# QM: Finite Potential Square Well Solved without Symmetry Assumption

as the title suggests I'm trying to figure out the solution of the finite potential well, without using the odd/even symmetry of the potential.

$$V(x) = \begin{cases} 0 & \text{otherwise} \\ -V_{0} & \text{-a \leq x \leq a} \\ \end{cases}$$

which has solutions $$\psi(x) = \begin{cases} \psi_{1}(x) = Ae^{\kappa x} + Be^{-\kappa x} & \text{x < -a} \\ \psi_{2}(x) = Ee^{ikx} + Fe^{-ikx} & \text{-a \leq x \leq a} \\ \psi_{3}(x) = Ce^{\kappa x} + De^{-\kappa x} & \text{x > a}\\ \end{cases},$$ with $\kappa = \sqrt{\frac{-2mE}{\hbar^{2}}}$ and $k = \sqrt{\frac{2m(E+V_{0})}{\hbar^{2}}}$.

As usually, we set $B=0$ and $C=0$, so that the function can still be normalized. Then we use continuity at $-a$, $a$ of $\psi(x)$ and its derivative $\psi'(x)$, yielding: \begin{align} \psi_{1}(-a) &= \psi_{2}(-a)\quad \rightarrow\quad Ae^{-\kappa a} = Ee^{-ika} + Fe^{ika} \\ \psi_{1}'(-a) &= \psi_{2}'(-a)\quad \rightarrow\quad \kappa Ae^{-\kappa a} = ik(Ee^{-ika} - Fe^{ika}) \end{align} and \begin{align} \psi_{2}(a) &= \psi_{3}(a)\quad \rightarrow \quad Ee^{ika} + Fe^{-ika} = De^{-\kappa a} \\ \psi_{2}'(a) &= \psi_{3}'(a)\quad \rightarrow \quad ik(Ee^{ika} - Fe^{-ika}) = -\kappa De^{-\kappa a} \end{align}

From here is basically where my struggle begins. I have tried several ways of adding, subtracting and dividing the equations but never come up with $\kappa = k\tan(ka)$ (or $\kappa = -k\cot(ka)$).

Can someone give me a hint as to how to solve this set of equations?

EDIT:

If I wanted to solve for the normalization, i.e.

$$\int_{-\infty}^{\infty}{|\Psi(x)|^{2}}dx=1 ,$$

how would I do that? Because it seems that, after solving the equations as far as possible, I'm still stuck with the TWO constants $$E$$ and $$F.$$

Take the equations at $x=-a$ and divide them. This gets rid of $A$ and $\text e^{-\kappa a}$. Then you can look at the real and imaginary parts of this equation separately and arrive at the tangent equations.
(I previously misunderstood the question $\to$ new answer)