Properties of the Linear Schrodinger Equation

Nonlinear PDE's Course
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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 \ dx = 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}dx=1. }[/math]

Therefore

[math]\displaystyle{ 2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}dx=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{ \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{ \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]