Particle in a Box

Chem 3240 · Lecture 3.3

Davit Potoyan

Classical vs Quantum

  • A particle trapped in a region \([0, L]\)
  • Classical: bounces freely, found everywhere with equal probability
  • Quantum: described by a wavefunction \(\psi(x)\), found in some regions with high probability, others with zero
  • Wavefunctions are standing waves, but with a probabilistic meaning

The Model

Infinite walls confine the particle:

\[V(x) = \begin{cases} \infty & x = 0 \text{ or } x = L \\ 0 & 0 < x < L \end{cases}\]

Boundary conditions: \[\psi(0) = \psi(L) = 0\]

Inside, only kinetic energy: \[\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\]

Solving the Schrödinger Equation

The eigenvalue problem \(\hat{H}\psi = E\psi\) becomes

\[\psi''(x) = -k^2\psi(x), \qquad k^2 = \frac{2mE}{\hbar^2}\]

General solution: \[\psi(x) = A\cos(kx) + B\sin(kx)\]

  • \(\psi(0)=0 \Rightarrow A = 0\)
  • \(\psi(L)=0 \Rightarrow \sin(kL) = 0\)

Quantization Emerges

The wall condition forces

\[kL = n\pi \quad\Rightarrow\quad k = \frac{n\pi}{L}, \quad n = 1, 2, 3, \dots\]

Energy is quantized: \[E_n = \frac{n^2 h^2}{8mL^2}\]

Confining a wave to a finite space forces discrete energy levels. Atoms, molecules, and solids inherit their quantized levels this way.

Normalization Fixes the Amplitude

Require total probability of \(1\):

\[\int_0^L \psi_n(x)^2\, dx = 1\]

Using \(\sin^2\theta = \tfrac{1}{2}(1-\cos 2\theta)\) gives \(\tfrac{B_n^2}{2}L = 1\), so

\[B_n = \sqrt{\frac{2}{L}}\]

The 1D Eigenstates

\[\psi_n(x) = \left(\frac{2}{L}\right)^{1/2}\sin\frac{n\pi x}{L}\]

\[E_n = \frac{n^2 h^2}{8mL^2}\]

  • Higher \(n\): more curvature, more energy
  • State \(n\) has \(n-1\) nodes where \(|\psi_n|^2 = 0\)

Zero-Point Energy

The lowest state \(n=1\) has nonzero energy:

\[E_1 = \frac{h^2}{8mL^2} \neq 0\]

  • Energy is purely kinetic, so a quantum particle is never at rest
  • Level spacing grows with \(n\): \[E_{n+1} - E_n = (2n+1)\frac{h^2}{8mL^2}\]

Smaller box means larger spacing. Quantum effects dominate when a particle is tightly confined.

Nodes and Non-Uniform Probability

  • \(|\psi_n|^2\) is not uniform, unlike the classical prediction
  • Nodes: points where probability is exactly zero
  • State \(n\) has \(n-1\) nodes at \(x = \frac{L}{n}, \frac{2L}{n}, \dots, \frac{(n-1)L}{n}\)
  • The existence of nodes is the signature of wave-like behavior
  • Yet for any \(n\): \(\langle x \rangle = \frac{L}{2}\) and \(\langle p \rangle = 0\) by symmetry

Pure vs Mixed States

Pure state (single eigenstate): probability is time-independent

\[|\psi_n(x,t)|^2 = |\psi_n(x)|^2\]

Mixed state (superposition): probability oscillates in time

\[\psi(x,t) = c_1\psi_1 e^{-iE_1 t/\hbar} + c_2\psi_2 e^{-iE_2 t/\hbar}\]

  • Interference term beats at frequency \((E_1 - E_2)/\hbar\)

The 3D Box

Separable solution \(\psi = X(x)Y(y)Z(z)\):

\[\psi = \sqrt{\frac{8}{abc}}\sin\frac{n_x\pi x}{a}\sin\frac{n_y\pi y}{b}\sin\frac{n_z\pi z}{c}\]

\[E_{n_x,n_y,n_z} = \frac{h^2}{8m}\left(\frac{n_x^2}{a^2} + \frac{n_y^2}{b^2} + \frac{n_z^2}{c^2}\right)\]

  • When \(a = b = c\), different \((n_x, n_y, n_z)\) can share one energy: degeneracy
  • Degeneracy arises from symmetry

Takeaway

Confining a quantum wave to a box quantizes its energy as \(E_n = \frac{n^2 h^2}{8mL^2}\), forbids rest through zero-point energy, and carves the probability into nodes, with symmetry producing degeneracy in higher dimensions.