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I'm stuck on a 2D square potential well:$$ \frac{-\hbar^2}{2m}(\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2})\Psi(x, y) + U(x, y)\Psi(x, y) = E\Psi(x, y)$$ $$U(x, y) =\begin{cases}0 & 0 < x < a, 0 < y < a\\ \infty & otherwise\end{cases} $$

I want to find all independent solutions of that problem. All that i have is a pretty nice solution based on the assumption that wave function can be factored as $\Psi(x, y) = \Psi_x(x)\Psi_y(y)$. But i haven't found anywhere the proof that that there are no other independent solutions of that problem. I'll be glad if somebody help me with the proof or another independent solution.

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You got a Fourier series in 2 variable. And since Fourier series are complete any function in 2 variable can be written as a Fourier series in 2 variable To see this in action see Ex 3.3 in Griffiths electrodynamics. Also take a look at the image as well. There might be a copyright issue.enter image description here

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  • $\begingroup$ This isn't even true in one variable. Kolmogorov constructed a continuous function in one variable whose Fourier series diverges everywhere (math.stackexchange.com/questions/608394/…). $\endgroup$ Mar 29, 2018 at 10:45
  • $\begingroup$ A wavefunction can't be such kind of pathological function. $\endgroup$
    – aitfel
    Mar 29, 2018 at 12:13
  • $\begingroup$ Why not? The above function is continuous. Wavefunctions need not be differentiable; for example, see the eigenfunction of the Dirac-delta potential. $\endgroup$ Mar 29, 2018 at 12:14
  • $\begingroup$ Is it square integrable? More technical does it even lie in Hilbert space? $\endgroup$
    – aitfel
    Mar 29, 2018 at 12:18
  • $\begingroup$ Ah, I see. It's in $L^1$, but I don't know if it's in $L^2$. From the Math SE post I linked to, apparently the function itself is integrable, but maybe it's not square-integrable. $\endgroup$ Mar 29, 2018 at 12:25

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