View source for Properties of the Linear Schrodinger Equation

{{nonlinear waves course
 | chapter title = Properties of the Linear Schrodinger Equation
 | next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
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}}

The linear Schrodinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>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>u\rightarrow0</math>
as <math>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>\delta</math> function potential==

We consider here the case when <math>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
<center><math>
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></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left(  x\right)  =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\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
<center><math>\begin{align}
\int_{0^{-}}^{0^{+}} \left(\partial_x^2 w +u_0\delta(x) w + \lambda w \right) \ \mathrm{d}x = 0.
\end{align}
</math></center>

This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left(  w\left(  x\right)  \right)  ^{2}\mathrm{d}x=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left(  ae^{-u_{0}x/2}\right)  ^{2}\mathrm{d}x=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left(  x\right)  =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left(  x\right)  =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
t\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left(  0^{+}\right)  -\partial_{x}w\left(  0^{-}\right)
+u_{0}w\left(  0\right)  =0.</math> This gives us
<center><math>\begin{matrix}
1+r  &  =t\\
-ikt+ik-ikr  &  =-tu_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r  &  =\frac{u_{0}}{2ik-u_{0}}\\
t  &  =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example 2: Hat Function Potential==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
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></center>
where <math>b>0.</math>

===Case when <math>\lambda<0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left(  x\right)  =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
a_{2}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varsigma</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equations, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
w\left(  x\right)  =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x, & -\varsigma< x <\varsigma,\\
a_{1}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
If we impose the condition that the function and its derivative are continuous at
<math>x=\pm\varsigma</math> we obtain the following equation
<center><math>
\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></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right)  =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\kappa\varsigma=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\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
<center><math>
w\left(  x\right)  =\left\{
\begin{matrix}
a_{1}e^{kx}, & x <-\varsigma,\\
b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
-a_{1}e^{-kx} & x > \varsigma,
\end{matrix}
\right.
</math></center>
and again imposing the condition that the solution and its derivative is continuous
at <math>x=\pm\varsigma</math> gives
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\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></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
\end{matrix}
\right)  =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k\varsigma}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\varsigma\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left(  x\right)  =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x <-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma< x <\varsigma\\
t\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\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}\\
t
\end{matrix}
\right)  =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

== Lecture Videos == 

=== Part 1 ===

{{#ev:youtube|anAThvCcpNw}}

=== Part 2 ===

{{#ev:youtube|SDPIx42VjLQ}}

=== Part 3 ===

{{#ev:youtube|OUmjeLZWr3M}}

=== Part 4 ===

{{#ev:youtube|hIfcO3a8_XU}}

=== Part 5 ===

{{#ev:youtube|z13lKSTficA}}

=== Part 6 ===

{{#ev:youtube|2XlQpEscxE4}}

=== Part 7 ===

{{#ev:youtube|iMMQ4NUdXNc}}

=== Part 8 ===

{{#ev:youtube|0F_dINNxMlw}}