Make the walls of the box finite and the wavefunction no longer stops dead at them: it seeps into the walls and decays there. The energies come out a little lower than in the infinite box, there are only finitely many bound states, and however shallow the well there is always at least one. Finding them means solving an equation that has no closed form, so it is solved by drawing two curves and reading off where they cross.

A plucked string fixed loosely at its ends, able to move a little at the pegs, vibrates slightly lower than one clamped hard: the loose ends let each shape stretch a little further.

The even states of a finite well with $z_{0}=a\sqrt{2mV_{0}}/\hbar=4$: the solutions of $\tan z=\sqrt{z_{0}^{2}/z^{2}-1}$ with $z=ka$. The curves cross twice, at the dots, so this well has two even bound states; with one odd state between them it has three in all.

Take $V=-V_{0}$ for $|x|<a$ and $0$ outside. A bound state has : it oscillates inside with wavenumber and decays outside with rate . Matching the logarithmic derivative at gives one transcendental equation for the even states and one for the odd states.

• In the variables $z=ka$ and $z_{0}=a\sqrt{2mV_{0}}/\hbar$ the even condition is .

Match the wavefunction and its derivative at the edge $x=a$, where the potential is finite. Dividing the two matching equations eliminates the unknown amplitude and leaves one equation between $k$ and $\kappa$ — equality of the logarithmic derivatives on the two sides.

$\psi$ is continuous at $x=a$.

$\psi'$ is continuous at $x=a$, since the step is finite.

Divide the second by the first: $A$ drops out.

Both come from the same energy: $k^{2}=2m(E+V_{0})/\hbar^{2}$ and .

A one-dimensional attractive well, however shallow or narrow, always has at least one bound state: the two curves of the even condition always cross once.

The number of bound states of a finite well is the smallest whole number that is at least $2z_{0}/\pi$.

Letting the wavefunction spread into the walls lets it bend less sharply, which lowers its kinetic energy compared with the infinite well of the same width.

Outside a finite well the bound wavefunction decays as $e^{-\kappa d}$ at a distance $d$. The particle has a small probability of being found in the classically forbidden region.

Great work! You now know how to:

• set up the finite well and match at its edges
• solve the quantisation condition graphically
• count bound states from $z_{0}$ and explain penetration into the walls