Step function potential scattering
introduction
\[E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi\]
step potential 1:
- From Classical to Quantum Mechanics: An Introduction to the Formalism, Foundations and Applications http://goo.gl/VN3OG 184p
- Let the potential is given by
\[\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}\]
- solution\[ \begin{cases} \psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 \\ \psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\ \psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a \end{cases} \] where $k_0=\sqrt{2m E/\hbar^{2}}$ if $x<0$ or $x>a$ and $k_1=\sqrt{2m (E-V_0)/\hbar^{2}}$ if $0<x<a$
- we impose two conditions on the wave function
- the wave function be continuous in the origin
- the first derivative of the wave function be continuous in the origin
- the coefficient must satisfy\[ \begin{cases} A_r + A_l - B_r - B_l = 0 \\ ik_0(A_r - A_l) = ik_1(B_r - B_l) \\ B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0} \\ ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0}) \end{cases} \]
scattering
- special case of scattering problem \(A_r=1, A_l=r, C_r=t , C_l = 0\)
- wave function
\[ \begin{cases} \psi_L(x)= e^{i k_0 x} + r e^{-i k_0x}\quad x<0 \\ \psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \\ \psi_R(x)= t e^{i k_0 x} \quad x>a \end{cases} \]
- coefficient satisfy
\[ \begin{cases} 1 + r - B_r - B_l = 0 \\ ik_0 (1-r)= ik_1(B_r - B_l) \\ B_re^{iak_1}+B_le^{-iak_1}=t e^{iak_0} \\ ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(te^{iak_0}) \end{cases} \]
- \(t-r=1\)\[t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!\]\[r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!\]\[R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!\]\[T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!\]
step potential 2 : not corrected
- 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
\[ \psi(x) = \begin{cases} \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ik_{0}x} + A_{\mathrm l}e^{-ik_{0}x}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ik_{1}x} + B_{\mathrm l}e^{-ik_{1}x}, & \text{ if } x>0, \end{cases} \]
- 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\[\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}\]
- \(t-r=1\)\[t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!\]\[r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!\]\[R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!\]\[T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!\]
history
encyclopedia
메타데이터
위키데이터
- ID : Q2279049
Spacy 패턴 목록
- [{'LOWER': 'solution'}, {'LOWER': 'of'}, {'LOWER': 'schrödinger'}, {'LOWER': 'equation'}, {'LOWER': 'for'}, {'LOWER': 'a'}, {'LOWER': 'step'}, {'LEMMA': 'potential'}]