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Some quantum mechanics derivation gave me the expectation values of some operator. I got some clean analytical result which makes me think that there is an analytical answer for the probability distribution.The odd moments are all $0$ and the even moments are given by:

$$\langle p^{2n}\rangle=2^{-2n}\binom{2n}{n}(k)^{2n}$$

$ghk$ is just some parameter. The most significant result for me will happen if this will result in some probability distribution based on Bessel $J$ functions but anything is good.

Ron Ruimy
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  • $ghk$ is just a scaling factor and can be largely ignored. So too is $2^{-2n}$. So the interesting question is what distribution gives even moments of ${2n \choose n}$ – Henry Feb 17 '21 at 01:02

3 Answers3

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Let us denote $q := ghk$ and your variable by $X.$ You can consider the moment-generating function of the variable $X,$ $M_X(t) = \mathbb{E}[e^{tX}] = \sum_{0 \leq k} \mathbb{E}[X^k] \frac{t^k}{k!},$ which in your case would be $$M_X(t) = \sum_{0 \leq k} 2^{-2k} \binom{2k}{k} \frac{(tq)^{2k}}{(2 k)!} = \sum_{k \geq 0} 2^{-2 k} \frac{(2 k)!}{(k!)^2} \frac{(t q)^{2k}}{(2k)!} = \sum_{k \geq 0} (\frac{t^k q^k}{2^k k!})^2.$$ Note that we already get that $M_X(t) \leq e^{tq},$ so this series is actually well defined in this case (it is not always the case with moment generating functions),. In fact, this is just a modified Bessel function of the first kind. More precisely, $M_X(t) = I_0(t q),$ so your intuition was right that we would get some kind of Bessel function somehow.

P.S.: We can determine the PDF of $X$ by using the inversion formula. Note that by the same reasoning as above, we deduce that the characteristic function of $X$ is given by $\varphi_X(t) = J_0(q t),$ a Bessel function of the first kind. It follows from the Inversion Theorem that $X$ admits a probability density function, which is given simply as the Fourier Transform of the characteristic function of $X.$ This has been computed before on this very site (see for example here or here). We have that $$\mathcal{F}(J_0)(\xi) = \chi_{(-1, 1)}(\xi)\frac1\pi \frac{1}{\sqrt{1 - \xi^2}}.$$ However, we are interested in a scaling of this function, so performing a simple substitution in the Fourier Transform integral, we get the expression of the PDF of $X:$ $$f_X(\xi) = \chi_{(-q, q)}(\xi) \frac{1}{\pi} \frac{1}{\sqrt{q^2 - \xi^2}}.$$

Actually Fritz
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    Wow that's cool! thanks :) Is there a way to get the probability distribution out of it? – Ron Ruimy Feb 17 '21 at 01:26
  • Yes, I shall add it in my answer. – Actually Fritz Feb 17 '21 at 02:05
  • I have added the derivation of the PDF from the moment-generating function/ the characteristic function of $X.$ It is not always the case that these Fourier Transforms are usually computed, we are kind of lucky that it works in this case, thank the gods of Physics for that haha. I hope this helps. :) Btw, what calculations led you to this variable? – Actually Fritz Feb 17 '21 at 02:29
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    It's kind of a weird thing - I'm experimenting with phase space techniques in quantum mechanics. Specifically something called truncated Wigner approximation in which you threat the quantum system as a classical statistical ensamble and try to solve problems for which an analytical solution is already known and compare. In this case this will be the spectrum of a free electron interacting with a driving field in a phase matched manner :)

    Thanks a lot for all your help!

     – Ron Ruimy Feb 17 '21 at 02:57
  • Hmmm, that sounds very cool, I’m into Statistical Physics myself. I will search the topic you mentioned, hopefully I will find something on that online, but if you already have some good references, do not hesitate to leave them in a comment or whatever. You are most welcome, it’s my pleasure! :) – Actually Fritz Feb 17 '21 at 03:09
  • https://arxiv.org/abs/0905.3384 This is the most comprehensive review of the subject I know of :) – Ron Ruimy Feb 17 '21 at 03:18
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The arcsine distribution stretched to have support on $[-ghk, ghk]$ has the moments you are looking for. So, with $-ghk < x < ghk$, the probability density function is $$f(x)=\frac{1}{\pi \sqrt{{{(ghk)}^{2}}-{{x}^{2}}}}$$ and the cumulative distribution function is $$F(x)=\mathbb P(X\le x) = \frac1\pi \cos^{-1}\left(\frac{-x}{ghk}\right)$$

As a quick check in R of the first twelve moments, setting $ghk=2$ gives the central binomial coefficients:

qstretchedarcsine <- function(prob, ghk){ - ghk * cos(prob * pi) }
thisdata <- qstretchedarcsine(ppoints(10^6), ghk=2)
round(colMeans(outer(thisdata, 1:12, "^")), 12)
#    0   2   0   6   0  20   0  70   0 252   0 924

See https://math.stackexchange.com/a/2629048/6460 for the relationship between the arcsine distribution and Bessel functions

Henry
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Let $f(x)$ be the (unknown) density function. Then the characteristic function is $h(t)=\int_R e^{itx}f(x)dx=\sum_{n=0}^\infty \frac{(it)^nM_n}{n!}$ Then the density function is given by $f(x)=\frac{1}{2\pi}\int_Re^{-itx}h(t)dt$.

Knowing the moments, $h(t)$ can be computed and then $f(x)$ using the inverse Fourier transform.

herb steinberg
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