Using a Toy Model to Improve the Quantization of Gravity and Field Theories

A half-harmonic oscillator, which gets its name because the coordinate is strictly positive, has been quantized and determined that it was a physically correct quantization. This positive result was found using a ﬃ ne quantization (AQ). The main purpose of this paper is to com-pare results of this new quantization procedure with those of canonical quantization (CQ). Using Ashtekar-like classical variables and CQ, we quantize the same toy model. While these two quantizations lead to di ﬀ erent results, they both would reduce to the same classical Hamiltonian if ¯ h → 0. Since these two quantizations have di ﬀ ering results, only one of the quantizations can be physically correct. Two brief sections illustrate how AQ can correctly help quantum gravity and the quantization of most ﬁeld theory problems.

The basic classical affine variables are the dilation d = pq and q ∕ = 0 because if q = 0, then d = 0 and p can not help. This leads to the quantum affine variables for an affine quantization (AQ) which are the dilation operator D = (P Q + QP )/2 and Q ∕ = 0, because if Q = 0 is allowed, then, just like the classical case, D = 0 and P can not help. 1 The half-harmonic oscillator has the same classical Hamiltonian, (p 2 + q 2 )/2 = (d 2 /q 2 + q 2 )/2 with q > 0, and the quantum Hamiltonian becomes (DQ −2 D + Q 2 )/2. That model has now been solved using affine quantization (AQ), and its eigenvalues are 2h(n + 1), with n = 0, 1, 2, ... [2,3]. Beyond that, the coordinate stopping point can be moved from q > 0 to q > −b, with b > 0, which, through computer calculations, have found that the eigenvalues continue to be equally spaced for all b > 0, and effectively found them to be the original eigenvalues (along with the original eigenfunctions) for the full harmonic oscillator when b → ∞ [4]. It is recognized that all of this is part of a valid quantization.

A Different Quantization Procedure for the
Half-Harmonic Oscillator

A change of classical variables
We start again with the classical Hamiltonian for the half-harmonic oscillator which is still H = (p 2 + q 2 )/2 and q > 0, but this time we will use different coordinates. To let our new coordinate variables span the whole real line, which makes them 'Ashtekar-like' [5], we choose q = s 2 , where −∞ < s < ∞. Thus, s is the new coordinate. For the new momentum we choose The classical Hamiltonian now becomes H = (p 2 + q 2 )/2 = (r 2 /4s 2 + s 4 )/2.

Quantization with the new quantum operators
For quantization, the new variables use canonical quantum operators, r → R and s → S, with [S, R] = ih1 1. Following the CQ rules, this leads to

A missing point
An issue that arose in Sec. 2.1 asked if q = s 2 > 0 or q = s 2 ≥ 0 should be adopted for the half-harmonic oscillator in order to achieve a valid halfharmonic oscillator quantization.
Should we care if a single point in a whole space is missing, i.e., does s ∕ = 0 really matter? In classical mechanics it might only be nothing but a nuisance.
However, as we will find out, in quantum mechanics it really can matter. We start by using many half-harmonic oscillators, leading to H = ! N n=1 (p 2 n + q 2 n )/2, with q n > 0 for all n. This equation can be interpreted differently as an N -dimensional vector, such as − → p for which − → p 2 = p 2 1 +p 2 2 +... and − → q for which − → q 2 = q 2 1 + q 2 2 + ..., with q n > 0 for all n. This implies that − → q 2 > 0, in which a single point in an N dimensional space has been removed (imagine that for N = 10, 000 ! ). How could that matter? It matters because the quantum theory of this 'toy model' is ]/2, which is dramatically sensitive to a single missing point where − → Q 2 = 0. By missing just one point in the entire coordinate region can lead to an incorrect quantization, as was shown by our toy model. Could that be likely to have any influence on the quantization of gravity?
A possible reply to that question may be found in the next section.

A Valid Quantum Gravity in a Nutshell
Before trying to solve a problem you should correctly formulate it! Physics says that the distance between two different, but very close, points in space is given by ds(x) 2 = g ab (x) dx a dx b > 0. This requirement ensures that g(x) ≡ det[g ab (x)] > 0. Likewise, from a purely mathematical view, and in preparation for an AQ quantization, the proper configuration space comes from J(x) = C a (x) g ab (x) C b (x), and all non-identically zero 'vectors' with components C a (x). While mathematically J(x) could be positive, zero, or negative, we choose only those metrics g ab (x) for which J(x) > 0. That leads to the desired physically correct configuration space.
Instead of choosing the classical functions π ab (x) and g cd (x), we choose the dilation field (also known as the momentric field) π a b (x) ≡ π ac (x) g bc (x) along with the metric field g ab (x). The Hamiltonian function (ignoring the cosmological constant) is given by is the 3-dimensional coordinate Ricci scalar [7].
For an AQ quantization, which uses Schrödinger's representation, the basic quantum operators are the metric quantum operatorĝ ab (x) = g ab (x) and the dilation quantum operatorκ a b (x) = [π ac (x) † g bc (x)+g bc (x)π ac (x)]/2, in whichπ ac (x) † g bc (x) =π ac (x)g bc (x). These operators lead to the commutation rules [g ab (x), g cd (x ′ )] = 0 , and the quantum Hamiltonian operator is While the quantum Hamiltonian of gravity is only part of the overall task, an incorrect version of that aspect is unlikely to use any further elements, like constraints, etc., to render a correct final quantization . Several articles pertaining to gravity by the author are [2,8,9,10].

Multiple Valid Quantum Field Theories in a Nutshell
Let us consider a scalar field ϕ(x), where x refers to an s-dimensional spatial field, as well as a momentum field π(x). These two classical fields lead to the dilation field κ(x) = π(x)ϕ(x), which is used instead of π(x), along with ϕ(x). Since κ(x) would vanish if ϕ(x) = 0, it is essential to require that ϕ(x) ∕ = 0. Moreover, if π(x) or ϕ(x) were infinite, they could not properly render κ(x), and therefore we require that |π(x)| + |ϕ(x)| < ∞, which then implies that |κ(x)| < ∞ as well. The presence of ( − → ∇ϕ(x)) 2 as part of the Hamiltonian density allows us to accept both positive and negative signs of ϕ(x) ∕ = 0, as if the field was still continuous. Using affine variables, the classical Hamiltonian density is given by Now we find that if 1/ϕ(x) 2 = 0, then κ(x) = 0. To prevent that from happening, we require that |ϕ(x)| < ∞, which reinforces that it has already been adopted. So not only must 0 < ϕ(x) −2 < ∞, it automatically requires that 0 < |ϕ(x)| p < ∞. Hence, it is fair to say that 0 ≤ H(x) < ∞. Finally, the classical Hamiltonian becomes which still must require that H(κ, ϕ) < ∞ in order to eliminate fields like ϕ(x) = 1 over any infinite space.
Using AQ and Schrödinger's representation for the dilation operator , and the quantum Hamiltonian becomes Using Monte Carlo, this last expression has already given positive results for the models ϕ 12 3 and ϕ 4 4 , where ϕ p n refers to the interaction power p and the spacetime number n = s + 1 [11,12].

Summary
This paper has largely been devoted to see if 'Ashtekar-like' coordinates can lead to physically correct quantizations by examining a toy model in order to find whether or not such variables might also lead to a physically correct quantization of gravity. It appears that a canonical quantization of the toy model, made possible by using Ashtekar-like variables, lead to a different quantum Hamiltonian from the known correct canonical equation. Thus, using canonical quantization and Ashtekar-like coordinates, have failed to lead to the physically correct quantization of the toy model.
In addition, using affine quantization procedures, and with various coordinate space removals, we have also given a highly realistic quantum version of quantum gravity, as well as for many realistic quantum examples of conventual quantum field theory.
If you use the right tools, you may solve a problem fairly easily. If you use the wrong tools, you may never solve the problem.