Delta potential scattering
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'}]