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I want to solve bound states (in fact only base state is needed) of time-independent Schrodinger equation with a 2D finite rectangular square well \begin{equation}V(x,y)=\cases{0,&$ |x|\le a \text{ and } |y|\le b$ \\ V_0,&\text{otherwise}}.\tag{1}\end{equation} $$\Big[-\frac{\hbar^2}{2m}(\partial_x^2+\partial_y^2)+V(x,y)\Big]\psi(x,y)=E\psi(x,y)$$ At first glance, this problem is simple. It seems that the solution is variable-separable and can be written as $\psi(x,y)=f(x)g(y)$. Then $$ \frac{f''(x)}{f(x)}+\frac{g''(y)}{g(y)}+\frac{2m}{\hbar^2}(E-V)=0.$$ Let $E=E_x+E_y$ and $V=V_x+V_y$, then the problem is reduced to two 1D problems $$\cases{f''(x)+\frac{2m}{\hbar^2}(E_x-V_x)f(x)=0\\g''(y)+\frac{2m}{\hbar^2}(E_y-V_y)g(y)=0}.$$

However, how to determine $V_x$ and $V_y$ in the 2D space? A definitely wrong method is making $$ V_x=\cases{0,&$|x|\le a$\\V_1,&$|x|>a$}\text{ and }V_y=\cases{0,&$|y|\le b$\\V_2,&$|y|>b$}\tag{2}.$$ In fact, the potential Eq. (2) is equivalent to two independent "1D finite square well" problems in $x$ and $y$ direction respectively. However, a careful reader will note that the potential Eq(2) is DIFFERENT from Eq(1), which means that the potential Eq(2) is NOT what we want. It's not a rectangular well, but as following Potential of Eq(2), but NOT a 2D square well..

Then, I find that a variable-separable bound state for finite 2D square well does not exist. Although analytical solutions exist in each region with a constant potential, problems occur when matching boundary conditions to keep the continuity of $\psi(x,y)$. Unlike matching boundary condition at descrete points in 1D, in 2D we have to match boundary conditions along lines, e.g., $$ f_1(a)g_1(y)=f_2(a)g_2(y)$$ in the boundary between $x<a$(region 1) and $x>a$ (region 2). This leads to $$ g_1(y)/g_2(y)=f_2(a)/f_1(a)=constant.$$ Matching all boundaries this way will lead to that $\psi(x,y)$ have to be 0 outside the well. But this cronsponds to the case of INFINITE well. It's not the solution of finite well. Then I think no solutions exist under the separating-variable method.

Then, the question is, beyond separating-variable method, how to solve this problem?

BTW: Does anyone know that what kind (shape) of 2D well is solvable for bound states and how? (Potential with circular symmetry is excluded, because I know how to solve it. I want to find another shape of 2D well which is solvable.)

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A potential with elliptical symmetry is solvable by separation of variables - in terms of Mathieu functions, for some details see this article. It talks about Helmholtz equation, but it can be extended to finite well analogously to circular symmetry case. –  Ruslan Dec 4 at 8:36

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I think this problem is similar to the problem of finding modes of rectangular dielectric waveguide. In this case, you can use the effective-index method for finding the approximated solution (For your problem, we can call it effective-potential method). For more information about effective-index method see the following articles:

  • Effective-index analysis of optical waveguides: Link

  • Analysis of integrated optical waveguides: variational method and effective-index method with built-in perturbation correction: Link

The basis of this method is that the mode of a waveguide can be separated into products of two functions, one in $x$ direction which is dependent only on $x$ and one in $y$ direction which is dependent only on $y$. These can be solved independently and combined to produce the mode structure. In this way, the 2D waveguide structure can be separated into two single structures, one being a step index planar waveguide in $x$ direction and other in $y$ direction. In fact, this is same as your suggestion for introducing $V_x$ and $V_y$, but in a special way that the solution is very closed to the exact solution

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Could you provide more details on this method than a link as links can be broken? –  Kyle Kanos Jul 18 at 17:21
    
The basis of this method is that the mode of a waveguide can be separated into products of two functions, one in x direction which is dependent only on x and one in y direction which is dependent only on y. These can be solved independently and combined to produce the mode structure. In this way, the 2D waveguide structure can be separated into two single structures, one being a step index planar waveguide in x direction and other in y direction. In fact, this is same as your suggestion for introducing $V_x$ and $V_y$, but in a special way that the solution is very closed to the exact solution –  Mojtaba Golshani Jul 19 at 6:55

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