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How to Solve a Wave-Mechanics Problem

A wave-mechanics problem asks for the quantum behavior of a system represented in coordinates. The input is usually a physical setup: a particle in a region, a potential energy function, a boundary condition, or an initial wave packet. The answer may be an energy spectrum, an evolving packet, or a probability for a specified measurement.

The central habit is to translate the words of the problem into a precise Hilbert-space and differential-equation problem before solving equations.

A physical problem before a differential equation

Section titled “A physical problem before a differential equation”

First identify the degrees of freedom, where the coordinate lives, and the quantity you want to predict. Then specify the Hamiltonian, inner product, and boundary conditions. Choose stationary states for an energy question or an initial wavefunction for a time-evolution question. After solving, normalize with the correct measure and calculate the requested observable.

The infinite-well example below follows these steps. You can read it first, then return to the individual choices as needed.

Start by naming the degrees of freedom. A one-dimensional particle has a coordinate xx. A central-potential problem has a radial coordinate rr and angular coordinates (θ,ϕ)(\theta,\phi). A two-level system may not have a coordinate-space wavefunction at all; it is included in this volume only as a canonical Hamiltonian model, not as a differential-equation problem.

For a nonrelativistic particle in one dimension, a common starting point is

H^=−ℏ22md2dx2+V(x).\hat H =-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x).

The symbols are not enough. One must also say where xx lives: the real line, a finite interval, a ring, a half-line, or a union of regions with matching conditions.

Coordinates determine the inner product. On a line,

⟨ψ∣ϕ⟩=∫−∞∞ψ∗(x)ϕ(x) dx.\langle \psi\vert\phi\rangle =\int_{-\infty}^{\infty}\psi^*(x)\phi(x)\,dx.

In three-dimensional Cartesian coordinates,

⟨ψ∣ϕ⟩=∫R3ψ∗(r)ϕ(r) d3r.\langle \psi\vert\phi\rangle =\int_{\mathbb R^3}\psi^*(\mathbf r)\phi(\mathbf r)\,d^3r.

In spherical coordinates the same volume element becomes

d3r=r2sin⁡θ dr dθ dϕ.d^3r=r^2\sin\theta\,dr\,d\theta\,d\phi.

Forgetting the measure changes normalization, expectation values, and sometimes the apparent form of differential operators.

The Hamiltonian encodes kinetic energy, potential energy, and external fields. For a time-independent potential in one dimension, stationary states solve

H^ψn(x)=Enψn(x).\hat H\psi_n(x)=E_n\psi_n(x).

For a time-dependent state, the equation of motion is

iℏ∂ψ(x,t)∂t=H^ψ(x,t).i\hbar\frac{\partial \psi(x,t)}{\partial t} =\hat H\psi(x,t).

The stationary equation finds energy eigenstates. The time-dependent equation evolves arbitrary initial states. When H^\hat H is time independent, these are linked by the expansion

ψ(x,t)=∑ncnψn(x)e−iEnt/ℏ\psi(x,t) =\sum_n c_n\psi_n(x)e^{-iE_n t/\hbar}

for a purely discrete spectrum, with integrals added when continuous spectra are present.

Boundary conditions are physical input. They may represent impenetrable walls, periodic geometry, regularity at the origin, square integrability at infinity, matching across finite potential jumps, or derivative jumps across singular potentials.

For example, an infinite square well on 0<x<L0<x<L imposes

ψ(0)=0,ψ(L)=0.\psi(0)=0,\qquad \psi(L)=0.

A particle on a ring of circumference LL instead imposes periodicity:

ψ(x+L)=ψ(x).\psi(x+L)=\psi(x).

For the Hamiltonian on a ring, the derivative is periodic as well: ψ′(0)=ψ′(L)\psi'(0)=\psi'(L). Both systems use the same kinetic-energy differential expression, but their spectra and eigenfunctions differ.

Solving usually means finding eigenvalues and eigenfunctions, then using them as a basis for dynamics. For bound states, normalization typically means

∫∣ψn(x)∣2 dx=1.\int \lvert\psi_n(x)\rvert^2\,dx=1.

For scattering states and plane waves, ordinary normalization may be impossible. One then uses box normalization, flux normalization, or delta normalization. The correct convention depends on the physical question and should be stated explicitly.

When a stationary equation is not analytically solvable, the numerical route must still respect the same boundary conditions and normalization. ODE Solvers covers shooting methods, while Matrix Diagonalization covers finite Hamiltonian eigenproblems.

Consider a particle of mass mm confined to 0<x<L0<x<L by infinite walls.

The Hilbert space is L2([0,L])L^2([0,L]) with inner product

⟨ψ∣ϕ⟩=∫0Lψ∗(x)ϕ(x) dx.\langle \psi\vert\phi\rangle =\int_0^L \psi^*(x)\phi(x)\,dx.

The Hamiltonian inside the well is

H^=−ℏ22md2dx2,\hat H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2},

with boundary conditions ψ(0)=ψ(L)=0\psi(0)=\psi(L)=0. The stationary Schrödinger equation is

−ℏ22md2ψdx2=Eψ.-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi.

The allowed normalized eigenfunctions and energies are

ψn(x)=2Lsin⁡nπxL,En=n2π2ℏ22mL2,n=1,2,3,…\psi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}, \qquad E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}, \qquad n=1,2,3,\ldots

The physical checks are immediate: energies scale as 1/L21/L^2, the ground-state energy is nonzero, and the nnth eigenfunction has n−1n-1 interior nodes. These checks connect the formula to confinement: a narrower region raises the kinetic-energy scale, and higher energies permit more oscillations.

Before trusting an answer, ask:

  • Does the wavefunction satisfy the boundary conditions?
  • Is it normalized with the correct measure?
  • Are the dimensions of the energy and wavefunction correct?
  • Does the result behave sensibly in a stated limit, such as removing a finite barrier or varying a well width? Specify which quantities are held fixed; enlarging a box at fixed level number is different from keeping its momentum fixed.
  • Are probability currents used where scattering probabilities are claimed?
  • Is the answer a bound state, scattering state, or generalized eigenstate?

These checks catch many wrong solutions before any sophisticated mathematics is needed.

  1. A particle on the interval 0<x<L0<x<L has Hamiltonian −ℏ2d2/(2mdx2)-\hbar^2 d^2/(2m dx^2). List two different boundary-condition choices and explain why they define different physical systems.
Solution

Hard-wall boundary conditions impose ψ(0)=ψ(L)=0\psi(0)=\psi(L)=0 and model an infinite square well. Periodic boundary conditions impose ψ(0)=ψ(L)\psi(0)=\psi(L) and ψ′(0)=ψ′(L)\psi'(0)=\psi'(L) and model a particle on a ring. The same differential expression has different domains, eigenfunctions, and spectra in the two cases.

  1. For the infinite square well, verify the normalization of ψn(x)=2/Lsin⁡(nπx/L)\psi_n(x)=\sqrt{2/L}\sin(n\pi x/L).
Solution

Use

∫0Lsin⁡2nπxL dx=L2.\int_0^L \sin^2\frac{n\pi x}{L}\,dx=\frac{L}{2}.

Then

∫0L∣ψn(x)∣2 dx=2L∫0Lsin⁡2nπxL dx=1.\int_0^L \lvert\psi_n(x)\rvert^2\,dx =\frac{2}{L}\int_0^L \sin^2\frac{n\pi x}{L}\,dx =1.
  1. You know the harmonic-oscillator spectrum but want to calculate transmission through a rectangular barrier. Which additional tools matter most?
Solution

Free plane waves and evanescent solutions, matching at both interfaces, and probability-current ratios. Ladder operators do not solve this piecewise-constant potential. Start with the potential step and reflection/transmission coefficients.

  1. Why is one-dimensional normalization insufficient preparation for radial hydrogen wavefunctions?
Solution

The radial factor R(r)R(r) is normalized with r2drr^2dr after angular normalization. The reduced function u(r)=rR(r)u(r)=rR(r) uses drdr. Using the wrong measure changes probabilities and expectation values even if the differential equation was solved correctly.

  • C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
  • D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
  • E. Merzbacher, Quantum Mechanics, 3rd ed., Wiley, 1998.
  • R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.