# 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 $$F.$$

## 1 Answer

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.

In particular, the real part gives you the equation with the tangent function and the imaginary part gives you the equation with the cotangent.

(I previously misunderstood the question $\to$ new answer)