Particle in a Box

A particle of mass m confined to 0 ≤ x ≤ L by walls it cannot pass: the simplest system in quantum mechanics, and the one where everything that makes it quantum is already visible. Everything below follows from the time-independent Schrödinger equation:

$$-\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2} + V(x)\Psi = E\Psi$$

1. Intuitive Explanation at Increasing Levels

1.1 Grade School Level

Pluck a guitar string. It is pinned at both ends, so it can only ring at certain notes: the open string, then the overtones you get by touching it lightly at the middle, at a third of the way along, and so on. Nothing in between. A particle in a box is the same instrument. The vibration is the particle, the walls are the two fixed ends, and the allowed energies are the notes. The lowest note is the ground state, and there is no silence: the string can never stop vibrating.

1.2 High School Level

A string fixed at both ends must fit a whole number of half-wavelengths between them: one hump for the fundamental, two for the octave, three for the next harmonic. Call that number n. More humps means a shorter wavelength, which by de Broglie means more momentum, which means more energy:

$$\lambda_n = \frac{2L}{n}, \qquad p_n = \frac{h}{\lambda_n} = \frac{n\pi\hbar}{L}, \qquad E_n = \frac{p_n^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2}.$$

Here the instrument differs from the string. A string's harmonics climb as n; the particle's energies climb as $n^2$. The second level has four times the energy of the first, the third has nine: a harmonic series, stretched. That is the whole spectrum, and it needed no calculus. The Play button at the bottom of the page sounds it.

1.3 College Level

Now ask what the vibration actually is. Inside the box the Schrödinger equation says $\Psi'' = -k^2\Psi$, whose solutions are sines and cosines. The walls kill the cosine and force $kL = n\pi$. Normalizing so the total probability is one fixes the amplitude. Section 2 does this step by step. The result is a standing wave $\Psi_n(x)$, and $|\Psi_n(x)|^2$ is where the particle is likely to be found: never at the walls, and for $n = 2$, never in the middle either, exactly where the octave harmonic has its node.

1.4 Grad Level

Drop the coordinates. The state is a vector $|\psi\rangle$ in a Hilbert space, the energy is an operator $\hat H$, and the allowed levels are its eigenvectors:

$$\hat H\,|n\rangle = E_n\,|n\rangle, \qquad \Psi_n(x) = \langle x | n \rangle.$$

The wavefunction is just the state written out in the position basis. The levels are orthonormal and complete,

$$\langle m | n \rangle = \delta_{mn}, \qquad \sum_n |n\rangle\langle n| = 1,$$

so any state is a sum of levels, with coefficients given by inner products:

$$|\psi\rangle = \sum_n c_n\,|n\rangle, \qquad c_n = \langle n | \psi \rangle, \qquad \sum_n |c_n|^2 = 1.$$

$$|\psi(t)\rangle = \sum_n c_n\, e^{-iE_n t/\hbar}\,|n\rangle.$$

A trivial example. Put the particle in an equal mix of the first two levels:

$$|\psi\rangle = \frac{|1\rangle + |2\rangle}{\sqrt 2}.$$

Each energy shows up half the time, the average energy is $\langle\hat H\rangle = \tfrac{1}{2}(E_1 + E_2) = \tfrac{5}{2}E_1$, and the position density

$$|\langle x|\psi(t)\rangle|^2 = \tfrac{1}{2}\left[\Psi_1^2 + \Psi_2^2 + 2\Psi_1\Psi_2\cos\!\left(\tfrac{(E_2 - E_1)\,t}{\hbar}\right)\right]$$

sloshes from one side of the box to the other with period $T = 2\pi\hbar/(E_2 - E_1) = 2\pi\hbar/3E_1$: two notes sounding at once, beating at their difference frequency. That beat is the only thing that ever moves in a box. A single level $|n\rangle$ is a stationary state, a pure tone, and $|\Psi_n|^2$ never changes.

2. Solution

2.1 Inside the Box

Inside the box $V(x) = 0$, so the Schrödinger equation becomes

$$\frac{d^2\Psi}{dx^2} + k^2\Psi = 0, \quad k^2 = \frac{2mE}{\hbar^2}$$

2.2 General Solution

$$\Psi(x) = A\sin(kx) + B\cos(kx)$$

2.3 Boundary Conditions

The walls impose $\Psi(0) = \Psi(L) = 0$, which kills the cosine and quantizes k:

$$\Psi_n(x) = A\sin\left(\frac{n\pi x}{L}\right), \quad n = 1, 2, 3, \ldots$$

2.4 Normalization and Energy

Normalizing over the box gives the stationary states and their energies:

$$\Psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right), \qquad E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$$

n = 1

Figure 1. Ψ₁ (top) and |Ψ₁|² (bottom); E₁ = 1 E₁. Play sounds the first 8 levels as f_n = n² · 55 Hz, since E_n ∝ n².