Operators and Observables | Physics

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.

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 $\hat{S}_x = \frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}$.

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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.

• 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.

• Calculus

• Linear Algebra

• Quantum Computing