"Step function potential scattering"의 두 판 사이의 차이
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| 3번째 줄: | 3번째 줄: | ||
* 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> | * 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^{ | + | * 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^{ikx} + B_{\mathrm l}e^{-ikx}, & \text{ if } x>0, \end{cases}</math><br> |
* we impose two conditions on the wave function<br> | * we impose two conditions on the wave function<br> | ||
** the wave function be continuous in the origin | ** the wave function be continuous in the origin | ||
| 10번째 줄: | 10번째 줄: | ||
* 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> | * 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> | * 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> | ||
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| 50번째 줄: | 48번째 줄: | ||
<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> | <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/Rectangular_potential_barrier | ||
* http://en.wikipedia.org/wiki/ | * http://en.wikipedia.org/wiki/ | ||
* http://www.scholarpedia.org/ | * http://www.scholarpedia.org/ | ||
2011년 2월 11일 (금) 02:58 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
\(\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ik_{0}x} + A_{\mathrm l}e^{-ik_{0}x}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ikx} + B_{\mathrm l}e^{-ikx}, & \text{ if } x>0, \end{cases}\) - 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
encyclopedia
- http://en.wikipedia.org/wiki/Rectangular_potential_barrier
- http://en.wikipedia.org/wiki/
- http://www.scholarpedia.org/
- http://eom.springer.de
- http://www.proofwiki.org/wiki/
- Princeton companion to mathematics(Companion_to_Mathematics.pdf)
books
expositions
articles
- 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/~reb/papers/index.html
- http://dx.doi.org/
question and answers(Math Overflow)
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experts on the field