Wednesday, December 21, 2016

quantum mechanics - 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)={0,|x|a and |y|b V0,otherwise.

[22m(2x+2y)+V(x,y)]ψ(x,y)=Eψ(x,y)
At first glance, this problem is simple. It seems that the solution is variable-separable and can be written as ψ(x,y)=f(x)g(y). Then f(x)f(x)+g(y)g(y)+2m2(EV)=0.
Let E=Ex+Ey and V=Vx+Vy, then the problem is reduced to two 1D problems {f(x)+2m2(ExVx)f(x)=0g(y)+2m2(EyVy)g(y)=0.


However, how to determine Vx and Vy in the 2D space? A definitely wrong method is making Vx={0,|x|aV1,|x|>a and Vy={0,|y|bV2,|y|>b.

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 ψ(x,y). Unlike matching boundary condition at descrete points in 1D, in 2D we have to match boundary conditions along lines, e.g., f1(a)g1(y)=f2(a)g2(y)

in the boundary between $xa(region2).Thisleadstog1(y)/g2(y)=f2(a)/f1(a)=constant.Matchingallboundariesthiswaywillleadtothat\psi(x,y)$ 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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