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I'm trying to solve the Schrodinger equation $$ -\frac{\hbar^2}{2m}\nabla^2 \psi+V(x,y)\psi(x,y)=E\psi(x,y) \tag{1}$$ for the finite two dimensional potential square well, that is, where

$$V(x,y)=\left\{\begin{matrix} 0 & if & \left |x \right |<a/2, \left |y \right |<a/2 \\ V_0 & if & \left |x \right |>a/2, \left |y \right |>a/2 \end{matrix}\right.\tag{2}$$

using separation of variables. My question arise when I make the suposition that $$V(x,y)=V_x(x)+V_y(y)\tag{3}$$ so $$\psi(x,y)=X(x) Y(y)\tag{4}$$

When I apply the separation of variable, I get: $$ \frac{-\hbar^2}{2m}\frac{X''(x)}{X(x)}+V_x(x)+\frac{-\hbar^2}{2m}\frac{Y''}{Y}+V_y(y)=E\tag{5} $$ So $$ \frac{-\hbar^2}{2m}X''(x)+V_x(x)X(x)=E_x X(x)\tag{6}$$ $$ \frac{-\hbar^2}{2m}Y''(y)+V_y(y)Y(y)=E_y Y(y)\tag{7}$$

where $E_x+E_y=E$

What actually happens with $V(x,y)$? Should I suppose that it's 'split' equally between $x$ and $y$? Or it can be 'split' in an arbritrary way? (as long as $V_x(x) + V_y(y) = V(x,y)$)

For example,

  1. inside the well, $V(x,y)=0$, so

    • Could be $V_x(x)=-5.24V_0$ and $V_y(y)=+5.24V_0$? (arbitrary partition)
    • Or should I suppose that $V_x(x)=0$ and $V_y(y)=0$ (equal partition)
  2. outside the well, $V(x,y)=V_0$, so

    • Could be $V_x(x)=\frac{1}{4}V_0$ and $V_y(y)=\frac{3}{4}V_0$? (arbitrary partition)
    • Or should I suppose that $V_x(x)=\frac{1}{2}V_0$ and $V_y(y)=\frac{1}{2}V_0$ (equal partition)

Edit:

This potential can't be solved by separation of variables. See this answer for more info: How to solve bound states of 2D finite rectangular square well?

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closed as unclear what you're asking by ACuriousMind, user36790, Sebastian Riese, Qmechanic Dec 15 '15 at 0:49

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ ...I don't think you can split that particular $V(x,y)$ in that way. What would be your candidate functions for $V(x)$ and $V(y)$, anyway? $\endgroup$ – ACuriousMind Dec 13 '15 at 23:23
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    $\begingroup$ Duplicate of: physics.stackexchange.com/q/73571 $\endgroup$ – Gert Dec 13 '15 at 23:39
  • $\begingroup$ They could be both $\frac{1}{2}V_0$ outside the well, and both 0 inside the well $\endgroup$ – user246185 Dec 13 '15 at 23:39
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    $\begingroup$ Oh, I see now you wrote it there as example, but that just doesn't make sense. When you write $V(x,y) = V(x) + V(y)$ you mean that you can split $V(x,y)$ into a function solely of $x$ and solely of $y$. By giving them "outside" and "inside" you elude the crucial question of how to include the information that $V(x,y)$ is zero inside the well into that split. In functional form, $V(x,y) = V_0 (\theta(x+a/2)-\theta(x-a/2))(\theta(y+a/2)-\theta(y-a/2))$ for $\theta$ the Heaviside function. That cannot be written as a sum $V(x)+V(y)$. $\endgroup$ – ACuriousMind Dec 13 '15 at 23:41
  • $\begingroup$ Comment to the question (v4): The if-structure in Definition (2) is incomplete/ambiguous. Consider clarifying it. $\endgroup$ – Qmechanic Dec 14 '15 at 0:07
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I think there is no adequate answer in the duplicate, so I tell you one way, which I believe resolve this problem.

Try to use a perturbation approach. Consider two one-dimensional finite wells for $x$ and $y$ respectfully, sum them up and take the difference between two-dimensional well, which you want to have, and the potential, which you get from summing two one-dimensional, as a perturbation. I don't remember if the second order perturbation correction coincide with an exact solution, but it is enough to see the degeneracy in energy levels.

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