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:
$$ V(x) = \begin{cases} \infty & x < -\tfrac{L}{2} \\ 0 & - \frac{L}{2} \leq x \leq \frac{L}{2} \\ \infty & x > \frac{L}{2} \end{cases} $$
The problems arise with the boundary conditions. We have
$$ \frac{d^2\Psi (x)}{dx^2} = -k^2 \Psi (x) $$
where $k^2 = \frac{2mE}{\hbar ^2}$. The general solution is
$$ \Psi (x) = Ae^{ikx} + Be^{-ikx} $$
Due to continuity and the nature of the potential, we must have
$$ \psi (-\tfrac{L}{2}) = \Psi (\tfrac{L}{2}) = 0 $$
Plugging in:
$$ \psi (-\tfrac{L}{2}) = Ae^{-ikL/2} + Be^{ikL/2} = 0 \\ \psi (\tfrac{L}{2}) = Ae^{ikL/2} + Be^{-ikL/2} = 0 $$
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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