Operators and Observables

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