Is there a closed form of this series?
$$\sum_{n=0}^{\infty} \frac{\left(\frac{-r}{\sqrt{a}}\right)^n}{n!} \sum_{\rho=0}^{n/2} \frac{\frac{n}{2}!}{\rho! \left(\frac{n}{2}-\rho\right)!}Q^{\frac{n}{2}-\rho}\sum_{k=0}^{\infty} \frac{(-Q)^k}{\Gamma(\alpha k+\frac{3}{2})}\times \frac{\Gamma(\alpha k+\frac{3}{2}-\frac{\alpha}{2})}{\Gamma(\alpha k+\frac{3}{2}-\frac{\alpha}{2}-\alpha \rho)}t^{\alpha k+\frac{1}{2}-\frac{\alpha}{2}-\alpha \rho}$$ Is there a closed form in the form of Wright or Mittag-Leffler functions or the product of these two? r is infinitely negative to infinite positivity. $a=\frac{i\hbar}{2m}$, $Q=\frac{A}{\hbar}\sqrt{\frac{\pi}{2}}-\frac{\pi B^{2} m}{4 \hbar^{2}}$, A and B are Coefficients and m is mass. $\alpha$ is between zero and one.
(This is not currently a complete answer.)
Call the OP's function $f(t)$. (Why does no one name the functions they want to talk about?) Let the operator $D^\mu$ be the fractional derivative operator with respect to $t$. For $\mu \geq 0$ and $z >-1$, $$ D^\mu t^z= \frac{\Gamma(z+1)}{\Gamma(z+1-\mu)}t^{z-\mu} \text{,} $$ so, with the parameter constraints mostly revealed in comments, we have $\alpha \rho \in [0,n/2]$ and $\alpha k + \frac{1}{2} - \frac{\alpha}{2} \in (0,\infty)$, so $$ D^{\alpha \rho} t^{\alpha k+\frac{1}{2}-\frac{\alpha}{2}} = \frac{\Gamma(\alpha k+\frac{3}{2}-\frac{\alpha}{2})}{\Gamma(\alpha k+\frac{3}{2}-\frac{\alpha}{2}-\alpha \rho)}t^{\alpha k+\frac{1}{2}-\frac{\alpha}{2}-\alpha \rho} $$ and we obtain $$ f(t) = \sum_{n=0}^{\infty} \frac{\left(\frac{-r}{\sqrt{a}}\right)^n}{n!} \sum_{\rho=0}^{n/2} \frac{\frac{n}{2}!}{\rho! \left(\frac{n}{2}-\rho\right)!}Q^{\frac{n}{2}-\rho}\sum_{k=0}^{\infty} \frac{(-Q)^k}{\Gamma(\alpha k+\frac{3}{2})} D^{\alpha \rho} t^{\alpha k+\frac{1}{2}-\frac{\alpha}{2}} \text{.} $$ The series in $k$ is a power series in $t^\alpha$, so we exchange the derivative and the sum (keeping the same radius of convergence). \begin{align*} \sum_{k=0}^{\infty} \frac{(-Q)^k}{\Gamma(\alpha k+\frac{3}{2})} D^{\alpha \rho} t^{\alpha k+\frac{1}{2}-\frac{\alpha}{2}} &= D^{\alpha \rho} \sum_{k=0}^{\infty} \frac{(-Q)^k}{\Gamma(\alpha k+\frac{3}{2})} t^{\alpha k+\frac{1}{2}-\frac{\alpha}{2}} \\ &= D^{\alpha \rho} t^{\frac{1}{2}-\frac{\alpha}{2}} \sum_{k=0}^{\infty} \frac{(-Q t^\alpha)^k}{\Gamma(\alpha k+\frac{3}{2})} \\ &= D^{\alpha \rho} t^{\frac{1}{2}-\frac{\alpha}{2}} E_{\alpha, \frac{3}{2}}(-Q t^\alpha) \text{,} \end{align*} where $E_{\alpha, \beta}(z) = \sum_{k=0}^\infty \frac{z^k}{\Gamma(\alpha k + \beta)}$ is (one of) the Mittag-Leffler function(s). Since $\alpha$ and $3/2$ are both positive, our $E$ is an entire function. So now we have \begin{align*} f(t) &= \sum_{n=0}^{\infty} \left(\frac{-r}{\sqrt{a}}\right)^n \frac{\frac{n}{2}!}{n!} \sum_{\rho=0}^{n/2} \frac{1}{\rho!}(D^{\alpha})^{\rho} \frac{Q^{\frac{n}{2}-\rho}}{\left(\frac{n}{2}-\rho\right)!} \cdot t^{\frac{1}{2}-\frac{\alpha}{2}} E_{\alpha, \frac{3}{2}}(-Q t^\alpha) \\ &= \sum_{n=0}^{\infty} \left(\frac{-r}{\sqrt{a}}\right)^n \frac{2^{-n}\sqrt{\pi}}{\Gamma(\frac{n}{2} + \frac{1}{2})} \sum_{\rho=0}^{n/2} \frac{1}{\rho!}(D^{\alpha})^{\rho} \frac{Q^{\frac{n}{2}-\rho}}{\left(\frac{n}{2}-\rho\right)!} \cdot t^{\frac{1}{2}-\frac{\alpha}{2}} E_{\alpha, \frac{3}{2}}(-Q t^\alpha) \end{align*}
... and it's too late here for me to continue. If anyone else wants to continue, be my guest. (The $\rho$ summation is rather similar to something like a truncated Maclaurin series.)