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Finite Well and Tunneling Probability Tool

Modern Physics • Quantum Mechanics

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Solve for bound states in a 1D finite square well and estimate transmission through a rectangular barrier. Enter the particle mass, well or barrier depth \(V_0\), width \(L\), and particle energy \(E\) to inspect bound levels, penetration depth, and tunneling probability.

Inputs
For the symmetric finite square well, bound states satisfy the transcendental conditions \[ z\tan z = \sqrt{z_0^2-z^2}\quad\text{(even)}, \qquad -z\cot z = \sqrt{z_0^2-z^2}\quad\text{(odd)}, \] where \[ z_0 = \frac{L}{2}\frac{\sqrt{2mV_0}}{\hbar}. \] For tunneling through a thick rectangular barrier with \(E
Animation and graph controls
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Interactive finite-well and tunneling preview
The left panel shows either the bound-state ladder or transmission probability versus energy. The right panel shows a bound-state wave function in a finite well and a tunneling wave through a rectangular barrier. Drag inside either panel to pan after zooming.
Left panel: bound-state or transmission graph. Right panel: finite-well and barrier concepts. Mouse-wheel zoom affects only the hovered panel.
Enter values and click “Calculate”.

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Frequently Asked Questions

What is the main formula for tunneling probability in this calculator?

For a thick rectangular barrier with E below V0, the calculator uses T ≈ 16E(V0 − E) / V0² × exp(−2κL), where κ = √(2m(V0 − E)) / ħ.

Why is the tunneling probability often extremely small?

Because the dominant factor is the exponential term exp(−2κL). Even moderate increases in barrier width, barrier height, or particle mass can make the transmission probability shrink very quickly.

What is the difference between a finite well and a barrier?

A finite well can support bound states with discrete energies below the outside potential, while a barrier describes a classically forbidden region that a particle may still cross by quantum tunneling.

Why are finite-well bound states not given by a simple formula like the infinite well?

Because matching the wave function and its derivative at the boundaries leads to transcendental equations such as k tan(ka) = α and −k cot(ka) = α, which generally require numerical solution.