"Step function potential scattering"의 두 판 사이의 차이

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*  Let the potential is given by<br><math>\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}</math><br>
 
*  Let the potential is given by<br><math>\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}</math><br>
 
*  solution<br><math>\psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 </math><br><math>\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a </math><br><math>\psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a</math><br> where<br><math>k_0=\sqrt{2m E/\hbar^{2}}\quad\quad\quad\quad x<0\quad or\quad x>a </math><br><math>k_1=\sqrt{2m (E-V_0)/\hbar^{2}}\quad 0<x<a </math><br>
 
*  solution<br><math>\psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 </math><br><math>\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a </math><br><math>\psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a</math><br> where<br><math>k_0=\sqrt{2m E/\hbar^{2}}\quad\quad\quad\quad x<0\quad or\quad x>a </math><br><math>k_1=\sqrt{2m (E-V_0)/\hbar^{2}}\quad 0<x<a </math><br>
 
 
*  we impose two conditions on the wave function<br>
 
*  we impose two conditions on the wave function<br>
 
** the wave function be continuous in the origin
 
** the wave function be continuous in the origin
 
** the first derivative of the wave function be continuous in the origin
 
** the first derivative of the wave function be continuous in the origin
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*  the coefficient must satisfy<br><math>A_r + A_l - B_r - B_l = 0</math><br><em><math>k_0(A_r - A_l) = ik_1(B_r - B_l)</math></em><br><math>B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0}</math><br><math>ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0})</math><br>
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 +
 
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<h5>scattering</h5>
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* special case of scattering problem <math>A_r=1, A_l=r, C_r=t , C_l = 0</math>
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*  wave function<br><math>\psi_L(x)= A_r e^{i k_0 x} + A_l e^{-i k_0x}\quad x<0 </math><br><math>\psi_C(x)= B_r e^{i k_1 x} + B_l e^{-i k_1x}\quad 0<x<a </math><br><math>\psi_R(x)= C_r e^{i k_0 x} + C_l e^{-i k_0x}\quad x>a</math><br>
 +
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* <math>t-r=1</math><br><math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math><br><math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math><br><math>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}}.\,\!</math><br><math>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}}\,\!</math><br>
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2011년 2월 11일 (금) 03:26 판

introduction

 

 

step potential 1:
  • 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
    \(A_r + A_l - B_r - B_l = 0\)
    \(k_0(A_r - A_l) = ik_1(B_r - B_l)\)
    \(B_re^{iak_1}+B_le^{-iak_1}=C_re^{iak_0}+C_le^{-iak_0}\)
    \(ik_1(B_re^{iak_1}-B_le^{-iak_1})=ik_0(C_re^{iak_0}-C_le^{-iak_0})\)

 

 

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

 

 

 

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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