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\heading{E7110: Sudden movement of a wall}
\auname{Shiran Nabha  \& Oron Landau }


{\bf The problem:}

\Dn

A particle is placed in potential well $0<x<L$, in his ground state.\\
Suddenly, the wall is moved, so the wide of the well is $-L<x<L$.\\
Find the probability $P(n)$ to find the particle in energy level n.\\
\Dn


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{\bf The Solution:}

\Dn

The particle was placed in the ground state of a potential well $0< x < L$, therefore his wave function is:
\Dn

\qquad
$\psi_0=\frac{1}{\sqrt{\frac{L}{2}}} \sin\left(\frac{\pi}{L}x\right)=\sqrt{\frac{2}{L}} \sin\left(\frac{\pi}{L}x\right) \ \ \ \ for \ \   0<x<L $

\Dn

After the sudden movement of the wall of the well, the particle remains in the same wave function, but the basic functions had changed.
\Dn


$\qquad {\psi _n} = \left\{ {\begin{array}{*{20}{l}}
{\frac{1}{{\sqrt L }}\sin\left(\frac{{\pi n}}{{2L}}x\right)}&{for\;n - even}\\
{\frac{1}{{\sqrt L }}\cos\left(\frac{{\pi n}}{{2L}}x\right)}&{for\;n - odd}
\end{array}} \right.$

\Dn

\qquad
$\psi_0=\sum_{n}^{}C_n\cdot \psi_n$\\

$\qquad C_n=<\psi_n|\psi_0 >=\int\limits_{-L}^{L} \psi_n^*\cdot\psi_0dx=\int\limits_{0}^{L} \psi_n^*\cdot\sqrt{\frac{2}{L}} \sin\left(\frac{\pi}{L}x\right)dx$\\
\Dn

\textbf{for n - even}\\
\begin{align}
C_n &=\int\limits_{0}^{L} \frac{1}{\sqrt{L}} \sin\left(\frac{\pi n}{2L}x\right)\cdot\sqrt{\frac{2}{L}} \sin\left(\frac{\pi}{L}x\right)dx=\frac{\sqrt{2}}{L} \int\limits_{0}^{L} \sin\left(\frac{\pi n}{2L}x\right)\cdot \sin\left(\frac{\pi}{L}x\right)dx=\nonumber\\
 &=\left[\frac{\sqrt{2}}{L} \frac{\frac{\pi(n-2)}{2L} \sin\left(\frac{\pi(n+2)}{2L}x\right) + \frac{\pi(n+2)}{2L} \sin\left(\frac{\pi(n-2)}{2L}x\right)}{2\cdot\left({\left(\frac{\pi}{L}\right)}^2-{\left(\frac{\pi \ n}{2L}\right)}^2\right)} \right]_{x=0}^{x=L}=\nonumber\\
 &=\frac{\sqrt{2}}{L} \frac{\frac{\pi(n-2)}{2L} \sin\left(\frac{\pi(n+2)}{2}\right) + \frac{\pi(n+2)}{2L} \sin\left(\frac{\pi(n-2)}{2}\right)}{2\cdot\left({\left(\frac{\pi}{L}\right)}^2-{\left(\frac{\pi \ n}{2L}\right)}^2\right)}\nonumber
 \end{align}

n is an even number, therfore, ${\frac{n\pm 2}{2}}=m$, where m is integer.\\

\qquad$\sin(m\pi)=0$\\

therefore,\\

$\qquad C_n=0$,\ \ \  for n - even.\\

\clearpage

\textbf{for n - odd}\\
\Dn
\begin{align}
C_n &=\int\limits_{0}^{L} \frac{1}{\sqrt{L}} \cos\left(\frac{\pi n}{2L}x\right)\cdot\sqrt{\frac{2}{L}} \sin\left(\frac{\pi}{L}x\right)dx=\frac{\sqrt{2}}{L} \int\limits_{0}^{L} \cos\left(\frac{\pi n}{2L}x\right)\cdot \sin\left(\frac{\pi}{L}x\right)dx=\nonumber\\
 &=\left[-\frac{\sqrt{2}}{L} \frac{\frac{\pi(2-n)}{2L} \cos\left(\frac{\pi(2+n)}{2L}x\right) + \frac{\pi(2+n)}{2L} \cos\left(\frac{\pi(2-n)}{2L}x\right)}{2\cdot\left({\left(\frac{\pi}{L}\right)}^2-{\left(\frac{\pi \ n}{2L}\right)}^2\right)} \right]_{x=0}^{x=L}=\nonumber\\
 &=-\frac{\sqrt{2}}{L} \frac{\frac{\pi(2-n)}{2L} \cos\left(\frac{\pi(2+n)}{2}\right) + \frac{\pi(2+n)}{2L} \cos\left(\frac{\pi(2-n)}{2}\right)}{2\cdot\left({\left(\frac{\pi}{L}\right)}^2-{\left(\frac{\pi \ n}{2L}\right)}^2\right)} +
\frac{\sqrt{2}}{L} \frac{\frac{\pi(2-n)}{2L} + \frac{\pi(2+n)}{2L}}{2\cdot\left({\left(\frac{\pi}{L}\right)}^2-{\left(\frac{\pi \ n}{2L}\right)}^2\right)}\nonumber
\end{align}
\Dn

n is an odd number, therefore, ${\frac{n\pm 2}{2}}=m+\frac{1}{2}$, where m is integer.\\

$\qquad\cos\left(\left(m+\frac{1}{2}\right)\pi\right)=0$\\

therefore,\\

$\qquad {C_n} = \frac{{\sqrt 2 }}{L}\frac{{\frac{{\pi \left( {2 - n} \right)}}{{2L}} + \frac{{\pi \left( {2 + n} \right)}}{{2L}}}}{{2\cdot\left( {{{\left( {\frac{\pi }{L}} \right)}^2} - {{\left( {\frac{{\pi \;n}}{{2L}}} \right)}^2}} \right)}} = \frac{{4\sqrt 2 }}{{\pi \left( {4 - {n^2}} \right)}}$\ \ \ for n - odd

\Dn

\textbf{finally:}\\
\Dn
$\qquad P(n) = {\left| {{C_n}} \right|^2} =  \left\{ {\begin{array}{*{20}{l}}
{\ \ \ \ \ 0}&{for\;n - even}\\
{\frac{{32}}{{{\pi ^2}{{(4 - {n^2})}^2}}}}&{for\;n - odd}
\end{array}} \right.$




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