# Finite two-dimensional potential square well [closed]

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?

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

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• ...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? – ACuriousMind Dec 13 '15 at 23:23
• Duplicate of: physics.stackexchange.com/q/73571 – Gert Dec 13 '15 at 23:39
• They could be both $\frac{1}{2}V_0$ outside the well, and both 0 inside the well – user246185 Dec 13 '15 at 23:39
• 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 $V(x)+V(y)$. – ACuriousMind Dec 13 '15 at 23:41
• Comment to the question (v4): The if-structure in Definition (2) is incomplete/ambiguous. Consider clarifying it. – Qmechanic Dec 14 '15 at 0:07

## 1 Answer

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.