The linear Schrodinger equation
[math]\displaystyle{
\partial_{x}^{2}w+uw=-\lambda w
}[/math]
has two kinds of solutions for [math]\displaystyle{ u\rightarrow0 }[/math] as [math]\displaystyle{ x\rightarrow\pm\infty. }[/math] The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided [math]\displaystyle{ u\rightarrow0 }[/math]
as [math]\displaystyle{ x\pm\infty }[/math] sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples
Example 1: [math]\displaystyle{ \delta }[/math] function potential
We consider here the case when [math]\displaystyle{ u\left( x,0\right) = u_0 \delta\left( x\right)
. }[/math] Note that this function can be thought of as the limit as of the potential
[math]\displaystyle{
u\left( x\right) =\left\{
\begin{matrix}
0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
}[/math]
In this case we need to solve
[math]\displaystyle{
\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
}[/math]
We consider the case of [math]\displaystyle{ \lambda\lt 0 }[/math] and [math]\displaystyle{ \lambda\gt 0 }[/math] separately. For the first
case we write [math]\displaystyle{ \lambda=-k^{2} }[/math] and we obtain
[math]\displaystyle{
w\left( x\right) =\left\{
\begin{matrix}
ae^{kx}, & x\lt 0\\
be^{-kx}, & x\gt 0
\end{matrix}
\right.
}[/math]
We have two conditions at [math]\displaystyle{ x=0, }[/math] [math]\displaystyle{ w }[/math] must be continuous at [math]\displaystyle{ 0 }[/math] and
[math]\displaystyle{ \partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_0 w\left( 0\right) =0. }[/math] This final condition is obtained by integrating `across' zero as follows
[math]\displaystyle{ \begin{align}
\int_{0^{-}}^{0^{+}} \partial_x^2 w +\delta(x) w + \lambda w \ \mathrm{d}x = 0.
\end{align}
}[/math]
This gives the condition that [math]\displaystyle{ a=b }[/math] and
[math]\displaystyle{ k=u_{0}/2. }[/math] We need to normalise the eigenfunctions so that
[math]\displaystyle{
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}\mathrm{d}x=1.
}[/math]
Therefore
[math]\displaystyle{
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1
}[/math]
which means that [math]\displaystyle{ a=\sqrt{u_{0}/2}. }[/math] Therefore, there is only one discrete
spectral point which we denote by [math]\displaystyle{ k_{1}=u_{0}/2 }[/math]
[math]\displaystyle{
w_{1}\left( x\right) =\left\{
\begin{matrix}
\sqrt{k_{1}}e^{k_{1}x}, & x\lt 0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x\gt 0
\end{matrix}
\right.
}[/math]
The continuous eigenfunctions correspond to [math]\displaystyle{ \lambda=k^{2}\gt 0 }[/math] are of the form
[math]\displaystyle{
w\left( x\right) =\left\{
\begin{matrix}
\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x\lt 0\\
a\mathrm{e}^{-\mathrm{i}kx}, & x\gt 0
\end{matrix}
\right.
}[/math]
Again we have the conditions that [math]\displaystyle{ w }[/math] must be continuous at [math]\displaystyle{ 0 }[/math] and
[math]\displaystyle{ \partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0. }[/math] This gives us
[math]\displaystyle{ \begin{matrix}
1+r & =a\\
-ika+ik-ikr & =-au_{0}
\end{matrix} }[/math]
which has solution
[math]\displaystyle{ \begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
a & =\frac{2ik}{2ik-u_{0}}
\end{matrix} }[/math]
Example 2: Hat Function Potential
The properties of the eigenfunction is perhaps seem most easily through the
following example
[math]\displaystyle{
u\left( x\right) =\left\{
\begin{matrix}
0 & x\notin\left[ -\varsigma,\varsigma\right] \\
b & x\in\left[ -\varsigma,\varsigma\right]
\end{matrix}
\right.
}[/math]
where [math]\displaystyle{ b\gt 0. }[/math]
Case when [math]\displaystyle{ \lambda\lt 0 }[/math]
If we solve this equation for the case when [math]\displaystyle{ \lambda\lt 0, }[/math] [math]\displaystyle{ \lambda=-k^{2} }[/math] we
get
[math]\displaystyle{
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x\lt -\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma\lt x\lt \varsigma\\
a_{2}e^{-kx} & x\gt \varsigma
\end{matrix}
\right.
}[/math]
where [math]\displaystyle{ \kappa=\sqrt{b-k^{2}} }[/math] which means that [math]\displaystyle{ 0\leq k\leq\sqrt{b} }[/math] (there is
no solution for [math]\displaystyle{ k\gt \sqrt{b}). }[/math] We then match [math]\displaystyle{ w }[/math] and its derivative at
[math]\displaystyle{ x=\pm\varsigma }[/math] to solve for [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math]. This leads to two system of
equation, one for the even ([math]\displaystyle{ a_{1}=a_{2} }[/math] and [math]\displaystyle{ b_{2}=0 }[/math] ) and one for the odd
solutions ([math]\displaystyle{ a_{1}=-a_{2} }[/math] and [math]\displaystyle{ b_{1}=0) }[/math]. The solution for the even solutions
is
[math]\displaystyle{
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x\lt -\varsigma\\
b_{1}\cos\kappa x & -\varsigma\lt x\lt \varsigma\\
a_{1}e^{-kx} & x\gt \varsigma
\end{matrix}
\right.
}[/math]
[math]\displaystyle{
\left(
\begin{matrix}
e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}
a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}
0\\
0
\end{matrix}
\right)
}[/math]
This has non trivial solutions when
[math]\displaystyle{
\det\left(
\begin{matrix}
e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) =0
}[/math]
which gives us the equation
[math]\displaystyle{
-\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
e^{-k\varsigma}=0
}[/math]
or
[math]\displaystyle{
\tan\kappa\varsigma=\frac{k}{\kappa}
}[/math]
We know that [math]\displaystyle{ 0\lt \kappa\lt \sqrt{b} }[/math] and if we plot this we see that we obtain a
finite number of solutions.
The solution for the odd solutions is
[math]\displaystyle{
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x\lt -\varsigma\\
+b_{2}\sin\kappa x & -\varsigma\lt x\lt \varsigma\\
-a_{1}e^{-kx} & x\gt \varsigma
\end{matrix}
\right.
}[/math]
[math]\displaystyle{
\left(
\begin{matrix}
e^{-k\varsigma} & -\sin\kappa\varsigma\\
ke^{-k\varsigma} & \cos\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}
a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}
0\\
0
\end{matrix}
\right)
}[/math]
This can non trivial solutions when
[math]\displaystyle{
\det\left(
\begin{matrix}
e^{-k\varsigma} & -\sin\kappa\varsigma\\
ke^{-k\varsigma} & \kappa\cos\kappa\varsigma
\end{matrix}
\right) =0
}[/math]
which gives us the equation
[math]\displaystyle{
\kappa e^{-k\varsigma}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k\varsigma}=0
}[/math]
or
[math]\displaystyle{
\tan\varsigma\kappa=-\frac{\kappa}{k}
}[/math]
Case when [math]\displaystyle{ \lambda\gt 0 }[/math]
When [math]\displaystyle{ \lambda\gt 0 }[/math] we write [math]\displaystyle{ \lambda=k^{2} }[/math] and we obtain solution
[math]\displaystyle{
w\left( x\right) =\left\{
\begin{matrix}
\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x\lt -\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma\lt x\lt \varsigma\\
a\mathrm{e}^{-\mathrm{i}kx} & x\gt \varsigma
\end{matrix}
\right.
}[/math]
where [math]\displaystyle{ \kappa=\sqrt{b+k^{2}}. }[/math] Matching [math]\displaystyle{ w }[/math] and its derivaties at [math]\displaystyle{ x=\pm1 }[/math] we
obtain
[math]\displaystyle{
\left(
\begin{matrix}
-\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa
\varsigma & 0\\
0 & \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\
0 & -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma &
ik\mathrm{e}^{-\mathrm{i}k\varsigma}
\end{matrix}
\right) \left(
\begin{matrix}
r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}
\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
}[/math]
