6.2: Potential step
- Page ID
- 14782
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consider a potential step
\[ V(x) = \left\{\begin{aligned}
& V_0 && x < a \\
& V_1 && x > a
\end{aligned}
\right. \label{6.3}\]
Figure \(\PageIndex{1}\): The step potential discussed in the text
Let me define
\[\begin{align} k_0 &= \sqrt{ \dfrac{2 m}{ ℏ^ 2} ( E − V_0 )} , \label{6.4} \\[5pt] k_1 &= \sqrt{ \dfrac{2 m}{ℏ^ 2} ( E − V_1 )} . \label{6.5} \end{align} \]
I assume a beam of particles comes in from the left,
\[ϕ(x) = A_0 e^{ i k_0 x} \, \, \text{for } x < 0 . \label{6.6}\]
At the potential step the particles either get reflected back to region I, or are transmitted to region II. There can thus only be a wave moving to the right in region II, but in region I we have both the incoming and a reflected wave,
\[\begin{align} ϕ_I (x) &= A 0 e^{ i k_0 x} + B_0 e^{ − i k_0 x} , \label{6.7} \\[5pt] ϕ_{II} (x) &= A_1 e^{ i k_1 x} . \label{6.8} \end{align} \]
We define a transmission (\(T\)) and reflection (\(R\)) coefficient as the ratio of currents between reflected or transmitted wave and the incoming wave, where we have canceled a common factor
\[\begin{align} R &= \dfrac{| B_0 |^2}{ | A_0 |^2} \\[5pt] T &= \dfrac{k_1 | A_1 |^2 }{ k_0 | A_0 |^2} . \label{6.9} \end{align} \]
Even though we have given up normalisability, we still have the two continuity conditions. At \(x= 0\) these imply, using continuity of \(ϕ\) and \(\dfrac{d}{dx}ϕ\),
\[\begin{align} A_0 + B_0 &= A_1 , \label{6.10} \\[5pt] i k_0 ( A_0 − B_0 ) &= i k_1 A_1 . \label{6.11} \end{align} \]
We thus find
\[\begin{align} A_1 &= \dfrac{2 k_0}{ k_0 + k_1 A_0} \label{6.12} \\[5pt] B_0 &= \dfrac{k_0 − k_1 }{k_0 + k_1 A_0} , \label{6.13} \end{align} \]
and the reflection and transmission coefficients can thus be expressed as
\[\begin{align} R &= \left( \dfrac{k_0 − k_1}{k_0 + k_1} \right)^2 , \label{6.14} \\[5pt] T &= \dfrac{ 4 k_1 k_0}{ ( k_0 + k_1 )^2} . \label{6.15} \end{align} \]
Notice that \(R+ T= 1\)!
Figure \(\PageIndex{2}\): The transmission and reflection coefficients for a square barrier.
In Figure \(\PageIndex{2}\) we have plotted the behaviour of the transmission and reflection of a beam of Hydrogen atoms impinging on a barrier of height 2 meV.