule sometimes leads to nonsense. That is corrected with new rule. Briefly stated, my essay offers a new and much better way to.assign a Hamiltonian operator to a classical Hamiltonian.

John Cox

Professor Klauder,

Thank-you very much for your time in reply, it's Christmas and I consider it a tribute that you would trouble to teach. That is truly a giving spirit.

Choosing a Hamiltonian might sound strange, as it would seem arbitrary. But not to my mind. Any physical experiment ultimately resolves to our limitation of being able to only know how a detector reacts, and we can only conject on the source. In classical, the Bohr quantum leap is contentious yet we have the known atomic spectra from which Bohr, Schrodinger et.al. evolve. To promote a c number to a quantum number would mean that we are choosing to simplify to integers. And many if not most times in classicism a value goes to a mathematical singularity rather than a finite conclusion. So a c number would become skewed in transforms and a corrected canonical could be expected. Also, many times lack of rigor comes into play.

e.g. 'beam diameter' relating the intensity across a laser beam cross-section, employs an exponential root, rather than e as the base. This would violate conventions in linear algebra where the natural exponential function can only be used as the base, not the index. But it is argued that such a usage is a non-linear function. If one were to input into computation, a truncated numerical value for e, while using the full numerical value of c; results will rapidly diverge from a true function line. But if the algebraic algorithm to obtain the transcendental number's value is employed in calculation, the result for;

[c(c)^1/e] will become a finite value to only three decimal points, = 2.143^14 cm/sec

and we can see that this would be non-linear per light second on a single pole in a spherical boundary.

So a Hamiltonian expressing the point energy value of a co-ordinate in classical mechanics, can and must be compatible with its quantum number co-ordinate. I really should read more on your esteemed work in Quantum Mechanics, before making a fool of myself, here, but as with those whom I've had the pleasure to learn from on the Relativistic side, I can tell when someone knows what there talking about. And thank-you, jrc

Lawrence Crowell

I will comment more on your paper later. It is in many ways quite interesting. At this point I would tend to think that your quantization is based on conformal sympletic transformations.

Cheers LC

John Klauder

Please fire away!

John

Lawrence Crowell

I keep having trouble finding time for this. The one little point I can make is that it seem odd that one has (p,q| that acts on a state covector |ψ). I am not sure how to interpret (p,q|ψ). This little bit I am having a bit of difficulty understanding. (Note I use parentheses because carrot signs don't work in this format)

Cheers LC

John Klauder

I gather you are familiar with ((I have trouble with Psi) where say Q |x> = x |x> and int |x> that can be integrated or differentiated, etc. They have the property that the vectors |x> are mutually orthogonal in the form = delta(x - x') which is zero if x=/=x' yet integrate to 1. The important fact is they make a functional representation rather than an abstract vector like |u>..

But there can be other representations and that is what provides. The vectors |p,q> are continuously labeled which makes these representations only involve continuous functions unlike . They are NOT,mutually orthogonal, namely < p,q|p',q'> is often never zero and this inner product is normally bounded by one, but these vectors do lead to the important relation that int |p,q>

John Klauder

Somehow several symbols failed to service being posted. Terms like (x|u) and (p,q|u) (I now understand the () situation).

John Klauder

A new try to get my prior post clear

But there can be other representations and that is what (p,q|u) provides. The vectors |p,q) are continuously labeled which makes these representations only involve continuous functions unlike (x|u). They are NOT mutually orthogonal, namely (p,q|p',q') is often never zero and this inner product is normally bounded by one, but these vectors do lead to the important relation that int |p,q)(p,q| dp dq/2 pi = I the identity operator. In brief, (p,q|u) is just another representation of |u).

I hope this helps.

Lawrence Crowell

It is an odd notation. To say (p,q| is then a sort of "quantum and" that is a form of logical or. It is apparently a part of the argument for (в€‚П€/в€‚t)dt в†' . The total derivative

dП€ = [в€‚П€/в€‚t + (dq/dt)В·в€‡ П€]dt = (в€‚П€/в€‚t)dt + (i/Д§)pВ·dq П€.

If dП€/dt = 0 this gives the connection between the Lagrangian and the Schrodinger equation

iД§в€‚П€/в€‚t - HП€ в†' pВ·dq - H = L

The Lagrangian is in configuration variables, which is a way of casting physics according to one set of variables, or in the case of QM according to position variables and momentum expressed as operators.

So there seems to be something very subtle with the notation (p, q|. This appears to point to some subtle issue of quantum logic. I am still pondering this some for there seems to be something potentially subtle or deep going on here.

My next question will involve the Wheeler DeWitt equation.

Cheers LC

Lawrence Crowell

There is a problem with carrot signs, I meant to write: It is apparently a part of the argument for (в€‚П€/в€‚t)dt в†' pdq

John Klauder

I am unclear about your comments. Like (x| spans the Hilbert space means that if (x|u) = 0 for all x means that |u) = 0, there are other state families that do the same. (p,q| for all p and all q also span Hilbert space. This implies that if (p,q|u) = 0 for all p,q, then |u) = 0. An important property about Hilbert space representations is their inner product like (u|v) = int (u|x)(x|v) dx , where "int" means "integral". A very similar integral leads to (u|v) = int (u|p,q)(p,q|v) dp dq/2 pi h-bar. My previous relation like this one also had an h-bar, but units where h-bar = 1 were implicit. Thus these are two different functional representations, I.e., (x|v) and (p,q|v), that serve the same purpose.

The reduced action functional is given by

A = int (p(t),q(t)| [ i h-bar d/dt - £ ] |p(t),q(t)) dt = int { p(t) q-dot(t) - H(p(t),q(t)) } dt

where £ is the Hamiltonian operator. There is no need to introduce only position coordinates to build a traditional Lagrangian.

Georgina Woodward

Hi John, thank you for sharing your improvement, and for replies to comments that help explain why you have chosen to present it as an entry to the competition. As well as the function the improvement fulfills in improving the consistency the specific kinds of transformations (preventing the peculiar outcomes).

The question 'what is fundamental?' is very open. If founational is considered fundamental, and quantum physics to be the modelling of that foundation, then getting it to work consistently is fundamental to that specific area of physics attempting to model the fundamental foundation from which physics arises.

Section 1 were you talk about the desirability and problems of adding extra dimensions to phase space was interesting to me. Something for me to ponder. I wish I could understand more of the presentation but see why you have written it. Kind regards Georgina