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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

2 Answers 2

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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Well, if I'm right, then this problem has no solution - at least no finite analytical one. And btw. the problem is more general than one might think. Look at the upper right quadrant ($x\gt0$ and $y\gt0$). Then the boundary conditions are symmetry boundary conditions (scalar product of normal vector and gradient of the solution is zero at $x=0$ and $y=0$). Furthermore, at x=a and x=b, the solution must be continuous (and the gradient, too, but this is not important). And, of course, the solution must vanish at infinity. These boundaries actually generate four regions: 0 - ($0\lt x \lt a, 0\lt y\lt b$), 1 - ($0\lt x\lt a, y\gt b$), 2 - ($x\gt a, 0\lt y\lt b$), and 3 - ($x\gt a, y\gt b$). In each of these regions, the solution is separable, as you correctly mentioned. The Schrodinger equation now makes the general statement. The total curvature in each region is a sum of two curvatures along the respective cartesian directon. $B_i^2=B_{i,x}^2+B_{i,y}^2$ with $i=0,1,2,3$. So, you have four equations for eight curvatures. Another four equations are added if you use the continuity boundary conditions. Namely, along the boundaries, the parallel curvatures at the adjacent regions must be equal. For instance, between region 0 and 1, $B_{0,x}^2=B_{1,x}^2$, etc. This, in total gives you a linear equation with a 8 by 8 square matrix, and a right-hand side (RHS) vector containing only the total curvatures $B_i^2$. First of all, this matrix is exactly singular (i.e. it does not have the full rank). To have any non-trivial solution at all, you have to make the RHS of the same rank. This gives you a condition for the total curvatures (which are related to the potential and the eigenvalue). And this condition is that the curvatures must be the same. Essentally, this means that the potential is everywhere zero - a clear contradiction to what we wanted. Please try this approach and report me your findings, as I might be wrong, as I outlined at the beginning ;)

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What is curvature in this context? Also, if your solution gives a "clear contradiction," is it really a solution? –  Kyle Kanos Mar 5 at 13:05
    
The curvature can be considered to be $\frac{1}{f}\frac{d^2f}{dx^2}$ in one direction, of generally, as $\frac{1}{f}\nabla^2f$. (Belongs to the last post by A.Friend - but I don't have 50 reputation - whatever this means ;) ) –  Johnny Doepp Mar 12 at 9:02
    
Hmm, I am not so sure about the 8 equations as for symmetry we know that in a square potential $f=g$. Even in a rectangular case I guess that re-scaling would work. –  mikuszefski Mar 12 at 14:15

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