- #1

etotheipi

- Homework Statement
- Given the particle is in its ground state with ##E <0##, calculate ##\mathbb{P}(x > \xi)## for the following potential:\begin{align*} V(x) =

\begin{cases}

\infty & x < 0 \\

- \lambda & 0\leq x\leq \xi \\

0 & x > \xi

\end{cases}

\end{align*}where ##\lambda > 0##.

- Relevant Equations
- N/A

I think I made an error somewhere. In ##[0,a]## I let ##\varphi(x) = \varphi_1(x) := p\sin{kx}## whilst in ##(\xi, \infty)## I let ##\varphi(x) = \varphi_2 (x) := re^{-\gamma x}##, and the constraints at ##x=\xi## are \begin{align*}

\varphi_1'(\xi) = \varphi_2'(\xi) &\implies pk\cos{k\xi} = -\gamma re^{-\gamma \xi} \\

\varphi_1(\xi) = \varphi_2(\xi) &\implies p\sin{k\xi} = re^{-\gamma \xi}

\end{align*}which implies ##\gamma = -k/\tan{k\xi}##. Then we can normalise,\begin{align*}

\int_0^\xi dx \, \varphi_1^2(x) + \int_\xi^{\infty} dx \, \varphi_2^2(x) &= \frac{(2k \xi - \sin{2 k \xi})p^2}{4k} + \frac{r^2 e^{-2\gamma \xi}}{2\gamma} \overset{!}{=} 1

\end{align*}and, since ##p^2 \sin^2{k \xi} = r^2 e^{-2\gamma \xi}## it follows that\begin{align*}

\mathbb{P}(x > \xi) = \frac{r^2 e^{-2\gamma \xi}}{2\gamma} = \frac{2\sin^2{k \xi} \tan{k \xi}}{2\sin^2{k \xi} \tan{k \xi} + \sin{2k \xi} - 2k \xi}\end{align*}That the particle is in its ground state implies ##k \xi = \pi / 2##, such that the sinusoidal part joins nicely onto the decaying exponential solution in the ##x > \xi## region. But that gives a non-sense answer; so I wonder where I made my mistake(s). Thanks!

\varphi_1'(\xi) = \varphi_2'(\xi) &\implies pk\cos{k\xi} = -\gamma re^{-\gamma \xi} \\

\varphi_1(\xi) = \varphi_2(\xi) &\implies p\sin{k\xi} = re^{-\gamma \xi}

\end{align*}which implies ##\gamma = -k/\tan{k\xi}##. Then we can normalise,\begin{align*}

\int_0^\xi dx \, \varphi_1^2(x) + \int_\xi^{\infty} dx \, \varphi_2^2(x) &= \frac{(2k \xi - \sin{2 k \xi})p^2}{4k} + \frac{r^2 e^{-2\gamma \xi}}{2\gamma} \overset{!}{=} 1

\end{align*}and, since ##p^2 \sin^2{k \xi} = r^2 e^{-2\gamma \xi}## it follows that\begin{align*}

\mathbb{P}(x > \xi) = \frac{r^2 e^{-2\gamma \xi}}{2\gamma} = \frac{2\sin^2{k \xi} \tan{k \xi}}{2\sin^2{k \xi} \tan{k \xi} + \sin{2k \xi} - 2k \xi}\end{align*}That the particle is in its ground state implies ##k \xi = \pi / 2##, such that the sinusoidal part joins nicely onto the decaying exponential solution in the ##x > \xi## region. But that gives a non-sense answer; so I wonder where I made my mistake(s). Thanks!

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