Solution of Schrödinger Equation for Finite Square Well Potential
Overview
Let’s examine how a particle moves when the potential takes the shape of a finite square well as shown in the figure above. The potential $U$ is
$$ U(x) = \begin{cases} 0 & x<-a \\ U_{0} & -a < x <a \\ 0 &a<x \end{cases} $$
When the potential is $U(x)$, the time-independent Schrödinger equation is
$$ \dfrac{d^2 u(x)}{dx^2}+\frac{2m}{\hbar ^2} \Big[ E-U(x) \Big]u(x)=0 $$
Solution
$E<-U_{0}$
If the energy is less than the potential, then a solution does not exist and therefore does not need to be considered.
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$-U_{0} < E < 0$
Part 2-1. $x<-a$
In this region, the time-independent Schrödinger equation is $$ \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}Eu=0 $$ Since $\frac{2m}{\hbar^2}E$ is negative, if we substitute with $-\kappa ^2$, $$ \dfrac{d^2 u}{dx^2}-\kappa^2u=0 $$ It is a very simple second-order ordinary differential equation. Solving the differential equation, the solution is $$ u_{1}(x)=A_{+}e^{\kappa x} + A_{-}e^{-\kappa x} $$ Since the wave function must be square-integrable, it is $A_{-}=0$.
Part 2-2. $-a<x<a$
In this region, the time-independent Schrödinger equation is $$ \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}(E+U_{0})u=0 $$ Since $E+U_{0}>0$, substituting with $\frac{2m}{\hbar^2}(E+U_{0})=k ^2$, $$ \dfrac{d^2 u}{dx^2}+k^2 u=0 $$ Solving the equation, $$ u_{2}(x) = B_{+}e^{i k x}+B_{-}e^{-ik x} $$
Part 2-3. $a<x$
In this region, the time-independent Schrödinger equation is $$ \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}Eu=0 $$ The solution has the same form as in Part 2-1. $$ u_{3}(x)=C_{+}e^{\kappa x} + C_{-}e^{-\kappa x} $$ Since the wave function must be square-integrable, it is $C_{+}=0$.
Part 2-4. Boundary Conditions
Assuming the wave function is smooth, it is continuous at $x=-a$ and $x=a$, and the derivative (slope) of the wave function is also continuous at $x=-a$ and $x=a$. Therefore, $$ \begin{cases}u_{1}(-a)=u_{2}(-a) \\ u_{2}(a)=u_{3}(a) \end{cases} \quad \implies \begin{cases} A_{+}e^{-\kappa a}+A_{-}e^{\kappa a} = B_{+}e^{-ik a}+B_{-}e^{ ik a} \quad \cdots (1) \\ B_{+}e^{ik a}+B_{-}e^{-ik a} = C_{+}e^{\kappa a}+C_{-} e^{-\kappa a} \ \quad \cdots (2) \end{cases} $$
$$ \begin{cases}u_{1}^{\prime}(-a)=u_{2}^{\prime}(-a) \\ u_{2}^{\prime}(a)=u_{3}^{\prime}(a) \end{cases} \quad \implies \begin{cases} \kappa A_{+}e^{-\kappa a}-\kappa A_{-}e^{\kappa a} = ik B_{+}e^{-ik a}-ik B_{-}e^{ik a} \quad \cdots (3) \\ ik B_{+}e^{ik a}-ik B_{-}e^{-ik a} = \kappa C_{+}e^{\kappa a}-\kappa C_{-} e^{\kappa a} \quad \cdots (4) \end{cases} $$
Expressing $(1)$ and $(3)$ in matrix form,
$$ \begin{pmatrix} e^{-\kappa a} & e^{ika} \\ ike^{-ika} & -ike^{ika} \end{pmatrix} \begin{pmatrix} A_{+} \\ A_{-} \end{pmatrix} = \begin{pmatrix} e^{-\kappa a} & e^{\kappa a} \\ \kappa e^{-\kappa a} & -\kappa e^{\kappa a} \end{pmatrix} \begin{pmatrix} B_{+} \\ B_{-} \end{pmatrix}\quad \cdots (5) $$
Expressing $(2)$ and $(4)$ in matrix form,
$$ \begin{pmatrix} e^{\kappa a} & e^{-\kappa a} \\ \kappa e^{\kappa a} & -\kappa e^{-\kappa a} \end{pmatrix} \begin{pmatrix} B_{+} \\ B_{-} \end{pmatrix} = \begin{pmatrix} e^{ik a} & e^{-ik a} \\ ik e^{ik a} & -ik e^{-ika } \end{pmatrix} \begin{pmatrix} C_{+} \\ 0 \end{pmatrix}\quad \cdots (6) $$