Properties of the Linear Schrodinger Equation: Difference between revisions

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Revision as of 08:44, 28 September 2025

The linear Schrödinger 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\rightarrow\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]

Case when [math]\displaystyle{ \lambda\lt 0 }[/math]

We consider the cases 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 (as [math]\displaystyle{ w\rightarrow0 }[/math] as [math]\displaystyle{ x\rightarrow\pm\infty }[/math])

[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^{+}} \left(\partial_x^2 w +u_0\delta(x) w + \lambda w \right) \ \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]

[math]\displaystyle{ 2\left( a^2\dfrac{e^{-u_{0}x}}{-u_{0}}\right)\Bigg|_{0}^{\infty}=1 }[/math]

[math]\displaystyle{ \dfrac{2a^2}{u_{0}}=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]

Case when [math]\displaystyle{ \lambda\gt 0 }[/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\\ t\mathrm{e}^{-\mathrm{i}kx}, & x\gt 0 \end{matrix} \right. }[/math]

where [math]\displaystyle{ \mathrm{e}^{-\mathrm{i}kx} }[/math] is the incident wave, [math]\displaystyle{ r\mathrm{e}^{\mathrm{i}kx} }[/math] is the reflected wave, and [math]\displaystyle{ t\mathrm{e}^{-\mathrm{i}kx} }[/math] is the transmitted wave.

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 & =t\\ -ikt+ik-ikr & =-tu_{0} \end{matrix} }[/math]

which has solution

[math]\displaystyle{ \begin{matrix} r & =\frac{u_{0}}{2ik-u_{0}}\\ t & =\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[ -\zeta,\zeta\right] \\ b & x\in\left[ -\zeta,\zeta\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 -\zeta,\\ b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\zeta\lt x \lt \zeta,\\ a_{2}e^{-kx}, & x\gt \zeta, \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\zeta }[/math] to solve for [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math]. This leads to two system of equations, 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 -\zeta,\\ b_{1}\cos\kappa x, & -\zeta\lt x \lt \zeta,\\ a_{1}e^{-kx}, & x\gt \zeta, \end{matrix} \right. }[/math]

If we impose the condition that the function and its derivative are continuous at [math]\displaystyle{ x=\pm\zeta }[/math] we obtain the following equation

[math]\displaystyle{ \left( \begin{matrix} e^{-k\zeta} & -\cos\kappa\zeta\\ ke^{-k\zeta} & -\kappa\sin\kappa\zeta \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\zeta} & -\cos\kappa\zeta\\ ke^{-k\zeta} & -\kappa\sin\kappa\zeta \end{matrix} \right) =0 }[/math]

which gives us the equation

[math]\displaystyle{ -\kappa e^{-k\zeta}\sin\kappa\zeta+k\cos\kappa\zeta e^{-k\zeta}=0 }[/math]

or

[math]\displaystyle{ \tan\kappa\zeta=\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 -\zeta,\\ b_{2}\sin\kappa x, & -\zeta\lt x \lt \zeta,\\ -a_{1}e^{-kx} & x \gt \zeta, \end{matrix} \right. }[/math]

and again imposing the condition that the solution and its derivative is continuous at [math]\displaystyle{ x=\pm\zeta }[/math] gives

[math]\displaystyle{ \left( \begin{matrix} e^{-k\zeta} & \sin\kappa\zeta\\ ke^{-k\zeta} & -\kappa\cos\kappa\zeta \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\zeta} & \sin\kappa\zeta\\ ke^{-k\zeta} & -\kappa\cos\kappa\zeta \end{matrix} \right) =0 }[/math]

which gives us the equation

[math]\displaystyle{ \kappa e^{-k\zeta}a\cos\kappa\zeta+k\sin\kappa\zeta e^{-k\zeta}=0 }[/math]

or

[math]\displaystyle{ \tan\zeta\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 -\zeta\\ b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\zeta\lt x \lt \zeta\\ t\mathrm{e}^{-\mathrm{i}kx} & x\gt \zeta \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=\pm\zeta }[/math] we obtain

[math]\displaystyle{ \left( \begin{matrix} -\mathrm{e}^{-\mathrm{i}k\zeta} & \cos\kappa\zeta & -\sin\kappa\zeta & 0\\ ik\mathrm{e}^{-\mathrm{i}k\zeta} & \kappa\sin\kappa\zeta & \kappa\cos\kappa \zeta & 0\\ 0 & \cos\kappa\zeta & \sin\kappa\zeta & -\mathrm{e}^{-\mathrm{i}k\zeta}\\ 0 & -\kappa\sin\kappa\zeta & \kappa\cos\kappa\zeta & ik\mathrm{e}^{-\mathrm{i}k\zeta} \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]

Lecture Videos

Part 1

Part 2

Part 3

Part 4

Part 5

Part 6

Part 7

Part 8

@@ Line 114: / Line 114: @@
 \begin{matrix}
-& x\notin\left[  -\varsigma,\varsigma\right] \\
+& x\notin\left[  -\zeta,\zeta\right] \\
-b & x\in\left[  -\varsigma,\varsigma\right]
+b & x\in\left[  -\zeta,\zeta\right]
 \end{matrix}
 \right.
@@ Line 128: / Line 128: @@
 w\left(  x\right)  =\left\{
 \begin{matrix}
-a_{1}e^{kx}, & x<-\varsigma,\\
+a_{1}e^{kx}, & x<-\zeta,\\
-b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
+b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\zeta< x <\zeta,\\
-a_{2}e^{-kx}, & x>\varsigma,
+a_{2}e^{-kx}, & x>\zeta,
 \end{matrix}
 \right.
@@ Line 136: / Line 136: @@
 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
+<math>x=\pm\zeta</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
@@ Line 143: / Line 143: @@
 w\left(  x\right)  =\left\{
 \begin{matrix}
-a_{1}e^{kx}, & x<-\varsigma,\\
+a_{1}e^{kx}, & x<-\zeta,\\
-b_{1}\cos\kappa x, & -\varsigma< x <\varsigma,\\
+b_{1}\cos\kappa x, & -\zeta< x <\zeta,\\
-a_{1}e^{-kx}, & x>\varsigma,
+a_{1}e^{-kx}, & x>\zeta,
 \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
+<math>x=\pm\zeta</math> we obtain the following equation
 <center><math>
 \left(
 \begin{matrix}
-e^{-k\varsigma} & -\cos\kappa\varsigma\\
+e^{-k\zeta} & -\cos\kappa\zeta\\
-ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
+ke^{-k\zeta} & -\kappa\sin\kappa\zeta
 \end{matrix}
 \right)  \left(
@@ Line 177: / Line 177: @@
 \begin{matrix}
-e^{-k\varsigma} & -\cos\kappa\varsigma\\
+e^{-k\zeta} & -\cos\kappa\zeta\\
-ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
+ke^{-k\zeta} & -\kappa\sin\kappa\zeta
 \end{matrix}
 \right)  =0
@@ Line 184: / Line 184: @@
 which gives us the equation
 <center><math>
--\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
+-\kappa e^{-k\zeta}\sin\kappa\zeta+k\cos\kappa\zeta
-e^{-k\varsigma}=0
+e^{-k\zeta}=0
 </math></center>
 or
 <center><math>
-\tan\kappa\varsigma=\frac{k}{\kappa}
+\tan\kappa\zeta=\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
@@ Line 198: / Line 198: @@
 w\left(  x\right)  =\left\{
 \begin{matrix}
-a_{1}e^{kx}, & x <-\varsigma,\\
+a_{1}e^{kx}, & x <-\zeta,\\
-b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
+b_{2}\sin\kappa x, & -\zeta< x <\zeta,\\
--a_{1}e^{-kx} & x > \varsigma,
+-a_{1}e^{-kx} & x > \zeta,
 \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
+at <math>x=\pm\zeta</math> gives
 <center><math>
 \left(
 \begin{matrix}
-e^{-k\varsigma} & \sin\kappa\varsigma\\
+e^{-k\zeta} & \sin\kappa\zeta\\
-ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
+ke^{-k\zeta} & -\kappa\cos\kappa\zeta
 \end{matrix}
 \right)  \left(
@@ Line 232: / Line 232: @@
 \begin{matrix}
-e^{-k\varsigma} & \sin\kappa\varsigma\\
+e^{-k\zeta} & \sin\kappa\zeta\\
-ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
+ke^{-k\zeta} & -\kappa\cos\kappa\zeta
 \end{matrix}
 \right)  =0
@@ Line 239: / Line 239: @@
 which gives us the equation
 <center><math>
-\kappa e^{-k\varsigma}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k\varsigma}=0
+\kappa e^{-k\zeta}a\cos\kappa\zeta+k\sin\kappa\zeta e^{-k\zeta}=0
 </math></center>
 or
 <center><math>
-\tan\varsigma\kappa=-\frac{\kappa}{k}
+\tan\zeta\kappa=-\frac{\kappa}{k}
 </math></center>
@@ Line 253: / Line 253: @@
 \begin{matrix}
-\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x <-\varsigma\\
+\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x <-\zeta\\
-b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma< x <\varsigma\\
+b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\zeta< x <\zeta\\
-t\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
+t\mathrm{e}^{-\mathrm{i}kx} & x>\zeta
 \end{matrix}
 \right.
 </math></center>
-where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm\varsigma</math> we
+where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm\zeta</math> we
 obtain
 <center><math>
@@ Line 265: / Line 265: @@
 \begin{matrix}
--\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\
+-\mathrm{e}^{-\mathrm{i}k\zeta} & \cos\kappa\zeta & -\sin\kappa\zeta & 0\\
-ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa
+ik\mathrm{e}^{-\mathrm{i}k\zeta} & \kappa\sin\kappa\zeta & \kappa\cos\kappa
-\varsigma & 0\\
+\zeta & 0\\
-& \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\
+& \cos\kappa\zeta & \sin\kappa\zeta & -\mathrm{e}^{-\mathrm{i}k\zeta}\\
-& -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma &
+& -\kappa\sin\kappa\zeta & \kappa\cos\kappa\zeta &
-ik\mathrm{e}^{-\mathrm{i}k\varsigma}
+ik\mathrm{e}^{-\mathrm{i}k\zeta}
 \end{matrix}
 \right)  \left(

Nonlinear PDE's Course
Current Topic	Properties of the Linear Schrodinger Equation
Next Topic	Connection betwen KdV and the Schrodinger Equation
Previous Topic	Introduction to the Inverse Scattering Transform

Properties of the Linear Schrodinger Equation: Difference between revisions

Revision as of 08:44, 28 September 2025

Contents

Example 1: [math]\displaystyle{ \delta }[/math] function potential

Case when [math]\displaystyle{ \lambda\lt 0 }[/math]

Case when [math]\displaystyle{ \lambda\gt 0 }[/math]

Example 2: Hat Function Potential

Case when [math]\displaystyle{ \lambda\lt 0 }[/math]

Case when [math]\displaystyle{ \lambda\gt 0 }[/math]

Lecture Videos

Part 1

Part 2

Part 3

Part 4

Part 5

Part 6

Part 7

Part 8

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