I am working on a problem in which I shall find the normalised solution to the 1D particle in a box. Solving for the particle in an asymmetric potential is quite straight forward, but I run into trouble when the potential is symmetric:
V(x)={∞x<−L20−L2≤x≤L2∞x>L2
The problems arise with the boundary conditions. We have
d2Ψ(x)dx2=−k2Ψ(x)
where k2=2mEℏ2. The general solution is
Ψ(x)=Aeikx+Be−ikx
Due to continuity and the nature of the potential, we must have
ψ(−L2)=Ψ(L2)=0
Plugging in:
ψ(−L2)=Ae−ikL/2+BeikL/2=0ψ(L2)=AeikL/2+Be−ikL/2=0
I know that when the potential is symmetric, we will find even (A=B) and odd (A=−B) wave functions. We will see that for even functions, n has to be odd whole numbers, and for odd functions n has to be even whole numbers. This leads to a sequence of sine and cosine curves as n increases by 1.
I am trouble getting there, however, from those boundary conditions, and I would really appreciate pointers and help.
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