- #1

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- Homework Statement:
- please see below

- Relevant Equations:
- please see below

To find the energy states of the particle, we define the wave function over three discrete domains defined by the sets ##\left\{x<-L\right\}##, ##\left\{L<x\right\}## and ##\left\{|x|<L\right\}##. The time independent Schr\"odinder equation is

\begin{equation}\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=[E-V(x)]\psi

\end{equation}

Outside of the finite well, the ##V(x)## term vanishes to zero to give the following linear second order ODE

\begin{equation}\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}|E|\psi

\end{equation}

whose characteristic polynomial is

\begin{equation}r^2-\frac{2m|E|}{\hbar^2}=0\end{equation} with the corresponding roots \begin{equation}r=\pm \sqrt{\frac{2m|E|}{\hbar^2}}=\pm \alpha\end{equation}

The general solution to the ODE is

\begin{equation}

\psi=C_1e^{\alpha x}+C_2e^{-\alpha x}

\end{equation}

To obtain the wave function inside of the finite well, we set ##V(x)=V_0## and solve the following time independent Schr\"odinder equation

\begin{equation}

\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}[V_0-|E|]\psi

\end{equation}

whose characteristic polynomial is

\begin{equation}

r^2+\frac{2m}{\hbar^2}[V_0-|E|]=0

\end{equation}

with the corresponding roots

\begin{equation}

r=\pm \sqrt{\frac{-2m}{\hbar^2}[V_0-|E|]}\Rightarrow\pm i\sqrt{q}

\end{equation}

The general solution to the ODE is

\begin{equation}

\psi=C_1e^{iqx}+C_2e^{-iqx}\Rightarrow \psi= C_3cos(qx)+iC_4sin(qx)

\end{equation}

Taking only the real parts of ##\psi## we have

\begin{equation}

\psi=A_1cos(qx)+A_2sin(qx)

\end{equation}

So the even parity wave functions are

\begin{equation}

\psi=Ce^{\alpha x}|_{\left\{x<-L\right\}}

\end{equation}

\begin{equation}

\psi=Acos(qx)|_{\left\{|x|<L\right\}}

\end{equation}

\begin{equation}

\psi=Ce^{-\alpha x}|_{\left\{L<x\right\}}

\end{equation}

and the odd parity wave functions are

\begin{equation}

\psi=-Ce^{\alpha x}|_{\left\{x<-L\right\}}

\end{equation}

\begin{equation}

\psi=Asin(qx)|_{\left\{|x|<L\right\}}

\end{equation}

\begin{equation}

\psi=Ce^{-\alpha x}|_{\left\{L<x\right\}}

\end{equation}

For even solutions, we impose the boundary condition ##\psi \in C^1(\pm L)## such that ##\psi## is continuous and once differentiable at ##x=\pm L##

\begin{equation}

\psi_{in}(\pm L)= \psi_{out}(\pm L)\Rightarrow Acos(qL)=Ce^{-\alpha L}

\end{equation}

\begin{equation}

\psi'_{in}(\pm L)= \psi'_{out}(\pm L)\Rightarrow Aqsin(qL)=\alpha Ce^{-\alpha L}

\end{equation}

We divide equation (16) by (15) and then make substitutions using $$y=qL$$ and $$r=\sqrt{\frac{2mV_0L^2}{\hbar^2}}$$ The solutions to the energy states are given implicitly by the following expression.

\begin{equation}

ytan(y)=\sqrt{r^2-y^2}

\end{equation}

We impose the same boundary condition on the odd functions

\begin{equation}

\psi_{in}(\pm L)= \psi_{out}(\pm L)\Rightarrow Asin(qL)=Ce^{-\alpha L}

\end{equation}

\begin{equation}

\psi'_{in}(\pm L)= \psi'_{out}(\pm L)\Rightarrow Aqcos(qL)=-\alpha Ce^{-\alpha L}

\end{equation}

We divide equation (20) by (21), make the same substitutions as above, and obtain the following expression.

\begin{equation}

-ycot(y)=\sqrt{r^2-y^2}

\end{equation}

The expressions do not have analytical solutions, so there are no closed forms of the energy levels of the even and odd states. However, given values for ##V_o## and ##L##, we can solve by graphing the curves given implicitly by the formulas.