Operators and Observables

Attempt these from memory before reading.

• What makes an operator an observable? Give the two conditions.
• Compute $[\hat{x}, \hat{p}] = \hat{x}\hat{p} - \hat{p}\hat{x}$ acting on a smooth test function.
• State the generalised uncertainty principle.
• Define the expectation value, and say what it is not.
• What is the difference between Hermitian and self-adjoint, in one sentence?

Check your answers against the text. The last prompt is the one that catches people, and it is the reason the page exists.

In finite-dimensional linear algebra, a Hermitian matrix is all you need. It has real eigenvalues and an orthonormal eigenbasis, so “measure $\hat{A}$” means “read off an eigenvalue and its weight” and the probabilities in the Born rule have somewhere to live.

Physics is not finite-dimensional. The position operator $\hat{x} = x$ and the momentum operator $\hat{p} = -i\hbar\,\mathrm{d}/\mathrm{d}x$ are defined on functions, and the set of functions on which a differential operator is defined is not incidental — it is part of the operator. Changing the domain changes the operator, and can change whether it has real eigenvalues at all.

This is not a technicality. The momentum operator on the half-line is Hermitian on its natural domain and is not self-adjoint, and self-adjointness is what the spectral theorem needs. The spectral theorem is what guarantees a complete family of spectral projections, and those projections are what “probability of outcome $a$” is defined in terms of. Lose self-adjointness and the Born rule has nothing to be a statement about.

So the two conditions are:

• Hermitian (symmetric): $\langle\phi, \hat{A}\psi\rangle = \langle \hat{A}\phi, \psi\rangle$ for all $\psi, \phi$ in the domain.
• Self-adjoint: Hermitian and the domain equals the domain of $\hat{A}^\dagger$.

On a finite-dimensional space, domains are everything, so the two coincide. On a Hilbert space of functions, they need not, and the difference is exactly where the interesting physics lives.

In the position representation:

$$\hat{x} = x, \quad \hat{p} = -i\hbar\frac{\partial}{\partial x}$$

These satisfy the canonical commutation relation:

$$[\hat{x}, \hat{p}] = i\hbar$$

Hermitian operators have real eigenvalues and orthogonal eigenstates — essential for observables.

Theorem 4.1. If $\hat{A}$ is Hermitian, then:

• All eigenvalues are real.
• Eigenstates corresponding to distinct eigenvalues are orthogonal.
• The eigenstates form a complete basis (for the space of physical states).

Proof that eigenvalues are real. Let $\hat{A}|a\rangle = a|a\rangle$ with $\langle a|a\rangle = 1$. Then:

$$\langle a|\hat{A}|a\rangle = a\langle a|a\rangle = a$$

Taking the complex conjugate:

$$\langle a|\hat{A}|a\rangle^* = \langle a|\hat{A}^\dagger|a\rangle = \langle a|\hat{A}|a\rangle = a^*$$

Where the second equality uses $\hat{A} = \hat{A}^\dagger$. Therefore $a = a^*$ So $a$ is real. $\blacksquare$

Proof that eigenstates are orthogonal. Let $\hat{A}|a\rangle = a|a\rangle$ and $\hat{A}|b\rangle = b|b\rangle$ With $a \neq b$:

$$\langle b|\hat{A}|a\rangle = a\langle b|a\rangle$$ $$\langle b|\hat{A}|a\rangle = \langle\hat{A}b|a\rangle = b^*\langle b|a\rangle = b\langle b|a\rangle$$

Where the last step uses $b^* = b$ (eigenvalues are real). Therefore:

$$(a - b)\langle b|a\rangle = 0$$

Since $a \neq b$ We must have $\langle b|a\rangle = 0$. $\blacksquare$

Theorem 4.2 (Spectral Theorem). Every Hermitian operator on a finite-dimensional Hilbert space Has a complete orthonormal set of eigenvectors. In infinite dimensions, this holds for Self-adjoint operators with a discrete spectrum; operators with continuous spectra require the Spectral theorem in its general form (resolution of the identity).

The commutator of two operators is $[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$.

Theorem 4.3 (Generalised Uncertainty Principle). For observables $\hat{A}$ and $\hat{B}$:

$$\sigma_A \sigma_B \geq \frac{1}{2}|\langle[\hat{A}, \hat{B}]\rangle|$$

Corollary 4.4 (Heisenberg Uncertainty Principle). $\sigma_x \sigma_p \geq \hbar/2$.

Proof. This follows from the generalised uncertainty principle with $[\hat{x}, \hat{p}] = i\hbar$:

$$\sigma_x \sigma_p \geq \frac{1}{2}|\langle i\hbar \rangle| = \frac{\hbar}{2}$$

$\blacksquare$

Theorem 4.5 (Robertson-Schrodinger inequality). For any state $|\psi\rangle$ and observables $\hat{A}$, $\hat{B}$:

$$\sigma_A^2\,\sigma_B^2 \geq \frac{1}{4}|\langle[\hat{A}, \hat{B}]\rangle|^2 + \frac{1}{4}\langle\{\Delta\hat{A}, \Delta\hat{B}\}\rangle^2$$

Where $\Delta\hat{A} = \hat{A} - \langle\hat{A}\rangle$ and $\sigma_A^2 = \langle\Delta\hat{A}^2\rangle$.

Proof. Define $|\alpha\rangle = (\Delta\hat{A} + i\lambda\Delta\hat{B})|\psi\rangle$ for a real Parameter $\lambda$. Since $\langle\alpha|\alpha\rangle \geq 0$:

$$\langle\psi|(\Delta\hat{A} - i\lambda\Delta\hat{B})(\Delta\hat{A} + i\lambda\Delta\hat{B})|\psi\rangle \geq 0$$ $$= \sigma_A^2 + i\lambda\langle[\Delta\hat{A}, \Delta\hat{B}]\rangle + \lambda^2\sigma_B^2 \geq 0$$

This is a quadratic in $\lambda$ that is non-negative for all $\lambda$ So its discriminant must be non-positive:

$$(\langle[\Delta\hat{A}, \Delta\hat{B}]\rangle)^2 - 4\sigma_A^2\sigma_B^2 \leq 0$$

Since $[\Delta\hat{A}, \Delta\hat{B}] = [\hat{A}, \hat{B}]$ (constants commute with everything):

$$\sigma_A^2\,\sigma_B^2 \geq \frac{1}{4}|\langle[\hat{A}, \hat{B}]\rangle|^2 \qquad \blacksquare$$

The stronger Robertson-Schrodinger form retains the anticommutator term $\langle\{\Delta\hat{A}, \Delta\hat{B}\}\rangle^2$ Which is always non-negative and provides a tighter bound.

Example 4.1. Show that the uncertainty principle is saturated for the harmonic oscillator ground state.

Solution

For the ground state $\psi_0(x) = (m\omega/\pi\hbar)^{1/4}\exp(-m\omega x^2/(2\hbar))$:

$$\langle x \rangle = 0, \quad \langle x^2 \rangle = \frac{\hbar}{2m\omega} \implies \sigma_x = \sqrt{\frac{\hbar}{2m\omega}}$$ $$\langle p \rangle = 0, \quad \langle p^2 \rangle = \frac{m\omega\hbar}{2} \implies \sigma_p = \sqrt{\frac{m\omega\hbar}{2}}$$ $$\sigma_x\,\sigma_p = \frac{\hbar}{2}$$

This saturates the Heisenberg bound, so the ground state is a minimum uncertainty state (Gaussian).

The expectation value of an observable $\hat{A}$ in state $|\psi\rangle$:

$$\langle A \rangle = \langle \psi | \hat{A} | \psi \rangle = \int \psi^* \hat{A} \psi\, dx$$

Theorem 4.6 (Ehrenfest’s Theorem). Quantum expectation values obey classical equations of motion:

$$\frac{d\langle \hat{x} \rangle}{dt} = \frac{\langle \hat{p} \rangle}{m}, \quad \frac{d\langle \hat{p} \rangle}{dt} = -\left\langle \frac{\partial V}{\partial x}\right\rangle$$

Proof of Ehrenfest’s Theorem. From the Schrodinger equation:

$$\frac{d\langle \hat{A} \rangle}{dt} = \frac{i}{\hbar}\langle[\hat{H}, \hat{A}]\rangle + \left\langle\frac{\partial \hat{A}}{\partial t}\right\rangle$$

For $\hat{A} = \hat{x}$ (no explicit time dependence), using $[\hat{p}^2, \hat{x}] = -2i\hbar\hat{p}$:

$$\frac{d\langle \hat{x} \rangle}{dt} = \frac{i}{\hbar}\!\left\langle\left[\frac{\hat{p}^2}{2m}, \hat{x}\right]\right\rangle = \frac{i}{\hbar}\cdot\frac{-2i\hbar}{2m}\langle\hat{p}\rangle = \frac{\langle\hat{p}\rangle}{m}$$

For $\hat{A} = \hat{p}$ Using $[V(\hat{x}), \hat{p}] = i\hbar\,V'(\hat{x})$:

$$\frac{d\langle \hat{p} \rangle}{dt} = \frac{i}{\hbar}\langle[V(\hat{x}), \hat{p}]\rangle = -\left\langle\frac{\partial V}{\partial x}\right\rangle$$

$\blacksquare$

Correspondence principle. Ehrenfest’s theorem embodies the correspondence principle: in the Classical limit (large quantum numbers or $\hbar \to 0$), quantum expectation values follow Classical trajectories. However, this is only exact for linear or quadratic potentials; for general Potentials, $\langle V'(x) \rangle \neq V'(\langle x \rangle)$ So quantum corrections persist even for large systems.

To find the eigenvalues and eigenvectors of an operator $\hat{A}$ Solve:

$$\hat{A}|\phi\rangle = a|\phi\rangle \implies \det(\hat{A} - a\hat{I}) = 0$$

The roots give the eigenvalues; substituting each back yields the eigenvectors.

Example 4.3. Find the eigenvalues and eigenvectors of $\hat{S}_x = \frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}$.

Solution

$$\det\!\left(\frac{\hbar}{2}\begin{pmatrix}-a & 1\\1 & -a\end{pmatrix}\right) = 0 \implies a^2 - 1 = 0 \implies a = \pm 1$$

Eigenvalues are $\pm\hbar/2$.

For $a = +1$: $\begin{pmatrix}-1 & 1\\1 & -1\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix} = 0 \implies c_1 = c_2$. Normalised: $|+\rangle_x = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$.

For $a = -1$: $c_1 = -c_2$. Normalised: $|-\rangle_x = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$.

These are equal superpositions of the $S_z$ eigenstates. Note that measuring $S_x$ on a state of Definite $S_z$ gives probabilistic outcomes, and vice versa.

Quantum operators are the bridge between abstract mathematical states and physical measurements. Each observable, such as position or momentum, is represented by an operator whose eigenvalues are the possible outcomes you could measure. Operators for incompatible observables do not commute, which means measuring one precisely forces the other to become uncertain. This is not a limitation of instruments but a fundamental feature of quantum reality. Ehrenfest’s theorem shows that quantum expectation values follow classical equations of motion on average, providing a smooth bridge from quantum to classical behaviour. Spin-half systems illustrate these ideas most vividly.

Mistake 1: Assuming $\langle V'(x) \rangle = V'(\langle x \rangle)$ in Ehrenfest’s theorem Ehrenfest’s theorem gives $d\langle p \rangle/dt = -\langle \partial V/\partial x \rangle$, not $-V'(\langle x \rangle)$. The expectation value of the derivative of $V$ is generally not the derivative of $V$ at the expectation value. This equality holds only for linear or quadratic potentials. Students frequently make this substitution when applying Ehrenfest’s theorem to anharmonic oscillators.

Mistake 2: Confusing Hermitian operators with self-adjoint operators Every self-adjoint operator is Hermitian, but the converse is not true for operators on infinite-dimensional spaces. The momentum operator $\hat{p} = -i\hbar\,d/dx$ is Hermitian on a dense domain but may fail to be self-adjoint depending on boundary conditions. Self-adjointness requires the domain of $\hat{A}$ and $\hat{A}^\dagger$ to coincide, which is essential for the spectral theorem.

Mistake 3: Assuming the uncertainty principle limits measurement precision The Heisenberg uncertainty principle $\sigma_x \sigma_p \geq \hbar/2$ is a property of quantum states, not a limitation of measurement apparatus. It reflects the inherent spread of a state in position and momentum space simultaneously. A minimum-uncertainty Gaussian state saturates the bound, showing it is a fundamental feature of quantum mechanics, not an instrumental error.

Quantum operators are the machinery that extracts measurable information from a quantum state. Each observable, like position or momentum, has an associated operator whose eigenvalues are the possible measurement outcomes. The expectation value of an operator gives the average result over many identical measurements. Commuting operators share eigenstates, meaning both observables can be known simultaneously. Non-commuting operators, like position and momentum, obey an uncertainty relation: measuring one precisely forces the other to become uncertain. This is not a limitation of instruments but a fundamental feature of quantum reality. Spin-half systems illustrate this vividly: measuring spin along one axis gives random results for the orthogonal axis.

The first row is the one that matters most, and it is also the one that feels like pedantry until you meet an example. The last row is its general form and is worth internalising: on an infinite-dimensional space, writing the operator down is only half the specification.

The third row deserves emphasis. The common statement is “commuting observables are simultaneously measurable”, and it is a theorem about a common eigenspace — which need not be the whole space. Two operators can commute and still leave you unable to assign both eigenvalues at once.

• Domains in quantum mechanics. Whether the Hamiltonian admits a self-adjoint extension is a genuine mathematical question with physical answers: boundary conditions at a singular endpoint can make the extension unique, absent, or a family parameterised by an angle, and that angle is the quantum number.
• Observables without spectra. Quantum field theory has no eigenvalues to read. The observable algebra is local and relativistic, and the theory of von Neumann algebras replaces “operator with spectrum” with “local algebra with representations”.
• Haag’s theorem. Local relativistic observables cannot be unitarily equivalent to the free field ones. It is a rigorous statement that the mathematical frame of canonical quantisation breaks down in QFT, and reconciling it remains open.
• Generalised measurement theory. POVMs generalise projective measurements; whether they describe real apparatus or are a mathematical convenience is still argued.
• Black-hole information and observables. Which operators count as observables inside a black hole, and what algebra they satisfy, is unsettled and empirically constrained.

Mix these with the postulates, wave functions, spin, Linear Algebra, Functional Analysis and Spectral Theory.

1. (Linear algebra.) Prove that a Hermitian operator on a finite-dimensional space has an orthonormal eigenbasis. Identify every step of the proof that fails in infinite dimensions, and say which hypothesis each one used.
2. (Functional analysis.) State the spectral theorem for self-adjoint operators, and explain why the theorem is false for merely Hermitian ones. Then take the first derivative on $L^2(0,1)$ and determine its adjoint, checking domains rather than formal expressions.
3. (Wave functions.) Compute $[\hat{x}, \hat{p}]$ acting on a smooth compactly supported function by differentiating, with no appeal to a formula. Then say why the answer is $i\hbar$ rather than the classical commutator $0$, and what that statement means physically.
4. (Spectral theory.) Show that the spectral measure of $\hat{p}$ is Lebesgue measure, using the Fourier transform: the projectors of $\hat{p}$ are the Fourier projections onto sets. What does it mean that position and momentum have continuous spectral measures rather than countable ones?
5. (Linear algebra.) Two Hermitian operators commute. Show that every eigenvector of one is an eigenvector of the other if the eigenvalues of the first are non-degenerate, and construct a two-dimensional example with degenerate eigenvalues where the conclusion fails.
6. (Research.) State Haag’s theorem in the form that local, covariant, time-ordered algebras cannot be unitarily equivalent to the free-field algebra. Explain what it implies about canonical quantisation, and why it is not regarded as showing QFT is inconsistent.

• Postulates of Quantum Mechanics: The postulates establish that observables are represented by Hermitian operators.

• Wave Functions and the Schrodinger Equation: Wave functions are the state vectors that operators act upon to extract physical information.

• Spin: Spin operators illustrate the algebraic properties of angular momentum in quantum mechanics.

• Linear Algebra

• Hilbert Spaces

• Bounded Linear Operators

• Angular Momentum and the Hydrogen Atom
• Approximation Methods
• Density Functional Theory: Conceptual Overview
• Historical Motivation