# How to solve bound states of 2D finite rectangular square well?

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 $$V(x,y)=\cases{0,& |x|\le a \text{ and } |y|\le b \\ V_0,&\text{otherwise}}.\tag{1}$$ $$\Big[-\frac{\hbar^2}{2m}(\partial_x^2+\partial_y^2)+V(x,y)\Big]\psi(x,y)=E\psi(x,y)$$ 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 .

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.)

-
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
It seems like it might be worth looking into edge effects when trying to determine if there is an analytic solution. Intuitively I imagine that on a very large scale, the solution near the center or centers of the edge will resemble the separable solution. – Shane Sep 4 '15 at 23:21
Maybe a change in variables by a conformal mapping of the square onto the circle helps in the case a = b : math.stackexchange.com/questions/1015205/… – jjcale Sep 5 '15 at 4:00
You might look at simpler problems with the same geometry, such as: "what is the electric field of a charged wire of rectangular cross section" or "what is the current from a wire of rectangular cross section into a surrounding resistive medium" (or heat flow problems). I don't know whether explicit solutions are known for these. – Keith McClary Sep 5 '15 at 5:42
Another thing one could do is divide by $V_0$ so your $V_x$=$V_y$=$V_xV_y$=1 but you still have the problem with matching bondry conditions – Shane Sep 6 '15 at 19:10

So far, the only thing that I can be sure of is that there is no solution of the form $\psi(x,y)=g(y)f(x)$, which would lead to an obvious contradiction:

If we have a solution $\psi(x,y)==f(x)g(y)$, denote $\frac{f''}{f}=F(x),\frac{g''}{g}=G(y)$, For any fixed x<-a, F(x)+G(y)=constant requires G(y) is constant, similarly, fixed y<-b, requires F(x) is constant, but F(x)+G(y) is not constant in the whole x-y plane , thus =><=.

-
@goodluck Another way to see that a separable solution is not possible is to note that separability requires $V(x,y) = V(x) + V(y)$. But the potential in the problem is $V(x,y) = V_0 \Pi(x/2a) \Pi(y/2a)$, where $\Pi(x)$ is the rectangular function: $\Pi(x) = 1$ for $-1/2 ≤ x ≤ 1/2$ and $\Pi(x) = 0$ otherwise. Equivalently we can express $V$ in terms of Heaviside step functions, but either way, it never separates into a sum $V(x) + V(y)$. – udrv Sep 9 '15 at 3:37

Electrons trapped in 2D circle is solvable, its solution is a Bessel function, I knew becaused i have solved it, it has also been detected by IBM scientist in their Lab prior.

With regrads to your problem, its pretty solvable by separation of variables, your problems has two regions, dont try to mix them cause you will get stuck as you are now. The two regions are simply IN-box and OUT-box.

IN-box:

At |x| < or = a and |y| < or = b, V=0

that means youre box has 'a' length on +/- x-axis, thats 2a overall, same as to with y. And in that region V=0 as your problem says, so your DE will reduce to:

$$-\frac{\hbar^2}{2m}(\partial_x^2+\partial_y^2)\psi (x,y) -E \psi (x,y) = 0$$

by this we can now employ separation of variables. Remember that your E is the a base energy (as you wish to solve) then it should therefore be the same anywhere inside, at its base state. Thus it is a constant and does not have any x or y component, no breaking or sum involved, you treat it the same as you treat a constant under the Technique of Separation.

OUT-box

if otherwise |x| is not < or = a that means greater than that, V = V_0, same goes with y-axis. Well V_0, as the name itself is a constant, no need separations for V_0, youre DE would then be

$$-\frac{\hbar^2}{2m}(\partial_x^2+\partial_y^2)\psi (x,y) +(V_0-E)\psi (x,y) = 0$$

again, this is also a separable equation, then therefore your Problem is solvable.

BTW: You seem to have problem with separations involving constats, kindly review it. I wish to discuss it, but that is not Physics anymore.

-
This is not called separation of variables. – Arek' Fu yesterday

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

-

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 ;)

-
What is curvature in this context? Also, if your solution gives a "clear contradiction," is it really a solution? – Kyle Kanos Mar 5 '15 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 ;) ) – user75205 Mar 12 '15 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 '15 at 14:15