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I just learned that derivatives can be represented as matrices. So, I was wondering if I could write the $\frac{d^2}{dx^2}$ and $V(x)$ as a matrix, maybe I could treat the eigenvalue problem, $${\bf H}\psi = E\psi$$ as an eigenvalue problem in matrices and find the values of E for which $$det({\bf H} - E {\bf I}) = 0$$ For example in a case where I consider the dimension of my vector space to be 3 (upto $x^2$), I would have the Hamiltonian matrix as $$ -\frac{\hbar^2}{2m}\left[ {\begin{array}{ccc} 0 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} } \right] + \left[ {\begin{array}{ccc} 0 & 0 & \frac{1}{2}k\\ 0 & \frac{1}{2}k & 0\\ \frac{1}{2}k & 0 & 0\\ \end{array} } \right] $$ I wanted to write a python code for finding the energy eigenstates this way. I would like to solve this for higher dimensions like upto terms containing $x^9$ or more, idk 10x10 matrices might be too much for my pc. Will my first few eigenvalues be accurate enough and is this even a plausible way to find the eigenvalues or is there some flaw in my method?

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  • $\begingroup$ It's not clear to me what you're actually trying to do here. If you represent the derivative as a "matrix", it's an infinite-dimensional matrix. It sounds as if you want to truncate this matrix (using powers of $x$) to approximate it by a finite-dimensional matrix, but it's not clear to me how you're actually doing this nor why you would expect this to be useful - if you consider that the non-truncated Hamiltonian has $\infty$ eigenstates, but the n-truncated matrix only can have $n$, why would you expect these $n$ eigenvalues to be useful approximations to the $\infty$ eigenvalues? $\endgroup$ – ACuriousMind Sep 29 '19 at 18:41
  • $\begingroup$ I don't expect them to be useful. I was just wondering if these n eigenvalues will be close to the first n eigenvalues of the real set of eigenvalues. $\endgroup$ – Brain Stroke Patient Sep 29 '19 at 18:50
  • $\begingroup$ What are the "first $n$ eigenvalues"? There is no natural order among them! (Unless you mean ordering them by their size as real numbers, but then I still don't see why you'd expect this truncation to produce anything close to these) $\endgroup$ – ACuriousMind Sep 29 '19 at 18:53
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    $\begingroup$ This numerical method is called FDM (Finite difference method). This is widely used to solve math problems including the eigenvalue problem for the Schrödinger equation. $\endgroup$ – Alex Trounev Sep 29 '19 at 19:33

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