Step function potential scattering

수학노트
imported>Pythagoras0님의 2012년 10월 29일 (월) 09:50 판
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introduction

 

 

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
    \(\psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 \)
    \(\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \)
    \(\psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a\)
    where
    \(k_0=\sqrt{2m E/\hbar^{2}}\quad\quad\quad\quad x<0\quad or\quad x>a \)
    \(k_1=\sqrt{2m (E-V_0)/\hbar^{2}}\quad 0<x<a \)
  • 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
    \(A_r + A_l - B_r - B_l = 0\)
    \(ik_0(A_r - A_l) = ik_1(B_r - B_l)\)
    \(B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0}\)
    \(ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0})\)

 

 

scattering

  • special case of scattering problem \(A_r=1, A_l=r, C_r=t , C_l = 0\)
  • wave function
    \(\psi_L(x)= e^{i k_0 x} + r e^{-i k_0x}\quad x<0 \)
    \(\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \)
    \(\psi_R(x)= t e^{i k_0 x} \quad x>a\)
  • coefficient satisfy
    \(1 + r - B_r - B_l = 0\)
    \(ik_0 (1-r)= ik_1(B_r - B_l)\)
    \(B_re^{iak_1}+B_le^{-iak_1}=t e^{iak_0}\)
    \(ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(te^{iak_0})\)
  • \(t-r=1\)
    \(t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!\)
    \(r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!\)
    \(R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!\)
    \(T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!\)

 

 

 

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
    \(\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = A_{\mathrm r}e^{ik_{0}x} + A_{\mathrm l}e^{-ik_{0}x}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) = B_{\mathrm r}e^{ik_{1}x} + B_{\mathrm l}e^{-ik_{1x}, & \text{ if } x>0, \end{cases}\)
  • 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
    \(\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}\)

 

  • \(t-r=1\)
    \(t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!\)
    \(r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!\)
    \(R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!\)
    \(T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!\)

 

 

 

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