Operators and Observables

Every electronic structure calculation run on a computer — Hartree–Fock, density functional theory, coupled cluster — begins by writing down the molecular Hamiltonian operator as a sum of one-electron kinetic-energy operators, one-electron nuclear-attraction operators, and two-electron repulsion operators, each built by exactly the substitution rule described here. Quantum chemistry software packages diagonalize matrix representations of $\hat H$ in a chosen basis; the resulting eigenvalues are the orbital and total electronic energies reported in a calculation, and the eigenvectors are the molecular orbital coefficients. NMR and EPR spectroscopy rely on constructing spin operators ($\hat S_x, \hat S_y, \hat S_z$) with their own commutation relations analogous to $[\hat x,\hat p_x]=i\hbar$, and the observed splitting patterns are eigenvalues of spin Hamiltonians built from these operators. Vibrational spectroscopy (IR and Raman) interprets absorption frequencies as energy gaps between eigenvalues of vibrational Hamiltonians constructed the same way as the harmonic-oscillator example above, now applied to real bond force constants extracted from measured spring stiffnesses. Scanning tunneling microscopy and photoelectron spectroscopy are direct experimental measurements of position-space electron density $|\psi|^2$ and orbital energy eigenvalues respectively — both quantities defined only because the position and energy operators tell us precisely what "density" and "energy level" mean at the quantum level. Even reaction-rate theory (transition-state theory, RRKM theory) implicitly uses momentum and kinetic-energy operators to define what "crossing a barrier" means quantum mechanically, feeding into modern computational catalysis and drug-binding free-energy calculations.

:::quiz[The commutator $[\hat x,\hat p_x]$ equals]

• $0$
• $\hbar$
• $i\hbar$ ✓
• $-i\hbar^2$

Ask the tutor: How do I construct the operator for angular momentum, and why does it end up not commuting with itself across different axes?

Classical mechanics describes a particle by a trajectory: at every instant it has a definite position $x(t)$ and a definite momentum $p(t)$, and any function of these, such as kinetic energy $p^2/2m$, is simply evaluated by plugging in the numbers. Quantum mechanics abandons trajectories. The state of a system is a wavefunction $\psi(x,t)$, and physical quantities are extracted from $\psi$ by a completely different mechanism: you act on $\psi$ with an operator. An operator is a rule that turns one function into another — "multiply by $x$," "take the second derivative," "multiply by $x$ and then differentiate." The postulate at the center of this article states that for every classically measurable quantity ("observable") $A$, quantum mechanics assigns a specific linear operator $\hat{A}$, constructed from the operators for position and momentum by the same algebraic formula that defines $A$ classically. Measuring $A$ on a system in state $\psi$ can only yield one of the eigenvalues of $\hat{A}$ — the numbers $a$ for which $\hat{A}\phi = a\phi$ for some function $\phi$. If $\psi$ itself equals one of these eigenfunctions $\phi$, the measurement returns $a$ with certainty. If $\psi$ is instead a superposition of several eigenfunctions, the measurement returns one of their eigenvalues at random, with probabilities set by how much of each eigenfunction is mixed into $\psi$, and only the average over many trials — the expectation value — is predictable in advance. The two operators from which everything else is built are position and momentum. The position operator is disarmingly simple: $\hat{x}$ just means "multiply the wavefunction by $x$." The momentum operator is the genuinely new ingredient of quantum theory:

Momentum, classically a number, becomes an instruction to differentiate and multiply by $-i\hbar$. This looks bizarre until you test it against the free-particle wavefunction $\psi(x) = e^{ikx}$, which describes a particle with classical momentum $p = \hbar k$ (the de Broglie relation). Applying $\hat{p}_x$:

The eigenvalue that pops out, $\hbar k$, is exactly the de Broglie momentum. The operator was not guessed; it was reverse-engineered so that plane waves — the wavefunctions of definite momentum — are its eigenfunctions with the correct eigenvalue. Once $\hat{x}$ and $\hat{p}_x$ exist, every other one-particle observable is built from them using the same functional form as its classical definition. Kinetic energy $T = p_x^2/2m$ becomes

Potential energy $V(x)$, already a function of position alone, becomes the multiplicative operator $\hat{V}(x)$ — multiply by whatever function of $x$ describes the potential (a Coulomb attraction, a harmonic spring, an infinite wall). Total energy, the Hamiltonian, is their sum:

This is precisely the operator appearing in the time-independent Schrödinger equation $\hat{H}\psi = E\psi$: that equation is nothing but the eigenvalue equation for the energy operator. Quantized energy levels — the discrete lines in an atomic spectrum, the shell structure of an atom, the vibrational ladder of a bond — are the eigenvalues of $\hat{H}$, and wavefunctions with sharp, definite energy are its eigenfunctions.

A crucial subtlety is order. Classically $xp_x$ and $p_xx$ are the same number, but $\hat{x}\hat{p}_x$ and $\hat{p}_x\hat{x}$ are different operators, because differentiation does not commute with multiplication by a variable. Acting on a test function $f(x)$:

so that

This nonzero commutator is not a mathematical curiosity — it is the algebraic origin of the position–momentum uncertainty relation and the reason position and momentum cannot simultaneously have sharp values in the same state. Any classical formula involving a product like $xp_x$ must be handled with care; the standard prescription is to symmetrize, replacing $xp_x$ by $\tfrac12(\hat{x}\hat{p}_x+\hat{p}_x\hat{x})$, which guarantees the resulting operator is Hermitian.

1. The operator postulate. For every measurable quantity $A$, there exists a linear operator $\hat{A}$, and measuring $A$ can only yield one of the eigenvalues of $\hat{A}$:

where: $\hat{A}$ is the operator for observable $A$; $\phi_n$ are its eigenfunctions; $a_n$ are the corresponding eigenvalues, i.e. the only numbers a measurement can produce. Reading: this equation says the entire menu of possible outcomes of any measurement is fixed the instant you know the operator — you never need to appeal to experiment to know what values are even allowed, only which one shows up.

2. Canonical operators.

where: $\hat{x}$ acting on $\psi$ means ordinary multiplication; $\hat{p}_x$ means differentiate with respect to $x$ and multiply by $-i\hbar$; $\hbar = h/2\pi$ is the reduced Planck constant. Reading: position keeps its classical, intuitive multiplicative character; momentum is promoted to a differential operator, and this asymmetry between $\hat x$ and $\hat p_x$ is the seed of everything nonclassical that follows, including uncertainty and quantization.

3. Constructing composite operators. Any classical observable $A(x,p_x)$ becomes an operator by direct substitution,

with symmetrization $xp_x \to \tfrac12(\hat x\hat p_x + \hat p_x \hat x)$ wherever position and momentum are multiplied together in the classical expression. where: $A(\hat x,\hat p_x)$ means "take the classical formula and replace every $x$ by $\hat x$, every $p_x$ by $\hat p_x$"; symmetrization is required only when $x$ and $p_x$ appear multiplied in the same term, since $\hat x$ and $\hat p_x$ do not commute. Reading: this is the master recipe. It is how $\hat H$, $\hat L_z$ (angular momentum), and every other operator in chemistry gets built — always starting from the classical mechanics you already know.

4. The Hamiltonian and the energy eigenvalue equation.

where: $\hat H$ is the total-energy operator; $E$ is an allowed (eigenvalue) energy; $\psi$ is the corresponding stationary-state wavefunction. Reading: this is exactly the Time-Independent Schrodinger Equation — not a separate law, but the operator postulate applied specifically to the energy observable.

5. The canonical commutator.

where: $[\hat A,\hat B] \equiv \hat A\hat B - \hat B\hat A$ is the commutator of two operators; a nonzero commutator signals that the two observables cannot both be given sharp, simultaneous values. Reading: whenever two operators fail to commute, no wavefunction can be a simultaneous eigenfunction of both (except in trivial special cases), and this single relation is the algebraic root of the Heisenberg uncertainty principle, foreshadowing the deeper structure of Hermitian Operators.

Example 1 (foundational): momentum eigenvalue of an asymmetric plane wave. Consider the one-dimensional wavefunction $\psi(x) = A\,e^{i(2.70\times10^{10}\,\text{m}^{-1})x}$, describing a free particle with wavevector $k = 2.70\times10^{10}\ \text{m}^{-1}$ (chosen off any round number deliberately). Find the momentum eigenvalue and the corresponding classical momentum and speed if this represents an electron. Apply the momentum operator directly:

So $\psi$ is an eigenfunction of $\hat p_x$ with eigenvalue $p = \hbar k$. Numerically, with $\hbar = 1.0546\times10^{-34}\ \text{J}\,\cdot\,\text{s}$:

For an electron ($m_e = 9.109\times10^{-31}\ \text{kg}$), the corresponding speed is

about 1% of the speed of light — comfortably nonrelativistic, consistent with using the nonrelativistic kinetic-energy operator. The associated de Broglie wavelength is $\lambda = 2\pi/k = 2.327\times10^{-10}\ \text{m}$, roughly 2.3 Å, a typical interatomic length scale — reassuring, since matter waves of atomic-scale wavelength are exactly where quantum mechanics becomes chemically relevant. Limiting case check: as $k\to 0$ the wavefunction becomes spatially uniform ($\psi \to A$, a constant) and $\hat p_x\psi = 0$ — zero momentum, exactly as expected for a "wave" with infinite wavelength, i.e., no spatial oscillation at all.

Example 2 (applied): energy eigenvalues of an electron confined to a conjugated chain (particle in a box). Treat a linear polyene's π-electron system as a particle confined to a one-dimensional box of length $L = 8.40\ \text{Å}$ (an asymmetric, non-round choice representing an actual short conjugated chain), with $V(x)=0$ inside and $V\to\infty$ at the walls. The Hamiltonian operator, per the postulate, is

Solving $\hat H \psi_n = E_n \psi_n$ with boundary conditions $\psi_n(0)=\psi_n(L)=0$ gives eigenfunctions $\psi_n(x) = \sqrt{2/L}\sin(n\pi x/L)$ and eigenvalues

Compute the energy gap between the 4th and 5th levels (a HOMO→LUMO-type transition for a chain holding 8 π-electrons, filling $n=1$–4):

Numerically, with $h = 6.626\times10^{-34}\ \text{J}\,\cdot\,\text{s}$, $m_e = 9.109\times10^{-31}\ \text{kg}$, $L = 8.40\times10^{-10}\ \text{m}$:

So

This corresponds to an absorption wavelength $\lambda = hc/\Delta E = (6.626\times10^{-34})(3.00\times10^8)/(7.66\times10^{-19}) = 2.60\times10^{-7}\ \text{m} = 260\ \text{nm}$, squarely in the UV region where conjugated polyenes actually absorb — the operator formalism, applied to a crude but honest model, reproduces the right order of magnitude and region of real spectroscopic behavior. Limiting case check: as $L\to\infty$ (an unconfined, "free" electron), $E_n \to 0$ for any fixed $n$ and the levels crowd together into a continuum — exactly the free-particle limit, where momentum (not a boxed quantum number) becomes the good quantum number and energy is no longer quantized.

Example 3 (exam/competition level): does a given trial function commute the kinetic and potential parts of $\hat H$, and what does that imply for simultaneous eigenfunctions? For a harmonic oscillator with $V(x) = \tfrac12 k_f x^2$ ($k_f$ the force constant, not to be confused with wavevector), compute the commutator $[\hat T,\hat V]$ and evaluate it acting on the specific (non-eigenfunction) trial function $f(x) = x^3$, then discuss what a nonzero commutator means for whether $\hat T$ and $\hat V$ can share eigenfunctions.

Setting up. $\hat T = -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}$ and $\hat V = \tfrac12 k_f x^2$. Compute $[\hat T,\hat V]f = \hat T(\hat V f) - \hat V(\hat T f)$ for $f(x)=x^3$.

First term, $\hat T(\hat V f)$. First $\hat V f = \tfrac12 k_f x^2\cdot x^3 = \tfrac12 k_f x^5$. Then

Second term, $\hat V(\hat T f)$. First $\hat T f = -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}(x^3) = -\dfrac{\hbar^2}{2m}(6x) = -\dfrac{3\hbar^2}{m}x$. Then

Subtract.

The result is nonzero (it is $-\tfrac{7}{2}(\hbar^2k_f/m)$ times the original function $x^3$, not the original eigenvalue equation being satisfied by $f$ itself — note $f=x^3$ was never claimed to be an eigenfunction of anything here, it's simply a test function used to evaluate the commutator operator).

Interpretation. Because $[\hat T,\hat V]\neq 0$ as an operator identity (confirmed by getting a nonzero result on this generic test function), $\hat T$ and $\hat V$ do not commute in general for the harmonic oscillator, so there is no reason to expect a common set of eigenfunctions for kinetic and potential energy separately. Only their sum, $\hat H = \hat T+\hat V$, has a well-defined discrete spectrum (the harmonic-oscillator energy ladder $E_n = (n+\tfrac12)\hbar\omega$) — individually, a state of sharp kinetic energy is never simultaneously a state of sharp potential energy, which is precisely why chemists speak of total energy levels and never "kinetic energy levels" as separately quantized, observable rungs. Limiting case check: if $k_f \to 0$ (no restoring force, i.e. $\hat V \to 0$), then $[\hat T,\hat V]\to 0$ trivially, and the operators commute vacuously — consistent with the free-particle case where $\hat H = \hat T$ and every eigenfunction of $\hat T$ is automatically an eigenfunction of $\hat H$.