Tunneling Through a Barrier

A particle with energy $E$ below a barrier of height $V_0$ has a non-zero probability of being found on the other side — quantum tunneling.

Why a wavepacket?

The plane wave $e^{ikx}$ normally used to derive a transmission coefficient carries a single, perfectly sharp energy $E = \hbar^2k^2/2m$ — but it also extends over all of space for all time. Nothing "arrives" at the barrier and nothing is ever "detected" on the other side; it's a steady-state idealization, not a particle.

A real particle is localized: it starts somewhere, travels toward the barrier, and eventually gets found on one side or the other. But localizing it in position forces its momentum — and therefore its energy — to spread out, by the uncertainty principle. So a real wavepacket isn't at energy $E$; it's a superposition of many energies, each with its own transmission probability $T(E)$.

That means the fraction of a real wavepacket that tunnels through isn't $T$ evaluated at some single "mean" energy — it's the energy-averaged transmission

where $|\varphi(k)|^2$ is the wavepacket's momentum distribution. The simulation further down checks this directly: it evolves a genuine wavepacket in time, measures how much ends up transmitted, and compares it to $T_{\text{eff}}$ computed from this formula.

Exact transmission coefficient

For a rectangular barrier of width $a$ and height $V_0$, with $\varepsilon = E/V_0$ and $\sigma = 2mV_0a^2/\hbar^2$:

for $\varepsilon < 1$ (tunneling below the barrier). For $\varepsilon > 1$ (above the barrier) the same formula holds with $\sinh \to \sin$. This is the exact result — valid at any energy, not just deep in the tunneling regime.

For the barrier used in the simulation below ($V_0 = 3$, $a = 1$) and a particle at the wavepacket's mean energy ($E = 2$, so $\varepsilon = 2/3$, $\sigma = 6$): $T \approx 0.192$. Watch for this number further down — it's not what the simulation measures.

The deep-tunneling approximation

When the barrier is opaque ($\sqrt{\sigma(1-\varepsilon)} \gg 1$), $\sinh(x) \approx \tfrac{1}{2} e^{x}$ and the exact formula above reduces to the familiar

— the approximation usually quoted for tunneling, and a good one once the barrier is thick or high enough. It breaks down as $E \to V_0$, where the exact formula must be used instead.

Watching it happen

Scanning tunneling microscopy

The exponential sensitivity of $T$ to the gap width $a$ — not just to its existence — is what gives the scanning tunneling microscope its atomic resolution: changing the tip-sample distance by about one atomic radius changes the tunneling current by an order of magnitude. A real sample is a surface, not a line: the gap is a function of two coordinates, $a(x, y)$, and the current is highest wherever that gap is smallest — typically directly over an atom.