Math Infinitum

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Intuition

On a line, finding the stationary states means solving one ordinary differential equation for each energy. Where the energy is above the potential the solution wiggles, like a wave; where it is below, the solution curves away from the axis and must die off exponentially. That single fact decides everything that follows: which energies are allowed, why bound states come in a discrete ladder, and how particles leak into places classical mechanics forbids.

A ball rolling in a valley turns back where its energy equals the height of the slope. A quantum particle does not turn back so sharply: its wavefunction slips a little way past the turning point, fading as it goes, before it gives up.

A potential $V(x)$ and an energy $E$ . Between the turning points, where $E>V$ , the region is classically allowed and a solution oscillates; outside them $E<V$ , the region is classically forbidden, and a solution grows or decays exponentially.

The equation for stationary states

For a particle on a line the time-independent Schrödinger equation is a second-order linear ordinary differential equation. Its character at each point depends on the sign of $E-V(x)$ , and the conditions at infinity decide which energies have acceptable solutions.

-\frac{\hbar^{2}}{2m}\,\psi''(x)+V(x)\,\psi(x)=E\,\psi(x)

Reading the equation

Where $E>V$ it reads $\psi''=-k^{2}\psi$ with : the solution curves back towards the axis and oscillates.

Every energy lies above the bottom of the potential

\hat{H}\psi=E\psi,\ \ \psi\ \text{normalisable}\ \Longrightarrow\ E>V_{\min}

Take the expectation value of the Hamiltonian in the eigenstate itself. The kinetic part, after an integration by parts, is an integral of $|\psi'|^{2}$ and so positive, and the potential part is at least $V_{\min}$ because integrates to one.

Proof steps

E=\bra{\psi}\hat{H}\ket{\psi}=\langle T\rangle+\langle V\rangle

In an eigenstate the energy is its own expectation value.

\langle T\rangle=-\frac{\hbar^{2}}{2m}\int\psi^{*}\psi''\,dx=\frac{\hbar^{2}}{2m}\int|\psi'|^{2}\,dx

Integrate by parts; the boundary term vanishes for a normalisable state.

\langle T\rangle>0

An integral of $|\psi'|^{2}$ is positive, since a normalisable $\psi$ cannot be constant.

\langle V\rangle=\int V|\psi|^{2}dx\ge V_{\min}\int|\psi|^{2}dx=V_{\min}

Replace $V$ by its minimum under an integral with weight one.

E=\langle T\rangle+\langle V\rangle>V_{\min}

Add the two.

Applications

Practice

Allowed Means Oscillating

Where the energy exceeds the potential, the equation says the second derivative has the opposite sign to the wavefunction, so it curves back towards the axis and oscillates.

\psi''=-k^{2}\psi,\qquad k^{2}=\frac{2m(E-V)}{\hbar^{2}}

Try it

In a region where $E>V(x)$ , what does a solution of the time-independent Schrödinger equation do?