# Difference between revisions of "Properties of the Linear Schrodinger Equation"

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Nonlinear PDE's Course
Current Topic Properties of the Linear Schrodinger Equation
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The linear Schrodinger equation

$\displaystyle{ \partial_{x}^{2}w+uw=-\lambda w }$

has two kinds of solutions for $\displaystyle{ u\rightarrow0 }$ as $\displaystyle{ x\rightarrow\pm\infty. }$ 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 $\displaystyle{ u\rightarrow0 }$ as $\displaystyle{ x\pm\infty }$ sufficiently rapidly) and a continum of solutions for the incident waves. This is easiest seen through the following examples

## Example 1: $\displaystyle{ \delta }$ function potential

We consider here the case when $\displaystyle{ u\left( x,0\right) = u_0 \delta\left( x\right) . }$ Note that this function can be thought of as the limit as of the potential

$\displaystyle{ u\left( x\right) =\left\{ \begin{matrix} 0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\ \frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right] \end{matrix} \right. }$

In this case we need to solve

$\displaystyle{ \partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w }$

We consider the case of $\displaystyle{ \lambda\lt 0 }$ and $\displaystyle{ \lambda\gt 0 }$ separately. For the first case we write $\displaystyle{ \lambda=-k^{2} }$ and we obtain

$\displaystyle{ w\left( x\right) =\left\{ \begin{matrix} ae^{kx}, & x\lt 0\\ be^{-kx}, & x\gt 0 \end{matrix} \right. }$

We have two conditions at $\displaystyle{ x=0, }$ $\displaystyle{ w }$ must be continuous at $\displaystyle{ 0 }$ and $\displaystyle{ \partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right) +u_0 w\left( 0\right) =0. }$ This final condition is obtained by integrating `across' zero as follows

\displaystyle{ \begin{align} \int_{0^{-}}^{0^{+}} \left(\partial_x^2 w +u_0\delta(x) w + \lambda w \right) \ \mathrm{d}x = 0. \end{align} }

This gives the condition that $\displaystyle{ a=b }$ and $\displaystyle{ k=u_{0}/2. }$ We need to normalise the eigenfunctions so that

$\displaystyle{ \int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}\mathrm{d}x=1. }$

Therefore

$\displaystyle{ 2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1 }$

which means that $\displaystyle{ a=\sqrt{u_{0}/2}. }$ Therefore, there is only one discrete spectral point which we denote by $\displaystyle{ k_{1}=u_{0}/2 }$

$\displaystyle{ w_{1}\left( x\right) =\left\{ \begin{matrix} \sqrt{k_{1}}e^{k_{1}x}, & x\lt 0\\ \sqrt{k_{1}}e^{-k_{1}x}, & x\gt 0 \end{matrix} \right. }$

The continuous eigenfunctions correspond to $\displaystyle{ \lambda=k^{2}\gt 0 }$ are of the form

$\displaystyle{ w\left( x\right) =\left\{ \begin{matrix} \mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x\lt 0\\ t\mathrm{e}^{-\mathrm{i}kx}, & x\gt 0 \end{matrix} \right. }$

Again we have the conditions that $\displaystyle{ w }$ must be continuous at $\displaystyle{ 0 }$ and $\displaystyle{ \partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right) +u_{0}w\left( 0\right) =0. }$ This gives us

$\displaystyle{ \begin{matrix} 1+r & =t\\ -ikt+ik-ikr & =-tu_{0} \end{matrix} }$

which has solution

$\displaystyle{ \begin{matrix} r & =\frac{u_{0}}{2ik-u_{0}}\\ t & =\frac{2ik}{2ik-u_{0}} \end{matrix} }$

## Example 2: Hat Function Potential

The properties of the eigenfunction is perhaps seem most easily through the following example

$\displaystyle{ u\left( x\right) =\left\{ \begin{matrix} 0 & x\notin\left[ -\varsigma,\varsigma\right] \\ b & x\in\left[ -\varsigma,\varsigma\right] \end{matrix} \right. }$

where $\displaystyle{ b\gt 0. }$

### Case when $\displaystyle{ \lambda\lt 0 }$

If we solve this equation for the case when $\displaystyle{ \lambda\lt 0, }$ $\displaystyle{ \lambda=-k^{2} }$ we get

$\displaystyle{ w\left( x\right) =\left\{ \begin{matrix} a_{1}e^{kx}, & x\lt -\varsigma,\\ b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\varsigma\lt x \lt \varsigma,\\ a_{2}e^{-kx}, & x\gt \varsigma, \end{matrix} \right. }$

where $\displaystyle{ \kappa=\sqrt{b-k^{2}} }$ which means that $\displaystyle{ 0\leq k\leq\sqrt{b} }$ (there is no solution for $\displaystyle{ k\gt \sqrt{b}). }$ We then match $\displaystyle{ w }$ and its derivative at $\displaystyle{ x=\pm\varsigma }$ to solve for $\displaystyle{ a }$ and $\displaystyle{ b }$. This leads to two system of equations, one for the even ($\displaystyle{ a_{1}=a_{2} }$ and $\displaystyle{ b_{2}=0 }$ ) and one for the odd solutions ($\displaystyle{ a_{1}=-a_{2} }$ and $\displaystyle{ b_{1}=0) }$. The solution for the even solutions is

$\displaystyle{ w\left( x\right) =\left\{ \begin{matrix} a_{1}e^{kx}, & x\lt -\varsigma,\\ b_{1}\cos\kappa x, & -\varsigma\lt x \lt \varsigma,\\ a_{1}e^{-kx}, & x\gt \varsigma, \end{matrix} \right. }$

If we impose the condition that the function and its derivative are continuous at $\displaystyle{ x=\pm\varsigma }$ we obtain the following equation

$\displaystyle{ \left( \begin{matrix} e^{-k\varsigma} & -\cos\kappa\varsigma\\ ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma \end{matrix} \right) \left( \begin{matrix} a_{1}\\ b_{1} \end{matrix} \right) =\left( \begin{matrix} 0\\ 0 \end{matrix} \right) }$

This has non trivial solutions when

$\displaystyle{ \det\left( \begin{matrix} e^{-k\varsigma} & -\cos\kappa\varsigma\\ ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma \end{matrix} \right) =0 }$

which gives us the equation

$\displaystyle{ -\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma e^{-k\varsigma}=0 }$

or

$\displaystyle{ \tan\kappa\varsigma=\frac{k}{\kappa} }$

We know that $\displaystyle{ 0\lt \kappa\lt \sqrt{b} }$ and if we plot this we see that we obtain a finite number of solutions.

The solution for the odd solutions is

$\displaystyle{ w\left( x\right) =\left\{ \begin{matrix} a_{1}e^{kx}, & x \lt -\varsigma,\\ b_{2}\sin\kappa x, & -\varsigma\lt x \lt \varsigma,\\ -a_{1}e^{-kx} & x \gt \varsigma, \end{matrix} \right. }$

and again imposing the condition that the solution and its derivative is continuous at $\displaystyle{ x=\pm\varsigma }$ gives

$\displaystyle{ \left( \begin{matrix} e^{-k\varsigma} & \sin\kappa\varsigma\\ ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma \end{matrix} \right) \left( \begin{matrix} a_{1}\\ b_{1} \end{matrix} \right) =\left( \begin{matrix} 0\\ 0 \end{matrix} \right) }$

This can non trivial solutions when

$\displaystyle{ \det\left( \begin{matrix} e^{-k\varsigma} & \sin\kappa\varsigma\\ ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma \end{matrix} \right) =0 }$

which gives us the equation

$\displaystyle{ \kappa e^{-k\varsigma}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k\varsigma}=0 }$

or

$\displaystyle{ \tan\varsigma\kappa=-\frac{\kappa}{k} }$

### Case when $\displaystyle{ \lambda\gt 0 }$

When $\displaystyle{ \lambda\gt 0 }$ we write $\displaystyle{ \lambda=k^{2} }$ and we obtain solution

$\displaystyle{ w\left( x\right) =\left\{ \begin{matrix} \mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x \lt -\varsigma\\ b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma\lt x \lt \varsigma\\ t\mathrm{e}^{-\mathrm{i}kx} & x\gt \varsigma \end{matrix} \right. }$

where $\displaystyle{ \kappa=\sqrt{b+k^{2}}. }$ Matching $\displaystyle{ w }$ and its derivaties at $\displaystyle{ x=\pm1 }$ we obtain

$\displaystyle{ \left( \begin{matrix} -\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\ ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa \varsigma & 0\\ 0 & \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\ 0 & -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma & ik\mathrm{e}^{-\mathrm{i}k\varsigma} \end{matrix} \right) \left( \begin{matrix} r\\ b_{1}\\ b_{2}\\ t \end{matrix} \right) =\left( \begin{matrix} \mathrm{e}^{\mathrm{i}k}\\ ik\mathrm{e}^{-\mathrm{i}k}\\ 0\\ 0 \end{matrix} \right) }$