"Step function potential scattering"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
(피타고라스님이 이 페이지를 개설하였습니다.)
 
1번째 줄: 1번째 줄:
 +
<h5>introduction</h5>
  
 +
*  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>
 +
 +
*  solution of the stationary S<br><math>\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ikx} + A_{\mathrm l}e^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ikx} + B_{\mathrm l}e^{-ikx}, & \text{ if } x>0, \end{cases}</math><br>
 +
*  we impose two conditions on the wave function<br>
 +
**  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<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>
 +
*  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>
 +
*  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>
 +
 +
 
 +
 +
 
 +
 +
 
 +
 +
<h5>delta potential scattering</h5>
 +
 +
 
 +
 +
* 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>
 +
 +
 
 +
 +
* <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>
 +
 +
 
 +
 +
 
 +
 +
 
 +
 +
<h5>history</h5>
 +
 +
* http://www.google.com/search?hl=en&tbs=tl:1&q=
 +
 +
 
 +
 +
 
 +
 +
<h5>related items</h5>
 +
 +
 
 +
 +
 
 +
 +
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">encyclopedia</h5>
 +
 +
* http://en.wikipedia.org/wiki/
 +
* http://www.scholarpedia.org/
 +
* [http://eom.springer.de/ http://eom.springer.de]
 +
* http://www.proofwiki.org/wiki/
 +
* Princeton companion to mathematics([[2910610/attachments/2250873|Companion_to_Mathematics.pdf]])
 +
 +
 
 +
 +
 
 +
 +
<h5>books</h5>
 +
 +
 
 +
 +
* [[2011년 books and articles]]
 +
* http://library.nu/search?q=
 +
* http://library.nu/search?q=
 +
 +
 
 +
 +
 
 +
 +
<h5>expositions</h5>
 +
 +
 
 +
 +
 
 +
 +
 
 +
 +
<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">articles</h5>
 +
 +
 
 +
 +
* http://www.ams.org/mathscinet
 +
* http://www.zentralblatt-math.org/zmath/en/
 +
* http://arxiv.org/
 +
* http://www.pdf-search.org/
 +
* http://pythagoras0.springnote.com/
 +
* [http://math.berkeley.edu/%7Ereb/papers/index.html http://math.berkeley.edu/~reb/papers/index.html]
 +
* http://dx.doi.org/
 +
 +
 
 +
 +
 
 +
 +
<h5>question and answers(Math Overflow)</h5>
 +
 +
* http://mathoverflow.net/search?q=
 +
* http://mathoverflow.net/search?q=
 +
 +
 
 +
 +
 
 +
 +
<h5>blogs</h5>
 +
 +
*  구글 블로그 검색<br>
 +
**  http://blogsearch.google.com/blogsearch?q=<br>
 +
** http://blogsearch.google.com/blogsearch?q=
 +
* http://ncatlab.org/nlab/show/HomePage
 +
 +
 
 +
 +
 
 +
 +
<h5>experts on the field</h5>
 +
 +
* http://arxiv.org/
 +
 +
 
 +
 +
 
 +
 +
<h5>links</h5>
 +
 +
* [http://detexify.kirelabs.org/classify.html Detexify2 - LaTeX symbol classifier]
 +
* [http://pythagoras0.springnote.com/pages/1947378 수식표현 안내]
 +
* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
 +
* http://functions.wolfram.com/

2011년 2월 10일 (목) 04:58 판

introduction
  • 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
    \(\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ikx} + A_{\mathrm l}e^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ikx} + B_{\mathrm l}e^{-ikx}, & \text{ if } x>0, \end{cases}\)
  • 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
    \(A_r + A_l - B_r - B_l = 0\)
    \(-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)\)

 

 

 

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

 

 

books

 

 

 

expositions

 

 

 

articles

 

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links