"Step function potential scattering"의 두 판 사이의 차이

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-==introduction==
-* [[Schrodinger equation]]
-:<math>E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi</math>
-==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
-:<math>\nu = \begin{cases} V_{0} & \text{ if } 0<x<a; \\ 0& \text{ otherwise } \end{cases}</math>
-*  solution:
-<math>
-\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}
-</math> 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:
-<math>
-\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}
-</math>
-===scattering===
-* special case of scattering problem <math>A_r=1, A_l=r, C_r=t , C_l = 0</math>
-*  wave function
-:<math>
-\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}
-</math>
-*  coefficient satisfy
-:<math>
-\begin{cases}
-+ 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}
-</math>
-* <math>t-r=1</math>:<math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math>:<math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math>:<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>:<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>
-==step potential 2 : not corrected==
-*  Let the potential is given by:<math>\psi(x) = \begin{cases} V_{-} & \text{ if } x<0; \\ V_{+}, & \text{ if } x>0, \end{cases}</math>
-*  solution of the stationary S
-:<math>
-\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}
-</math>
-*  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:<math>\psi(0) =\psi_L(0) = \psi_R(0) = A_r + A_l = B_r + B_l</math>:<math>A_r + A_l - B_r - B_l = 0</math>
-*  second condition:<math> -\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</math> LHS becomes <math>-\frac{\hbar^2}{2m}[\psi_R'(0)-\psi_L'(0)] +\lambda\psi(0)</math> RHS becomes 0:<math>-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)</math>
-*  the coefficient must satisfy
-:<math>
-\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}
-</math>
-===delta potential scattering===
-* special case of scattering problem <math>A_r=1,  A_l=r,  B_r=t , B_l = 0</math>
-* wave function:<math>\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}</math>
-* <math>t-r=1</math>:<math>t=\cfrac{1}{1-\cfrac{m\lambda}{i\hbar^2k}}\,\!</math>:<math>r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!</math>:<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>:<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>
-==history==
-* http://www.google.com/search?hl=en&tbs=tl:1&q=
-==related items==
-==encyclopedia==
-* http://en.wikipedia.org/wiki/Rectangular_potential_barrier
-[[분류:개인노트]]
-[[분류:physics]]
-[[분류:math and physics]]

"Step function potential scattering"의 두 판 사이의 차이

2020년 11월 14일 (토) 01:49 판

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