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I am trying to numerically find eigenstates of an Hamiltonian. Let $V(x)$ be some potential. Suppose my space is between $-L,L$ and that the allowed positions are every $\Delta x$, such that $\frac{1}{\Delta x}$ is some natural number $N$. I want to write the Hamiltonian, in this case $$ H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) $$ discretely, and then represent it as an $N\times N$ matrix in the discrete position basis. How can I transfer the continuous Hamiltonian to a discrete one?

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  • $\begingroup$ Do you want to calculate eigenfunctions of the differential operator $H$ by sampling the space and discretizing in computer, or do you want to calculate eigenfunctions of analogous Hamiltonian for a discrete lattice space? $\endgroup$ – Ján Lalinský May 28 '16 at 11:28
  • $\begingroup$ @JánLalinský The latter $\endgroup$ – JonTrav1 May 28 '16 at 11:41
  • $\begingroup$ Look at this 1997 paper which has done it, and references therein. This represents an enormous cottage industry flooding the literature. $\endgroup$ – Cosmas Zachos May 28 '16 at 13:35
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You're looking for a set of wavefunctions $\psi_n(x)$ that are solutions to $H\psi_n=E\psi_n$. The index $n$ does not refer to a discretized position. For numerical solutions, you will have to discretize $x$, for example with constant steps $h$ (you call them $\Delta x$); then the second derivative will be approximated as $$ {\partial^2\psi\over\partial x^2} \approx {\psi(x+h)+\psi(x-h)-2\psi(x) \over h^2}. $$ You can write $\psi_{nm}\equiv \psi_n(mh)$ and express the partial derivative above as a matrix multiplication. Make sure that $h$ is much smaller than the smallest expected wavelength of your wavefunctions. So your discretized Schrödinger equation becomes $$ -{\hbar^2\over 2m} \sum_m \mathcal D_{jm}\psi_{nm} + V_j\psi_{nj} = E_n\psi_{nj}, $$ where $\mathcal D$ is the discretized differential operator. For example, if you have discretized your $x$ space into four possible $x$ values, $\{0,h,2h,3h\}$, then the differential operator is $$ \mathcal D = \frac{1}{h^2} \left( \begin{array}{cccc} -2 & 1 & 0 & 0 \\ 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & -2 \end{array}\right). $$ In this representation, you have to deal carefully with the edges of the domain. You'd better choose your domain large enough that $\psi$ is very close to zero at those edges. The potential operator is a diagonal matrix, $$ V = \left( \begin{array}{cccc} V(0) & 0 & 0 & 0 \\ 0 & V(h) & 0 & 0 \\ 0 & 0 & V(2h) & 0 \\ 0 & 0 & 0 & V(3h) \end{array}\right) $$ And the total Hamiltonian is $$H_{jm}=-(h^2/2m)\mathcal D_{jm} + V_{jm},$$ of which you want to find eigenvectors.

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  • $\begingroup$ Could you elaborate please? I didn't fully understand. $\endgroup$ – JonTrav1 May 28 '16 at 11:42
  • $\begingroup$ Answer updated. $\endgroup$ – Han-Kwang Nienhuys May 28 '16 at 15:20

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