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Double Causality: A Framework for Physics


Summary of results



Draft-in-progress


An example of a more sophisticated framework, an example of an investigation, and an example of how one conceptual framework can have wide reaching implications, can be the foundation for all of physics.

Fivesome

The fivesome refers to the division of everything into five perspectives, that which is needed for decision making, and which yield two representations, one in terms of time and the other in terms of space.

Here are two presentations that I gave about the fivesome:

The main idea of this conceptual framework: Every effect has had its cause. Not every cause has had its effect.

  • Our minds can move backwards from effect to cause, but also forwards from cause to effect. And a fifth perspective (the present or the boundary) relates the two causal directions.

We conceive this framework by way of two conceptions, in terms of time and in terms of space.

  • In time, the cause is in the past, the effect is in the future, and the present relates them in both directions.
  • In space, the cause is outside of a system, the effect is inside of a system, and a boundary separates them.

Examples of double causality

  • Jesus offers communion. "He took bread, and when he had given thanks, he broke and gave it to them, saying, 'This is my body which is given for you. Do this in memory of me.'"
  • Bob Dylan, "Don't think twice". "I wish that there was something you could do or say to try and make me change my mind and stay but we never did too much talking, anyway, so don't think twice, it's all right."
  • Beatles, "She loves you". "You think you've lost your love. Well, I saw her yesterday. It's you she's thinking of and she told me what to say. She say's she loves you."
  • Beatles, "Yesterday". "Yesterday love was such an easy game to play. Now I need a place to hide away. Oh, I believe in yesterday."
  • John Lennon, "God". "I don't believe in Jesus, ... I don't believe in Zimmerman, I don't believe in Beatles. I just believe in me Yoko and me. And that's reality. The dream is over. What can I say? The dream is over. Yesterday. I was the dream weaver, but now I'm reborn. I was the Walrus, but now I'm John. And so dear friends, you'll just have to carry on. The dream is over."

Combinatorial Interpretation of Schroedinger's equation

{$i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi$}

{$i\hbar\frac{\partial\Psi}{\partial t} + \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}=V\Psi$}

Solutions of Schroedinger's equation are orthogonal Sheffer polynomials

  • A wave function which is a solution to Schroedinger's equation can be thought of as consisting of an orthogonal polynomial (the information state) and a weight function (the space-time wrapper).
  • Question to investigate: In what sense can I claim that the only mathematical solutions to Schroedinger's equation are those which have this form?
  • Question to investigate: Are there examples of mathematical solutions or physical solutions that do not have this form?
  • Solutions for the quantum harmonic oscillator are based on Hermite polynomials, and solutions for the hydrogen atom are based on Laguerre polynomials.
  • These are orthogonal polynomials and their weight functions provide the inner product that is the basis for the Schroedinger equation.
  • The orthogonal polynomials and weight function depend on the potential specified.
  • Question to investigate: Assuming the solutions are orthogonal polynomials, then what are the possible potentials?

{$\langle Q\rangle =\int\Psi^*\hat{Q}\Psi \textrm{dx}=\langle\Psi |\hat{Q}\Psi\rangle$}

A combinatorial interpretation of a physical observable relates three elements.

  • The equation is satisfied by orthogonal polynomials.
    • Their coefficients are integers that have combinatorial interpretations.
    • The polynomials are in terms of a variable (such as x) which can be given a combinatorial interpretation (such as cells of free space).
  • Orthogonality is expressed with regard to a weight function that can be thought of as a space-time wrapper.
    • The combinatorics can be interpreted in terms of the moments, thus bypassing the wrapper, the sum or integral.
    • However, the weight function can also be interpreted combinatorially.
  • The physical operator, the observable, can be interpreted as a shift of orthogonal polynomials, by way of the relevant Rodrigues' formula, which is then multiplied with the conjugates of the originals and then submitted to the orthogonality relations.

Working hypothesis: In physical situations, the orthogonal polynomials must be Sheffer polynomials.

  • When other polynomials occur, such as Chebyshev polynomials for the infinite well, then we only get toy models which do not actually occur in physics.

Question to investigate: What is special about Sheffer polynomials?

answered by Tom Copeland, Shadows of Simplicity

Hidden structure, hidden symmetry, that is waiting to be discovered.

  • Maxwell's theory of the electromagnetic field, as we know, contains two hidden symmetries that will rock twentieth century physics. One is relativistic invariance and the other is gauge invariance. Now, both of these symmetries were already contained in Maxwell's theory, but nobody including Maxwell in the 19th century noticed it. And so is it conceivable that our present day theory also contains some kind of hidden structure? To me, that's a possibility. Hidden structure that's waiting for young people to discover just like Einstein discovered relativistic invariance hidden in Maxwell's equation and Hermann Weyl discovers gauge invariance in Maxwell's theory. – Anthony Zee, October 14, 2020, Quantum Field Theory: An Overview, 29:15.


Sheffer polynomials

Sheffer polynomial sequences {$\{P_n(x)\}$} are those with respect to which the shift operator {$T_a$} commutes with the operator Q. The shift operator is defined by {$T_af(x)=f(x+a)$}, and the operator Q is defined by {$QP_n(x)=nP_{n-1}(x)$}. The operator Q acts like a derivative on the highest power, and the remaining terms can be thought of as echo terms, which are dependent on the highest power of each polynomial in the sequence. Thus we have {$(T_aQ-QT_a)P_n(x)=0$}.

Sheffer sequences form a group under the operation of umbral composition, which is defined as follows. Given polynomial sequences {$p_n(x)=\sum_{k=0}^{n=\infty}a_{n,k}x^k$} and {$q_n(x)=\sum_{k=0}^{n=\infty}b_{n,k}x^k$}, we define {$(p_n\circ q)(x)=\sum_{k=0}^na_{n,k}q_k(x)x^k=\sum_{n\geq k>l\geq 0}a_{n,k}b_{k,l}x^l$}. [Investigation: Interpret this combinatorially.]

Sheffer polynomials {$\{P_n(x)\}$} have the form {$\sum_{n=0}^{\infty}P_n(x)t^n=A(t)e^{xu(t)}$}.

It is helpful to construct a generating sequence {$J(y)$}, a formal inverse to {$u(t)$}, such that {$J(u(t)) = u(J(t)) = t$}.

{$u(t)=\sum_{k=1}^{\infty}u_kt^k=t+u_2t^2+u_3t^3+\cdots$}

{$J(y)=\sum_{l=1}^{\infty}j_ly^l=y+j_2y^2+j_3y^3+\cdots$}

{$J(u(t))=\sum_{l=1}^{\infty}j_l(\sum_{k=1}^{\infty}u_kt^k)^l$}

{$u(J(t))=\sum_{k=1}^{\infty}u_k(\sum_{l=1}^{\infty}j_lt^l)^k$}

By examining {$J(u(t)) = u(J(t)) = t$} for each power {$t$}, e can get {$j_l$} in terms of the {$u_k$} and vice versa.

{$j_1=u_1=1$}

{$j_2+u_2=0$} thus {$j_2=-u_2$}

{$j_3+2j_2u_2+u_3=0$} thus {$j_3=-u_3+2u_2^2$}

{$j_4+ 2j_3u_2 +2j_2u_3 +j_2^2u_2 +j_2u_2^2 + 3j_3u_3 + u_4 = 0$} thus {$j_4=-u_4 + 3u_3^2 - 6u_2^2u_3 - 4u_2^3 + 4u_2u_3$}

Note that the equations are symmetric.

{$\frac{\textrm{d}}{\textrm{dt}}J(u(t))=\frac{\textrm{d}}{\textrm{dt}}t=1$}

{$\frac{\textrm{dJ}}{\textrm{du}}\frac{\textrm{du}}{\textrm{dt}}=1$}

{$J'(u)\equiv\frac{\textrm{dJ}}{\textrm{du}}=\frac{1}{u'(t)}$}

Let {$D=\frac{\textrm{d}}{\textrm{dx}}$}.

{$$DA(t)e^{xu(t)}=A(t)(u(t))e^{xu(t)}$$}

{$$D^kA(t)e^{xu(t)}=A(t)(u(t))^ke^{xu(t)}$$}

{$$J(D)A(t)e^{xu(t)}=\sum_{k=1}^{\infty}j_kA(t)(u(t))^ke^{xu(t)}$$}

{$$=A(t)e^{xu(t)}\sum_{k=1}^{\infty}J(u(t))$$}

{$$=A(t)e^{xu(t)}\sum_{k=1}^{\infty}j_k(u(t))^k$$}

{$$=A(t)e^{xu(t)}t$$}

{$$J(D)\sum_{n=0}^{\infty}P_n(x)t^n = \sum_{n=0}^{\infty}P_n(x)t^{n+1}$$}

By considering the coefficient for {$t^n$} we see that {$J(D)P_n(x)=P_{n-1}(x)$}.



Orthogonal Sheffer polynomials

All orthogonal polynomials satisfy a recurrence relation of the form {$xP_n(x) = P_{n+1}(x) + A_n P_n(x) + B_n P_{n-1}(x)$}.

More specifically, Sheffer polynomials satisfy a recurrence relation of the form {$P_{n+1}(x)=[x-(ln+f)]P_n(x)-[kn(n-1)-\gamma n]P_{n-1}(x)$}. See: Sheffer classification, Sheffer polynomials and the paper by Galiffa and Riston.

Here we can assume that {$f$} = 0 because that's equivalent to translating {$x$} to {$x-f$}.

{$$A'(t) = - \gamma t + \textrm{echo terms}$$}

{$$u'(t) = 1 - lt + (l^2-k)t^2 + \textrm{echo terms}$$}

I mean by "echo terms" that the higher order terms are completely determined by the initial terms and have no independent significance.

We can choose to write

{$$u'(t) = \frac{1}{(1-\alpha t)(1-\beta t)} = 1 + (\alpha + \beta) t + (\alpha^2 + \alpha\beta + \beta^2)t^2 + \textrm{echo terms}$$} {$$=(1+\alpha t+\alpha^2t^2+...)(1+\beta t+\beta^2t^2+...) = \sum_{n=0}^{\infty}t^n\sum_{i=0}^n \alpha^i\beta^{n-i}$$}

We have {$l=(-\alpha)+(-\beta)$} and {$k=(-\alpha)(-\beta)$} and {$(1-\alpha t)(1-\beta t) = 1+lt+kt^2$}. Thus we have

{$$u'(t)=\frac{1}{(1-\alpha t)(1-\beta t)}=\frac{1}{1+lt+kt^2}=\frac{A'(t)}{\gamma tA(t)}$$}

In the most general case, the solution gives the Meixner polynomials.

{$u(t)$}{$\textrm{ln}A(t)$}{$A(t)$}{$e^{x(u(t))}$}
{$\frac{1}{\beta - \alpha}\textrm{ln}(1-\alpha t)+\frac{1}{\alpha - \beta}\textrm{ln}(1-\beta t)$}{$\frac{\gamma}{\alpha \beta}[\frac{\alpha}{\alpha - \beta} \textrm{ln}(1-\beta t) - \frac{\beta}{\alpha - \beta} \textrm{ln}(1 - \alpha t)]$}{$(\frac{(1-\beta t)^\alpha}{(1-\alpha t)^\beta})^{\frac{\gamma}{\alpha\beta(\alpha - \beta)}}$}{$(\frac{1-\beta t}{1-\alpha t})^{\frac{x}{\alpha-\beta}}$}

Fivefold classification

Various specializations of the Meixner polynomials yield a fivefold classification.

polynomials{$\alpha$}, {$\beta$}{$u'(t)$}{$u(t)$}
Meixner{$\alpha \neq \beta$}, both real, nonzero{$\frac{1}{(1-\alpha t)(1-\beta t)}$}{$\frac{1}{\beta - \alpha}\textrm{ln}(1-\alpha t)+\frac{1}{\alpha - \beta}\textrm{ln}(1-\beta t)$}
Charlier{$\alpha\neq\beta$}, {$\beta=0$}{$\frac{1}{1-\alpha t}$}{$-\frac{1}{\alpha}\textrm{ln}(1-\alpha t)$}
Laguerre{$\alpha=\beta$}, {$\beta\neq 0$}{$\frac{1}{(1-\alpha t)^2}$}{$\frac{t}{(1 - \alpha t)}$}
Hermite{$\alpha=\beta$}, {$\beta=0$}{$1$}{$t$}
Meixner-Pollaczek{$\alpha=\overline{\beta}$}, complex conjugates, nonzero{$\frac{1}{(1-\alpha t)(1-\overline{\alpha} t)}$}{$\frac{1}{\overline{\alpha} - \alpha}\textrm{ln}(1-\alpha t)+\frac{1}{\alpha - \overline{\alpha}}\textrm{ln}(1-\overline{\alpha} t)$}

Combinatorial interpretation

The Sheffer family satisfies the recurrence relation below, with initial polynomials {$P_{-1}=0$} and {$P_0=1$}:

{$$P_{n+1}(x) = (x - (nl + f)) P_n(x) - n(nk + c) P_{n-1}(x)$$}

Writing it like this helps remember that {$l$} is the weight for links, {$k$} for kinks, {$f$} for fixed points (self-entangled trees), {$\gamma$} for cycles (trees). Here {$x$} is taken without coefficient and so the resulting polynomials are all monic. A coefficient can always be introduced by substituting {$x=\gamma y$}. {$f$} is trivial in that it corresponds to a translation of {$x$} and thus can be set as zero.





Space-time wrapper

Each orthogonal polynomial has a corresponding weight function against which it is integrated against. The weight function serves as the inner product with regard to which it is orthogonal.

PolynomialWeight {$w(x)$}Combinatorial interpretation
Meixnerstep function {$\frac{a^k(\beta)_k}{k!}$} on interval {$[k,k+1)$} 
Charlierstep function {$\frac{a^k}{k!}$} on interval {$[k,k+1)$} 
Laguerre{$x^\alpha e^{-x}$}Tracy-Widom distribution upper end, weak coupling, lower temperature, particles
Hermite{$e^{-x^2}$}Tracy-Widom distribution lower end, strong coupling, higher temperature, plasma
Meixner-Pollaczek{$\frac{|\Gamma(\frac{n+ix}{2})|^2}{|\Gamma(\frac{n}{2})|^2}e^{-x\;\textrm{tan}^{-1}\delta} $} 

The {$k$}th moment {$\mu_k$} of an orthogonal polynomial is gotten by integrating {$x^k$} against the weight {$\textrm{dw}$}.

{$$\mu_k=\int x^k \textrm{dw}$$}

In the case of the various Sheffer polynomials, these moments are fundamental numbers in combinatorics which count the most basic objects.

See also: Corteel, Kim, Stanton. Moments of Orthogonal Polynomials and Combinatorics.

Relate these to the construction of families of Zeng trees or of cycle double paths.

PolynomialMoment {$\mu_n$}Combinatorial interpretationAbstract data structurePossibility functions {$a_k,q_k,s_k$}{$L(x^n)$}Example when n=4
Meixnerordered Bell numberordered set partitions of n (weak orderings)(made of links)linear list{$k+1,0,k$} 541 in all: W>X>Y>Z, W>X>Y=Z, W>X=Y>Z, W>X=Y=Z, W=X>Y>Z ... Y>X>W>Z ... Y=Z>X=W etc.>
CharlierBell number {$B_n$}set partitions of n (equivalence relations)symbol table{$k+1,k,1$} 15 in all: {1}{2}{3}{4}, {12}{3}{4}, {13}{2}{4}, {14}{2}{3}, {23}{1}{4}, {24}{1}{2}, {34}{1}{2},{12}{34},{13}{24},{14}{23},{123}{4},{124}{3},{134}{2},{234}{1},{1234}.
Laguerren!permutations of n (strict orderings)dictionary{$k+1,2k+1,k$} 24 in all: 1234=(), 1243=(34), 1324=(23), 1342=(234), 1423=(243), 1432=(24), 2134=(12), 2143=(12)(34), 2314=(123), etc.
Hermite(n-1)!! or 0permutations of n where each cycle is disjoint and has length 2priority queue{$k+1,0,1$} 3 in all: (12)(34), (13)(24), (14)(23)
Meixner-PollaczekSecant number {$A_{2n}$}zig zag permutations of 2n (like kinks)   61 in all: 1 < 3 > 2 < 4 > 6 < 8 > 7 and 2 < 6 > 3 < 8 > 1 < 7 > 4 < 5 etc.

Kim and Zeng have a general formula for the moment, which in our case specializes to

{$$L(x^n)=\sum_{\sigma\in S_n}(-\alpha)^{a\sigma} (-\beta)^{d\sigma} 0^{fix\sigma}(\frac{-\gamma}{\alpha\beta})^{cycle\sigma}$$}

Thus the moment {$\mu_n$} considers permutations of length n, where the steps {$\alpha$} indicate ascents, steps {$\beta$} indicate descents, and the translation steps {$f$} (if we had {$f\neq 0$}), would indicate fixed points. (Evidently, fixed points of permutations characterize translations in Minkowski space.)

Quantum Field Theory

The Schroedinger equation is only a preliminary model of quantum reality. We need to add special relativity and we also need to allow for the ongoing creation and annihilation of particles. Thus we need to consider quantum field theory.


Scattering Zones


The Invisible Bowling Ball

Modify the classical picture of fast moving marbles scattering off a bowling ball.

Think of the "interaction potential" as an invisible bowling ball, the sum of the relevant quantum fields.

If we suppose that the field drops off further away, for example, as {$\frac{1}{r^2}$}, then at a large enough distance, by the Heisenberg uncertainty principle, the field becomes too subtle to be meaningful in any way. Thus we can say that at a certain distance it does not exist, but then at a certain distance it does.

The two roots {$\{{\alpha, \beta}\}$} describe the particle's clock. The clock can be thought of as relating two coordinate systems by steps {$\alpha$} from one to the other and by steps {$\beta$} going back. The steps can be thought of as ticks by the clock. Setting {$\alpha$} and {$\beta$} equal synchronizes the clocks. Setting {$\alpha$} or {$\beta$} to {$0$} resets the clock.

There are 5 zones:

  • {$\{{\alpha, \beta}\}$} - distant past, asymptotically free - unentangled state
  • {$\{{\alpha, 0}\}$} - entering the interaction potential - wave packet
  • {$\{{\alpha, \alpha}\}$} - within the interaction potential - bound state
  • {$\{{0, 0}\}$} - leaving the interaction potential - free space
  • {$\{\alpha, \bar{\alpha}\}$} - distant future, asymptotically free - entangled state

Remarks

  • Currently, as far as I know, quantum field theory only considers the zone where particles are leaving the interaction potential. This is akin to free space.
  • In what sense do the clocks themselves establish Minkowski space-time? What need, if any, do we have for a continuum of space-time?
  • Starting with Feynman diagrams (the lowest layer of the QFT theoretical cake) work backwards with alternate versions of Wick's theorem corresponding to each of the five zones. What does this yield in place of the traditional quantum field theory (the Hermite case)?
  • All of physics should be understood in terms of pairs of reference frames which are related by particle clocks.

Understanding the 5 zones in terms of an event

Reference frame with step {$\beta$} measures with regard to the entering of an event and reference frame with step {$\alpha$} measures with regard to the leaving of an event.

  • Before the interaction, the particle clock has steps that are real and independent, {$\{{\alpha, \beta}\}$}, which express the grid from the perspective of the two reference frames.
  • Reference frame with step {$\beta$} enters the event and resets itself to zero so we have {$\{{\alpha, 0}\}$} while the other frame continues counting as before.
  • Then the reset frame (in the event) now counts along {$\beta = \alpha$} with the frame which has yet to leave the event. This is the bound state, {$\{{\alpha, \alpha}\}$}.
  • Then the frame with step {$\alpha$} leaves the event and resets itself to zero, and so they are both zero, as they were both equal: {$\{{0, 0}\}$}
  • Then the two frames are counting as conjugates {$\{\alpha, \bar{\alpha}\}$} where the real component is the same and the imaginary component is opposite.

What happens if we reverse time and enter the event from the opposite direction? What happens to nonentanglement and entanglement?

Initial state and final state

Creation and annihilation operators

Creation and annihilation operators, Ladder operator

{$a\mid\! n\rangle= \sqrt{n} \ |n-1\rangle$}

The eigenvalue {$sqrt{n}$} of the annihilation operator is based on the state {$n$} it is leaving.

{$a^\dagger \mid\! n\rangle= \sqrt{n+1}\mid\! n+1\rangle$}

The eigenvalue {$\sqrt{n+1}$} of the creation operator is based on the state {$n+1$} it is going to.

Contraction

Contraction expresses the movement of a particle because it relates its creation and annihilation.

Commutator

{$[a,a^\dagger]\mid\! n\rangle = aa^\dagger\mid\! n\rangle - a^\dagger a\mid\! n\rangle = a\sqrt{n+1}\mid\! n+1\rangle - a^\dagger\sqrt{n}\mid\! n-1\rangle = \sqrt{n+1}\sqrt{n+1}\mid\! n\rangle - \sqrt{n}\sqrt{n}\mid\! n\rangle = 1\mid\! n\rangle$}

{$[a,a^\dagger]=1$}

The creation operator (raising operator) and annihilation operator (lowering operator) do not commute. This means that they cannot be measured at the same time.

Gaining knowledge of the eigenvalues of the raising operator changes knowledge of the eigenvalues of the lowering operator and vice versa.

Noncommutativity expresses how the definition of the operators changes based on their order.

  • A helpful example. If you start on Earth on the equator at zero longitude, then compare 1) moving 50 km North and then 50 km East, 2) moving 50 km East and then 50 km North. You end up in different locations. The reason is that at higher latitudes the motion East takes you further out around the sphere. Which can be understood to mean that the definition of East has changed.
  • In the example, consider what it means at the South Pole (not obviously defined). Then move one step away and they become well defined. Then going 50 km changes the definition of going East radically.

Wick's theorem

Wick's theorem

What's up with Wick's theorem with references in the comments.

Relate to the Hermite polynomials.

Normal-ordered terms acting on vacuum state give a null contribution to the sum. We conclude that m is even and only completely contracted terms remain.



Feynman diagrams

Generate the Feynman diagrams.

Label the diagrams with the relevant weights.

interpretation

  • 2 additional operators (or modifications of creation and annihilation) to represent the particle's internal clock (and/or causal state - how is it relating to single and double causality)
  • cycle - sidestepping forward and back - are ticks in a clock - measure time
  • alpha, beta relate two
  • one tree, one particle? (create and annihilate?)
  • trees express causality in space
  • creates space and time like gravity - no gravity as you leave (synchronized) - but gravity plays a role in the other four states
  • particle clock should take care of space-time
  • energy not too relevant - Lagrangian, Lagrangian density not that relevant
  • quadratic because of second order relationship (two inputs, one output) of infinite polynomials
  • The reset of a clock (to zero) is the collapse of the wave function.

Wick's theorem

  • normal ordered terms acting on null space give zero contribution
  • m is even and only completely contracted terms remain
  • fermions - add negative sign with each swap
  • in general - add clock with each swap
  • My guess: swapping operators changes clock direction for fermions, but not for bosons
  • With orthogonal terms, the translation term, which can be removed, is a tree without a double root, and thus is a clock without a particle.

Not action at a distance. Velocity of transmission?

Commutator

  • Operators that commute can be measured at the same time.
  • Operators that do not commute cannot be measured at the same time.
  • If they are canonical conjugates, then one is the Fourier transform of the other, as with {$[\hat{x},\hat{p}_x]=i\hbar\mathbb{I}$}
  • If we multiplied the creation operator by {$i\hbar$} or {$-i\hbar$} (for bosons or fermions) then it would be a canonical conjugate to the annihilation operator. Would this mean that they simply have a phase difference?
  • Or we could multiply each operator, creation and annihilation, by {$\sqrt{i\hbar}$} Note that {$\sqrt{i}$} is an eighth root of unity and could be related to the eight-cycle of divisions of everything.
  • The basis for the commutator: You can't annihilate a particle if it is not there. You can't lower the null state. But this argument is not relevant because it would hold even if the created particle and the annihilated particle were of different kinds, but that just gives zero for the commutator.
  • Why is the negative sign added in the fermion case?
  • What is the physical meaning of commutators in quantum mechanics?
    • Self-adjoint operators are generators of unitary groups representing (strongly continuous) physical transformations of the physical system. Such a continuous transformation is represented by a unitary one-parameter group R∋a↦Ua. A celebrated theorem by Stone indeed establishes that Ua=eiaA for a unique self-adjoint operator A and all reals a.
    • The action of a symmetry group Ua on an observable B is made explicit by the well-known formula in Heisenberg picture: Ba:=U†aBUa For instance, if Ua describes rotations of the angle a around the z axis, Ba is the analog of the observable B measured with physical instruments rotated of a around z.
    • The commutator here is a first-order evaluation of the action of the transformation on the observable B, since (again up to mathematical subtleties especially regarding domains): Ba=B−ia[A,B]+O(a2).
    • Usually, information encompassed in commutation relations is very deep. When dealing with Lie groups of symmetries, it permits to reconstruct the whole representation (there is a wonderful theory by Nelson on this fundamental topic) under some quite mild mathematical hypotheses. Therefore commutators play a crucial role in the analysis of symmetries.
    • [A,B] quantifies the extent to which the action of B changes the value of the dynamical variable A, and vice versa.
    • The B(Δ) are "ladder operators" which act to increase the value of the variable A by an amount Δ. The commutator thus induces a natural decomposition of B into contributions that change the value of A by a given amount.
    • If [A,B]≠0, one can get a measure of how much B changes A by computing the Hilbert-Schmidt norm (squared) of the commutator: Tr{[A,B]†[A,B]}=∑a,a′(a−a′)2|⟨a|B|a′⟩|2. This is the sum of the (squared) matrix elements of B which link different eigenstates of A, weighted by the corresponding change in eigenvalues (squared). So this clearly quantifies the change in A brought about by applying B.
  • Quora: What is the physical signficance of a commutator?
    • If the commutator of two operators[1] is zero, it means that you can make simultaneous measurements corresponding to those operators. In other words, making a measurement corresponding to one operator does not alter the state of the particle if you previously made a measurement corresponding to the other operator.
    • If the commutator of two operators is nonzero, then an uncertainty principle applies and you cannot measure values of both operators simultaneously. The commutator [x,px]=iℏ - since it is nonzero, this means that, if you measure position first, it pushes the particle into a superposition of possible momentum states, which have to be collapsed again by a momentum measurement.
  • Measuring (gaining knowledge of) position changes knowledge of momentum; measuring momentum changes knowledge of position; measuring time changes knowledge of energy; measuring energy changes knowledge of time.

Look at the eigenvalues - creation operator's eigenvalue references the state to which it goes, annihilation operator's eigenvalue references the state from which it comes. Thus in one order (create, then annihilate) you get the (n+1) from the higher state, and in the opposite order (annihilate, then create) you get the n from the lower state.

  • Change in definition means that you have to use a different basis of eigenvectors?

Space-time


Understanding space-time as probability distribution.

Hermite.Standard normal distribution. Binomial distribution.  
Charlier.Poisson distribution{$\!f(k; \lambda)= \Pr(X{=}k)= \frac{\lambda^k e^{-\lambda}}{k!}$} where {$\lambda$} is the mean and also the variance.A discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. The Poisson distribution can also be used for the number of events in other specified interval types such as distance, area or volume.
Laguerre.Chi-squared distribution{$f(x;\,k) = \dfrac{x^{\frac k 2 -1} e^{-\frac x 2}}{2^{\frac k 2} \Gamma\left(\frac k 2 \right)}$} where {$0 < x < \infty$}The distribution of a sum of the squares of k independent standard normal random variables. The chi-squared distribution is a special case of the gamma distribution and is one of the most widely used probability distributions in inferential statistics, notably in hypothesis testing and in construction of confidence intervals.

The Gaussian distribution can be derived from the binomial distribution in the limit {$N → ∞, N p → ∞$}. The Poisson distribution is another possibility: {$N → ∞, p → 0, N p = λ$} where {$\lambda$} is constant, so that the probability p for a step is assumed to be very small. Thus moving from the Poisson distribution to the Gaussian distribution simply means redefining the circumstances so that the probability is larger.

Weight function for Meixner polynomials

Think of {$N=\frac{-\gamma}{\alpha\beta}$} as a positive integer. Then the Meixner polynomials are the same as the Kravchuk polynomials.

This weight {$N$} is the global quantum. It gets assigned to each particle-clock, which is to say, to each cycle. It is the cut-off from above. It could mean that you can't have cycles of length greater than {$N$}.

Suppose {$\alpha\neq\beta$} and think of {$\alpha - \beta$} as the unit of the one-dimensional space-time grid, which is to say, steps in space or ticks by the clock.

The weight function for the Meixner polynomials is {$w((\alpha-\beta)n)=(\frac{\alpha}{\beta})^n\binom{N}{n}$}. This means that you choose {$n$} steps and you turn them around. You replace {$n$} steps {$\beta$} with {$n$} steps {$\alpha$}.

An alternative way to write this is {$w((\alpha-\beta)n)=\frac{1}{n!}(\frac{\alpha}{\beta})^n(N)_n$}.

Adjoint probability

I think of {$p$} and {$p-1$} as adjoint probabilities. This is to say that {$p-1$} is the complementary probability {$1-p$} but in the negative direction. I think of {$p-1$} as expressing the complementary probability in the opposite direction in time. [Investigation: Consider what this might mean in terms of Bayesian and frequentist probabilities.]

Discriminant

The discriminant {$\sqrt{l^2-4k}$} is important for classifying orthogonal Sheffer polynomials. It can be positive (Meixner), negative (Meixner-Pollaczek) or zero (Laguerre, Hermite). There are also the Charlier polynomials. The discriminant arises from the quadratic equation.

It is helpful to realize that {$\alpha -\beta=\sqrt{l^2-4k}$}. For {$l=(-\alpha)+(-\beta)$} and {$k=(-\alpha)(-\beta)$}.

Thus the discrete unit of space-time is the discriminant!

The Meixner weight function should be the most general because it is in terms of both {$\alpha$} and {$\beta$}. However, we cannot have {$\beta=0$} because it is in the denominator.

Quantum-discrete vs. classical-continuous

If {$\alpha - \beta \neq 0$} then the entire system is moving with a mean {$pN$} and a variance {$p(p-1)N$}.

We have the extra bosonic condition that {$\alpha - \beta = 1$} which means that you can't have both {$\alpha$} and {$\beta$} equal to {$0$}.

{$\alpha = p$}, {$\beta = -(1-p)$}, {$l=-\alpha -\beta = 1-2p$}, {$k=(-\alpha)(-\beta)=p(p-1)$}, {$f=pN$}.

Meixner-Pollaczek

If we substitute {$\alpha = a+bi, \beta=\bar{\alpha}=a-bi$}, then we get the Meixner-Pollazcek polyomials. Note that in polar coordinates {$\frac{\alpha}{\bar{\alpha}}=\textrm{cos}2\theta + i\textrm{sin}2\theta$}. We can apply de Moivre's formula. Also, the unit of space-time is {$\alpha - \bar{\alpha}=2bi$}. The weight function is then {$w((2bi)n)=w(\frac{\alpha}{\bar{\alpha}}|\alpha|n) = (\textrm {cos}2n\theta + i \textrm{sin}2n\theta)\binom{N}{n}$}.


Calculating observables


Calculating energy

Consider how this all comes together in calculating an observable, namely, the energy.

The probability density (the space-time) is a distribution of a random process that converts the information state into an observational probability.

{$\hat{H}=\frac{1}{2m}(-i\hbar\frac{\partial}{\partial x})(-i\hbar\frac{\partial}{\partial x}) + \frac{1}{2m}(m\omega x)^2$}

{$\langle H\rangle =\int\psi^*\hat{H}\psi dx = \langle\psi | \hat{H}\psi\rangle $}

{$\psi_n(x) = (\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\frac{1}{\sqrt{2^nn!}}H_n(\xi)e^{\frac{-\xi^2}{2}}$} is the expression for the wave function in Griffiths (2.86) where {$\xi =\sqrt{\frac{m\omega}{\hbar}}x$} and {$H_n$} is the physicist Hermite polynomial.

We may write {$\psi_n(x) = H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}}$} where {$C=(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\frac{1}{\sqrt{2^nn!}}$}.

{$\frac{\partial H_n}{\partial\xi}=2nH_{n-1}(\xi)$}

{$\hat{H}\psi_n(x) = [\frac{1}{2m}(-i\hbar\frac{\partial}{\partial x})(-i\hbar\frac{\partial}{\partial x}) + \frac{1}{2m}(m\omega x)^2] H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}}$}

{$ = \frac{1}{2m}(-i\hbar\frac{\partial}{\partial x})(-i\hbar\frac{\partial}{\partial x}) H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}} + \frac{1}{2m}(m\omega)^2 x^2 H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}}$} and note that {$x^2=\frac{\hbar}{m\omega}\xi^2$}.

{$ = \frac{-1}{2m}\hbar^2\frac{\partial}{\partial x}\frac{\partial}{\partial x} H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}} + \frac{1}{2m}(m\omega)^2 \frac{\hbar}{m\omega} \xi^2 H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}}$}

{$ = \frac{-1}{2m}\hbar^2\frac{m\omega}{\hbar}\frac{\partial}{\partial\xi}\frac{\partial}{\partial\xi} H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}} + \frac{1}{2m}(m\omega)^2 \frac{\hbar}{m\omega} \xi^2 H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}}$}

{$ = \frac{-1}{2}\hbar\omega\frac{\partial}{\partial\xi}(2n H_{n-1}(\xi)\cdot Ce^{-\frac{\xi^2}{2}} - H_n(\xi)\cdot \xi Ce^{-\frac{\xi^2}{2}}) + \frac{1}{2}\hbar \omega \xi^2 H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}}$}

{$ = \frac{-1}{2}\hbar\omega(4n(n-1) H_{n-2}(\xi)\cdot Ce^{-\frac{\xi^2}{2}} - 2n H_{n-1}(\xi)\cdot \xi Ce^{-\frac{\xi^2}{2}} - 2n H_{n-1}(\xi)\cdot \xi Ce^{-\frac{\xi^2}{2}} - H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}} + H_n(\xi)\cdot \xi^2 Ce^{-\frac{\xi^2}{2}} ) + \frac{1}{2}\hbar \omega \xi^2 H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}}$}

{$ = \frac{-1}{2}\hbar\omega (4n(n-1) H_{n-2}(\xi) - 4n \xi H_{n-1}(\xi) + (-1 + \xi^2 + -\xi^2 )H_n(\xi))Ce^{-\frac{\xi^2}{2}} $}

{$ = \frac{-1}{2}\hbar\omega (4n(n-1) H_{n-2}(\xi) - 4n \xi H_{n-1}(\xi) - H_n(\xi))Ce^{-\frac{\xi^2}{2}} $} and make use of the recursion relation {$H_{n+1}(\xi)=2\xi H_n(\xi) - 2 n H_{n-1}(\xi)$} to identify {$4n \xi H_{n-1}(\xi)= 2n H_{n}(\xi) + 4 n^2 H_{n-2}(\xi)$}.

{$ = \frac{-1}{2}\hbar\omega (4n(n-1) H_{n-2}(\xi) - 2n H_{n}(\xi) - 4n^2 H_{n-2}(\xi) - H_n(\xi))Ce^{-\frac{\xi^2}{2}} $}

{$ = \frac{-1}{2}\hbar\omega ((-4n) H_{n-2}(\xi) - (2n+1) H_n(\xi))Ce^{-\frac{\xi^2}{2}} $}

Then by orthogonality

{$\langle H\rangle =\int\psi^*\hat{H}\psi dx = \int H_n(\xi)\cdot C^*e^{-\frac{\xi^2}{2}} \frac{-1}{2}\hbar\omega (-4n H_{n-2}(\xi) - (2n+1) H_n(\xi))Ce^{-\frac{\xi^2}{2}} dx$}

{$ = \int H_n(\xi)\cdot C^*e^{-\frac{\xi^2}{2}} \frac{-1}{2}\hbar\omega (-4n H_{n-2}(\xi) - (2n+1) H_n(\xi))Ce^{-\frac{\xi^2}{2}} \sqrt{\frac{\hbar}{m\omega}}d\xi$}

{$= - (2n+1)\frac{-1}{2}\hbar\omega \sqrt{\frac{\hbar}{m\omega}} C^2 \int H_n(\xi) H_n(\xi) e^{-\xi^2} d\xi $}

{$= \frac{2n-1}{2}\hbar\omega \sqrt{\frac{\hbar}{m\omega}} (\frac{m\omega}{\pi\hbar})^{\frac{1}{2}}\frac{1}{2^nn!} \int_{-\infty}^{\infty} H_n(\xi) H_n(\xi) Ce^{-\xi^2} d\xi $} and the integral by orthogonality yields the factor {$2^nn!\sqrt{\pi}$}

{$= \frac{2n+1}{2}\hbar\omega\frac{1}{\sqrt{\pi}}\frac{1}{2^nn!} 2^nn!\sqrt{\pi} $}

{$= \frac{2n+1}{2}\hbar\omega$} as required.

Note that because the Hermite polynomials are orthogonal under the relevant inner product, each of them is an eigenvector for an observable's operator. For whatever the operator does to the wave function, multiplying the output by the conjugate and integrating over the product will yield only the contribution by the original polynomial.

The same calculation in terms of probabilist Hermite polynomials

Our combinatorial interpretation is best made in terms of the probabilist Hermite polynomials and so we need to convert the expressions above.

The physicist Hermite polynomial {$H_n(x)$} is related to the probabilist Hermite polynomial {$\textrm{He}_n(y)$} by the equations {$H_n(x)=2^{\frac{n}{2}}\textrm{He}_n(\sqrt{2}x)$} and {$\textrm{He}_n(y)=2^{-\frac{n}{2}}H_n(\frac{x}{\sqrt{2}})$}.

Instead of {$\psi_n(x) = H_n(\xi)\cdot Ce^{-\frac{\xi^2}{2}}$} where {$\xi =\sqrt{\frac{m\omega}{\hbar}}x$} we should write {$\psi_n(x) = 2^{\frac{n}{2}}\textrm{He}_n(y)\cdot Ce^{-\frac{y^2}{4}}$} where {$y =\sqrt{\frac{2m\omega}{\hbar}}x$}.

{$\frac{\partial\textrm{He}_n}{\partial y}=n\textrm{He}_{n-1}(y)$}

{$\hat{H}\psi_n(x) = \frac{-1}{2m}\hbar^2\frac{\partial}{\partial x}\frac{\partial}{\partial x} 2^{\frac{n}{2}}\textrm{He}_n(y)\cdot Ce^{-\frac{y^2}{4}} + \frac{1}{2m}(m\omega)^2 \frac{\hbar}{m\omega} \frac{1}{2} y^2 2^{\frac{n}{2}}\textrm{He}_n(y)\cdot Ce^{-\frac{y^2}{4}}$}

{$ = \frac{-1}{2m}\hbar^2\frac{2m\omega}{\hbar}2^{\frac{n}{2}}\frac{\partial}{\partial y}\frac{\partial}{\partial y} \textrm{He}_n(y)\cdot Ce^{-\frac{y^2}{4}} + \frac{1}{2m}(m\omega)^2 \frac{\hbar}{m\omega}\frac{1}{2} y^2 2^{\frac{n}{2}} \textrm{He}_n(y)\cdot Ce^{-\frac{y^2}{4}}$}

{$ = -2^{\frac{n}{2}}\hbar\omega\frac{\partial}{\partial y}(n \textrm{He}_{n-1}(y)\cdot Ce^{-\frac{y^2}{4}} - \textrm{He}_n(y)\cdot \frac{1}{2}y Ce^{-\frac{y^2}{4}}) + \frac{1}{4}y^2 \textrm{He}_n(y)\cdot \hbar \omega 2^{\frac{n}{2}}Ce^{-\frac{y^2}{4}}$}

{$ = -2^{\frac{n}{2}}\hbar\omega(n(n-1) \textrm{He}_{n-2}(y)\cdot Ce^{-\frac{y^2}{4}} - n \textrm{He}_{n-1}(y)\cdot \frac{1}{2} y Ce^{-\frac{y^2}{4}} - n \textrm{He}_{n-1}(y)\cdot \frac{1}{2}y Ce^{-\frac{y^2}{4}} - \textrm{He}_n(y)\cdot \frac{1}{2} Ce^{-\frac{y^2}{4}} + \textrm{He}_n(y)\cdot \frac{1}{4} y^2 Ce^{-\frac{y^2}{4}} ) + \frac{1}{4}y^2 \textrm{He}_n(y)\cdot \hbar \omega 2^{\frac{n}{2}} Ce^{-\frac{y^2}{4}}$}

{$ = -2^{\frac{n}{2}}\hbar\omega (n(n-1) \textrm{He}_{n-2}(y) - ny \textrm{He}_{n-1}(y) + (-\frac{1}{2} + \frac{1}{4}y^2 + -\frac{1}{4}y^2 )\textrm{He}_n(y))Ce^{-\frac{y^2}{4}} $}

{$ = -2^{\frac{n}{2}}\hbar\omega (n(n-1) \textrm{He}_{n-2}(y) - ny \textrm{He}_{n-1}(y) - \frac{1}{2}\textrm{He}_n(y))Ce^{-\frac{y^2}{2}} $} and make use of the recursion relation {$\textrm{He}_{n+1}(y)=y \textrm{He}_n(y) - n \textrm{He}_{n-1}(y)$} to identify {$n y \textrm{He}_{n-1}(y)= n \textrm{He}_{n}(y) + n^2 \textrm{He}_{n-2}(y)$}.

{$ = -2^{\frac{n}{2}}\hbar\omega (n(n-1) \textrm{He}_{n-2}(y) - n \textrm{He}_{n}(y) - n^2 \textrm{He}_{n-2}(y) - \frac{1}{2}\textrm{He}_n(y))Ce^{-\frac{y^2}{4}} $}

{$ = -2^{\frac{n}{2}}\hbar\omega (-n\textrm{He}_{n-2}(y) - (n+\frac{1}{2}) \textrm{He}_n(y))Ce^{-\frac{y^2}{4}} $}

Then by orthogonality

{$\langle H\rangle =\int\psi^*\hat{H}\psi dx = \int 2^{\frac{n}{2}} \textrm{He}_n(y)\cdot C^*e^{-\frac{y^2}{4}}(-2^{\frac{n}{2}})\hbar\omega (-n \textrm{He}_{n-2}(y) - (n+\frac{1}{2}) \textrm{He}_n(y))Ce^{-\frac{y^2}{4}} dx$}

{$ = -2^n\int \textrm{He}_n(\xi)\cdot C^*e^{-\frac{y^2}{4}}\hbar\omega (-n \textrm{He}_{n-2}(y) - (n+\frac{1}{2}) \textrm{He}_n(y))Ce^{-\frac{y^2}{4}} \sqrt{\frac{\hbar}{2m\omega}}dy$}

{$= (n+\frac{1}{2})2^n\hbar\omega \sqrt{\frac{\hbar}{2m\omega}} C^2 \int \textrm{He}_n(y) \textrm{He}_n(y) e^{-\frac{y^2}{2}} dy $}

{$= (n+\frac{1}{2})2^n\hbar\omega \sqrt{\frac{\hbar}{2m\omega}} (\frac{m\omega}{\pi\hbar})^{\frac{1}{2}}\frac{1}{2^nn!} \int_{-\infty}^{\infty} \textrm{He}_n(y) \textrm{He}_n(y) Ce^{-\frac{y^2}{2}} dy $} and the integral by orthogonality yields the factor n!{$\sqrt{2\pi}$}

{$= (n+\frac{1}{2})\hbar\omega\frac{1}{\sqrt{2\pi}}\frac{1}{n!}n!\sqrt{2\pi}$}

{$= (n+\frac{1}{2})\hbar\omega$} as required.

Explain the combinatorics of the Hamiltonian in terms of the combinatorics of differentiation and multiplication.

Investigation: Calculate {$\langle H\rangle$} in the general case for {$\psi=\sum_{n=0}^{\infty}c_n\psi_n$}.


Overview of combinatorial interpretation


I need to provide combinatiorial interpretations of the following:

  • The definition of the Hermite polynomials. (Done)
  • The recursion relation for the Hermite polynomials. (Done)
  • The orthogonality relation for the Hermite polynomials. (I need to undertand the paper by Kim and Zeng.)
  • The Hamiltonian operator's action on the Hermite polynomial. (I need to interpret the action and also translate from physicist Hermite polynomials to probabilist Hermite polynomials.)

The x's (free space, position) are time events that stay fixed. The i's (momentum) are time events that switch order (they link two events, two interactions). What does it mean if we switch from position to momentum, if we take the Fourier transform?


Space builders


I have been trying to argue that the only solutions to Schroedinger's equation (and related quantum field theory equations) which are physical are those which are in terms of orthogonal Sheffer polynomials. I now share my idea of what that means physically.

Sheffer polynomials are those that build space incrementally

Combinatorially, Sheffer sequences are those which have the generating function {$\sum_{n=0}^{\infty} P_n(x)s^n = C(s) e^{x u(s)}$}. This means that the terms in the polynomial {$P_n(x)$} are assembled by pairing a component in {$C(s)$} with one or more components from {$x u(s)$}.

On Monday I told John that we have the equation {$n! P_n(x)$} = the {$n$}th derivative with respect to {$s$} of {$C(s) e^{x u(s)}$} evaluated at {$0$}. So I became interested to calculate that {$n$}th derivative with respect to {$s$}.

The number of terms in the derivatives grew as {$1, 2, 5, 15 \cdots$} So I looked it up at the online encyclopedia of integer sequences and I got the Bell numbers.

The Bell numbers count the number of ways of partitioning a set of size n+1. Namely:

for {$P_0(x)$}: {$[0]$}

for {$P_1(x)$}: {$[01]$} & {$[0][1]$}

for {$P_2(x)$}: {$[012]$} & {$[01][2]$} & {$[02][1]$} & {$[0][12]$} & {$[0][1][2]$}

The sequence grows as {$1, 2, 5, 15, 52 \dots$} It also counts the number of ways of setting up equivalence classes on n elements.

I realized that I can prove by induction straightforwardly that Sheffer sequences are, combinatorially, those that build up space in terms of partitions.

Differentation of {$C(s) e^{x u(s)}$} generates two kinds of components. On the one hand, we can differentiate {$C = C(s)$} to get {$C'$} and then {$C''$} and {$C'''$} and so on. On the other hand, the derivative of {$E = e^{x u(s)}= e^W$} is {$W' E$}. And then we simply use the product rule, and if we have a product of many factors, then we simply differentiate one of those factors and sum up the possibilities.

What this means, spatially, is that we start of with a space {$C E$}. I won't write the {$E$} because it is omnipresent.

{$C = [0]$} is the initial (empty) space.

Then we can add a new cell 1 in two ways.

We can add it in a new compartment: {$C W' = [0][1]$}

Or we can add it to an existing compartment by lengthening that compartment, which means by differentiating that compartment. {$C' = [01]$}.

Consider now adding another new cell 2. We can add it as a new compartment: {$C W' W' = [0][1][2]$} & {$C' W' = [01][2]$}.

Or we can add it to an existing comparment by differentiating: {$C' W' = [02][1]$} & {$C W'' = [0][12]$} & {$C'' = [012]$}.

The way to prove the claim is simply by induction to suppose that the terms in the nth derivative match the partitions of {$n$}, and then to show that there is a bijection between the growth of the partitions and the effects of differentiation. Adding the {$n+1$}st cell as a new compartment {$[n]$} matches the differentiation of E. Inserting the n+1st cell into an existing compartment matches the differentiation of some factor in the existing derivative. So the two ways of thinking, in terms of partitions and in terms of differentiation, match up.

So it is easy to get all of the terms of the nth derivative with regard to {$s$}. Simply write out all of the partitions of {$\{0,1,..., n-1\}$}. Then a part which contains {$0$} and has size {$k$} will be given weight {$C^(k)$}, which is to say, the {$k$}th derivative of {$C$}, with regard to {$s$}. And any part which does not contain {$0$} and has size {$k$} will be given weight {$W^(k+1)$}, the {$k$}th derivative of {$W$}, with regard to {$s$}. Multiply together the weights for the parts in a partition and then add up the resulting weights for the partitions.

Furthermore, each part (which does not contain {$0$}) gets a weight {$x$}. So the term {$x^d$} in {$P_n(x)$} gives the partitions with {$d+1$} parts. Thus the power of {$x$} is the measure of compartments. As the number {$n$} of cells grows, as given by the degree of {$s$}, and the index of {$P_n(x)$}, then so does the possible number of compartments, as given by the degree of {$x$}.

I have to remember to divide by {$n!$}.

Investigation: What does it mean to differentiate with regard to {$s$} ?

So I need to think why differentiating a compartment is the same as lengthening it. Or why integrating has to do with shortening.

I can also think of what differentation (with regard to {$s$}) means for a space {$s^n$}.

Compare with multiplying out

We can also simply multiply out the terms of the generating functions

{$$\sum_{n=0}^{\infty}P_n(x)t^n=(1+a_1t+a_2t^2+\cdots)(1+xh_1t+\frac{(xh_1t)^2}{2!}+\cdots)(1+xh_2t^2+\frac{(xh_2t^2)^2}{2!}+\cdots)(1+xh_3t^3+\frac{(xh_3t^3)^2}{2!}+...)\cdots$$}

We likewise get terms that correspond to configurations of parts. Here we ignore the initial element [0]. Each term contains a factor of the form {$a_i$} where {$i+1$} is the length of the compartment that contains [0]. (Note that {$a_0=1$}, in which case this compartment simply consists of [0].) The number of additional compartments is given by the degree of {$x$}. And then each term is a combination of compartments, with the additional compartments having weight {$h_j$} with length {$j$}.

This gives all of the terms generated. But to match them up with the space builders above, I need to figure out how to account for the frequencies. This is where the {$n!$} is important because it arises on the left hand side in differentiating {$n$} times.

For example, for {$n=1$} we have two configurations: {$a_1 \leftrightarrow [01], h_1x\leftrightarrow [0][1]$}. And for {$n=2$} we have four configurations: {$a_2 \leftrightarrow [012], a_1h_1x \leftrightarrow [01][2], [02][1], h_2x [0][12], h_1^2 [0][1][2]$}. So here we need to have that {$a_1 h_1 = 2C'W'$}.

The roles of the Lie group and Lie algebra

From the combinatorics we see that the Lie algebra {$xh(t))$} contributes the additional compartments whereas {$A(t)$}, which expresses the real form (the compactness or noncompactness), gives the original compartment, thus the nature of the spacetime. Is there a sense in which they constitute a pair of conjugate variables?

The compartmentalization is given by {$x$} and so that expresses the amount of energy but also any observable for which the powers in {$x$} are meaningful.

And the growth of the combinatorial objects suggests that it should be expressible as representations of a group.

Idea: Orthogonality of Sheffer polynomials makes space three-dimensional

Furthermore, if the Sheffer polynomials are orthogonal, then their recursion relation is even more limited. The generating functions for {$C(s)$} and {$u(s)$} depend only on the first few terms, and all the remaining terms can be thought of as "echo terms", depending only on the first few. Indeed, for x u(s) only the first three terms matter. The {$x$} terms only arise from differentiating {$e^{x u(s)}$}. Each of the first three terms is qualitatively different, the first term is constant, the second term is {$α + β$}, the third term is {$α^2+ αβ + β^2$}, and the "echo terms" are similar but with higher powers, {$α^n +... + β^n$}.

So I am intrigued that this might yield an explanation why space is three-dimensional.

Time is compatibility with space

Stone's theorem says that a wave function {$f(x,t)$} evolves as {$\sum_{n=0}^{\infty} a_n P_n(x) e^{i λ_n t} \sqrt{w(x)}$}.

Substituting in for {$P_n(x)$} as above this means that {$f(x,t) = \sum_{n=0}^{\infty} \frac{a_n}{n!} (\textrm{d}/\textrm{ds})^n (C(s) e^{u(s)x - i λ_n t}) |_{s=0} \sqrt w(x)$}

I think this means that the time evolution is simply compatible with all that I have written above. The key term is {$e^{u(s) x - i λ_n t}$ and differentiating it with regards to s brings down {$u'(s) x$} but does not bring down {$i λ_n t$}. I think this means that the time dimension is what is compatible with space. Note that what goes down is {$u'(s) x$} and the first three terms matter. So time is like the constant term in {$u(s)$} whose derivative with regard to {$s$} is simply zero.

There is a question of how to think about {$s$}. The powers of {$s$} are what link together the spatial dimensions. But only the first three dimensions are independent.

The five kinds of Sheffer polynomials are five interpretations of space building

The possible values for {$α$} and {$β$} are limited. The general case is {$α, β$} and the four special cases are {$α$}, {$0$} and {$α$}, {$α$} and {$0$}, {$0$} and {$α$}, {$\bar{α}$}.

These further interpret and specialize the meaning of the space building.

Investigation: Relate to the moments of the orthogonal Sheffer polynomials

The moments are the numbers {$μ_k = \int x^k \textrm{dw}$} where {$\int \textrm{dw }$} is the norm with regard to which the polynomials are orthogonal.

Sheffer polynomials are space builders, as dictated by the Bell numbers, and furthermore, if they are orthogonal, then their coefficients, considered as a vector, are orthogonal to their moments, considered as a vector.

{$\int P_n(x) \textrm{dw} = \int \sum p_{n,k} x^k \textrm{dw} = \sum p_{n,k} \int x^k \textrm{dw} = \sum_{n=0}^{\infty} p_{n,k} \mu_k$}

And note that {$P_0(x)$} is constant and orthogonal to {$P_n(x)$} for all {$n\neq 0$}, which means that {$\int P_0(x) P_n(x) \textrm{dw} = \int P_n(x) \textrm{dw} = 0$} and consequently {$0=\sum_{n=0}^{\infty} p_{n,k} \mu_k$} for all {$n\neq 0$}.

There are precisely five sequences of moments that are orthogonal in this way. And each of these has a combinatorial meaning.

The numbers {$μ_k$}, in the general case (the Meixner polynomials) {$α$}, {$β$}, are the ordered Bell numbers, which count the ordered set partitions of {$n$}.

In the special case {$α, 0$}, which is the Charlier polynomials, they are the Bell numbers.

In the special case {$α, α$}, the Laguerre polynomials, they are the number of permutations {$n!$}

In the special case {$0, 0$}, the Hermite polynomials, they are the number of involutions without fixed points, which is to say, permutations where each cycle is disjoint and has length {$2$}. (As with the links in Wick's theorem.)

In the special case {$α, \bar{α}$}, the Meixner-Pollaczek polynomials, they are zig-zag permutations of {$2n$}, which resemble "kinks".

So I need to understand how these moments relate to the Bell numbers as derived above and how we get specializations of space building.

Investigation: Why are these Lorentz invariant?

As I learn more I can try again to understand why the orthogonal Sheffer polynomials might be Lorentz invariant.

I will be reading Mattuck's book about Feynman diagrams because he is explicit in doing calculations.

I think that the invariance should relate to the effect of translating {$x$} to {$x-f$} which combinatorially has a nice interpretation in terms of objects going from having double roots to single roots.


Notes


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Šis puslapis paskutinį kartą keistas August 09, 2022, at 12:07 AM