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Derivation Of The Classification Of ShefferPolynomials, 20230109 Research Program Fivesome Double Causality: A Framework for Physics Contents
Hypothesis to explore
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Want to find a combinatorial interpretation for the measurement of the Hamiltonian {$\hat{H}$} given a wave function {$\psi$} {$\langle H\rangle =\int\psi^*\hat{H}\psi dx = \langle\psi  \hat{H}\psi\rangle $} where, for example, {$\psi_n(x) = D_n\textrm{He}_n(\sqrt{\frac{2m\omega}{\hbar}}x) e^{\frac{m\omega}{2\hbar}x^2}$} We are given the generating function for the orthogonal Sheffer polynomials {$P_n(x)$}. {$\sum_{n=0}^{\infty}P_n(x)\frac{t^n}{n!}=A(t)e^{xu(t)}$} I write the recurrence relation as follows: {$P_{n+1}(x)=[x(ln+f)]P_n(x)[kn(n1)\gamma n]P_{n1}(x)$} Combinatorially, we will see that the coefficients in the recurrence relation are weights for combinatorial components:
The weights {$\alpha$} and {$\beta$} are the basis for the fivefold classification. We have {$(\alpha\beta)^2=l^24k$}. {$\alpha,\beta = \frac{\alpha + \beta}{2}\pm \frac{\alpha  \beta}{2} = \frac{l}{2}\pm\frac{\sqrt{l^24k}}{2}$} {$u'(t)=\frac{1}{(1\alpha t)(1\beta t)} = \frac{1}{1 + lt + kt^2} = \frac{A'(t)}{\gamma t A(t)} = \frac{1}{\gamma t}\frac{\textrm{d}}{\textrm{dt}}\textrm{ln}A(t)$} {$u'(t)=1+t(\alpha + \beta)+t^2(\alpha^2+\alpha\beta+\beta^2)+\cdots$} interpreted as particle clocks {$u'(t)= \frac{1}{\alpha\beta}[(\alpha  \beta) + (\alpha^2  \beta^2)t + (\alpha^3\beta^3)t^2 + \cdots]$} {$u(t)=t+\frac{1}{2}t^2(\alpha + \beta)+\frac{1}{3}t^3(\alpha^2+\alpha\beta+\beta^2)+\cdots$} has an extra point in each circle which is placed and then removed (how does that relate to quantum perturbation?) The constraint of orthogonality yields the constraint {$A(t)=e^{\gamma v(t)}$} where {$v(t)=\frac{1}{2}t^2 + \frac{1}{3}t^3(\alpha + \beta)+\frac{1}{4}t^4(\alpha^2+\alpha\beta+\beta^2)+\cdots $} has an extra double point in each circle which is placed and then removed {$v'(t) = t +t^2(\alpha + \beta)+t^3(\alpha^2+\alpha\beta+\beta^2)+\cdots$} {$v'(t)=tu'(t)$} is the quantum perturbation of {$u(t)$}. {$A(t)=e^{\gamma (\frac{1}{2}t^2 + \frac{1}{3}t^3(\alpha + \beta)+\frac{1}{4}t^4(\alpha^2+\alpha\beta+\beta^2)+\cdots)}$} Thus {$\sum_{n=0}^{\infty}P_n(x)\frac{t^n}{n!}=e^{\gamma v(t)+xu(t)}$} {$\mathcal{L}[x^n]=\sum_{\sigma\in S_n}\alpha^{\textrm{asc}(\sigma)}\beta^{\textrm{desc}(\sigma)}(\frac{f}{\alpha\beta})^{\textrm{fix}(\sigma)}(\frac{\gamma}{\alpha\beta})^{\textrm{cyc}(\sigma)\textrm{fix}(\sigma)}$} {$\frac{\textrm{d}}{\textrm{dx}}\textrm{He}_n(x) = n\textrm{He}_{n1}(x)$} {$x\textrm{He}_n(x) = \textrm{He}_{n+1}(x) + n\textrm{He}_{n1}(x)$} because {$\textrm{He}_{n+1}(x) = x\textrm{He}_n(x)  n\textrm{He}_{n1}(x)$} {$t(u(s))=s$} {$t(D)=D+t_2D^2+t_3D^3+\cdots$} {$t(D)P_n(x)=nP_{n1}(x)$} {$t(D)=\frac{e^{(\alpha\beta)D}1}{e^{(\alpha\beta)D}\beta}$} Mathematically, specifically, I want to:
=============================== The fivesome refers to the division of everything into five perspectives, that which is needed for decision making, and which yield two representations, one in terms of time and the other in terms of space. Here are two presentations that I gave about the fivesome:
The main idea of this conceptual framework: Every effect has had its cause. Not every cause has had its effect.
We conceive this framework by way of two conceptions, in terms of time and in terms of space.
Examples of double causality
{$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$} Solutions of Schroedinger's equation are orthogonal Sheffer polynomials
{$\langle Q\rangle =\int\Psi^*\hat{Q}\Psi \textrm{dx}=\langle\Psi \hat{Q}\Psi\rangle$} A combinatorial interpretation of a physical observable relates three elements.
Working hypothesis: In physical situations, the orthogonal polynomials must be Sheffer polynomials.
Question to investigate: What is special about Sheffer polynomials? answered by Tom Copeland, Shadows of Simplicity Hidden structure, hidden symmetry, that is waiting to be discovered.
=============================== Sheffer polynomial sequences {$\{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_{n1}(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_aQQT_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$}, we define the {$n$}th term of {$p\circ q$} as {$(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$}. [Investigation: Interpret this combinatorially.] A scanned copy of an important article, Some properties of polynomial sets of type zero by I.M.Sheffer, Duke Mathematical Journal, September 1939, pages 590622, is available here through Archive.org It also available to subscribers here through Project Euclid. Sheffer polynomials {$\{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$} Note that the equations are symmetric in {$j$} and {$u$}. {$\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}}$}. {$$DA(t)e^{xu(t)}=A(t)(u(t))e^{xu(t)}$$} {$$D^kA(t)e^{xu(t)}=A(t)(u(t))^ke^{xu(t)}$$} {$$J(D)A(t)e^{xu(t)}=\sum_{k=1}^{\infty}j_kA(t)(u(t))^ke^{xu(t)}$$} {$$=A(t)e^{xu(t)}\sum_{k=1}^{\infty}j_k(u(t))^k$$} {$$=A(t)e^{xu(t)}J(u(t))$$} {$$=A(t)e^{xu(t)}t$$} {$$J(D)\sum_{n=0}^{\infty}P_n(x)t^n = \sum_{n=0}^{\infty}P_n(x)t^{n+1}$$} By considering the coefficient for {$t^n$} we see that {$J(D)P_n(x)=P_{n1}(x)$}. =============================== All orthogonal polynomials satisfy a recurrence relation of the form {$xP_n(x) = P_{n+1}(x) + A_n P_n(x) + B_n P_{n1}(x)$}. More specifically, Sheffer polynomials satisfy a recurrence relation written by various authors as follows:
I will write it in the following form to emphasize the combinatorial meaning of the terms {$P_{n+1}(x)=[x(ln+f)]P_n(x)[kn(n1)\gamma n]P_{n1}(x)$} Combinatorially, we will see that the variables are weights for components:
We can assume that {$f$} = 0 because that's equivalent to translating {$x$} to {$xf$}. We will be expressing the weights in terms of {$\alpha$} and {$\beta$}. Notably {$l=(\alpha)+(\beta)$} and {$k=(\alpha)(\beta)$}. This yields a discriminant {$$A'(t) =  \gamma t + \textrm{echo terms}$$} {$$u'(t) = 1  lt + (l^2k)t^2 + \textrm{echo terms}$$} I mean by "echo terms" that the higher order terms are completely determined by the initial terms and have no independent significance. Calculating {$u(t)$} I am trying to relate the notation in Sheffer with that in Kim & Zeng. Sheffer calls {$H(t)$} what I am writing {$u(t)$}. He sets {$ u'(t)=\sum_{n=0}^{\infty}q_{n+1,1}t^n$}. Thus {$u^{(k)}(0)=(k1)!q_{k,1}$}. But I know that {$u^{(k)}(0) = (k1)!(\alpha^{k1}+\alpha^{k2}\beta + \cdots + \alpha\beta^{k2}+\beta^{k1})$}. Thus {$q_{k,1} = \alpha^{k1}+\alpha^{k2}\beta + \cdots + \alpha\beta^{k2}+\beta^{k1} $}. Calculating {$A(t)$} Using Sheffer's notation... Define {$\frac{A'(t)}{A(t)} = \sum_{k=0}^{\infty}q_{k+1,0}t^k $}. {$$\frac{\textrm{d}}{\textrm{dt}}\textrm{ln}A(t)=\frac{A'(t)}{A(t)}$$} {$$ \textrm{ln}A(t) = \int\sum_{k=0}^{\infty}t^k\textrm{dt} = \sum_{k=0}^{\infty}\int t^k\textrm{dt} = \sum_{k=0}^{\infty}\frac{q_{k+1,0}}{k+1}t^{k+1} $$} {$$ A(t) = e^{\sum_{k=0}^{\infty}\frac{q_{k+1,0}}{k+1}t^{k+1}} = e^{q_{1,0}t + \frac{1}{2}q_{2,0}t^2 + \frac{1}{3}q_{3,0}t^3 + \cdots} $$} {$$ A(t) = 1 + (q_{1,0}t + \frac{1}{2}q_{2,0}t^2 + \frac{1}{3}q_{3,0}t^3 + \cdots) + \frac{1}{2}(q_{1,0}t + \frac{1}{2}q_{2,0}t^2 + \frac{1}{3}q_{3,0}t^3 + \cdots)^2 + \frac{1}{6}(q_{1,0}t + \frac{1}{2}q_{2,0}t^2 + \frac{1}{3}q_{3,0}t^3 + \cdots)^3 + \cdots $$} {$$ A(t) = 1 + t(q_{1,0}) + t^2(\frac{1}{2}q_{2,0} + \frac{1}{2}q_{1,0}^2) + t^3(\frac{1}{3}q_{3,0} + \frac{1}{2}q_{1,0}q_{2,0} + \frac{1}{6}q_{1,0}^3) + \cdots $$} Recurrence relation Sheffer gives the recurrence relation, in his notation: {$q_{k+2,1} = \gamma q_{k+1,1} + \epsilon q_{k,1}$} Here {$\gamma = \alpha + \beta =  (\alpha  \beta)$} and {$\epsilon =  \alpha\beta =  (\alpha)(\beta) $} in my notation. The recurrence relation yields the characteristic equation {$u^2 = \gamma u + \epsilon $}. This equation expresses the ansatz that solutions will have the form {$q_{k,1} = r^n$}. This equation yields three outcomes, see the article Linear recurrence with constant coefficients.
Relating {$P_n(x)$}, {$u(t)$} and {$A(t)$} From Sheffer: {${P'}_n(x)=u_1P_{n1}(x) + u_2P_{n2}(x) + \cdots + u_nP_0(x)$} As Sheffer notes, this equation shows that {$u(t)$} by itself determines {${P'}_n(x)$}. Thus {$A(t)$} simply contributes the constant term that must be determined in integrating {${P'}_n(x)$} to yield {$P_n(x)$}. In the form {$nP_n(x)=\sum_{k=1}^{\infty}(q_{k,0}+xq_{k,1})P_{nk}$} where {$\frac{A'(t)}{A(t)}=\sum_{k=0}^{\infty}q_{k+1,0}t^k$} and {$u'(t)=\sum_{k=0}^{\infty}q_{k,0}t^k$} it is apparent that {$A(t)$} contributes the constant terms and {$u(t)$} contributes the coefficients of {$x$}. Interpreting {$u(t)$} and {$A(t)$} We can choose to write {$$u'(t) = \frac{1}{(1\alpha t)(1\beta t)} = 1 + (\alpha + \beta) t + (\alpha^2 + \alpha\beta + \beta^2)t^2 + \textrm{echo terms}$$} {$$=(1+\alpha t+\alpha^2t^2+...)(1+\beta t+\beta^2t^2+...) = \sum_{n=0}^{\infty}t^n\sum_{i=0}^n \alpha^i\beta^{ni}$$} 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.
======================== Fivefold classification Various specializations of the Meixner polynomials yield a fivefold classification.
The Sheffer family satisfies the recurrence relation below, with initial polynomials {$P_{1}=0$} and {$P_0=1$}: {$$P_{n+1}(x) = (x  (nl + f)) P_n(x)  n(nk + c) P_{n1}(x)$$} Writing it like this helps remember that {$l$} is the weight for links, {$k$} for kinks, {$f$} for fixed points (selfentangled 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. ======================== Spacetime 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.
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.
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. ======================== 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:
Remarks
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.
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 {$a\mid\! n\rangle= \sqrt{n} \ n1\rangle$} 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 {$[a,a^\dagger]\mid\! n\rangle = aa^\dagger\mid\! n\rangle  a^\dagger a\mid\! n\rangle = a\sqrt{n+1}\mid\! n+1\rangle  a^\dagger\sqrt{n}\mid\! n1\rangle = \sqrt{n+1}\sqrt{n+1}\mid\! n\rangle  \sqrt{n}\sqrt{n}\mid\! n\rangle = 1\mid\! n\rangle$} {$[a,a^\dagger]=1$} 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.
Wick's theorem What's up with Wick's theorem with references in the comments. Relate to the Hermite polynomials. Normalordered 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
Wick's theorem
Not action at a distance. Velocity of transmission? Commutator
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.
======================== Understanding spacetime as probability distribution.
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 particleclock, which is to say, to each cycle. It is the cutoff 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 onedimensional spacetime 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 {$p1$} as adjoint probabilities. This is to say that {$p1$} is the complementary probability {$1p$} but in the negative direction. I think of {$p1$} 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 {$\sqrt{l^24k}$} is important for classifying orthogonal Sheffer polynomials. It can be positive (Meixner), negative (MeixnerPollaczek) or zero (Laguerre, Hermite). There are also the Charlier polynomials. The discriminant arises from the quadratic equation. It is helpful to realize that {$\alpha \beta=\sqrt{l^24k}$}. For {$l=(\alpha)+(\beta)$} and {$k=(\alpha)(\beta)$}. Thus the discrete unit of spacetime 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. Quantumdiscrete vs. classicalcontinuous If {$\alpha  \beta \neq 0$} then the entire system is moving with a mean {$pN$} and a variance {$p(p1)N$}. We have the extra bosonic condition that {$\alpha  \beta = 1$} which means that you can't have both {$\alpha$} and {$\beta$} equal to {$0$}. {$\alpha = p$}, {$\beta = (1p)$}, {$l=\alpha \beta = 12p$}, {$k=(\alpha)(\beta)=p(p1)$}, {$f=pN$}. MeixnerPollaczek If we substitute {$\alpha = a+bi, \beta=\bar{\alpha}=abi$}, then we get the MeixnerPollazcek 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 spacetime is {$\alpha  \bar{\alpha}=2bi$}. The weight function is then {$w((2bi)n)=w(\frac{\alpha}{\bar{\alpha}}\alphan) = (\textrm {cos}2n\theta + i \textrm{sin}2n\theta)\binom{N}{n}$}. ======================== Calculating energy Consider how this all comes together in calculating an observable, namely, the energy. The probability density (the spacetime) 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 C_ne^{\frac{\xi^2}{2}}$} where {$C_n=(\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}\frac{1}{\sqrt{2^nn!}}$}. {$\frac{\partial H_n}{\partial\xi}=2nH_{n1}(\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 C_ne^{\frac{\xi^2}{2}}$} {$ = \frac{1}{2m}(i\hbar\frac{\partial}{\partial x})(i\hbar\frac{\partial}{\partial x}) H_n(\xi)\cdot C_ne^{\frac{\xi^2}{2}} + \frac{1}{2m}(m\omega)^2 x^2 H_n(\xi)\cdot C_ne^{\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 C_ne^{\frac{\xi^2}{2}} + \frac{1}{2m}(m\omega)^2 \frac{\hbar}{m\omega} \xi^2 H_n(\xi)\cdot C_ne^{\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 C_ne^{\frac{\xi^2}{2}} + \frac{1}{2m}(m\omega)^2 \frac{\hbar}{m\omega} \xi^2 H_n(\xi)\cdot C_ne^{\frac{\xi^2}{2}}$} {$ = \frac{1}{2}\hbar\omega\frac{\partial}{\partial\xi}(2n H_{n1}(\xi)\cdot C_ne^{\frac{\xi^2}{2}}  H_n(\xi)\cdot \xi C_ne^{\frac{\xi^2}{2}}) + \frac{1}{2}\hbar \omega \xi^2 H_n(\xi)\cdot C_ne^{\frac{\xi^2}{2}}$} {$ = \frac{1}{2}\hbar\omega(4n(n1) H_{n2}(\xi)\cdot C_ne^{\frac{\xi^2}{2}}  2n H_{n1}(\xi)\cdot \xi C_ne^{\frac{\xi^2}{2}}  2n H_{n1}(\xi)\cdot \xi C_ne^{\frac{\xi^2}{2}}  H_n(\xi)\cdot C_ne^{\frac{\xi^2}{2}} + H_n(\xi)\cdot \xi^2 C_ne^{\frac{\xi^2}{2}} ) + \frac{1}{2}\hbar \omega \xi^2 H_n(\xi)\cdot C_ne^{\frac{\xi^2}{2}}$} {$ = \frac{1}{2}\hbar\omega (4n(n1) H_{n2}(\xi)  4n \xi H_{n1}(\xi) + (1 + \xi^2 + \xi^2 )H_n(\xi))C_ne^{\frac{\xi^2}{2}} $} {$ = \frac{1}{2}\hbar\omega (4n(n1) H_{n2}(\xi)  4n \xi H_{n1}(\xi)  H_n(\xi))C_ne^{\frac{\xi^2}{2}} $} and make use of the recursion relation {$H_{n+1}(\xi)=2\xi H_n(\xi)  2 n H_{n1}(\xi)$} to identify {$4n \xi H_{n1}(\xi)= 2n H_{n}(\xi) + 4 n^2 H_{n2}(\xi)$}. {$ = \frac{1}{2}\hbar\omega (4n(n1) H_{n2}(\xi)  2n H_{n}(\xi)  4n^2 H_{n2}(\xi)  H_n(\xi))C_ne^{\frac{\xi^2}{2}} $} {$ = \frac{1}{2}\hbar\omega ((4n) H_{n2}(\xi)  (2n+1) H_n(\xi))C_ne^{\frac{\xi^2}{2}} $} Then by orthogonality {$\langle H\rangle =\int\psi^*\hat{H}\psi dx = \int H_n(\xi)\cdot C_ne^{\frac{\xi^2}{2}} \frac{1}{2}\hbar\omega (4n H_{n2}(\xi)  (2n+1) H_n(\xi))C_ne^{\frac{\xi^2}{2}} dx$} {$ = \int H_n(\xi)\cdot C_ne^{\frac{\xi^2}{2}} \frac{1}{2}\hbar\omega (4n H_{n2}(\xi)  (2n+1) H_n(\xi))C_ne^{\frac{\xi^2}{2}} \sqrt{\frac{\hbar}{m\omega}}d\xi$} {$=  (2n+1)\frac{1}{2}\hbar\omega \sqrt{\frac{\hbar}{m\omega}} C_n^2 \int H_n(\xi) H_n(\xi) e^{\xi^2} d\xi $} {$= \frac{2n1}{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) C_ne^{\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 C_ne^{\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 C_ne^{\frac{y^2}{4}}$} where {$y =\sqrt{\frac{2m\omega}{\hbar}}x$}. {$\frac{\partial\textrm{He}_n}{\partial y}=n\textrm{He}_{n1}(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 C_ne^{\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 C_ne^{\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 C_ne^{\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 C_ne^{\frac{y^2}{4}}$} {$ = 2^{\frac{n}{2}}\hbar\omega\frac{\partial}{\partial y}(n \textrm{He}_{n1}(y)\cdot C_ne^{\frac{y^2}{4}}  \textrm{He}_n(y)\cdot \frac{1}{2}y C_ne^{\frac{y^2}{4}}) + \frac{1}{4}y^2 \textrm{He}_n(y)\cdot \hbar \omega 2^{\frac{n}{2}}C_ne^{\frac{y^2}{4}}$} {$ = 2^{\frac{n}{2}}\hbar\omega(n(n1) \textrm{He}_{n2}(y)\cdot C_ne^{\frac{y^2}{4}}  n \textrm{He}_{n1}(y)\cdot \frac{1}{2} y C_ne^{\frac{y^2}{4}}  n \textrm{He}_{n1}(y)\cdot \frac{1}{2}y C_ne^{\frac{y^2}{4}}  \textrm{He}_n(y)\cdot \frac{1}{2} C_ne^{\frac{y^2}{4}} + \textrm{He}_n(y)\cdot \frac{1}{4} y^2 C_ne^{\frac{y^2}{4}} ) + \frac{1}{4}y^2 \textrm{He}_n(y)\cdot \hbar \omega 2^{\frac{n}{2}} C_ne^{\frac{y^2}{4}}$} {$ = 2^{\frac{n}{2}}\hbar\omega (n(n1) \textrm{He}_{n2}(y)  ny \textrm{He}_{n1}(y) + (\frac{1}{2} + \frac{1}{4}y^2 + \frac{1}{4}y^2 )\textrm{He}_n(y))C_ne^{\frac{y^2}{4}} $} {$ = 2^{\frac{n}{2}}\hbar\omega (n(n1) \textrm{He}_{n2}(y)  ny \textrm{He}_{n1}(y)  \frac{1}{2}\textrm{He}_n(y))C_ne^{\frac{y^2}{2}} $} and make use of the recursion relation {$\textrm{He}_{n+1}(y)=y \textrm{He}_n(y)  n \textrm{He}_{n1}(y)$} to identify {$n y \textrm{He}_{n1}(y)= n \textrm{He}_{n}(y) + n^2 \textrm{He}_{n2}(y)$}. {$ = 2^{\frac{n}{2}}\hbar\omega (n(n1) \textrm{He}_{n2}(y)  n \textrm{He}_{n}(y)  n^2 \textrm{He}_{n2}(y)  \frac{1}{2}\textrm{He}_n(y))C_ne^{\frac{y^2}{4}} $} {$ = 2^{\frac{n}{2}}\hbar\omega (n\textrm{He}_{n2}(y)  (n+\frac{1}{2}) \textrm{He}_n(y))C_ne^{\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_ne^{\frac{y^2}{4}}(2^{\frac{n}{2}})\hbar\omega (n \textrm{He}_{n2}(y)  (n+\frac{1}{2}) \textrm{He}_n(y))C_ne^{\frac{y^2}{4}} dx$} {$ = 2^n\int \textrm{He}_n(\xi)\cdot C_ne^{\frac{y^2}{4}}\hbar\omega (n \textrm{He}_{n2}(y)  (n+\frac{1}{2}) \textrm{He}_n(y))C_ne^{\frac{y^2}{4}} \sqrt{\frac{\hbar}{2m\omega}}dy$} {$= (n+\frac{1}{2})2^n\hbar\omega \sqrt{\frac{\hbar}{2m\omega}} C_n^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) C_ne^{\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$}. ======================== I need to provide combinatiorial interpretations of the following:
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? Tom Leinster. The Categorical Origins Of Entropy.
======================== Here are key papers that I am studying.
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