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The reflection and transmission coefficients for a potential problem in one dimension are defined. For this purpose it is sufficient to know the behavior of the wave function at large distances. To set up the problem one needs to impose a suitable boundary condition on the wave function at large distances. It is shown that, for real potentials, the probability conservation implies that the reflection and transmission coefficients add to unity.
$\newcommand{\Label}[1]{\label{#1}}\newcommand{\Prime}{{^\prime}}
\newcommand{\pp}[2][]{\frac{\partial#1}{\partial #2}}\newcommand{\PP}[2][]{\frac{\partial^2#1}{\partial #2^2}}
\newcommand{\dd}[2][]{\frac{d#1}{d #2}} \newcommand{\DD}[2][]{\frac{d^2#1}{d #2^2}}$
Potential properties at $\pm \infty$
We shall define reflection and transmission coefficients for a particle incident on a potential which tends to constant values
\begin{equation}\label{EQ01}
V(x) \to \begin{cases}
V_1 & \text{as } x\to -\infty,\\
V_2 & \text{as } x\to + \infty.
\end{cases}
\end{equation}
\paragraph{Set up solution for $E > V_!, V_2$\\}
Since the potential becomes constant at large distances, the particle travels like a free particle with momenta \(p_{1,2}^2 = \sqrt{2m(E-V_{1,2})}\). The wave function for large distances is therefore given by \begin{equation}\label{EQ02} u_E (x) = \begin{cases} A e^{ikx} + B e^{-ikx}, & \text{ as } x\to -\infty,\\ C e^{iqx} + D e^{-iqx}, & \text{ as } x\to \infty. \end{cases} \end{equation} where \begin{equation} k=p_1/\hbar=\sqrt{\frac{2m(E-V_1)}{\hbar^2}}, \qquad q = p_2/\hbar =\sqrt{\frac{2m(E-V_2)}{\hbar^2}}. \end{equation} The equations \eqref{EQ02} follow from the Schr\"{o}dinger equation, by replacing the potential by constant \(V_{1,2}\) at large distances.
Assume incident wave comes from $-\infty$
For a beam incident from the left, we can identify the incident, reflected and transmitted waves as follows.
| Wave Type | Wave function | Flux |
| Incident beam | \(A e^{ikx}\) | \(\frac{\hbar k}{m}|A^2|\) |
| Reflected beam | \(B e^{-ikx}\) | \(\frac{\hbar k}{m}|B^2|\) |
| Transmitted beam | \(C e^{iqx}\) | \(\frac{\hbar q}{m}|C^2|\) |
Boundary condition for \(x\to\infty\)
We have not shown the term proportional to \(D\) in the table. There should be no wave travelling to the left in the transmission region. Thus for physical reasons,
we must set \(D=0\).
Define reflection and transmission coefficients
The last column in the table shows the flux as computed using probability current \begin{equation}\label{EQ04} J = \frac{\hbar}{2im}\left\{ \psi^*(x)\dd[\psi]{x} - \psi(x)\dd[\psi^*]{x} \right\} \end{equation}
The reflection and transmission coefficients are computed as ratios of corresponding fluxes with the flux of the incident beam.
\begin{eqnarray}
\text{Transmission coefficient } &=& \frac{\text{Flux of transmitted
wave}}{\text{Flux of incident wave}} = \frac{|B|^2}{|A|^2}\\
\text{Reflection coefficient } &=& \frac{\text{Flux of reflected
wave}}{\text{Flux of incident wave}} = \frac{q}{k}\frac{|C|^2}{|A|^2}
\end{eqnarray}
Probability conservation implies $T+R =1$
The state of system corresponding to wave function \(\psi_E(x)\), solution of time independent Schrodinger has time dependence given by
\begin{equation}
H u_E(x) = Eu_E(x) \Longrightarrow \text{ wave function at time } t =
\psi(x,t) = u_E(x)e^{-iEt/\hbar}.
\end{equation}
Hence the probability density, as given by \(\rho(x,t)= |\psi(x,t)|^2 =|u_E(x)|^2\), is independent of time. Thus the equation of continuity, in one dimension, takes the form,
\begin{equation}
\pp[\rho]{t} + \pp[j]{x} = 0 \Longrightarrow \pp[j]{x} = 0
\end{equation}
Therefore \(j(x,t)\) is independent of \(x\) also. We therefore compute the flux density at large distances using the wave function at large distance in
\eqref{EQ02}
\begin{eqnarray} j(-\infty) &=& \frac{\hbar k}{m}(|A|^2-|B|^2),\\ j(+\infty) &=& \frac{\hbar q}{m}(|C|^2-|D|^2) = \frac{\hbar q}{m}|C|^2,\qquad \because D=0 \end{eqnarray}
Setting \(D=0\), and equating the values at \(+\infty\) and \(-\infty\)
\begin{equation}
j(-\infty) = j(+\infty)\Longrightarrow \frac{\hbar k}{m}(|A|^2-|B|^2)
=\frac{\hbar q}{m}|C|^2
\end{equation}
The last relation can be rearranged to give \(T+R=1\).
