A potential built from pieces — steps, walls, wells — is solved piece by piece, and the pieces must be joined. The joining rules come from the equation itself: the wavefunction is always continuous, its slope is continuous wherever the potential is finite, it vanishes at an infinite wall, and at a delta-function spike its slope jumps by an amount proportional to the value there.

A rope tied to a wall cannot move at the wall; a rope passing a smooth post keeps both its position and its slope; a rope pulled sharply at one point by a hook bends there in a corner. Each rule for the wavefunction is one of these.

The wavefunction $e^{-|x|}$ of a delta-function well: continuous everywhere, with a corner at the origin where the delta sits. The slope jumps from $+1$ to $-1$, by an amount fixed by the strength of the delta.

Integrating the Schrödinger equation across a point shows how the wavefunction and its derivative may behave there. Where the potential is finite, both are continuous; where it has a delta function the derivative jumps; where it is infinite the wavefunction vanishes.

• $\psi$ is continuous everywhere: a jump in $\psi$ would make $\psi''$ contain the derivative of a delta function, which nothing in the equation can balance.
• $\psi'$ is continuous where is finite, even at a step: the integral over a vanishing interval is zero.

Integrate the Schrödinger equation over a short interval around the origin and let it shrink. The second derivative integrates to the jump in the first derivative; the energy term vanishes with the interval's width; the delta function survives and picks out $\psi(0)$.

Integrate each term of the equation over $[-\varepsilon,\varepsilon]$.

The derivative integrates to a difference, the delta picks out the value at 0.

A bounded function integrated over a vanishing interval gives zero.

Rearrange: the slope jumps by an amount proportional to the value.

The wavefunction itself is continuous at every point, whatever the potential does: a jump would make its second derivative too singular to balance.

At a delta-function potential of strength $\alpha$, the slope of the wavefunction jumps by $2m\alpha/\hbar^{2}$ times the value there.

The wavefunction cannot enter where the potential is infinite, so it vanishes at an infinite wall. Its slope may be anything there.

At a finite step the integral of the potential over a shrinking interval goes to zero, so the slope has no jump.

Great work! You now know how to:

• state the matching conditions at steps, walls and delta functions
• derive the jump in slope at a delta function
• explain what makes bound-state energies discrete