- #1
docnet
Gold Member
- 696
- 348
- 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.