Step function potential scattering

introduction

• Schrodinger equation
  \(E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi\)
  

 

 

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
  \(\psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 \)
  \(\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a \)
  \(\psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a\)
  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
  \(\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}\)
  

 

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

 

 

 

history

• http://www.google.com/search?hl=en&tbs=tl:1&q=