Operators, Expectation Values, and the Harmonic Oscillator

A classical solution reports the position of a particle as a function of time. The wave nature of matter makes that impossible: the most that can be known is the probability distribution $∣Ψ (x, t) ∣^{2}$ . Predictions about measurements are therefore statistical, and they are computed from the wave function as expectation values.¹

Expectation values

The expectation value of position is the average of $x$ weighted by the probability density, the value obtained by measuring $x$ on a large number of identically prepared particles:

⟨ x ⟩ = - \infty \int \infty Ψ^{*} (x, t) x Ψ (x, t) d x .

For a stationary state the time-dependent phase cancels in $Ψ^{*} Ψ$ , so

⟨ x ⟩ = - \infty \int \infty ψ^{*} (x) x ψ (x) d x .

For the infinite square well the density $∣ ψ_{n} ∣^{2}$ is symmetric about the midpoint, so $⟨ x ⟩ = L /2$ for every $n$ , whether by symmetry or direct integration.

The last clause matters. For even $n$ in the well, $ψ_{n} (L /2) = 0$ , so the probability of measuring $x = L /2$ is zero — yet $⟨ x ⟩ = L /2$ , because the density is symmetric about that point. The expectation value is a mean, not a prediction of one outcome.

The expectation value is the balance point of the probability density: each slice of x contributes weighted by the height of the modulus-squared curve, and the mean sits under the centroid.

Operators

Position is easy because it is just a number multiplying $ψ$ . Momentum is not: by the uncertainty principle $p$ cannot be written as a function of $x$ , so $⟨ p ⟩$ cannot be computed as $\int ψ^{*} p (x) ψ d x$ . Momentum is instead represented by a differential operator acting on the wave function.

The operator sits between $Ψ^{*}$ and $Ψ$ , acting on $Ψ$ to its right. For a multiplicative operator like $f (x)$ the placement is immaterial, but for a derivative it is essential: $Ψ^{*} \partial_{x} Ψ \neq = (\partial_{x} Ψ^{*}) Ψ$ in general. Squaring the operator gives $⟨ p^{2} ⟩$ :

⟨ p^{2} ⟩ = - \infty \int \infty Ψ^{*} (\frac{ℏ}{i} \frac{\partial}{\partial x}) (\frac{ℏ}{i} \frac{\partial}{\partial x}) Ψ d x = - ℏ^{2} - \infty \int \infty Ψ^{*} \frac{\partial ^{2} Ψ}{\partial x ^{2}} d x .

Momentum in the ground state. For the well's ground state $ψ_{1} = 2/ L sin (π x / L)$ ,

⟨ p ⟩ = \frac{ℏ}{i} \frac{2}{L} 0 \int L sin \frac{π x}{L} \frac{π}{L} cos \frac{π x}{L} d x = 0,

since the particle is equally likely to move in either direction. But $⟨ p^{2} ⟩$ is not zero:

p_{op}^{2} ψ_{1} = - ℏ^{2} \frac{d ^{2} ψ _{1}}{d x ^{2}} = ℏ^{2} \frac{π ^{2}}{L ^{2}} ψ_{1} ⟹ ⟨ p^{2} ⟩ = \frac{ℏ ^{2} π ^{2}}{L ^{2}} = 2 m E_{1} .

A vanishing mean momentum with a nonzero mean-square momentum is the signature of a standing wave: equal parts moving left and right, so the average cancels while the average of the square does not.²

The Hamiltonian operator

Writing the classical energy $H = p^{2} /2 m + V$ and replacing $p$ by $p_{op}$ turns the total energy into the Hamiltonian operator:

H_{op} = \frac{p _{op}^{2}}{2 m} + V (x) = - \frac{ℏ ^{2}}{2 m} \frac{\partial ^{2}}{\partial x ^{2}} + V (x) .

The time-independent Schrödinger equation compresses to an eigenvalue equation,

H_{op} ψ = E ψ,

read as: the stationary states are the eigenfunctions of the Hamiltonian, and the allowed energies are its eigenvalues. This form generalizes cleanly — write the classical energy in terms of position and momentum, replace each momentum by its operator, and the Hamiltonian for many particles in three dimensions follows the same way, as it does for hydrogen.

Physical quantity	Operator
Function of position $f (x)$	$f (x)$ (multiply)
$x$ component of momentum $p_{x}$	$\frac{ℏ}{i} \frac{\partial}{\partial x}$
Kinetic energy $E_{k}$	$- \frac{ℏ ^{2}}{2 m} \frac{\partial ^{2}}{\partial x ^{2}}$
Total energy (Hamiltonian)	$- \frac{ℏ ^{2}}{2 m} \frac{\partial ^{2}}{\partial x ^{2}} + V (x)$
Total energy (time form)	$i ℏ \frac{\partial}{\partial t}$

Each observable is a rule for acting on $ψ$ ; its expectation value is the sandwich $\int ψ^{*} (operator) ψ d x$ .

The simple harmonic oscillator

The next solvable potential is the harmonic oscillator, the parabolic well

V (x) = \frac{1}{2} K x^{2} = \frac{1}{2} m ω^{2} x^{2}, ω = K / m = 2 π f .

Its importance is that any smooth potential near a minimum is parabolic to leading order, so the oscillator describes the vibration of molecules in gases and solids and the normal modes of a crystal.³ Classically a particle of energy $E$ oscillates between the turning points $x = \pm A$ where the kinetic energy vanishes,

E = \frac{1}{2} m ω^{2} A^{2},

and any energy $E \geq 0$ is allowed, with the lowest being $E = 0$ (the particle at rest at the origin).

The Schrödinger equation for this potential is

- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} ψ}{d x ^{2}} + \frac{1}{2} m ω^{2} x^{2} ψ = E ψ .

The exact solution uses Hermite's differential equation, but the qualitative structure follows from the curvature rule alone. For $∣ x ∣ < A$ the energy exceeds the potential, so $ψ$ curves toward the axis and oscillates; for $∣ x ∣ > A$ the potential exceeds the energy, so $ψ$ curves away and must decay. Only special energies give a solution that decays on both sides. They are evenly spaced:

E_{n} = (n + \frac{1}{2}) ℏ ω, n = 0, 1, 2, \dots

Two features separate this from the square well. The spacing is constant, $ℏ ω$ , rather than growing as $n^{2}$ ; and the ground state $E_{0} = \frac{1}{2} ℏ ω$ is nonzero, just as the well has a nonzero $E_{1}$ .

The parabolic potential holds equally spaced levels separated by one quantum hbar omega; every allowed transition obeying Delta n equals one releases or absorbs the same energy, the classical oscillation frequency times h.

The eigenfunctions

The solutions are a Gaussian envelope multiplied by a Hermite polynomial $H_{n}$ of degree $n$ :

ψ_{n} (x) = C_{n} e^{- mω x^{2} /2ℏ} H_{n} (x),

with $C_{n}$ fixed by normalization. The lowest three, in the shorthand $A_{n}$ for the normalization constants, are

ψ_{0} = A_{0} e^{- mω x^{2} /2ℏ}, ψ_{1} = A_{1} \frac{mω}{ℏ} x e^{- mω x^{2} /2ℏ}, ψ_{2} = A_{2} (1 - \frac{2 mω x ^{2}}{ℏ}) e^{- mω x^{2} /2ℏ} .

The ground state is a pure Gaussian bell, the minimum-uncertainty wave packet; each higher state adds one node. The polynomial degree equals $n$ , so $ψ_{n}$ has $n$ zeros.

The first three oscillator states: a Gaussian ground state, an antisymmetric first excited state with one node, and a symmetric second state with two nodes, each under a Gaussian envelope.

At large $n$ the probability density $∣ ψ_{n} ∣^{2}$ develops $n$ peaks and, smeared over a detector window, approaches the classical distribution, which piles up near the turning points where the particle moves slowest. This is the correspondence principle again, now for the oscillator.

Parity

The oscillator potential is symmetric, $V (- x) = V (x)$ , so the Hamiltonian is unchanged by the reflection $x \to - x$ . This operation is parity, denoted $P$ . If $ψ (x)$ solves $H_{op} ψ = E ψ$ , then $ψ (- x)$ solves the same equation with the same energy. When the level is nondegenerate, $ψ (- x)$ can differ from $ψ (x)$ only by a constant $C$ , and applying $P$ twice gives $C^{2} = 1$ , so $C = \pm 1$ .

For the oscillator, states with even $n$ are even and states with odd $n$ are odd: $ψ_{0}$ and $ψ_{2}$ are symmetric, $ψ_{1}$ is antisymmetric. Parity organizes the states of every symmetric system and returns in the analysis of atomic and particle transitions.

An even state maps onto itself under reflection through the origin; an odd state maps onto its own negative, matching magnitude but flipping sign across x equals zero.

Transitions and the selection rule

A stationary state does not radiate; its density is time-independent. Emission or absorption of light requires a superposition of two states whose combined density oscillates. The strength of a dipole transition between states $n$ and $m$ is governed by the integral $\int ψ_{n} x ψ_{m} d x$ , which for the oscillator has the property

- \infty \int \infty ψ_{n} x ψ_{m} d x = 0 unless n - m = \pm 1.

Parity makes this immediate: $x$ is odd, so the product $ψ_{n} x ψ_{m}$ integrates to zero over the symmetric range unless $ψ_{n}$ and $ψ_{m}$ have opposite parity, and for the oscillator opposite parity plus a nonzero integral forces $∣ n - m ∣ = 1$ . The result is a selection rule.

The equal spacing and the $Δ n = \pm 1$ rule together give a single emission line at the classical frequency. This is the assumption Planck built into the blackbody derivation: an oscillator exchanges energy with the field only in units of $ℏ ω$ . When the potential is not symmetric or the barrier is finite, the states are no longer confined, which is the setting for reflection, transmission, and tunneling.

Tipler & Llewellyn, Modern Physics, §6-4 — expectation values as probability-weighted averages, the momentum operator $p_{op} = (ℏ/ i) \partial_{x}$ , operator ordering, and the Hamiltonian form $H_{op} ψ = E ψ$ (Table 6-1). ↩
Tipler & Llewellyn, Modern Physics, §6-4, Example 6-5 — $⟨ p ⟩ = 0$ and $⟨ p^{2} ⟩ = ℏ^{2} π^{2} / L^{2} = 2 m E_{1}$ for the infinite-well ground state. ↩
Tipler & Llewellyn, Modern Physics, §6-5 — the simple harmonic oscillator: classical turning points, the levels $E_{n} = (n + \frac{1}{2}) ℏ ω$ , the Gaussian-times-Hermite eigenfunctions, and the $Δ n = \pm 1$ selection rule. ↩