Step function potential scattering - 편집 역사

Pythagoras0: /* 메타데이터 */

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메타데이터

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imported>Pythagoras0: 찾아 바꾸기 – “* Princeton companion to mathematics(Companion_to_Mathematics.pdf)” 문자열을 “” 문자열로

2012-11-01T21:11:42Z

찾아 바꾸기 – “* Princeton companion to mathematics(Companion_to_Mathematics.pdf)” 문자열을 “” 문자열로

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@@ 127번째 줄: / 127번째 줄: @@
 [[분류:math and physics]]
 [[분류:migrate]]
 ==메타데이터==

@@ 130번째 줄: / 130번째 줄: @@
 == 메타데이터 ==
 ===위키데이터===
 * ID :  [https://www.wikidata.org/wiki/Q2279049 Q2279049]

← 이전 판		2020년 12월 26일 (토) 13:44 판
127번째 줄:		127번째 줄:
	[[분류:math and physics]]		[[분류:math and physics]]
	[[분류:migrate]]		[[분류:migrate]]
		+
		+	== 메타데이터 ==
		+
		+	===위키데이터===
		+	* ID : [https://www.wikidata.org/wiki/Q2279049 Q2279049]

← 이전 판	2020년 11월 14일 (토) 09:49 판
1번째 줄:	1번째 줄:
	+	==introduction==

	+	* [[Schrodinger equation]]
	+	:<math>E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi</math>
	+
	+
	+
	+
	+
	+	==step potential 1:==
	+
	+	* From Classical to Quantum Mechanics: An Introduction to the Formalism, Foundations and Applications http://goo.gl/VN3OG 184p
	+
	+	* Let the potential is given by
	+	:<math>\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}</math>
	+	* solution:
	+	<math>
	+	\begin{cases}
	+	\psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 \\
	+	\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\
	+	\psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a
	+	\end{cases}
	+	</math> where $k_0=\sqrt{2m E/\hbar^{2}}$ if $x<0$ or $x>a$ and $k_1=\sqrt{2m (E-V_0)/\hbar^{2}}$ if $0<x<a$
	+	* we impose two conditions on the wave function
	+	** the wave function be continuous in the origin
	+	** the first derivative of the wave function be continuous in the origin
	+	* the coefficient must satisfy:
	+	<math>
	+	\begin{cases}
	+	A_r + A_l - B_r - B_l = 0 \\
	+	ik_0(A_r - A_l) = ik_1(B_r - B_l) \\
	+	B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0} \\
	+	ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0})
	+	\end{cases}
	+	</math>
	+
	+
	+
	+	===scattering===
	+
	+	* special case of scattering problem <math>A_r=1, A_l=r, C_r=t , C_l = 0</math>
	+	* wave function
	+	:<math>
	+	\begin{cases}
	+	\psi_L(x)= e^{i k_0 x} + r e^{-i k_0x}\quad x<0 \\
	+	\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\
	+	\psi_R(x)= t e^{i k_0 x} \quad x>a
	+	\end{cases}
	+	</math>
	+	* coefficient satisfy
	+	:<math>
	+	\begin{cases}
	+	1 + r - B_r - B_l = 0 \\
	+	ik_0 (1-r)= ik_1(B_r - B_l) \\
	+	B_re^{iak_1}+B_le^{-iak_1}=t e^{iak_0} \\
	+	ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(te^{iak_0})
	+	\end{cases}
	+	</math>
	+
	+	* <math>t-r=1</math>:<math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math>:<math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math>:<math>R=\|r\|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!</math>:<math>T=\|t\|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!</math>
	+
	+
	+
	+
	+
	+
	+
	+	==step potential 2 : not corrected==
	+
	+	* Let the potential is given by:<math>\psi(x) = \begin{cases} V_{-} & \text{ if } x<0; \\ V_{+}, & \text{ if } x>0, \end{cases}</math>
	+
	+	* solution of the stationary S
	+	:<math>
	+	\psi(x) =
	+	\begin{cases}
	+	\psi_{\mathrm L}(x) = A_{\mathrm r}e^{ik_{0}x} + A_{\mathrm l}e^{-ik_{0}x}, & \text{ if } x<0; \\
	+	\psi_{\mathrm R}(x) = B_{\mathrm r}e^{ik_{1}x} + B_{\mathrm l}e^{-ik_{1}x}, & \text{ if } x>0,
	+	\end{cases}
	+	</math>
	+	* we impose two conditions on the wave function
	+	** the wave function be continuous in the origin
	+	** integrate the Schrödinger equation around x = 0, over an interval [−ε, +ε] and In the limit as ε → 0, the right-hand side of this equation vanishes; the left-hand side becomes
	+	* first condition:<math>\psi(0) =\psi_L(0) = \psi_R(0) = A_r + A_l = B_r + B_l</math>:<math>A_r + A_l - B_r - B_l = 0</math>
	+	* second condition:<math> -\frac{\hbar^2}{2 m} \int_{-\epsilon}^{+\epsilon} \psi''(x) \,dx + \int_{-\epsilon}^{+\epsilon} V(x)\psi(x) \,dx = E \int_{-\epsilon}^{+\epsilon} \psi(x) \,dx</math> LHS becomes <math>-\frac{\hbar^2}{2m}[\psi_R'(0)-\psi_L'(0)] +\lambda\psi(0)</math> RHS becomes 0:<math>-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)</math>
	+	* the coefficient must satisfy
	+	:<math>
	+	\begin{cases}
	+	A_r + A_l - B_r - B_l = 0 \\
	+	-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)
	+	\end{cases}
	+	</math>
	+
	+
	+
	+	===delta potential scattering===
	+	* special case of scattering problem <math>A_r=1, A_l=r, B_r=t , B_l = 0</math>
	+	* wave function:<math>\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}</math>
	+	* <math>t-r=1</math>:<math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math>:<math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math>:<math>R=\|r\|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!</math>:<math>T=\|t\|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!</math>
	+
	+
	+
	+
	+
	+
	+
	+	==history==
	+
	+	* http://www.google.com/search?hl=en&tbs=tl:1&q=
	+
	+
	+
	+
	+
	+	==related items==
	+
	+
	+
	+
	+
	+	==encyclopedia==
	+
	+	* http://en.wikipedia.org/wiki/Rectangular_potential_barrier
	+
	+
	+	[[분류:개인노트]]
	+	[[분류:physics]]
	+	[[분류:math and physics]]
	+	[[분류:migrate]]

← 이전 판		2020년 11월 14일 (토) 09:49 판
1번째 줄:		1번째 줄:
−	~~==introduction==~~

−	* [[Schrodinger equation]]
−	~~:<math>E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi</math>~~
−
−
−
−
−
−	~~==step potential 1:==~~
−
−	* From Classical to Quantum Mechanics: An Introduction to the Formalism, Foundations and Applications http://goo.gl/VN3OG 184p
−
−	* Let the potential is given by
−	~~:<math>\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}</math>~~
−	* solution:
−	~~<math>~~
−	~~\begin{cases}~~
−	~~\psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 \\~~
−	~~\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\~~
−	~~\psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a~~
−	~~\end{cases}~~
−	~~</math> where $k_0=\sqrt{2m E/\hbar^{2}}$ if $x<0$ or $x>a$ and $k_1=\sqrt{2m (E-V_0)/\hbar^{2}}$ if $0<x<a$~~
−	* we impose two conditions on the wave function
−	** the wave function be continuous in the origin
−	** the first derivative of the wave function be continuous in the origin
−	* the coefficient must satisfy:
−	~~<math>~~
−	~~\begin{cases}~~
−	~~A_r + A_l - B_r - B_l = 0 \\~~
−	~~ik_0(A_r - A_l) = ik_1(B_r - B_l) \\~~
−	~~B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0} \\~~
−	~~ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0})~~
−	~~\end{cases}~~
−	~~</math>~~
−
−
−
−	~~===scattering===~~
−
−	* special case of scattering problem <math>A_r=1, A_l=r, C_r=t , C_l = 0</math>
−	* wave function
−	~~:<math>~~
−	~~\begin{cases}~~
−	~~\psi_L(x)= e^{i k_0 x} + r e^{-i k_0x}\quad x<0 \\~~
−	~~\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\~~
−	~~\psi_R(x)= t e^{i k_0 x} \quad x>a~~
−	~~\end{cases}~~
−	~~</math>~~
−	* coefficient satisfy
−	~~:<math>~~
−	~~\begin{cases}~~
−	~~1 + r - B_r - B_l = 0 \\~~
−	~~ik_0 (1-r)= ik_1(B_r - B_l) \\~~
−	~~B_re^{iak_1}+B_le^{-iak_1}=t e^{iak_0} \\~~
−	~~ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(te^{iak_0})~~
−	~~\end{cases}~~
−	~~</math>~~
−
−	* <math>t-r=1</math>:<math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math>:<math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math>:<math>R=\|r\|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!</math>:<math>T=\|t\|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!</math>
−
−
−
−
−
−
−
−	~~==step potential 2 : not corrected==~~
−
−	* Let the potential is given by:<math>\psi(x) = \begin{cases} V_{-} & \text{ if } x<0; \\ V_{+}, & \text{ if } x>0, \end{cases}</math>
−
−	* solution of the stationary S
−	~~:<math>~~
−	~~\psi(x) =~~
−	~~\begin{cases}~~
−	~~\psi_{\mathrm L}(x) = A_{\mathrm r}e^{ik_{0}x} + A_{\mathrm l}e^{-ik_{0}x}, & \text{ if } x<0; \\~~
−	~~\psi_{\mathrm R}(x) = B_{\mathrm r}e^{ik_{1}x} + B_{\mathrm l}e^{-ik_{1}x}, & \text{ if } x>0,~~
−	~~\end{cases}~~
−	~~</math>~~
−	* we impose two conditions on the wave function
−	** the wave function be continuous in the origin
−	** integrate the Schrödinger equation around x = 0, over an interval [−ε, +ε] and In the limit as ε → 0, the right-hand side of this equation vanishes; the left-hand side becomes
−	* first condition:<math>\psi(0) =\psi_L(0) = \psi_R(0) = A_r + A_l = B_r + B_l</math>:<math>A_r + A_l - B_r - B_l = 0</math>
−	* second condition:<math> -\frac{\hbar^2}{2 m} \int_{-\epsilon}^{+\epsilon} \psi''(x) \,dx + \int_{-\epsilon}^{+\epsilon} V(x)\psi(x) \,dx = E \int_{-\epsilon}^{+\epsilon} \psi(x) \,dx</math> LHS becomes <math>-\frac{\hbar^2}{2m}[\psi_R'(0)-\psi_L'(0)] +\lambda\psi(0)</math> RHS becomes 0:<math>-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)</math>
−	* the coefficient must satisfy
−	~~:<math>~~
−	~~\begin{cases}~~
−	~~A_r + A_l - B_r - B_l = 0 \\~~
−	~~-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)~~
−	~~\end{cases}~~
−	~~</math>~~
−
−
−
−	~~===delta potential scattering===~~
−	* special case of scattering problem <math>A_r=1, A_l=r, B_r=t , B_l = 0</math>
−	* wave function:<math>\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}</math>
−	* <math>t-r=1</math>:<math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math>:<math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math>:<math>R=\|r\|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!</math>:<math>T=\|t\|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!</math>
−
−
−
−
−
−
−
−	~~==history==~~
−
−	* http://www.google.com/search?hl=en&tbs=tl:1&q=
−
−
−
−
−
−	~~==related items==~~
−
−
−
−
−
−	~~==encyclopedia==~~
−
−	* http://en.wikipedia.org/wiki/Rectangular_potential_barrier
−
−
−	~~[[분류:개인노트]]~~
−	~~[[분류:physics]]~~
−	~~[[분류:math and physics]]~~

← 이전 판	2020년 11월 16일 (월) 18:07 판
(차이 없음)

@@ 90번째 줄: / 90번째 줄: @@
 * [http://eom.springer.de/ http://eom.springer.de]
 * http://www.proofwiki.org/wiki/
-* Princeton companion to mathematics([[2910610/attachments/2250873|Companion_to_Mathematics.pdf]])

← 이전 판		2012년 10월 29일 (월) 17:55 판
168번째 줄:		168번째 줄:
	[[분류:개인노트]]		[[분류:개인노트]]
	[[분류:physics]]		[[분류:physics]]
		+	[[분류:math and physics]]

← 이전 판		2012년 10월 29일 (월) 17:50 판
167번째 줄:		167번째 줄:
	* http://functions.wolfram.com/		* http://functions.wolfram.com/
	[[분류:개인노트]]		[[분류:개인노트]]
		+	[[분류:physics]]