Properties of the Linear Schrodinger Equation - Revision history

Levi: /* Case when \lambda>0 */

2025-09-28T09:27:56Z

Case when \lambda>0

← Older revision		Revision as of 09:27, 28 September 2025
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	\right.		\right.
	</math></center>		</math></center>
	where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its ~~derivaties~~ at <math>x=\pm\zeta</math> we		where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivatives at <math>x=\pm\zeta</math> we
	obtain		obtain
	<center><math>		<center><math>

Levi: change varsigma to actual zeta

2025-09-28T08:44:25Z

change varsigma to actual zeta

Show changes

Levi: /* Case when \lambda>0 */

2025-09-28T08:43:39Z

Case when \lambda>0

← Older revision		Revision as of 08:43, 28 September 2025
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	\right.		\right.
	</math></center>		</math></center>
	where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\~~pm1~~</math> we		where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm\varsigma</math> we
	obtain		obtain
	<center><math>		<center><math>

Levi: a bit more explanations

2025-09-28T06:18:36Z

a bit more explanations

← Older revision		Revision as of 06:18, 28 September 2025
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	\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w		\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
	</math></center>		</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		===Case when <math>\lambda<0</math>===

			We consider the cases 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 (as <math>w\rightarrow0</math> as <math>x\rightarrow\pm\infty</math>)
	<center><math>		<center><math>
	w\left( x\right) =\left\{		w\left( x\right) =\left\{
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	<center><math>		<center><math>
	2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1		2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1
	</math></center>		</math>
			<math>
			2\left( a^2\dfrac{e^{-u_{0}x}}{-u_{0}}\right)\Bigg\|_{0}^{\infty}=1
			</math>
			<math>\dfrac{2a^2}{u_{0}}=1</math></center>
	which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete		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>		spectral point which we denote by <math>k_{1}=u_{0}/2</math>
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	\right.		\right.
	</math></center>		</math></center>

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

	The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form		The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
	<center><math>		<center><math>
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	\right.		\right.
	</math></center>		</math></center>
			where <math>\mathrm{e}^{-\mathrm{i}kx}</math> is the incident wave, <math>r\mathrm{e}^{\mathrm{i}kx}</math> is the reflected wave, and <math>t\mathrm{e}^{-\mathrm{i}kx}</math> is the transmitted wave.

	Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and		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)		<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)

Levi at 06:38, 27 September 2025

2025-09-27T06:38:07Z

← Older revision		Revision as of 06:38, 27 September 2025
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	first are waves and the second are bound solutions. It is well known that		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>		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		as <math>x\rightarrow\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
	incident waves. This is easiest seen through the following examples		incident waves. This is easiest seen through the following examples

Levi at 20:27, 25 September 2025

2025-09-25T20:27:31Z

← Older revision		Revision as of 20:27, 25 September 2025
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	}}		}}

	The linear ~~Schrodinger~~ equation		The linear Schrödinger equation
	<center><math>		<center><math>
	\partial_{x}^{2}w+uw=-\lambda w		\partial_{x}^{2}w+uw=-\lambda w

Meylan: /* Case when \lambda>0 */

2020-09-24T01:03:35Z

Case when \lambda>0

← Older revision		Revision as of 01:03, 24 September 2020
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	\right)		\right)
	</math></center>		</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}}

Meylan: /* Case when \lambda>0 */

2020-09-21T02:25:36Z

Case when \lambda>0

← Older revision		Revision as of 02:25, 21 September 2020
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	\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 <-\varsigma\\
	b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma< x <\varsigma\\		b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma< x <\varsigma\\
	a\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma		t\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
	\end{matrix}		\end{matrix}
	\right.		\right.
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	b_{1}\\		b_{1}\\
	b_{2}\\		b_{2}\\
	a		t
	\end{matrix}		\end{matrix}
	\right) =\left(		\right) =\left(

Meylan: /* Example 1: \delta function potential */

2020-09-21T02:24:08Z

Example 1: \delta function potential

← Older revision		Revision as of 02:24, 21 September 2020
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	\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\		\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
	a\mathrm{e}^{-\mathrm{i}kx}, & x>0		t\mathrm{e}^{-\mathrm{i}kx}, & x>0
	\end{matrix}		\end{matrix}
	\right.		\right.
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	+u_{0}w\left( 0\right) =0.</math> This gives us		+u_{0}w\left( 0\right) =0.</math> This gives us
	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	1+r & =a\\		1+r & =t\\
	-~~ika~~+ik-ikr & =-~~au_~~{0}		-ikt+ik-ikr & =-tu_{0}
	\end{matrix}</math></center>		\end{matrix}</math></center>
	which has solution		which has solution
	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	r & =\frac{u_{0}}{2ik-u_{0}}\\		r & =\frac{u_{0}}{2ik-u_{0}}\\
	a & =\frac{2ik}{2ik-u_{0}}		t & =\frac{2ik}{2ik-u_{0}}
	\end{matrix}</math></center>		\end{matrix}</math></center>

Meylan: /* Example 2: Hat Function Potential */

2019-05-24T05:18:55Z

Example 2: Hat Function Potential

← Older revision		Revision as of 05:18, 24 May 2019
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	\begin{matrix}		\begin{matrix}
	a_{1}e^{kx}, & x<-\varsigma,\\		a_{1}e^{kx}, & x<-\varsigma,\\
	b_{1}\cos\kappa, x & -\varsigma< x <\varsigma,\\		b_{1}\cos\kappa x, & -\varsigma< x <\varsigma,\\
	a_{1}e^{-kx}, & x>\varsigma,		a_{1}e^{-kx}, & x>\varsigma,
	\end{matrix}		\end{matrix}