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\heading{E212: Particle in an potential well, with scatterer} 
\auname{Yaakov Kleeorin}
{\bf The problem:}

\Dn
A particle is placed inside an infinite square well [0,2d] in the center there is a delta scatterer ${V(x)=-u\delta(x-d)}$:

\Dn

(1) What is the minimum u for which the energy states are negative?\\
(2) How does the wave function look when we decrease u to a point where the ground state E=0?\\
\Dn\Dn

{\bf The solution:}
\Dn

The time independent schrodinger equation is as follows:
\[[\frac{\hat{p}^2}{2m}+V(x)]\psi{(x)}=E\psi{(x)}\]
inserting ${\hat{p}=-i\frac{d}{d{x}}}$ and the potential we get:
\[[-\frac{1}{2m}{\nabla}^2-u\delta(x-d)]\psi{(x)}=E\psi{(x)}\]
with the following boundry conditions for the infinite walls and delta stitching:
\[[1]\psi|_{x=0}=\psi|_{x=2d}=0\]
\[[2]\psi|_{x=d-}=\psi|_{x=d+}\]
by means of integrating the Schrodinger equation over infinitysimal ${\epsilon}$ around x=d we get the a second stitching condition:
\[[3]\frac{d\psi}{dx}|_{x=d+}-\frac{d\psi}{dx}|_{x=d-}=2mu\psi|_{x=d}\]
using the equation and the boundry conditions we can introduce a wave function as follows (already satisfying condition [1]):
\[[0,d]\ \ \  \psi(x)=\left\{ \begin{array}{lcl}	
	A\sin(kx) & \mbox{for}	& E>0 \\
 	A\sinh(ax) & \mbox{for} & E<0 \\
	\end{array}\right.\]\\
\[[d,2d]\ \ \  \psi(x)=\left\{ \begin{array}{lcl}	
	A\sin(k(2d-x)) & \mbox{for}	& E>0 \\
 	A\sinh(a(2d-x)) & \mbox{for} & E<0 \\
	\end{array}\right.\]
from the scrodinger equation we get:
\[k=\sqrt{2mE}\ for \ E>0\]
\[a=\sqrt{-2mE}\ for \ E<0\]
using the stitching conditions:
\[\left\{ \begin{array}{lcl}	
	Ak\cos(kd)+Ak\cos(kd)=2muA\sin(kd) & \mbox{for}	& E>0 \\
 	Aa\cosh(ad)+Aa\cosh(ad)=2muA\sinh(ad) & \mbox{for} & E<0 \\
	\end{array}\right.\]\\
or finally:
\[\left\{ \begin{array}{lcl}	
	k\cot(kd)=mu & \mbox{for}	& E>0 \\
 	a\coth(ad)=mu & \mbox{for} & E<0 \\
	\end{array}\right.\] \\
\\

(1)
we use the second relation to find the minimum for which E is still negative which means ${E \rightarrow 0-}$. we remember:
\[E=-\frac{a^2}{2m}\]
if we take E to zero we get:
\[\sqrt{2mE}\coth(\sqrt{2mE}d)=mu\]
reminded that:
\[\coth(cx)=\frac{1}{\tanh(cx)}\rightarrow \frac{1}{cx}\]
we get:
\[\sqrt{2mE}\frac{1}{\sqrt{2mE}d}=mu\Rightarrow u_0=\frac{1}{md}\]
So energy levels will remain negative for ${u>\frac{1}{md}}$\\
\\
\\
(2)
Now let us take the limit [a goes to zero] for the wave function:


\[E<0\ \ \  \psi(x)=\left\{ \begin{array}{lcl}	
	lim_{a\rightarrow{0}}A\sinh(ax) & \mbox{for}	& [0,d] \\
 	lim_{a\rightarrow{0}}A\sinh(a(2d-x)) & \mbox{for} & [d,2d] \\
	\end{array}\right.\]\\
reminded that:
\[\sinh(cx)\rightarrow cx\]
we get:
\[E<0\ \ \  \psi(x)=\left\{ \begin{array}{lcl}	
	\tilde{A}x & \mbox{for}	& [0,d] \\
 	\tilde{A}(2d-x) & \mbox{for} & [d,2d] \\
	\end{array}\right.\]\\
\\
Another way of getting the wave function at E=0 is by inserting E into the schrodinger equation:
\[[-\frac{1}{2m}{\nabla}^2-u\delta(x-d)]\psi{(x)}=0\]
This gives rise to the form
\[\psi(x)=\left\{ \begin{array}{lcl}	
	A_1x+B_1 & \mbox{for}	& [0,d] \\
 	A_2x+B_2 & \mbox{for} & [d,2d] \\
	\end{array}\right.\]
Using the same boundery conditions and stitching 
\[[1]\psi|_{x=0}=\psi|_{x=2d}=0\]
\[[2]\psi|_{x=d-}=\psi|_{x=d+}\]
we get:
\[\psi(x)=\left\{ \begin{array}{lcl}	
	Ax & \mbox{for}	& [0,d] \\
 	A(2d-x) & \mbox{for} & [d,2d] \\
	\end{array}\right.\]\\
\\
And the delta stitching is automatically satisfied because we specified the u for E=0 (u=1/md):

\[[3]\frac{d\psi}{dx}|_{x=d+}-\frac{d\psi}{dx}|_{x=d-}=2mu\psi|_{x=d}\]
\[A+A=2Aumd=2A\]

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