LESSON NOTES · 02

Schrödinger dynamics and the box

Prerequisites: State and measurement; second derivatives and integration

1. Time evolution and stationary density

For an isolated nonrelativistic particle the Schrödinger equation gives unitary dynamics. A time-independent Hamiltonian permits eigenfunctions with phase evolution. The phase cancels from a single eigenstate’s density, but relative phases survive in a superposition. An infinite box is idealized confinement, not a literal molecular wall.

\[ i\hbar\partial_t\psi=\hat H\psi,\qquad\psi_n(x,t)=\psi_n(x)e^{-iE_nt/\hbar}. \]

2. Impose both boundaries

Inside \(0<x<L\), the potential is zero, and infinite outside. The general interior solution is \(A\sin kx+B\cos kx\). The left boundary eliminates \(B\); the right requires integer multiples of \(\pi\). The index \(n=0\) gives the zero function, which cannot normalize. Quantization comes from both boundaries, not an additional classical rule.

\[ \psi^{\prime\prime}+k^2\psi=0,\quad\psi(0)=\psi(L)=0\Rightarrow k_n=n\pi/L,\quad n=1,2,\ldots \]

3. Normalize and recover energies

The sine-square integral is \(L/2\), fixing the amplitude. Each wavefunction has \(n-1\) interior nodes. Energies scale as \(n^2/L^2\), so doubling the width quarters every energy. A stationary density is not spatially uniform.

\[ 1=|A|^2\int_0^L\sin^2(n\pi x/L)dx=|A|^2L/2,\quad\psi_n=\sqrt{2/L}\sin(n\pi x/L),\quad E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}. \]

4. Relative phase creates changing density

Use equal weights of two real eigenfunctions. Squaring the complex sum gives an interference term whose frequency is the energy difference divided by \(\hbar\). Orthonormality makes that term integrate to zero, so total probability remains one while the density changes locally.

\[ \psi=\frac{\psi_1e^{-iE_1t/\hbar}+\psi_2e^{-iE_2t/\hbar}}{\sqrt2},\quad|\psi|^2=\tfrac12(\psi_1^2+\psi_2^2)+\psi_1\psi_2\cos[(E_2-E_1)t/\hbar]. \]

Teaching schematic drawn from the equations or algorithm steps in this lesson; not measured data.

Worked probability integral

For the ground state, integrate over the middle half of the box. The result is larger than \(1/2\) because the ground-state density is concentrated near the center. This is an exact analytic toy model, not a simulated molecule.

\[ P(L/4<x<3L/4)=\frac2L\int_{L/4}^{3L/4}\sin^2(\pi x/L)dx=\frac12+\frac1\pi\simeq0.8183. \]

Check yourself

Why no zero-energy state?

Answer and reasoning

It would satisfy both boundaries only as the zero function.

Do spacings shrink at high \(n\)?

Answer and reasoning

No: \(E_{n+1}-E_n=(2n+1)E_1\).

Can density change in a stationary energy eigenstate?

Answer and reasoning

Its overall time phase cancels, so not under this time-independent Hamiltonian.

References and further reading

• MIT 5.73: Infinite box

Derivations and numerical examples here are original teaching constructions, not copied passages or reported research data.

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