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\heading{E255: finding the S matrix for a delta function in the origin:} 
\auname{Guy Brand}


{\bf The problem:}

\Dn

For the potential $V(x) = u\delta(x)$ write the wave function as
{[}sorry for the typo{]}
\[\Psi=Ae^{ikx}+Be^{-ikx} {[}x<0{]}\]

\[\Psi=Ce^{ikx}+De^{-ikx} {[}x>0{]}\]


\Dn

(1) find the S matrix

(2) represent the S matrix with the transmission probability g and three phases.

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) 
Because the wave amplitude at all points must be defined we require:
 \[\Psi^+(0)=\Psi^-(0)\]
 \[A+B=C+D\]

And because we require integrability at all points we get:

\[\lim_{\varepsilon\rightarrow0}\int\limits _{-\varepsilon}^{\varepsilon}\frac{-1}{2m}\frac{d^{2}}{dx^{2}}\Psi+u\delta(x)\Psi dx=\int\limits _{-\varepsilon}^{\varepsilon}i\frac{d}{dt}\Psi dx=0\]
which, in a simpler writing would be:
\[\frac{1}{2m}{[}\frac{d}{dx}\Psi(\varepsilon)-\frac{d}{dx}\Psi(-\varepsilon){]}=u\Psi(0)\]
which yields:
\[\frac{ik}{2m}C-\frac{ik}{2m}D-(\frac{ik}{2m}A-\frac{ik}{2m}B)=u(A+B)=u(C+D)\]
in terms of $u=\frac{U}{v_E}$ where $v_E=\sqrt{\frac{2E}{m}}=\frac{k}{m}$
we can simplify this expression:
\[C-D-(A-B)=-2ui(A+B)\] 

Let us assume the wave comes from the left: A is then the coming wave
amplitude, 

B is the reflected wave amplitude, C is the transmitted wave amplitude, 

and D does not exist:
we get:
\\I. \[A+B=C\]
and 
II. \[C-A+B=-2iuC\]

now, for doing the same process again for a wave coming from the right, $A=0$ for a wave coming from the right.

we get this set of equations:

III. \[D+C=B\]
and
IV. \[C-D+B=-2iuB\]

in this case we get a matrix which looks like this:

\[
\bm{S} = \left(
\amatrix{
r & t \cr
t & r
}\right)
\]


where r is the relation between B and A or C and D (both are equal
because the system is symetric around the Y axis), and in the same manner $t=\frac{B}{D}=\frac{C}{A}$.

by using the equation $(-A+B=C-2A)$ on II we get: \[2A=2(iu+1)C\]

\[t=\frac{1}{1-iu}\]

we can see that if we subtitude $C$ for $B-D$ on IV we get \[2D=2(iu+1)B\] which gives the same result for $t$. 

if we substitute $C$ for $A+B$ on II we get:

\[2B=-2iu(A+B)\]

from which we get the relation:

\[\frac{B}{A}=r=\frac{-iu}{1+iu}\]
which is the same as $\frac{C}{D}$.

now we can write $r$ and $t$:
\[t=\frac{1}{1+iu}=\frac{1-iu}{1+u^2}\]
\[r=\frac{-iu}{1+iu}=\frac{-u(i+u)}{1+u^2}\]

so

\[
\bm{S} = \left(
\amatrix{
\frac{-iu}{1+iu} & \frac{1}{1+iu} \cr
\frac{1}{1+iu} & \frac{-iu}{1+iu}
}\right)
\]


%%%%%%%%%%%%%%%
\Dn

(2)

the S matrix with the transmission probability g and three phases, would be written as such:

\[
\mathbf{S} = 
\eexp{i\gamma}
\left(\amatrix{i\sqrt{1-g}\eexp{i\alpha} & \sqrt{g}\eexp{-i\phi} \cr
\sqrt{g}\eexp{i\phi} &  i\sqrt{1-g}\eexp{-i\alpha}}\right)
\]

where

\[g=|t|^2=\frac{1+u^2}{(1+u^2)^2}=\frac{1}{1+u^2}\]

we get the phase of i*t from: 
\[\arctan{\frac{\Im(it)}{\Re(it)}}=\arctan\frac{1}{u}\]
and the same goes for r: \[\arctan{\frac{\Im(r)}{\Re(r)}}=\arctan\frac{1}{u}\]
in both cases we get the phase $\arctan{\frac{1}{u}}$, so both $\sqrt{1-g}$ and $i\sqrt{g}$ have the same phase: \[\gamma=\arctan{\frac{1}{u}}\]

because t and r for both waves coming from the left and from the right are the same (the problem is symetric), $\phi=\alpha=0$.
if this problem was not only one dimensional, we could have noticed the effect of the vector potential and get $\phi$.
and if the potential was a-symetric we would have gotten different r phases, and use $\alpha$ for it.


\ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\ 

{\bf Summarizing formula the way I like [D.C.]:} \\

The $\bm{S}$~matrix for a delta function 
scatterer $V(r) = u \delta(r)$ is 
%
\begin{eqnarray} 
\bm{S} = \left(
\amatrix{
r & t \cr
t & r
}\right)
 \,\,=\,\,
\eexp{i\gamma} \left(
\amatrix{
-i\sqrt{1{-}g}\eexp{i\alpha} & \sqrt{g}\eexp{-i\phi}   \cr
\sqrt{g} \eexp{i\phi}  & -i\sqrt{1{-}g}\eexp{-i\alpha}
}\right)
\end{eqnarray}
%
with
%
\begin{eqnarray}
v_E &\equiv& (2E/\mass)^{1/2} \\
u_E &\equiv& \frac{u}{\hbar v_E} \\
t &=& \frac{1}{1+iu_E} \\
r &=& t-1 \\
\gamma &=& \arg(t) \ \ = \ \ -\arctan(u_E) \\
g &=& \frac{1}{1+u_E^2} \ = \ (\cos(\gamma))^2 \\
\alpha &=& 0 \\
\phi &=& 0
\end{eqnarray}
%
Note that we use the common ad-hoc convention 
of writing the channel functions as 
${\psi(r)=A\exp(-ikr)+B\exp(-ikr)}$  
where $r=|x|$. In the alternate convention,  
which is used for s-scattering, $B \mapsto -B$, 
so as to have $\bm{S}=\bm{1}$ in the 
limit of zero coupling ($u=\infty$).  
 



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