A bit of theory first. If $t \in \Bbb C$ with $\text{Im } t > 0$ and if $z \in \Bbb C$, then one may consider the following series:

$$\vartheta (z, t) = \sum _{m \in \Bbb Z} \Bbb e ^{\pi \Bbb i \ m^2 t + 2 \pi \Bbb i \ n z} .$$

Remarkably, the series converges. Even more remarkably, is converges uniformly on compact subsets. Even more remarkably than that, if $| \text{Im } z | < c$ and $\text{Im } t > \varepsilon$, then the convergence is also *very fast*.

It is also obvious that $\vartheta$ can be extended to the case $\text{Im } t < 0$ by

$$\vartheta (t, z) = \overline {\vartheta (\bar t, \bar z)} .$$

Returning to your question, it is clear that if $t = 0$ then your series diverges. On the other hand, if $t<0$ then your series is

$$\sum _{m \in \Bbb Z} \exp \left( \pi \Bbb i \ m^2 \frac {-t} \pi + 2 \pi \Bbb i \ m \frac z {2 \pi} \right) = \vartheta \left( - \frac t \pi, \frac z {2 \pi} \right) .$$

If $t>0$ then use the above formula involving conjugates.

Extending the framework, your series makes sense even when $t=0$, provided that you agreed to view its sum as a distribution, in which case it is equal to $2 \pi \delta$ (with $\delta$ being Dirac's distribution).

One nice property of your distribution (let's call it $u$) is that for $t \in \Bbb R \setminus \{0\}$ and $z \in \Bbb R$, you have

$$\frac 1 {\Bbb i} \frac {\partial u} {\partial t} = \frac {\partial^2 u} {\partial x^2}, \quad \lim _{t \to 0} u = 2 \pi \delta$$

so $\frac 1 {2 \pi} u$ is the kernel of Schrödinger's equation.

Even more, notice that $u(t,z) = u(t,z + 2 \pi)$, so in fact $u$ "descends" naturally to a function defined on $\Bbb R \setminus \{ 0 \} \times S^1$, with $S^1$ being the unit circle.

If you don't want to view $u$ as a distribution, then $u$ may be made to verify the heat equation, but judging from your question it seems that this is not what you want.

You can find more information in the first chapter of David Mumford's "Tata Lectures on Theta", vol. 1.