Tunneling Through a Square Barrier

The reflection and transmission coefficients for a particle through a potential will be defined in terms of large distance asymptotic properties of the solutions of the Schr\"{o}dinger equation. We shall, therefore, consider the motion of a particle in a one dimensional potential $V(x)$. We assume that the potential is such that it approaches a constant value $V_1$ as  $x\to -\infty$ and a constant value $V_2$ as $x \to + \infty$.
\begin{equation}
  \lim_{x \to -\infty} = V_1, \qquad \lim_{x\to + \infty} = V_2. \Label{EQ01}
\end{equation}

The motion of the particle from $-\infty$ to $+\infty$ is possible only when  energy $E$ of the particle is greater than both $E_1$ and $E_2$. We assume this to be the case. The Schr\"{o}dinger equation for the particle is
\begin{equation}   -\frac{\hbar^2}{2m} \DD[\psi(x)]{x} + V(x) \psi(x) = E \psi(x) .\Label{EQ02} \end{equation}
We look for behaviour of solutions as $x \to -\infty$, so replace $V(x)$ with $V_1$ and solve the resulting equation

\begin{equation}    -\frac{\hbar^2}{2m} \DD[\psi(x)]{x} + V_1 \psi(x) = E \psi(x) .\Label{EQ03} \end{equation}
or
\begin{equation}
  \DD[\psi]{x} + \frac{2m(E-V_1)}{\hbar^2} \psi(x) =0 . \Label{EQ04}
\end{equation}
Defining $k_1$ by
\begin{equation}
    k_1 = \sqrt{\frac{2m(E-V_1)}{\hbar^2}} \Label{EQ05}
\end{equation}

we get, for $x\to -\infty$,
\begin{equation}
    \DD[\psi]{x} + k_1^2 \psi(x) =0 . \Label{EQ06}
\end{equation}
Therefore we have
\begin{equation}
  \psi(x) \to A e^{ik_1x} + B e^{-ik_1x} \Label{EQ07}
\end{equation}
as $x\to -\infty$. Similarly, the most general form of the solution for large $x \to \infty$ is
\begin{equation}
  \psi(x) \to C e^{ik_2x} + De^{-ik_2x} \Label{EQ08}
\end{equation}
where
\begin{equation}
   k_2 = \sqrt{\frac{2m(E-V_2)}{\hbar^2}}. \Label{EQ09}
\end{equation}
The four constants $A,B,C,D$ are not arbitrary and  have two relations to satisfy. Hence
two parameters remain undetermined and there will be two linearly independent solutions for energy $E > V_1, V_2$.

The physical interpretation of the wave function for the problem of reflection and  transmission, requires a suitable additional boundary condition that will restrict further the parameters. We will have only one (linearly independent) solution from which the reflection and transmission coefficients can be determined.

Let us now turn to analysis of scattering experiment. We consider a beam of  particles incident on a target whose effect will be modelled by the potential  $V(x)$ satisfying conditions \eqref{\LabelPrefix;EQ01}.  A part of the beam will get reflected and a part will will be transmitted. If the beam is incident from the left,  (from $=-\infty$), we shall have only transmitted beam on the right of the target. A part of the incident beam will get reflected and will be travelling to the left. Also the transmitted beam will be travelling to the
right.

Thus to the left of the target, we have superposition of plane waves travelling to the left and to the right, but on the right, as $x\to\infty$, we have wave travelling in the positive  $x$ direction only.This experimental situation will be described by a wave function satisfying the following asymptotic behaviour for arge distances.
\begin{eqnarray}
  \psi(x) &\stackrel{x \to -\infty}{\longrightarrow} &  A \exp(ik_1x) +
B\exp(-ik_1x)\Label{EQ10}\\
  \psi(x) &\stackrel{x \to + \infty} {\longrightarrow}&  C \exp(ik_2x)
\Label{EQ11}.
\end{eqnarray}
{\it i.e.} we should look for solutions of the Schr\"{o}dinger equation satisfying \Ref{EQ10} and \Ref{EQ11} ($D=0$ in $\EqRef{EQ08}$
The physical interpretation of the coefficients $A,B,C$ in the wave function is obtained by computing the current density
\begin{equation}
  j(x) =\frac{\hbar}{2im}\left[\psi^*(x)\dd{x} \psi(x) - \psi(x)
\dd{x}\psi^*(x) \right] \label{EQ12}
\end{equation}
for large distances. We have
\begin{equation}
\begin{array}{ll}
   \text{as } x \to -\infty, & j \to (|A|^2-|B|^2)\times \Big(\dfrac{\hbar
k_1}{m}\Big) \\
   \text{as } x \to + \infty & j \to |C|^2\Big( \dfrac{\hbar k_2}{m}\Big)
\end{array}\Label{EQ13}
\end{equation}
In the region $x\to -\infty$ we have a superposition of waves travelling to the right and to the left with fluxes $|A|^2\times \frac{\hbar
k_1}{m}$ and $|B|^2 \times \frac{\hbar k_1}{m}$ respectively. In the region $x\to +\infty$ we have a wave travelling
to the right with flux $|C|^2\frac{\hbar k_2}{m}$. Hence we are led to
define the reflection and transmission coefficients as
\begin{eqnarray}
   T &=& \frac{\text{Flux for the transmitted beam}}{\text{Incident flux}}
      = \left|\frac{C}{A}\right|^2\frac{k_2}{k_1} \\
   R &=& \frac{\text{Flux for the reflected beam}}{\text{Incident flux}}
      = \left|\frac{B}{A}\right|^2.
 \end{eqnarray}
We will now show that the  conservation of probability holds for a real local potential and that it implies
\begin{eqnarray}
    \pp[\rho]{t} &+& \nabla \cdot \vec{j} =0 \Label{EQ16}\\
    \rho(\vec{r},t)=|\psi(x)|^2, \quad \vec{j}(\vec{r},t) &=&
\frac{\hbar}{2im}\Big( \psi^*(\vec{r},t)         \nabla \psi(\vec{r},t) -
\psi(\vec{r},t) \nabla \psi^*(\vec{r},t)\Big).
\end{eqnarray}
For  one dimensional problems, $\vec{j}$ has only one component and
\begin{equation}
     j(x) =\frac{\hbar}{2im}\Big(\psi^*(x) \dd{x}\psi(x) - \psi(x)
           \dd{x}\psi^*(x)\Big).
\end{equation}
As $\psi(x,t)$ is corresponds to a definite energy $E$, its time dependence is given by
\begin{equation}
     \psi(x,t)= e^{-iEt/hbar} \psi(x,0),
\end{equation}
and hence the probability density
\begin{equation}
  \rho = |\psi(x,0)|^2
  \end{equation}
is independent of time and $\dd[\rho]{t}=0$. Thus for a stationary state in  one dimension $\EqRef{EQ16}$ implies that
\begin{equation}
  \dd[j]{x}  \Rightarrow j(x) = \text{constant},
\end{equation}
Thus $j(x)$ is independent of $x$ and
\begin{equation}
  j(+\infty)=j(-\infty)
\end{equation}
Using $\EqRef{EQ13}$        
\begin{equation}
  \frac{\hbar k_1}{m}(|A|^2-|A|^2) = \frac{\hbar k_2}{m}|C|^2.
\end{equation}
Divide by $\frac{\hbar k_1}{m} |A|^2$ to get
\begin{equation}
   1- \Big|\frac{B}{A}\Big|^2= \frac{k_2}{k_2}|C|^2 \Rightarrow 1-R =T
\end{equation}
Thus as a consequence of probability conservation we get the result that the transmission and reflection coefficients must add to unity, {\it i.e.} nothing is lost in transmission.

We wish to compute the reflection and transmission coefficients for a beam incident on  a target represented by a square barrier. The potential is assumed to be of the form
\begin{equation}
 V(x)  =\begin{cases}      0, & x < 0\\     V_0, & 0\le  x \le L \\      0,   & x > L   \end{cases}.
\end{equation}
where \(V_0 >0\).

\subsection*{Solve eigenvalue problem}
We assume $0< E < V_0$ and write the solutions in the three regions \(R_I,
R_{II}, R_{III}\)
\[\begin{array}{lll}
  R_I = \{x| x < 0\},\quad & R_{II} =\{x|0 \le x \le L\}, \quad & R_{III} =\{x|
x >L\}. \end{array}\]
as
\begin{eqnarray}
  \psi_{_\text{I}}(x) &=& A \exp(ikx) + B \exp(-ikx)\\
  \psi_{_\text{II}}(x) &=& C \exp(\alpha x) + B \exp(-\alpha x)\\
  \psi_{_\text{III}}(x) &=& F \exp(ikx) + G \exp(-ikx)
\end{eqnarray}
where
\begin{equation}
     k =\sqrt{\frac{2mE}{\hbar^2}}, \qquad \alpha =
\sqrt{\frac{2m(V_0-E)}{\hbar^2}}.
\end{equation}
Now we ask what boundary conditions and matching conditions are to be imposed on the solution of the Schr\"{o}dinger equation to fix the unknown constants.
The wave function and its derivative must be continuous at $x=0$ and at $x=L$.

Apart from above general requirements,  a boundary condition specific to the problem of finding the reflection and transmission coefficients is to be imposed. Let us assume that the incident beam is coming from the left. In general some particles will be reflected and some will be transmitted. Thus we should expect particles travelling both ways, to the left and to the right in region $R_{_\text{I}}$, but in  $R_{_\text{III}}$ there are only transmitted particles travelling to the right. Thus the solution in $R_{_\text{III}}$ should have only $\exp(ikx)$ term and we must set $G=0$.

Setting $G=0$ and requiring continuity of wave function and its derivative at $x=0$ and at $x=L$ gives
\begin{equation}
  \begin{array}{lc}
  \underline{x=0} & A+B = C+D \\
                   & ik(A-B) = \alpha(C-D)\\
  \underline{x=L}  & C \exp(\alpha L) + D\exp(-\alpha L) = F \exp(ikL)\\
                   & i\alpha C \exp(\alpha L) - i\alpha D\exp(-\alpha L) =ik F
\exp(ikL)
  \end{array}
\end{equation}
Solving these equations for $F/A$ we get the transmission amplitude as
\begin{equation}
  S(E) = \frac{F}{A} = \frac{2i\alpha k}{2i\alpha k \cosh \alpha L +
(k^2-L^2)\sinh \alpha L}
\end{equation}
and the transmission coefficient is
\begin{equation}
  T = |S(E)|^2 = \left[ 1+ \frac{\sinh^2 \alpha L}{4(E/V_0)(1-E/V_0)}\right]^2
\end{equation}
A classical particle with energy $E< V_0$ cannot cross the barrier. However,  quantum mechanically there is nonzero probability that the particle will be transmitted and we say that the particle 'tunnels through the barrier'. For $E\approx V_0, \alpha \approx 0$, and one has
\begin{equation}
  T(E) \to \left(1+ \frac{mV_0L^2}{2\hbar^2} \right)^{-1}.
\end{equation}
For an opaque barrier $\alpha L >> 1$ the transmission coefficient  is given by
\begin{equation}   T(V_0) \approx \Big(\frac{16E(E-V_0)}{V_0^2}\Big)\exp(-2\alpha L). \end{equation}