"Step function potential scattering"의 두 판 사이의 차이

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1번째 줄: 1번째 줄:
 
==introduction==
 
==introduction==
  
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* [[Schrodinger equation]]<br><math>E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi</math><br>
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* [[Schrodinger equation]]
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:<math>E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi</math>
  
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==step potential 1:==
 
==step potential 1:==
11번째 줄: 12번째 줄:
 
* From Classical to Quantum Mechanics: An Introduction to the Formalism, Foundations and Applications http://goo.gl/VN3OG 184p
 
* From Classical to Quantum Mechanics: An Introduction to the Formalism, Foundations and Applications http://goo.gl/VN3OG 184p
  
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*  Let the potential is given by<br><math>\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}</math><br>
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*  Let the potential is given by
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*  solution<br><math>\psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 </math><br><math>\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a </math><br><math>\psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a</math><br> where<br><math>k_0=\sqrt{2m E/\hbar^{2}}\quad\quad\quad\quad x<0\quad or\quad x>a </math><br><math>k_1=\sqrt{2m (E-V_0)/\hbar^{2}}\quad 0<x<a </math><br>
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:<math>\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}</math>
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*  we impose two conditions on the wave function<br>
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*  solution:
 +
<math>
 +
\begin{cases}
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\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}
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</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 wave function be continuous in the origin
 
** the first derivative of the wave function be continuous in the origin
 
** the first derivative of the wave function be continuous in the origin
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*  the coefficient must satisfy<br><math>A_r + A_l - B_r - B_l = 0</math><br><em><math>ik_0(A_r - A_l) = ik_1(B_r - B_l)</math></em><br><math>B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0}</math><br><math>ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0})</math><br>
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*  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>
  
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+
  
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===scattering===
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==scattering==
 
  
 
* special case of scattering problem <math>A_r=1, A_l=r, C_r=t , C_l = 0</math>
 
* special case of scattering problem <math>A_r=1, A_l=r, C_r=t , C_l = 0</math>
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*  wave function<br><math>\psi_L(x)= e^{i k_0 x} + r e^{-i k_0x}\quad x<0 </math><br><math>\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a </math><br><math>\psi_R(x)= t e^{i k_0 x} \quad x>a</math><br>
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*  wave function
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*  coefficient satisfy<br><math>1 + r - B_r - B_l = 0</math><br><em><math>ik_0 (1-r)= ik_1(B_r - B_l)</math></em><br><math>B_re^{iak_1}+B_le^{-iak_1}=t e^{iak_0}</math><br><math>ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(te^{iak_0})</math><br>
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:<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>
  
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* <math>t-r=1</math><br><math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math><br><math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math><br><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><br><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><br>
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* <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>
  
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==step potential 2 : not corrected==
 
==step potential 2 : not corrected==
  
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*  Let the potential is given by<br><math>\psi(x) = \begin{cases} V_{-} & \text{ if } x<0; \\ V_{+}, & \text{ if } x>0, \end{cases}</math><br>
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*  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<br><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_{1x}, & \text{ if } x>0, \end{cases}</math><br>
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*  solution of the stationary S
−
*  we impose two conditions on the wave function<br>
+
:<math>
−
**  the wave function be continuous in the origin
+
\psi(x) =  
−
**  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
+
\begin{cases}  
−
*  first condition<br><math>\psi(0) =\psi_L(0) = \psi_R(0) = A_r + A_l = B_r + B_l</math><br><math>A_r + A_l - B_r - B_l = 0</math><br>
+
\psi_{\mathrm L}(x) = A_{\mathrm r}e^{ik_{0}x} + A_{\mathrm l}e^{-ik_{0}x}, & \text{ if } x<0; \\  
−
*  second condition<br><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><br> LHS becomes <math>-\frac{\hbar^2}{2m}[\psi_R'(0)-\psi_L'(0)] +\lambda\psi(0)</math><br> RHS becomes 0<br><math>-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)</math><br>
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\psi_{\mathrm R}(x) = B_{\mathrm r}e^{ik_{1}x} + B_{\mathrm l}e^{-ik_{1}x}, & \text{ if } x>0,  
−
*  the coefficient must satisfy<br><math>A_r + A_l - B_r - B_l = 0</math><br><math>-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)</math><br>
+
\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>
  
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==delta potential scattering==
 
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 +
===delta potential scattering===
 
* special case of scattering problem <math>A_r=1,  A_l=r,  B_r=t , B_l = 0</math>
 
* special case of scattering problem <math>A_r=1,  A_l=r,  B_r=t , B_l = 0</math>
−
* wave function<br><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><br>
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* 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>
  
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* <math>t-r=1</math><br><math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math><br><math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math><br><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><br><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><br>
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==history==
 
==history==
73번째 줄: 108번째 줄:
 
* http://www.google.com/search?hl=en&tbs=tl:1&q=
 
* http://www.google.com/search?hl=en&tbs=tl:1&q=
  
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==related items==
 
==related items==
  
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==encyclopedia==
 
==encyclopedia==
  
 
* http://en.wikipedia.org/wiki/Rectangular_potential_barrier
 
* http://en.wikipedia.org/wiki/Rectangular_potential_barrier
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* http://en.wikipedia.org/wiki/
 
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* http://www.scholarpedia.org/
 
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* [http://eom.springer.de/ http://eom.springer.de]
 
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* http://www.proofwiki.org/wiki/
 
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* Princeton companion to mathematics([[2910610/attachments/2250873|Companion_to_Mathematics.pdf]])
 
  
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[[분류:개인노트]]
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[[분류:physics]]
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==books==
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[[분류:math and physics]]
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[[분류:migrate]]
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* [[2011년 books and articles]]
 
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* http://library.nu/search?q=
 
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* http://library.nu/search?q=
 
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==expositions==
 
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==articles==
 
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* http://www.ams.org/mathscinet
 
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* http://www.zentralblatt-math.org/zmath/en/
 
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* http://arxiv.org/
 
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* http://www.pdf-search.org/
 
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* http://pythagoras0.springnote.com/
 
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* [http://math.berkeley.edu/%7Ereb/papers/index.html http://math.berkeley.edu/~reb/papers/index.html]
 
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* http://dx.doi.org/
 
  
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==메타데이터==
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===위키데이터===
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* ID : [https://www.wikidata.org/wiki/Q2279049 Q2279049]
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===Spacy 패턴 목록===
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==question and answers(Math Overflow)==
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* [{'LOWER': 'solution'}, {'LOWER': 'of'}, {'LOWER': 'schrödinger'}, {'LOWER': 'equation'}, {'LOWER': 'for'}, {'LOWER': 'a'}, {'LOWER': 'step'}, {'LEMMA': 'potential'}]
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* http://mathoverflow.net/search?q=
 
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* http://mathoverflow.net/search?q=
 
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==blogs==
 
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*  구글 블로그 검색<br>
 
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**  http://blogsearch.google.com/blogsearch?q=<br>
 
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** http://blogsearch.google.com/blogsearch?q=
 
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* http://ncatlab.org/nlab/show/HomePage
 
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==experts on the field==
 
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* http://arxiv.org/
 
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==links==
 
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* [http://detexify.kirelabs.org/classify.html Detexify2 - LaTeX symbol classifier]
 
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* [http://pythagoras0.springnote.com/pages/1947378 수식표현 안내]
 
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
 
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* http://functions.wolfram.com/
 
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[[분류:개인노트]]
 

2021년 2월 17일 (수) 01:04 기준 최신판

introduction

\[E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi\]



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

\[\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}\]

  • solution\[ \begin{cases} \psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 \\ \psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\ \psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a \end{cases} \] 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\[ \begin{cases} A_r + A_l - B_r - B_l = 0 \\ ik_0(A_r - A_l) = ik_1(B_r - B_l) \\ B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0} \\ ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0}) \end{cases} \]


scattering

  • special case of scattering problem \(A_r=1, A_l=r, C_r=t , C_l = 0\)
  • wave function

\[ \begin{cases} \psi_L(x)= e^{i k_0 x} + r e^{-i k_0x}\quad x<0 \\ \psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\ \psi_R(x)= t e^{i k_0 x} \quad x>a \end{cases} \]

  • coefficient satisfy

\[ \begin{cases} 1 + r - B_r - B_l = 0 \\ ik_0 (1-r)= ik_1(B_r - B_l) \\ B_re^{iak_1}+B_le^{-iak_1}=t e^{iak_0} \\ ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(te^{iak_0}) \end{cases} \]

  • \(t-r=1\)\[t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!\]\[r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!\]\[R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!\]\[T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!\]




step potential 2 : not corrected

  • Let the potential is given by\[\psi(x) = \begin{cases} V_{-} & \text{ if } x<0; \\ V_{+}, & \text{ if } x>0, \end{cases}\]
  • solution of the stationary S

\[ \psi(x) = \begin{cases} \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ik_{0}x} + A_{\mathrm l}e^{-ik_{0}x}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ik_{1}x} + B_{\mathrm l}e^{-ik_{1}x}, & \text{ if } x>0, \end{cases} \]

  • 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\[\psi(0) =\psi_L(0) = \psi_R(0) = A_r + A_l = B_r + B_l\]\[A_r + A_l - B_r - B_l = 0\]
  • second condition\[ -\frac{\hbar^2}{2 m} \int_{-\epsilon}^{+\epsilon} \psi''(x) \,dx + \int_{-\epsilon}^{+\epsilon} V(x)\psi(x) \,dx = E \int_{-\epsilon}^{+\epsilon} \psi(x) \,dx\] LHS becomes \(-\frac{\hbar^2}{2m}[\psi_R'(0)-\psi_L'(0)] +\lambda\psi(0)\) RHS becomes 0\[-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)\]
  • the coefficient must satisfy

\[ \begin{cases} A_r + A_l - B_r - B_l = 0 \\ -A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l) \end{cases} \]


delta potential scattering

  • special case of scattering problem \(A_r=1, A_l=r, B_r=t , B_l = 0\)
  • wave function\[\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}\]
  • \(t-r=1\)\[t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!\]\[r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!\]\[R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!\]\[T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!\]




history



related items

encyclopedia

메타데이터

위키데이터

Spacy 패턴 목록

  • [{'LOWER': 'solution'}, {'LOWER': 'of'}, {'LOWER': 'schrödinger'}, {'LOWER': 'equation'}, {'LOWER': 'for'}, {'LOWER': 'a'}, {'LOWER': 'step'}, {'LEMMA': 'potential'}]
원본 주소 "https://wiki.mathnt.net/index.php?title=Step_function_potential_scattering&oldid=51669"