A particle passing over a well, or over a barrier with energy to spare, usually reflects a little. But at special energies the reflection vanishes completely and the particle passes as if the obstacle were not there. At those energies a whole number of half-wavelengths fits across the obstacle, and the waves reflected from its two edges cancel exactly.

A thin film of soap or oil reflects some colours strongly and others not at all: where the film is a whole number of half-wavelengths thick, the reflections from its two surfaces cancel. Matter waves crossing a well do the same.

The transmission probability over a square well of depth 5 and width 2, against the energy, in units with $\hbar^{2}/2m=1$. It reaches exactly 1 at the resonances, where the wavelength inside the well fits a whole number of times into twice its width, and between them dips.

For a particle of energy $E>0$ crossing a square well of depth $V_{0}$ and width $a$, the transmission probability has the same form as for a barrier with $\sinh$ replaced by . It reaches one whenever , with the wavenumber inside the well.

• $k'=\sqrt{2m(E+V_{0})}/\hbar$; resonance means the width holds a whole number of half-wavelengths of the wave inside.

The second term of the bracket is proportional to $\sin^{2}k'a$, so it vanishes exactly when $k'a$ is a whole multiple of , and then . Solving for the energy gives the resonant energies, the levels of an infinite well of width shifted down by the depth.

The exact transmission of the well.

The correction vanishes with the sine.

Where the sine is zero.

Use $k'^{2}=2m(E+V_{0})/\hbar^{2}$.

Transmission over a well is perfect when the width of the well holds a whole number of half-wavelengths of the wave inside it.

The resonant energies of a well are the levels of an infinite well of the same width, shifted down by the depth.

Resonances also occur for a particle passing over a barrier with energy above its top, when the wave above the barrier fits a whole number of half-wavelengths into it.

At resonance the waves reflected from the two edges of the well are equal in size and opposite in phase, so they cancel.

Great work! You now know how to:

• find the resonant energies of a well
• explain perfect transmission as cancelling reflections
• connect sharp resonances to nearly bound states