"Delta potential scattering"의 두 판 사이의 차이
둘러보기로 가기
검색하러 가기
Pythagoras0 (토론 | 기여) |
|||
| (사용자 2명의 중간 판 16개는 보이지 않습니다) | |||
| 1번째 줄: | 1번째 줄: | ||
| − | + | ==introduction== | |
| − | * Let the potential is given by <math>V(x) = \lambda\delta(x)</math> | + | * Let the potential is given by |
| − | * we impose two conditions on the wave function | + | :<math>V(x) = \lambda\delta(x)</math> |
| − | ** | + | * wave function |
| − | ** | + | :<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> |
| − | * first condition | + | * we impose two conditions on the wave function |
| − | * second condition | + | ** the wave function be continuous in the origin |
| − | * the | + | ** 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 coefficients <math>A_r, A_l,B_r,B_l</math> 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> | * special case of scattering problem <math>A_r=1, A_l=r, B_r=t , B_l = 0</math> | ||
| − | * wave function | + | * 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> | ||
| − | + | ||
| − | + | ==계산 리소스== | |
| + | * https://docs.google.com/file/d/0B8XXo8Tve1cxZjdDY3BfRFEwM0E/edit | ||
| − | + | ||
| − | + | ==encyclopedia== | |
| − | + | * [http://en.wikipedia.org/wiki/Delta_potential_barrier_%28QM%29 http://en.wikipedia.org/wiki/Delta_potential_barrier_(QM)] | |
| + | [[분류:physics]] | ||
| + | [[분류:math and physics]] | ||
| + | [[분류:migrate]] | ||
| − | * [ | + | ==메타데이터== |
| + | ===위키데이터=== | ||
| + | * ID : [https://www.wikidata.org/wiki/Q2381860 Q2381860] | ||
| + | ===Spacy 패턴 목록=== | ||
| + | * [{'LOWER': 'delta'}, {'LEMMA': 'potential'}] | ||
2021년 2월 17일 (수) 02:30 기준 최신판
introduction
- Let the potential is given by
\[V(x) = \lambda\delta(x)\]
- wave function
\[\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ikx} + A_{\mathrm l}e^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ikx} + B_{\mathrm l}e^{-ikx}, & \text{ if } x>0, \end{cases}\]
- 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 coefficients \(A_r, A_l,B_r,B_l\) 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
\[\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}\]
- \(t-r=1\)
\[t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!\] \[r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!\] \[R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!\] \[T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!\]
계산 리소스
encyclopedia
메타데이터
위키데이터
- ID : Q2381860
Spacy 패턴 목록
- [{'LOWER': 'delta'}, {'LEMMA': 'potential'}]