I'm doing quantum mechanics and I have an eigenfunction which is a theta function. I then discretised it, since I want see if I can find the eigenvalues for the discrete case by finding the eigenfunctions for the continuum and discretising them then applying translation operators for derivatives.

I have a theta function of the form $$\theta_3(z|\tau)$$ where $$z=\frac{ivL}{N}-\frac{ik_1}{N}+\frac{k_2}{N};$$ $$\tau=\frac{i}{N},$$ and $k_1$, $k_2=0,1,..., N-1$ with $v$ and $L$ being constants. Using the periodicity of the theta function

$$\theta_3(z+\tau|\tau)=\exp(-\pi i\tau-2\pi iz)\theta_3(z|\tau);$$ and $$\theta_3(z-\tau|\tau)=\exp(-\pi i\tau+2\pi iz)\theta_3(z|\tau)$$ where the above condition come from $k_1\mapsto k_1\pm 1$. My question is thus: for the other translation I acquire the resulting theta functions $$\theta_3(z\pm i\tau|\tau),$$ is there a rule for this such that $$\theta_3(z\pm i\tau|\tau)=A(z,\tau_\pm)\theta_3(z|\tau)?$$

I've tried looking at the series representation and computing it explicitly, but no luck. Am I just being hopeful, when in fact, there is no way to find the neat form of the $z\pm i\tau$ periodicity condition?

This leads to another question: how does one use an eigenfunction which is theta function in the Schrödinger equation to find eigenvalues? The derivatives aren't defined since it creates a sum like $\sum_{m\in\mathbb{Z}}\alpha m\hspace{1mm}\mathrm{e}^{\hspace{1mm}\beta m^2+m\gamma}$ where $\alpha$, $\beta$ and $\gamma$ are constants, which as far as I know can't be computed...

  • $\begingroup$ See if this is useful: reed.edu/physics/faculty/wheeler/documents/Quantum%20Mechanics/… $\endgroup$ – Alex R. Jun 2 '17 at 18:27
  • $\begingroup$ $\displaystyle f(z,\tau) = \sum_n e^{- \pi n^2 \tau - 2i \pi n z} = f(z+1,\tau) = f(z,\tau+2i)$ This function is interesting because from the Poisson summation formula $ \displaystyle f(z,1/\tau) = \tau^{1/2} \sum_n e^{- \pi (n-z)^2 / \tau - 2i \pi n z}=\tau^{1/2} \sum_n e^{- \pi (n^2+z^2-2nz)/ \tau - 2i \pi n z} = \tau^{1/2} e^{- \pi z^2 \tau} f(1/\tau, z+i z/\tau )$ and so on for the derivative $\displaystyle\frac{\partial}{\partial z} f(z,\tau) = -\sum_n 2i \pi n e^{- \pi n^2 \tau - 2i \pi n z} = \frac{\partial}{\partial z} f(z+1,\tau)=\frac{\partial}{\partial z} f(z+1,\tau+2i)$ $\endgroup$ – reuns Jun 2 '17 at 18:50
  • $\begingroup$ Sorry I don't quite understand what you are trying to say. Are you saying that the periodicity condition $\theta(z\pm i\tau|\tau)$ doesn't have the form I suggested and that only the $\tau$ has a complex periodicity? and that the derivatives are related via the expression you gave? Incidentally, isn't $\theta(z+1|\tau)=\theta(z|\tau)$? So the derivative you gave says $\frac{\partial}{\partial z}\theta=\frac{\partial}{\partial z}\theta$? $\endgroup$ – Lewis Proctor Jun 6 '17 at 11:53

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