Math Infinitum

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Intuition

Shut a particle in a box with perfectly hard walls. Its wavefunction must vanish at both walls, so it must fit a whole number of half-wavelengths inside, exactly like a guitar string. Each fit gives one energy, growing as the square of the number of half-waves, and even the lowest is not zero: squeezing a particle into a box costs energy.

A guitar string pinned at both ends can only vibrate in whole numbers of half-waves: the fundamental, the octave, and so on. The particle in a box has exactly the same set of shapes; its energies go up as the squares of the harmonics.

The first three states of a particle in a box of width $L$ , each drawn on its own energy level, in units of the ground energy $E_{1}$ : 1, 4 and 9. The $n$ -th state has $n-1$ nodes inside the box.

Energies and states of the box

For $V=0$ on $0<x<L$ and infinite outside, the stationary states are sines that vanish at both walls. The energies grow as $n^{2}$ ; the states are orthonormal and complete, so any state in the box is a series of them.

\begin{aligned}E_{n}&=\frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}\\ \psi_{n}(x)&=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L},\quad n=1,2,3,\dots\end{aligned}

Properties

The ground energy $E_{1}$ is not zero: a particle confined to width $L$ has $\Delta x\lesssim L$ and so $Δ p ≳ ℏ /$ , which costs kinetic energy.

Solving the box

\psi(0)=\psi(L)=0\ \Longrightarrow\ k=\frac{n\pi}{L},\quad E=\frac{\hbar^{2}k^{2}}{2m}

Inside the box the equation is $\psi''=-k^{2}\psi$ , solved by sines and cosines. The wall at 0 removes the cosine; the wall at $L$ forces the sine to vanish there, which only whole numbers of half-waves can do. Normalising fixes the constant.

Proof steps

\psi=A\sin kx+B\cos kx,\qquad k^{2}=\frac{2mE}{\hbar^{2}}

The general solution where $V=0$ , for $E>0$ .

\psi(0)=B=0

The wall at $x=0$ removes the cosine.

\psi(L)=A\sin kL=0\ \Rightarrow\ kL=n\pi

The wall at $L$ : $A\neq0$ , so $\sin kL=0$ with $n$ ; gives nothing.

\int_{0}^{L}A^{2}\sin^{2}\frac{n\pi x}{L}dx=\frac{A^{2}L}{2}=1\ \Rightarrow\ A=\sqrt{\frac{2}{L}}

The square of a sine averages to one half over whole half-periods.

E_{n}=\frac{\hbar^{2}}{2m}\Bigl(\frac{n\pi}{L}\Bigr)^{2}

Substitute $k=n\pi/L$ .

Applications

Practice

Energies Grow as n^{2}

The energy of the $n$ -th state of a box is $n^{2}$ times the ground energy.

E_{n}=n^{2}E_{1}

Try it

In an infinite well, what is $E_{3}/E_{1}$ ?