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I have a summation of the form:

$$\sum_{m\in\mathbb{Z}}m\exp\left\{-\frac{B}{2}\left(z-\frac{2\pi iv}{B}-\frac{2\pi m}{BL_2}\right)^2\right\}$$ where $B, v, L_2$ are constants and $z$ is a variable. I've been trying to look for a solution via Gauss sums, but have found anything of help yet (my background is in theoretical physics so I have to read and learn number theory as I go). I was maybe thinking about using a Fourier expansion of a function which would produce the series above, but haven't found anything yet...

Any suggestions?

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This is related to a derivative of a Jacobi theta-function: if we write $$ \theta(z,\tau) = \sum_{m = -\infty}^{\infty} \exp{(\pi i m^2 \tau + 2\pi i n z)}, $$ then $$ \sum_{m = -\infty}^{\infty} \exp{\left(-\frac{B}{2}(z-2\pi i \nu/B)^2+\frac{2\pi m}{L_2}(z-2\pi i \nu/B) - \frac{2\pi^2}{BL_2^2}m^2 \right)} = e^{-B(z-2\pi i \nu/B)^2/2} \theta\left(\frac{z-2\pi i \nu/B}{iL_2},\frac{2\pi i}{BL_2^2}\right), $$ and then your sum is $$ e^{-B(z-2\pi i \nu/B)^2/2} \frac{L_2}{2\pi}\frac{\partial}{\partial z}\theta\left(-i\frac{z-2\pi i \nu/B}{L_2},\frac{2\pi i}{BL_2^2}\right) $$ (Notice that the derivative brings down a $2\pi m/L_2$ inside the sum.) There is in general no way to write this more simply.

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  • $\begingroup$ Thank you for your reply. The problem I have been looking at indeed involves theta functions. The theta function I have been looking at is a wave function, and the sum I'm looking at comes from the derivative of this wave function. Is there anyway to re-arrange the derivative to pull out a term and leave a theta function, so in effect, find an eigenvalue? $\endgroup$ – Lewis Proctor May 19 '17 at 11:57
  • $\begingroup$ No, I don't think so. The logarithmic derivatives of the theta-functions are given in dlmf.nist.gov/20.5#ii , and are decidedly not constant. Quotients of theta-functions satisfy nonlinear first-order differential equations, and there is of course the heat equation relating the second $z$-derivative to the $\tau$-derivative, but I think those are the only ones possible (note the derivative of a theta-function has additive quasi-periodicity, while the function itself is multiplicatively quasi-periodic). $\endgroup$ – Chappers May 19 '17 at 13:09
  • $\begingroup$ Yeah I thought that might be the case. This means then that a quantum system with a theta function may possibly not yield an eigenvalue. Which seems bizarre to me. Thank you anyway. $\endgroup$ – Lewis Proctor May 19 '17 at 14:41

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