I'm going through this paper.
The article defines function function $\phi_n^\mu(x)$ that is orthonormal on $L^2$ with measure $dm = dx$: \begin{equation} \phi^\mu_n =\left(\frac{\gamma_\mu(n)}{\Gamma(\mu+\frac{1}{2})}\right)^\frac{1}{2}\frac{1}{\sqrt{2^n}n!}e^{-\frac{x^2}{2}}H^\mu_n(x)x^\mu; \mu,n\in{N}, \end{equation}
Also, it defines differential operator (one-dimensional case, $d=1$, $\mu_1=\mu$)
\begin{equation} \label{eq:hcal} \mathcal{H_\mu} = \frac{1}{2}\left( \sum^d_{k=1}\left[ \mathcal{D}^2_{\mu_k}-\frac{2\mu_k}{x} \left( \frac{\partial}{\partial x}+\mu_k-1 \right) \right] -|x|^2 \right), \end{equation} where \begin{equation} \label{eq:dunkl} \mathcal{D}_\mu\phi(x)=\phi'(x) + \frac{\mu}{x}\left( \phi(x)-\phi(-x) \right), \end{equation} \begin{equation}\label{eq:dunkl2} (\mathcal{D}^2_\mu\phi)(x)=\phi''(x) + \frac{2\mu}{x}\phi'(x) - \frac{\mu}{x^2}\left( \phi(x)-\phi(-x) \right), \end{equation}
It is said (in the paper) that $\phi$ is the eigenfunction of $\mathcal{H}^\mu$ operator, i.e. $-2(\mathcal{H}^\mu \phi_n^\mu) (x) = 2(n+\mu+1/2)\phi_n^\mu(x)$ for one-dimensional case.
I have the functions written in Python: dunkl2 for $\mathcal{D}_\mu$, and calh for $\mathcal{H}_\mu$ operator:
def genhermite(n,m,x):
if(n==0):
return 1
if(n%2==0):
return (-1)**(n//2) * 2**n * misc.factorial(n//2) * special.eval_genlaguerre(n//2,m-1/2,x**2) / genfactorial(n,m)* misc.factorial(n)
else:
return (-1)**(n//2) * 2**n * misc.factorial(n//2) * x * special.eval_genlaguerre(n//2,m+1/2,x**2) / genfactorial(n,m) * misc.factorial(n)
def genphi(n,m,x):
return np.sqrt(genfactorial(n,m)/special.gamma(m+1/2)) / (2**(n/2)*misc.factorial(n)) * np.exp(-x**2/2) * genhermite(n,m,x) * x**m
def dunkl2(n,m,func,x):
return misc.derivative(func,x,0.001,2) + 2*m*(misc.derivative(func,x,0.001))/x - m*(func(x)-func(-x))/x**2
def calh(n,m,func,x):
return -dunkl2(n,m,lambda x: func(n,m,x),x) + 2*m/x*misc.derivative(lambda x: func(n,m,x),x,0.001) + (2*m/x*(m-1)+x**2)*func(n,m,x)
for i in range(0,10):
print("n =", i, "; m =", 10-i, "; x =", 1, "; (Cal.H * f)(x) / f(x) =", calh(i,10-i,genphi,1)/genphi(i,10-i,1)
I test my Python functions by simply dividing both sides of the differential equation by $\phi_n^\mu(x)$ and checking if the result is $2(n+\mu+1/2)$, and this gives me the set of numbers below. But, as you can see, there's something wrong, at least because the resulting coefficient depends not only on $n$ and $\mu$, but on $x$ also.
Where is my mistake? I can assure you that this phi I use is indeed orthonormal.
Data:
n = 0 ; m = 9 ; x = 1 ; (Cal.H * f)(x) / f(x) = 108.999910168
n = 1 ; m = 8 ; x = 1 ; (Cal.H * f)(x) / f(x) = 74.9999101676
n = 2 ; m = 7 ; x = 1 ; (Cal.H * f)(x) / f(x) = 75.0000086295
n = 3 ; m = 6 ; x = 1 ; (Cal.H * f)(x) / f(x) = 49.0000086295
n = 4 ; m = 5 ; x = 1 ; (Cal.H * f)(x) / f(x) = 49.0000089668
n = 5 ; m = 4 ; x = 1 ; (Cal.H * f)(x) / f(x) = 31.0000089668
n = 6 ; m = 3 ; x = 1 ; (Cal.H * f)(x) / f(x) = 30.9999828753
n = 7 ; m = 2 ; x = 1 ; (Cal.H * f)(x) / f(x) = 20.9999828754
n = 8 ; m = 1 ; x = 1 ; (Cal.H * f)(x) / f(x) = 20.9999722442
n = 9 ; m = 0 ; x = 1 ; (Cal.H * f)(x) / f(x) = 18.9999722445
n = 0 ; m = 9 ; x = 2 ; (Cal.H * f)(x) / f(x) = 77.5000026668
n = 1 ; m = 8 ; x = 2 ; (Cal.H * f)(x) / f(x) = 61.0000026667
n = 2 ; m = 7 ; x = 2 ; (Cal.H * f)(x) / f(x) = 53.9999972207
n = 3 ; m = 6 ; x = 2 ; (Cal.H * f)(x) / f(x) = 41.4999972198
n = 4 ; m = 5 ; x = 2 ; (Cal.H * f)(x) / f(x) = 36.5000043004
n = 5 ; m = 4 ; x = 2 ; (Cal.H * f)(x) / f(x) = 28.0000043004
n = 6 ; m = 3 ; x = 2 ; (Cal.H * f)(x) / f(x) = 24.9999862338
n = 7 ; m = 2 ; x = 2 ; (Cal.H * f)(x) / f(x) = 20.4999862334
n = 8 ; m = 1 ; x = 2 ; (Cal.H * f)(x) / f(x) = 19.4999826454
n = 9 ; m = 0 ; x = 2 ; (Cal.H * f)(x) / f(x) = 18.9999826456
I'm no Python expert, but I believe your implementation of the operator $\mathcal{D}_\mu$ is \begin{equation} \mathcal{D}_\mu \phi(x) = \phi''(x) +2 \frac{\mu}{x} \phi'(x) - \frac{\mu}{x^2} (\phi(x) - \phi(-x)), \end{equation} which is not the same as the operator description for $\mathcal{D}_\mu$ in your question.
Edit based on updated question: Another issue might be that there are two types of (generalised) Hermite polynomials, see here. From the article you cited (and the references therein), it's clear that they use the 'physicists' ' definition; the function $f(x)$ you test might use the 'probabilists' ' definition.
Addition: I used the definition from the original paper, where $\Phi$ therein is the Kummer confluent hypergeometric function $_1F_1$, to test in Mathematica whether $\phi_n^\mu$ is an eigenfunction of $\mathcal{H}_\mu$, and it turns out it's not. So, there might be something wrong with the operator expression for $\mathcal{H}_\mu$, which would mean that this paper contains a typo in equation (1.8).
More precisely, I obtain \begin{equation} -2 \mathcal{H}_\mu \phi_{2m}^\mu = 2\left(2m+2\mu+\frac{1}{2}+\frac{\mu(\mu-1)}{x}\right)\phi_{2m}^\mu -2 (1+2m+2\mu)\frac{\mu}{(1+2m)x} \phi_{2m+1}^\mu \end{equation}