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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 '14 at 8:36

1 Answer 1

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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