Schrödinger Equation Solutions

Solving the time-independent Schrödinger equation for analytically tractable potentials is the cornerstone of quantum mechanics. These solutions are not mere mathematical exercises; they provide the fundamental language for describing quantized energy levels, atomic structure, and the behavior of particles in confinement. Mastering the techniques and results for a few canonical systems unlocks your ability to model a vast array of physical phenomena, from electrons in nanostructures to molecular vibrations.

The Framework: Time-Independent Schrödinger Equation

The central eigenvalue problem for a particle of mass $m$ in a potential $V (r)$ is the time-independent Schrödinger equation:

$\hat{H} ψ (r) = E ψ (r)$

where the Hamiltonian operator is $\hat{H} = - \frac{ℏ ^{2}}{2 m} \nabla^{2} + V (r)$ . Here, $E$ represents the possible energy eigenvalues, and $ψ (r)$ are the corresponding stationary-state eigenfunctions, or wavefunctions. A bound state is characterized by a wavefunction that decays to zero at large distances, confining the particle to a region of space. The solutions we examine—the infinite square well, harmonic oscillator, and hydrogen atom—are the prototypical examples of bound systems, each introducing essential mathematical methods and physical concepts.

The Infinite Square Well: Quantization from Boundary Conditions

The infinite square well (or particle in a box) models a particle perfectly confined to a one-dimensional region, say from $x = 0$ to $x = L$ . The potential is $V (x) = 0$ inside the well and infinite outside. The infinite potential imposes the rigid boundary condition that the wavefunction must be zero at the walls: $ψ (0) = ψ (L) = 0$ .

Inside the well, the Schrödinger equation simplifies to: $- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} ψ}{d x ^{2}} = E ψ$ The general solution is $ψ (x) = A sin (k x) + B cos (k x)$ , where $k = 2 m E /ℏ$ . Applying $ψ (0) = 0$ forces $B = 0$ . Applying $ψ (L) = 0$ gives $A sin (k L) = 0$ , which requires $k L = nπ$ , for $n = 1, 2, 3, ...$ . This integer $n$ is the quantum number for this system.

The quantization condition leads directly to the discrete bound state energies: $E_{n} = \frac{ℏ ^{2} k _{n}^{2}}{2 m} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m L ^{2}}$ The corresponding normalized wavefunctions are: $ψ_{n} (x) = \frac{2}{L} sin (\frac{nπ x}{L})$ This system showcases how boundary conditions alone force energy quantization. The wavefunctions are standing waves, with $n$ corresponding to the number of half-wavelengths fitting in the box. The ground state ( $n = 1$ ) has no nodes inside the well, and the number of nodes increases with $n$ .

The Harmonic Oscillator: Power Series and Ladder Operators

The harmonic oscillator potential, $V (x) = \frac{1}{2} m ω^{2} x^{2}$ , is paramount for modeling vibrations, from diatomic molecules to quantum field theory. The Schrödinger equation becomes: $- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} ψ}{d x ^{2}} + \frac{1}{2} m ω^{2} x^{2} ψ = E ψ$ This equation is typically solved via the power series method. By introducing dimensionless variables $ξ = \frac{mω}{ℏ} x$ and $ϵ = \frac{2 E}{ℏ ω}$ , the equation simplifies to $\frac{d ^{2} ψ}{d ξ ^{2}} + (ϵ - ξ^{2}) ψ = 0$ . A detailed series solution reveals that for the wavefunction to be normalizable (i.e., not blow up at infinity), the series must terminate, becoming a polynomial. This termination condition requires $ϵ = 2 n + 1$ , where $n = 0, 1, 2, ...$ .

This yields the famous quantized energies: $E_{n} = ℏ ω (n + \frac{1}{2})$ The corresponding wavefunctions are given by $ψ_{n} (x) = N_{n} H_{n} (ξ) e^{- ξ^{2} /2}$ , where $H_{n}$ are the Hermite polynomials and $N_{n}$ is a normalization constant. The zero-point energy $E_{0} = ℏ ω /2$ is a purely quantum mechanical feature. An elegant alternative approach uses ladder operators ( $\overset{a}{^}_{+}$ and $\overset{a}{^}_{-}$ ) to algebraically derive the spectrum without solving differential equations, highlighting the power of operator methods.

The Hydrogen Atom: Separation of Variables in 3D

The hydrogen atom consists of an electron in the Coulomb potential of a proton, $V (r) = - \frac{e ^{2}}{4 π ϵ _{0} r}$ . This is a central force problem, making it ideal for separation of variables in spherical coordinates $(r, θ, ϕ)$ . We assume a product solution for the wavefunction: $ψ (r, θ, ϕ) = R (r) Y (θ, ϕ)$ .

The angular part $Y (θ, ϕ)$ is solved by the spherical harmonics, $Y_{l}^{m} (θ, ϕ)$ . This introduces two quantum numbers:

The azimuthal quantum number $l = 0, 1, 2, ...$ (orbital angular momentum).
The magnetic quantum number $m = - l, - l + 1, ..., l - 1, l$ (angular momentum projection).

Substituting the separated form back leads to the radial equation for $R (r)$ : $- \frac{ℏ ^{2}}{2 m r ^{2}} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + [\frac{ℏ ^{2} l ( l + 1 )}{2 m r ^{2}} - \frac{e ^{2}}{4 π ϵ _{0} r}] R = ER$ Solving this radial equation (e.g., via a power series or other methods) yields the final quantum condition and the principal quantum number $n = 1, 2, 3, ...$ , with the constraint $l < n$ .

The resulting bound state energies depend only on $n$ : $E_{n} = - \frac{m e ^{4}}{2 ( 4 π ϵ _{0} ) ^{2} ℏ ^{2} n ^{2}} = - \frac{13.6 eV}{n ^{2}}$ The wavefunctions $ψ_{n l m} (r, θ, ϕ) = R_{n l} (r) Y_{l}^{m} (θ, ϕ)$ are characterized by all three quantum numbers. The radial functions $R_{n l} (r)$ involve associated Laguerre polynomials and determine the electron's radial probability distribution. The ground state (1s orbital) has $n = 1, l = 0, m = 0$ and is spherically symmetric.

Common Pitfalls

Misapplying Boundary Conditions: In the infinite square well, a common error is to include the $n = 0$ solution from $sin (nπ x / L)$ . This gives a wavefunction that is identically zero everywhere, which is not a physically valid state. Always check that your final wavefunction is normalizable and non-trivial.

Confusing Quantum Number Roles: In the hydrogen atom, remember that the energy depends only on $n$ , but the full state is labeled by $n, l,$ and $m$ . The quantum number $l$ determines the shape and angular momentum, while $m$ determines the orientation. Stating that "the 2s and 2p states have different energies" is incorrect—they are degenerate for the pure Coulomb potential (though this degeneracy is lifted in more complex atoms).

Mishandling the Series Solution for the Harmonic Oscillator: The power series method requires analyzing the asymptotic behavior for large $ξ$ . A frequent mistake is to force the series to terminate for any arbitrary $E$ , not realizing that only specific energies allow the solution to remain normalizable. The termination condition is the source of quantization.

Neglecting Normalization and Interpretation: After solving for the functional form, always normalize the wavefunction. Furthermore, remember that $∣ ψ (x) ∣^{2}$ is a probability density. For the hydrogen atom, $R_{n l} (r)$ alone is not the radial probability density; you must consider the volume element $4 π r^{2} ∣ R_{n l} (r) ∣^{2}$ .

Summary

The infinite square well demonstrates that boundary conditions lead directly to energy quantization $E_{n} \propto n^{2}$ , with wavefunctions as standing sine waves.
The harmonic oscillator, solved via the power series method, yields equally spaced energy levels $E_{n} = ℏ ω (n + 1/2)$ and introduces the vital concept of zero-point energy. Ladder operators provide an elegant algebraic solution.
The hydrogen atom requires separation of variables in spherical coordinates, yielding wavefunctions labeled by three quantum numbers $(n, l, m)$ . The energy depends only on $n$ , resulting in the $E_{n} \propto - 1/ n^{2}$ Rydberg formula.
The mathematical techniques of separation of variables (for problems with symmetric potentials) and the power series method (for equations with variable coefficients) are essential tools for solving the Schrödinger equation analytically.
Understanding the origin and role of quantum numbers in each system is crucial for describing the state of a quantum particle and its properties, such as energy, angular momentum, and spatial distribution.

Schrodinger Equation Solutions