- #1
docnet
- 796
- 486
- Homework Statement
- psb
- Relevant Equations
- psb
Thank you for reading
Section 1
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<L\right\}##, and ##\left\{L<x\right\}##. The time independent Schrodinder equation is
\begin{equation}\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}[V_(x)-|E|]\psi
\end{equation}
Outside of the well, setting ##V(x)=V_0## gives
\begin{equation}\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}[V_0-|E|]\psi
\end{equation}
The characteristic polynomial is
\begin{equation}\lambda^2-\frac{2m[V_0-|E|]}{\hbar^2}=0\end{equation} with the roots \begin{equation}
\lambda=\pm \sqrt{\frac{2m[V_0-|E|]}{\hbar^2}}\Rightarrow \pm \alpha\end{equation}
and the general solution is
\begin{equation}
\psi=C_1e^{\alpha x}+C_2e^{-\alpha x}
\end{equation}
Inside of the finite well, setting ##V(x)=0## gives
\begin{equation}
\frac{d^2\psi}{dx^2}=\frac{-2m}{\hbar^2}|E|\psi
\end{equation}
which has the following characteristic polynomial
\begin{equation}
\lambda^2+\frac{2m}{\hbar^2}|E|=0
\end{equation}
with the roots
\begin{equation}
\lambda=\pm i\sqrt{\frac{2m}{\hbar^2}|E|}\Rightarrow\pm i\sqrt{q}
\end{equation}
and the general solution is
\begin{equation}
\psi=C_1e^{iqx}+C_2e^{-iqx}\Rightarrow \psi= C_3cos(qx)+iC_4sin(qx)
\end{equation}
The real and imaginary parts of a complex valued solution form a fundamental set of solutions of the second order linear ODE, giving
\begin{equation}
\psi=A_1cos(qx)+A_2sin(qx)
\end{equation}
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}
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}
Section 2
For the even solution, we require ##\psi \in C^1(\pm L)\Rightarrow \psi## is once continuously 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 (18) by (17) and do a change of variables using $$x=qL$$ and $$r^2=\frac{2mV_0L^2}{\hbar^2}$$ The even solutions are given implicitly by
\begin{equation}
xtan(x)=\sqrt{r^2-x^2}
\end{equation}
For the odd functions, we again require that ##\psi \in C^1(\pm L)##
\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}
and divide equation (20) by (21). Following a change of variables for ##x## and ##r^2## we have
\begin{equation}
-x cot(x)=\sqrt{r^2-x^2}
\end{equation}
Section 3
(19) and (22) are transcendental equations whose solutions are obtained using numerical approximations. The even solutions are given by
\begin{equation}
\{ \{y=xtan(x)\}\cap \{y=\sqrt{r^2-x^2}\}|x\in [0,\infty)\}
\end{equation}
and the odd solutions are given by
\begin{equation}
\{\{y=-xcot(x)\}\cap\{y=\sqrt{r^2-x^2}\}|x\in [0,\infty)\}
\end{equation}
It is worth mentioning the solutions exist in the domain \begin{equation}
\{(x,y)\in {\rm I\!R} \times[0,\infty)\}
\end{equation}
and the solution space is symmetric in the ##x## and the ##-x##. For wells with
\begin{equation}
V_0<\frac{(\hbar \pi)^2}{8mL^2}\Rightarrow L<\frac{\hbar \pi}{\sqrt{8mV_0}}\Rightarrow r<\frac{\pi}{2}
\end{equation}
we use the fact ##r=\frac{\pi}{2}## is in the exceptional set defined by $$\varepsilon(f)=\Big\{\alpha\in\bar{\mathbb{Q}}:f(\alpha)\in\bar{\mathbb{Q}}\Big\}$$ where ##Q## denotes the field of rational numbers.
$$0=-\Big(\frac{\pi}{2}\Big)cot\Big(\frac{\pi}{2}\Big)=\sqrt{\Big(\frac{\pi}{2}\Big)^2-x^2}$$
It follows that
\begin{equation}
\{\{y=-xcot(x)\}\cap\{y=\sqrt{r^2-x^2}\}|(x,r)\in [0,\infty)\times{[0,\frac{\pi}{2}})\}=\{\emptyset\}
\end{equation}
so there are no odd parity energy states. In the meanwhile, we find
\begin{equation}
\{ \{y=xtan(x)\}\cap \{y=\sqrt{r^2-x^2}\}|(x,r)\in [0,\infty)\times[0,\infty)\}\neq\{\emptyset\}
\end{equation}
so there will always be an even parity solution for any ##V_o##. In particular, we find that ##\{y=xtan(x)\}\cap \{y=\sqrt{r^2-x^2}\}\}## where ##r\in[0,\frac{\pi}{2})## is the graph of ##y=xtan(x)## in the region ##\{|x|\leq.934\}##. So, there exists only one energy state of even parity for
$$V_0<\frac{(\hbar \pi)^2}{8mL^2}$$
Section 4
Setting ##L=1nm## and ##V_0=1eV##, we compute
$$
r^2=\frac{2mV_0L^2}{\hbar^2}=\frac{2(9.1\times10^{-31}kg)(1.6\times10^{-19}J)(10^{-9})^2m}{(1.054\times10^{-34}Js)^2}\approx26.2729J
$$
we plot
\begin{equation}
y=xtan(x)
\end{equation}
\begin{equation}
y=-xcot(x)
\end{equation}
\begin{equation}
y=\sqrt{26.2729-x^2}
\end{equation}
and find
$$x_{even}=\{1.312, 3.86\}$$
and
$$x_{odd}=\{2.608, 4.964\}$$
We compute the energy using
\begin{equation}
E=\frac{x^2\hbar^2}{2mL^2}
\end{equation}
giving
$$E_{1even}\approx1.05\times10^{-20}J$$
$$E_{2even}\approx9.09\times10^{-20}J$$
and
$$E_{1odd}\approx4.15\times10^{-20}J$$
$$E_{2odd}\approx1.50\times10^{-19}J$$
The expression for the energy levels of the infinite well is
\begin{equation}
E_n=\frac{n^2\pi^2\hbar^2}{8mL^2}
\end{equation}
giving
$$E_1\approx1.50\times10^{-20}J$$
$$E_2\approx6.02\times10^{-20}J$$
so the ground state energy level for the finite well is lower.
Last edited by a moderator:
