"Delta potential scattering"의 두 판 사이의 차이
둘러보기로 가기
검색하러 가기
imported>Pythagoras0 |
imported>Pythagoras0 |
||
1번째 줄: | 1번째 줄: | ||
==introduction== | ==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> |
− | ** | + | :<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> |
− | ** | + | * we impose two conditions on the wave function |
− | * first condition | + | ** the wave function be continuous in the origin |
− | * second condition | + | ** 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 |
− | * the coefficient must satisfy | + | * 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== | ==delta potential scattering== |
2014년 10월 5일 (일) 18:15 판
introduction
- Let the potential is given by
\[V(x) = \lambda\delta(x)\] \[\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ikx} + A_{\mathrm l}e^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ikx} + B_{\mathrm l}e^{-ikx}, & \text{ if } x>0, \end{cases}\]
- 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
\[ \begin{cases} A_r + A_l - B_r - B_l = 0 \\ -A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l) \end{cases} \]
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