"Step function potential scattering"의 두 판 사이의 차이

2011년 2월 11일 (금) 03:10 판

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

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
\(A_r + A_l - B_r - B_l = 0\)
\(-A_r + A_l + B_r - B_l =\frac{2m\lambda}{ik\hbar^2}(A_r + A_l)\)

special case of scattering problem \(A_r=1, A_l=r, B_r=t , B_l = 0\)
wave function
\(\psi(x) = \begin{cases} \psi_{\mathrm L}(x) = e^{ikx} + re^{-ikx}, & \text{ if } x<0; \\ \psi_{\mathrm R}(x) =te^{ikx} , & \text{ if } x>0, \end{cases}\)

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

@@ 1번째 줄: / 1번째 줄: @@
 <h5>introduction</h5>
+* [[#]]
+<math>E \psi = -\frac{\hbar^2}{2m}{\partial^2 \psi \over \partial x^2} + V(x)\psi</math>
 *  Let the potential is given by<br><math>\psi(x) = \begin{cases} V_{-} & \text{ if } x<0; \\ V_{+}, & \text{ if } x>0, \end{cases}</math><br>