11.3: Scattering in one dimension- Step function
- Page ID
- 28678
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Firstly, we review the problem of scattering by a step function in one dimension. Consider a particle moving from a region \((x < 0)\) where the potential is \(V = 0\) to a region \((x > 0)\) where the potential is \(V = V_0\).
Assuming the particle energy \(E > V_0\), this is simply the free particle problem, the spatial solution to which is:
\[\Phi = A \text{ exp}(ikx) + B \text{ exp}(−ikx) \quad (x < 0); \quad \Phi = C \text{ exp}(ik'x) + D \text{ exp}(−ik'x) \quad (x > 0) \nonumber\]
where \(k = \sqrt{2mE/}\hbar\) and \(k' = \sqrt{2m(E − V_0)}/\hbar\)
Figure \(\PageIndex{1}\): Scattering at a step function.
From the boundary condition that all particles start from \(x = −\infty\), we can immediately set \(D=0\).
From the condition of continuity of \(\Phi\) and \(d\Phi /dx\) at \(x = 0\) we also require \(A + B = C\) and \(k(A − B) = k' C\)
This gives the reflected amplitude \(B/A = (k − k')/(k + k')\) and the transmitted amplitude \(C/A = 2k/(k + k' )\)
The reflected flux is thus
\[\frac{\hbar k}{m} A^2 \left( \frac{k − k'}{k + k'}\right)^2 \nonumber\]
and the transmitted flux is
\[\frac{\hbar k'}{m} A^2 \left(\frac{2k}{k + k'}\right)^2 \nonumber\]
Note that \(A^2 \neq B^2 + C^2\). The conserved quantity is the flux of particles, not the probability density. In this case the transmitted particles are moving more slowly than the incident ones.
Notice that if \(V_0\) is negative, the transmitted flux gets smaller as \(|V_0|\) gets larger: it is difficult to fall off a big cliff! This anomaly is due to the unphysical potential - the discontinuous first derivative at \(x = 0\).
We have not considered the case of \(E < V_0\). Now the square root is imaginary and \(\Phi (x > 0) = Ce^{-\kappa' x}\) where we define a real quantity \(\kappa' = ik' = \sqrt{2m(V_0 − E)}/\hbar\). The boundary conditions are then \(A+B = C\) and \(ik(A−B) = \kappa' C\), which gives the reflected amplitude \(B/A = (ik−\kappa' )/(ik+\kappa')\) and the transmitted amplitude \(C\kappa' /Ak = 2ik/(ik + \kappa' )\).
Now the reflected flux is equal to the incident flux, and although the wavefunction penetrates the region \(x > 0\), it decays exponentially and there is no propagating wave.