Barrier Width and Probability

A quantum particle with energy $E$ approaching a rectangular potential barrier of height $V_0 > E$ and width $a$ has no classical path through the barrier: classically the particle bounces back every time. Quantum mechanics, however, permits the wavefunction to extend into the barrier region as an exponentially decaying (evanescent) wave — and if the barrier is thin enough, the wavefunction emerges on the other side with nonzero amplitude. This is quantum tunneling.

Wavefunction inside the barrier

For $0 \leq x \leq a$ the time-independent Schrödinger equation becomes

\frac{d^2\psi}{dx^2} = \kappa^2\,\psi, \qquad \kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar},

whose general solution is a real exponential (not a complex oscillatory wave):

\psi_{\text{barrier}}(x) = A e^{-\kappa x} + B e^{+\kappa x}.

The growing term $e^{+\kappa x}$ is kept in the full calculation but contributes negligibly when $\kappa a \gg 1$ (the opaque or "thick barrier" limit). In that limit only the decaying piece survives across the barrier, and the probability amplitude at $x = a$ is suppressed by $e^{-\kappa a}$ relative to its value at $x = 0$ .

The transmission probability

The exact transmission coefficient for a rectangular barrier requires matching boundary conditions at both walls $x = 0$ and $x = a$ . The exact result is:

T = \left[1 + \frac{(k^2 + \kappa^2)^2}{4\,k^2\kappa^2}\sinh^2(\kappa a)\right]^{-1}, \qquad k = \frac{\sqrt{2mE}}{\hbar},

where $k$ is the wave number in the free region and $\sinh(\kappa a) = \tfrac{1}{2}(e^{\kappa a} - e^{-\kappa a})$ .

When $\kappa a \gg 1$ the factor $\sinh^2(\kappa a) \approx \tfrac{1}{4} e^{2\kappa a}$ grows rapidly, so $T$ simplifies to the opaque-barrier approximation:

T \approx \frac{16\,k^2\kappa^2}{(k^2+\kappa^2)^2}\,e^{-2\kappa a} \approx e^{-2\kappa a}.

The prefactor is of order unity and varies slowly, so the dominant dependence on barrier width is the exponential $e^{-2\kappa a}$ .

Why width matters so much

Doubling the barrier width replaces $e^{-2\kappa a}$ with $e^{-4\kappa a} = (e^{-2\kappa a})^2$ , squaring the (already small) transmission probability. For typical atomic-scale values ( $m = m_e$ , $V_0 - E = 1\,\text{eV}$ ) the decay constant is

\kappa = \sqrt{\frac{2m_e \cdot 1\,\text{eV}}{\hbar^2}} \approx 5.1\,\text{nm}^{-1},

derived using $\hbar^2/(2m_e) \approx 3.81\,\text{eV·Å}^2$ . A barrier only $1\,\text{Å}$ wider reduces $T$ by a factor of $e^{-2\kappa\cdot 0.1\,\text{nm}} = e^{-1.02} \approx 0.36$ . This extreme sensitivity to width is what makes scanning tunneling microscopes (STMs) capable of resolving individual atoms: the tunneling current changes by roughly an order of magnitude per angstrom of tip-to-surface distance.

A worked example

An electron with $E = 1.0\,\text{eV}$ meets a barrier of $V_0 = 3.0\,\text{eV}$ and width $a = 2.0\,\text{Å}$ . Using $\hbar^2/(2m_e) = 3.81\,\text{eV·Å}^2$ :

\kappa = \sqrt{\frac{3.0 - 1.0}{3.81}}\,\text{Å}^{-1} = \sqrt{\frac{2.0}{3.81}}\,\text{Å}^{-1} \approx 0.724\,\text{Å}^{-1}.

2\kappa a = 2 \times 0.724 \times 2.0 \approx 2.90.

T \approx e^{-2.90} \approx 0.055 \quad (5.5\%).

Extending the barrier to $a = 3.0\,\text{Å}$ gives $2\kappa a \approx 4.35$ , so $T \approx e^{-4.35} \approx 0.013$ — the probability drops to about $1.3\%$ , a factor of four reduction for just one extra angstrom.

Try it

This is a numerical exercise — return a number. An electron (use $\hbar^2/(2m) = 3.81\,\text{eV·Å}^2$ ) with energy $E = 2.0\,\text{eV}$ faces a barrier of height $V_0 = 4.0\,\text{eV}$ and width $a = 3.0\,\text{Å}$ . Using the opaque-barrier approximation $T \approx e^{-2\kappa a}$ , compute the transmission probability.

Run your code to see the quantum state.