To express $f(x,z)=\sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{n!}$ as known functions $\alpha,\beta >0$
$$f(x,z)=\sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{n!}$$
$$\frac{\partial{f(x,z)}}{\partial z}=\beta \sum \limits_{n=1}^\infty \frac{e^{-\alpha n^2 x+\beta n z}}{(n-1)!}$$ 
$$\frac{\partial{f(x,z)}}{\partial z}|_{z=2 \frac{ \alpha}{\beta} x+ z_1}=\beta \sum \limits_{n=1}^\infty \frac{e^{-\alpha n^2 x+\beta n (2 \frac{ \alpha}{\beta} x+  z_1)}}{(n-1)!}$$ 
$$\frac{\partial{f(x,z)}}{\partial z}|_{z=2 \frac{ \alpha}{\beta} x+ z_1}=\beta e^{\alpha x+ \beta z_1} \sum \limits_{n=1}^\infty \frac{e^{-\alpha (n-1)^2 x+\beta (n-1) z_1}}{(n-1)!}$$ 
$$\frac{\partial{f(x,z)}}{\partial z}|_{z=2 \frac{ \alpha}{\beta} x+ z_1}=\beta e^{\alpha x+ \beta z_1} \sum \limits_{n=0}^\infty \frac{e^{-\alpha n^2 x+\beta n z_1}}{n!}$$
$$\frac{\partial{f(x,z)}}{\partial z}|_{z=2 \frac{ \alpha}{\beta} x+ z_1}=\beta e^{\alpha x+ \beta z_1} f(x,z_1)$$
I do not know how to solve this kind  differential equations. 
Do you know how to solve that?
Can we express the function as known functions such as Jacobi Theta Functions etc?
Also  could you please share your knowledge about the function if you know it.
Thanks a lot for answers
EDIT:
Another property is:
$$-\alpha\frac{\partial^2{f(x,z)}}{\partial z^2}=\beta^2 \frac{\partial{f(x,z)}}{\partial x} $$
 A: First note that $\alpha$ and $x$ , $\beta$ and $z$ are both taking the same roles in $f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-\alpha n^2x+\beta nz}}{n!}$ and should be burdensome. So you should simplify your definition to $f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+zn}}{n!}$ , where $x\geq0$ .
$f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+zn}}{n!}$ can create the following differential equation:
$\dfrac{df(x,z)}{dz}=\sum\limits_{n=0}^\infty\dfrac{ne^{-xn^2+zn}}{n!}=\sum\limits_{n=1}^\infty\dfrac{ne^{-xn^2+zn}}{n!}=\sum\limits_{n=1}^\infty\dfrac{e^{-xn^2+zn}}{(n-1)!}=\sum\limits_{n=0}^\infty\dfrac{e^{-x(n+1)^2+z(n+1)}}{n!}=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+(z-2x)n+z-x}}{n!}=e^{z-x}\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+(z-2x)n}}{n!}=e^{z-x}f(x,z-2x)$
Note that this is a DDE. Unfortunately even the forms of the general solutions of DDEs we still can't know well, we can't know the number of I.C.s should be required. So this approach fails.
$f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+zn}}{n!}$ can create the following PDE:
$\dfrac{\partial^2f(x,z)}{\partial z^2}=\sum\limits_{n=0}^\infty\dfrac{n^2e^{-xn^2+zn}}{n!}$
$\dfrac{\partial f(x,z)}{\partial x}=\sum\limits_{n=0}^\infty\dfrac{-n^2e^{-xn^2+zn}}{n!}$
$\therefore\dfrac{\partial f(x,z)}{\partial x}+\dfrac{\partial^2f(x,z)}{\partial z^2}=0$
Note that the form of the general solution of this PDE is as follows:
Let $f(x,z)=X(x)Z(z)$ ,
Then $X'(x)Z(z)+X(x)Z''(z)=0$
$X'(x)Z(z)=-X(x)Z''(z)$
$-\dfrac{X'(x)}{X(x)}=\dfrac{Z''(z)}{Z(z)}=s^2$
$\begin{cases}\dfrac{X'(x)}{X(x)}=-s^2\\Z''(z)-s^2Z(z)=0\end{cases}$
$\begin{cases}X(x)=c_3(s)e^{-xs^2}\\Z(z)=\begin{cases}c_1(s)\sinh zs+c_2(s)\cosh zs&\text{when}~s\neq0\\c_1z+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore f(x,z)=C_1z+C_2+\int_sC_3(s)e^{-xs^2}\sinh zs~ds+\int_sC_4(s)e^{-xs^2}\cosh zs~ds$ or $C_1z+C_2+\sum\limits_sC_3(s)e^{-xs^2}\sinh zs+\sum\limits_sC_4(s)e^{-xs^2}\cosh zs~ds$
It is clear that two I.C.s should be required to solve this PDE uniquely, however from the functional property itself, only $f(0,z)=\sum\limits_{n=0}^\infty\dfrac{e^{zn}}{n!}=\sum\limits_{n=0}^\infty\dfrac{(e^z)^n}{n!}=e^{e^z}$ is trivially known, others for example $f(x,0)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2}}{n!}$ cannot be trivially known.
So we still no hope to express $f(x,z)=\sum\limits_{n=0}^\infty\dfrac{e^{-xn^2+zn}}{n!}$ as known functions.
