Physics / Advanced Theoretical Physics Quantum Mechanics II 100% Free Open Access
Chapter 1 • Theory & Derivations

Matrix Mechanics, Hilbert Space & Operator Representations

Comprehensive mathematical and physical foundations of modern quantum state space: the physical significance of two-slit interference, state vectors in abstract complex Hilbert space, Dirac notation, adjoints, projection operators, continuous coordinate and momentum representations, transformation theory, spatial inversion and parity, pure versus mixed state ensembles with the density matrix formalism, and the complete algebraic ladder operator matrix mechanics of the quantum harmonic oscillator.

§1.1Foundations of Quantum State Space: Slit Experiments, Superposition & Dirac Bra-Ket Algebra

1. Physical Impetus: The Double-Slit Experiment & Quantum Probability Amplitudes

The conceptual foundation of quantum mechanics rests upon the breakdown of classical probability theory, most dramatically demonstrated by electron diffraction through two closely spaced slits. In classical statistical mechanics, if an event can occur via two mutually exclusive classical alternatives (passing through slit 1 or slit 2), the resultant probability distribution on a detection screen is the direct sum of the individual probabilities:

$$\mathcal{P}_{\text{classical}}(x) = \mathcal{P}_1(x) + \mathcal{P}_2(x)$$

In quantum mechanics, when no measurement is made to determine the specific path traversed by the electron, the detector records an interference pattern characteristic of wave phenomena. Quantum kinematics resolves this paradox by associating with every physical event a complex-valued probability amplitude $\psi(x) \in \mathbb{C}$, such that the total state is a coherent linear superposition of the two path alternatives:

$$\psi(x) = \psi_1(x) + \psi_2(x)$$

The observable probability density $\mathcal{P}(x)$ is the modulus squared of the total amplitude:

$$\mathcal{P}(x) = |\psi(x)|^2 = |\psi_1(x) + \psi_2(x)|^2 = |\psi_1(x)|^2 + |\psi_2(x)|^2 + 2\,\text{Re}\left[\psi_1^*(x)\,\psi_2(x)\right]$$

The final term, $2\sqrt{\mathcal{P}_1\mathcal{P}_2}\cos(\phi_1 - \phi_2)$, is the quantum interference cross-term. If a non-destructive measurement is introduced to register which slit the particle traversed, the relative phase coherence is irrevocably destroyed by quantum backaction (decoherence), collapsing the interference pattern into the classical additive distribution $\mathcal{P}_1 + \mathcal{P}_2$.

2. The Dirac Bra and Ket Formalism

Paul Dirac synthesized the wave mechanics of Schrödinger and the matrix mechanics of Heisenberg into an abstract vector space representation. A physical state of an isolated quantum system is represented by a ray in a complex vector space $\mathcal{H}$, termed a ket vector and denoted by $|\psi\rangle$.

Associated with the vector space $\mathcal{H}$ is its dual space $\mathcal{H}^*$, consisting of all continuous linear functionals mapping $\mathcal{H} \to \mathbb{C}$. Dirac denoted elements of the dual space as bra vectors, $\langle \phi| \in \mathcal{H}^*$. By the Riesz representation theorem, for every ket $|\psi\rangle \in \mathcal{H}$, there exists a unique conjugate dual bra $\langle \psi| \in \mathcal{H}^*$ under an anti-linear bijective map termed the Hermitian adjoint:

$$(c_1|\psi_1\rangle + c_2|\psi_2\rangle)^\dagger = c_1^*\langle \psi_1| + c_2^*\langle \psi_2|$$

The action of a functional $\langle \phi|$ on a state ket $|\psi\rangle$ forms the Dirac bracket, representing the inner product:

$$\langle \phi | \psi \rangle = \int_{\Omega} \phi^*(x)\,\psi(x)\,dx \in \mathbb{C}$$

The inner product satisfies three defining axioms:

  1. Skew-symmetry (Hermitian conjugate property): $\langle \phi | \psi \rangle = \langle \psi | \phi \rangle^*$
  2. Linearity in the ket argument: $\langle \phi | (c_1 |\psi_1\rangle + c_2 |\psi_2\rangle) = c_1 \langle \phi | \psi_1\rangle + c_2 \langle \phi | \psi_2\rangle$
  3. Positive-definiteness: $\langle \psi | \psi \rangle \ge 0$, with equality holding if and only if $|\psi\rangle = 0$.

§1.2Hilbert Space Geometry: Inner Products, Norms, Cauchy-Schwarz Inequality & Bases

1. Definition of Complex Hilbert Space

An abstract Hilbert space $\mathcal{H}$ is a complete complex inner product space. Completeness ensures that every Cauchy sequence of state vectors $\{|\psi_n\rangle\}_{n=1}^\infty$ converges to an element within $\mathcal{H}$ with respect to the induced norm:

$$\|\psi\| = \sqrt{\langle \psi | \psi \rangle}$$ $$\lim_{n,m\to\infty} \|\psi_n - \psi_m\| = 0 \implies \exists |\psi\rangle \in \mathcal{H} \text{ such that } \lim_{n\to\infty} \|\psi_n - \psi\| = 0$$

2. The Cauchy-Schwarz and Triangle Inequalities

For any two state kets $|\psi\rangle, |\phi\rangle \in \mathcal{H}$, consider the arbitrary real or complex parameter $\lambda$. Since the norm of the vector $|\chi\rangle = |\psi\rangle + \lambda |\phi\rangle$ is strictly non-negative:

$$\langle \chi | \chi \rangle = \langle \psi | \psi \rangle + \lambda \langle \psi | \phi \rangle + \lambda^* \langle \phi | \psi \rangle + |\lambda|^2 \langle \phi | \phi \rangle \ge 0$$

Choosing the specific variation $\lambda = -\frac{\langle \phi | \psi \rangle}{\langle \phi | \phi \rangle}$ (assuming $\langle \phi | \phi \rangle \ne 0$) yields:

$$\langle \psi | \psi \rangle - \frac{|\langle \phi | \psi \rangle|^2}{\langle \phi | \phi \rangle} \ge 0 \implies |\langle \phi | \psi \rangle|^2 \le \langle \psi | \psi \rangle \langle \phi | \phi \rangle$$

This is the foundational Cauchy-Schwarz inequality:

$$|\langle \phi | \psi \rangle| \le \|\phi\| \, \|\psi\|$$

Equality holds if and only if the kets are linearly dependent: $|\psi\rangle = c|\phi\rangle$. Using this result, the Minkowski (triangle) inequality directly follows:

$$\| |\psi\rangle + |\phi\rangle \| \le \|\psi\| + \|\phi\|$$

3. Orthonormal Bases and Completeness (Closure) Relations

Let $\{|u_n\rangle\}_{n=1}^N$ (where $N$ may be finite or countably infinite) be an orthonormal set spanning $\mathcal{H}$, satisfying the orthonormality condition:

$$\langle u_n | u_m \rangle = \delta_{nm}$$

Any arbitrary state $|\psi\rangle \in \mathcal{H}$ can be expanded uniquely as:

$$|\psi\rangle = \sum_{n} c_n |u_n\rangle, \quad c_n = \langle u_n | \psi \rangle$$

Substituting $c_n$ into the expansion reveals the identity operator $\hat{I}$:

$$|\psi\rangle = \sum_{n} |u_n\rangle \langle u_n | \psi \rangle = \left( \sum_{n} |u_n\rangle \langle u_n| \right) |\psi\rangle \implies \sum_{n} |u_n\rangle \langle u_n| = \hat{I}$$

This equation is the completeness relation (or resolution of the identity). The norm of $|\psi\rangle$ expressed through basis coordinates gives Parseval's identity:

$$\langle \psi | \psi \rangle = \sum_{n} |c_n|^2 = \sum_n |\langle u_n | \psi \rangle|^2 = 1$$

§1.3Linear Operators, Matrix Representations, Adjoints & Spectral Decomposition

1. Linear Operators and Dyadic Outer Products

A linear operator $\hat{A}: \mathcal{H} \to \mathcal{H}$ maps kets to kets while preserving linear combinations:

$$\hat{A}\left(c_1|\psi_1\rangle + c_2|\psi_2\rangle\right) = c_1 \hat{A}|\psi_1\rangle + c_2 \hat{A}|\psi_2\rangle$$

The outer product between a ket $|\phi\rangle$ and a bra $\langle \chi|$ forms a rank-1 linear operator (a dyad):

$$\hat{M} = |\phi\rangle \langle \chi| \implies \hat{M}|\psi\rangle = |\phi\rangle \langle \chi | \psi \rangle = (\langle \chi | \psi \rangle) |\phi\rangle$$

2. Matrix Representation in a Discrete Basis

By inserting the completeness relation on both sides of an operator $\hat{A}$, we express $\hat{A}$ in terms of its matrix elements:

$$\hat{A} = \hat{I} \hat{A} \hat{I} = \sum_{n} \sum_{m} |u_n\rangle \langle u_n | \hat{A} | u_m \rangle \langle u_m|$$

Defining the matrix element $A_{nm} \equiv \langle u_n | \hat{A} | u_m \rangle$, the operator acts as a matrix acting upon coordinate column vectors:

$$\hat{A} \doteq \begin{pmatrix} A_{11} & A_{12} & \cdots \\ A_{21} & A_{22} & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix}, \quad |\psi\rangle \doteq \begin{pmatrix} c_1 \\ c_2 \\ \vdots \end{pmatrix}$$

3. Hermitian Adjoint and Observable Operators

The Hermitian adjoint $\hat{A}^\dagger$ of an operator $\hat{A}$ is defined by the condition:

$$\langle \phi | \hat{A} | \psi \rangle^* = \langle \psi | \hat{A}^\dagger | \phi \rangle \quad \forall |\psi\rangle, |\phi\rangle \in \mathcal{H}$$

In matrix terms, $(A^\dagger)_{nm} = (A_{mn})^*$, representing the conjugate transpose. An operator represents a physical observable if and only if it is self-adjoint (Hermitian):

$$\hat{A}^\dagger = \hat{A} \iff \langle u_n | \hat{A} | u_m \rangle = \langle u_m | \hat{A} | u_n \rangle^*$$

Hermitian operators possess two vital physical properties:

  1. Real Eigenvalues: If $\hat{A}|a_n\rangle = a_n |a_n\rangle$, then $a_n = a_n^* \in \mathbb{R}$.
  2. Orthogonality of Eigenvectors: If $\hat{A}|a_n\rangle = a_n|a_n\rangle$ and $\hat{A}|a_m\rangle = a_m|a_m\rangle$ with $a_n \ne a_m$, then $\langle a_n | a_m \rangle = 0$.

4. Projection Operators and the Spectral Decomposition Theorem

A projection operator $\hat{P}_n$ onto the one-dimensional subspace spanned by the normalized ket $|u_n\rangle$ is defined as:

$$\hat{P}_n \equiv |u_n\rangle \langle u_n|$$

Projection operators satisfy the algebraic properties of idempotency and Hermiticity:

$$\hat{P}_n^2 = (|u_n\rangle \langle u_n|)(|u_n\rangle \langle u_n|) = |u_n\rangle \langle u_n | u_n \rangle \langle u_n| = |u_n\rangle \langle u_n| = \hat{P}_n$$ $$\hat{P}_n^\dagger = \hat{P}_n, \quad \hat{P}_n \hat{P}_m = \delta_{nm}\hat{P}_n$$

According to the Spectral Decomposition Theorem, any Hermitian observable $\hat{A}$ with a non-degenerate discrete spectrum can be represented as a weighted sum of its projection operators:

$$\hat{A} = \sum_n a_n |a_n\rangle \langle a_n| = \sum_n a_n \hat{P}_n$$

For any analytic function $f(\hat{A})$, the spectral representation evaluates directly to:

$$f(\hat{A}) = \sum_n f(a_n) |a_n\rangle \langle a_n|$$

§1.4Unitary Transformation Theory, Change of Basis & Continuous Representations

1. Change of Basis and Unitary Operators

Consider two distinct complete orthonormal bases $\{|u_n\rangle\}$ and $\{|v_k\rangle\}$ spanning the same Hilbert space $\mathcal{H}$. The transformation connecting the two bases is mediated by a linear operator $\hat{U}$:

$$|v_k\rangle = \sum_n |u_n\rangle \langle u_n | v_k\rangle \equiv \hat{U} |u_k\rangle$$ $$\hat{U} = \sum_k |v_k\rangle \langle u_k|$$

Evaluating the adjoint operator:

$$\hat{U}^\dagger = \sum_j |u_j\rangle \langle v_j| \implies \hat{U} \hat{U}^\dagger = \sum_{k,j} |v_k\rangle \langle u_k | u_j\rangle \langle v_j| = \sum_k |v_k\rangle \langle v_k| = \hat{I}$$ $$\hat{U}^\dagger \hat{U} = \sum_{j,k} |u_j\rangle \langle v_j | v_k\rangle \langle u_k| = \sum_k |u_k\rangle \langle u_k| = \hat{I}$$

Thus, $\hat{U}^\dagger = \hat{U}^{-1}$, meaning $\hat{U}$ is a unitary operator. Unitary transformations preserve inner products, vector norms, and the algebraic spectra of operators:

$$\langle \phi' | \psi' \rangle = \langle \phi | \hat{U}^\dagger \hat{U} | \psi \rangle = \langle \phi | \psi \rangle$$ $$\hat{A}' = \hat{U}^\dagger \hat{A} \hat{U}$$

2. Continuous Spectra: Position and Momentum Representations

When an observable exhibits a continuous spectrum, such as position $\hat{x}$ or momentum $\hat{p}$, the Kronecker delta is replaced by the Dirac delta distribution:

$$\hat{x}|x\rangle = x|x\rangle, \quad \langle x | x' \rangle = \delta(x - x'), \quad \int_{-\infty}^\infty |x\rangle \langle x| \, dx = \hat{I}$$ $$\hat{p}|p\rangle = p|p\rangle, \quad \langle p | p' \rangle = \delta(p - p'), \quad \int_{-\infty}^\infty |p\rangle \langle p| \, dp = \hat{I}$$

The Schrödinger wave function $\psi(x)$ is the coordinate-basis projection of the state ket $|\psi\rangle$:

$$\psi(x) = \langle x | \psi \rangle, \quad \phi(p) = \langle p | \psi \rangle$$

The transformation kernel connecting coordinate and momentum space is the overlap bracket $\langle x | p \rangle$. Using the fundamental commutator $[\hat{x}, \hat{p}] = i\hbar\hat{I}$ and the differential representation $\langle x | \hat{p} | \psi \rangle = -i\hbar \frac{\partial}{\partial x}\psi(x)$:

$$\langle x | \hat{p} | p \rangle = p \langle x | p \rangle = -i\hbar \frac{\partial}{\partial x} \langle x | p \rangle$$ $$\implies \langle x | p \rangle = \frac{1}{\sqrt{2\pi\hbar}} \exp\left(\frac{i p x}{\hbar}\right)$$

Consequently, transforming from coordinate to momentum wavefunctions is mathematically identical to a continuous Fourier transformation:

$$\phi(p) = \langle p | \psi \rangle = \int_{-\infty}^\infty \langle p | x \rangle \langle x | \psi \rangle \, dx = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^\infty e^{-ipx/\hbar} \psi(x) \, dx$$

§1.5Parity and Symmetry Operators: Spatial Inversion, Selection Rules & Invariance

1. The Spatial Inversion (Parity) Operator

The parity operator $\hat{\Pi}$ acts on spatial coordinate eigenkets by inverting all Cartesian axes through the origin:

$$\hat{\Pi} |x, y, z\rangle = |-x, -y, -z\rangle$$

Acting twice upon any arbitrary state returns the original spatial coordinates:

$$\hat{\Pi}^2 |x\rangle = \hat{\Pi} |-x\rangle = |x\rangle \implies \hat{\Pi}^2 = \hat{I}$$

Since $\hat{\Pi}$ is also unitary ($\hat{\Pi}^\dagger \hat{\Pi} = \hat{I}$), it is simultaneously Hermitian:

$$\hat{\Pi}^\dagger = \hat{\Pi}^{-1} = \hat{\Pi}$$

The eigenvalues $\pi_k$ of the parity operator must satisfy:

$$\hat{\Pi}|\psi\rangle = \pi_k |\psi\rangle \implies \hat{\Pi}^2|\psi\rangle = \pi_k^2 |\psi\rangle = |\psi\rangle \implies \pi_k = \pm 1$$

States with eigenvalue $+1$ are termed even parity states ($\psi(-x) = +\psi(x)$), while states with eigenvalue $-1$ are termed odd parity states ($\psi(-x) = -\psi(x)$).

2. Transformation of Dynamical Observables Under Parity

Under coordinate inversion, the position and momentum operators change sign:

$$\hat{\Pi}^\dagger \hat{x} \hat{\Pi} = -\hat{x} \iff \{\hat{\Pi}, \hat{x}\} = \hat{\Pi}\hat{x} + \hat{x}\hat{\Pi} = 0$$ $$\hat{\Pi}^\dagger \hat{p} \hat{\Pi} = -\hat{p} \iff \{\hat{\Pi}, \hat{p}\} = 0$$

Conversely, the orbital angular momentum operator $\hat{\vec{L}} = \hat{\vec{r}} \times \hat{\vec{p}}$ is an axial vector (pseudovector) and commutes with parity:

$$\hat{\Pi}^\dagger \hat{\vec{L}} \hat{\Pi} = (\hat{\Pi}^\dagger \hat{\vec{r}} \hat{\Pi}) \times (\hat{\Pi}^\dagger \hat{\vec{p}} \hat{\Pi}) = (-\hat{\vec{r}}) \times (-\hat{\vec{p}}) = \hat{\vec{r}} \times \hat{\vec{p}} = \hat{\vec{L}} \implies [\hat{\Pi}, \hat{\vec{L}}] = 0$$

3. Parity Conservation and Laporte's Selection Rule

If the Hamiltonian of a system is invariant under spatial inversion, $V(-\vec{r}) = V(\vec{r})$, then:

$$[\hat{\Pi}, \hat{H}] = 0$$

Consequently, stationary states can be chosen as simultaneous eigenstates of $\hat{H}$ and $\hat{\Pi}$. Furthermore, parity is a constant of motion:

$$\frac{d}{dt}\langle \hat{\Pi} \rangle = \frac{1}{i\hbar}\langle [\hat{\Pi}, \hat{H}] \rangle = 0$$

For electric dipole transitions mediated by the vector dipole operator $\hat{\vec{d}} = q\hat{\vec{r}}$, the transition matrix element between states $|\psi_i\rangle$ and $|\psi_f\rangle$ with definite parities $\pi_i$ and $\pi_f$ satisfies:

$$\langle \psi_f | \hat{\vec{r}} | \psi_i \rangle = \langle \psi_f | \hat{\Pi}^\dagger \hat{\Pi} \hat{\vec{r}} \hat{\Pi}^\dagger \hat{\Pi} | \psi_i \rangle = \pi_f \pi_i \langle \psi_f | (-\hat{\vec{r}}) | \psi_i \rangle = -\pi_f \pi_i \langle \psi_f | \hat{\vec{r}} | \psi_i \rangle$$

Thus, the matrix element is identically zero unless $\pi_f \pi_i = -1$. This establishes Laporte's selection rule: electric dipole transitions can only occur between states of opposite parity ($\Delta l = \pm 1$).

§1.6The Density Matrix Formalism: Pure vs Mixed States, Ensembles & von Neumann Entropy

1. Limitations of Pure State Vectors and Statistical Ensembles

A single state ket $|\psi\rangle$ describes a pure state, in which complete maximal knowledge about the quantum preparation is available. However, in realistic experimental scenarios (such as an unpolarized thermal beam of particles, or a subsystem entangled with an unobserved environment), the system is described by a statistical ensemble: it has classical probability $p_k \ge 0$ of being in state $|\psi_k\rangle$, where $\sum_k p_k = 1$. Note that the kets $\{|\psi_k\rangle\}$ need not be mutually orthogonal.

2. The Density Operator

To describe such a statistical mixture, John von Neumann introduced the density operator (or density matrix) $\hat{\rho}$:

$$\hat{\rho} \equiv \sum_k p_k |\psi_k\rangle \langle \psi_k|$$

The ensemble average expectation value of any physical observable $\hat{A}$ is given by:

$$\langle \hat{A} \rangle = \sum_k p_k \langle \psi_k | \hat{A} | \psi_k \rangle = \sum_k p_k \sum_n \langle \psi_k | u_n \rangle \langle u_n | \hat{A} | \psi_k \rangle = \sum_n \langle u_n | \hat{A} \left( \sum_k p_k |\psi_k\rangle \langle \psi_k| \right) | u_n \rangle$$ $$\langle \hat{A} \rangle = \text{Tr}(\hat{\rho}\hat{A})$$

3. Mathematical Properties of the Density Operator

  1. Hermiticity: $\hat{\rho}^\dagger = \sum_k p_k (|\psi_k\rangle \langle \psi_k|)^\dagger = \hat{\rho}$
  2. Unit Trace (Conservation of Total Probability): $\text{Tr}(\hat{\rho}) = \sum_k p_k \langle \psi_k | \psi_k \rangle = \sum_k p_k = 1$
  3. Positive Semi-Definiteness: For any ket $|\phi\rangle$, $\langle \phi | \hat{\rho} | \phi \rangle = \sum_k p_k |\langle \phi | \psi_k \rangle|^2 \ge 0$
  4. Purity Criterion: $$\text{Tr}(\hat{\rho}^2) \le 1$$ Equality $\text{Tr}(\hat{\rho}^2) = 1 \iff \hat{\rho}^2 = \hat{\rho}$ holds if and only if the state is pure ($\hat{\rho} = |\psi\rangle \langle \psi|$). For a strictly mixed state in an $N$-dimensional Hilbert space, $\frac{1}{N} \le \text{Tr}(\hat{\rho}^2) < 1$.

4. von Neumann Entropy and Thermal Density Operators

The quantum mechanical generalization of Gibbs-Shannon entropy is the von Neumann entropy:

$$S(\hat{\rho}) = -k_B \text{Tr}(\hat{\rho} \ln \hat{\rho}) = -k_B \sum_j \lambda_j \ln \lambda_j$$

where $\{\lambda_j\}$ are the eigenvalues of $\hat{\rho}$. For any pure state, $S(\hat{\rho}) = 0$. For a maximally mixed state in $N$ dimensions ($\hat{\rho} = \frac{1}{N}\hat{I}$), $S = k_B \ln N$.

For a system in canonical thermal equilibrium at temperature $T = (k_B \beta)^{-1}$, the thermal density operator is:

$$\hat{\rho}_{\text{th}} = \frac{e^{-\beta \hat{H}}}{\mathcal{Z}}, \quad \mathcal{Z} = \text{Tr}\left(e^{-\beta \hat{H}}\right)$$

§1.7Algebraic Solution of the Harmonic Oscillator: Ladder Operators & Matrix Mechanics

1. Factorization of the Harmonic Oscillator Hamiltonian

Consider the one-dimensional quantum harmonic oscillator Hamiltonian:

$$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 \hat{x}^2$$

We define the dimensionless non-Hermitian creation (raising) operator $\hat{a}^\dagger$ and annihilation (lowering) operator $\hat{a}$:

$$\hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} + \frac{i}{m\omega}\hat{p}\right), \quad \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} - \frac{i}{m\omega}\hat{p}\right)$$

Evaluating the commutator of $\hat{a}$ and $\hat{a}^\dagger$ using $[\hat{x}, \hat{p}] = i\hbar$:

$$[\hat{a}, \hat{a}^\dagger] = \frac{m\omega}{2\hbar}\left[\hat{x} + \frac{i}{m\omega}\hat{p}, \hat{x} - \frac{i}{m\omega}\hat{p}\right] = \frac{m\omega}{2\hbar}\left(-\frac{i}{m\omega}[\hat{x}, \hat{p}] + \frac{i}{m\omega}[\hat{p}, \hat{x}]\right) = \frac{m\omega}{2\hbar}\left(2\frac{\hbar}{m\omega}\right) = 1$$ $$[\hat{a}, \hat{a}^\dagger] = \hat{I}$$

Multiplying $\hat{a}^\dagger \hat{a}$:

$$\hat{a}^\dagger \hat{a} = \frac{m\omega}{2\hbar}\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} - \frac{i}{m\omega}[\hat{x}, \hat{p}]\right) = \frac{1}{\hbar\omega}\hat{H} - \frac{1}{2}$$

Defining the Hermitian number operator $\hat{N} \equiv \hat{a}^\dagger \hat{a}$, the Hamiltonian takes the canonical diagonal form:

$$\hat{H} = \hbar\omega\left(\hat{N} + \frac{1}{2}\right)$$

2. The Fock State Eigenvalue Spectrum

The commutators of $\hat{N}$ with $\hat{a}$ and $\hat{a}^\dagger$ are:

$$[\hat{N}, \hat{a}] = [\hat{a}^\dagger \hat{a}, \hat{a}] = [\hat{a}^\dagger, \hat{a}]\hat{a} = -\hat{a}$$ $$[\hat{N}, \hat{a}^\dagger] = [\hat{a}^\dagger \hat{a}, \hat{a}^\dagger] = \hat{a}^\dagger[\hat{a}, \hat{a}^\dagger] = +\hat{a}^\dagger$$

Let $|n\rangle$ denote an eigenstate of $\hat{N}$ with eigenvalue $n$: $\hat{N}|n\rangle = n|n\rangle$. Then:

$$\hat{N}(\hat{a}|n\rangle) = (\hat{a}\hat{N} + [\hat{N}, \hat{a}])|n\rangle = (\hat{a}n - \hat{a})|n\rangle = (n - 1)(\hat{a}|n\rangle)$$ $$\hat{N}(\hat{a}^\dagger|n\rangle) = (\hat{a}^\dagger\hat{N} + [\hat{N}, \hat{a}^\dagger])|n\rangle = (\hat{a}^\dagger n + \hat{a}^\dagger)|n\rangle = (n + 1)(\hat{a}^\dagger|n\rangle)$$

Thus, $\hat{a}$ decreases $n$ by 1, and $\hat{a}^\dagger$ increases $n$ by 1. Since $\langle n | \hat{N} | n \rangle = \|\hat{a}|n\rangle\|^2 \ge 0$, the spectrum must be bounded from below. There must exist a unique ground state $|0\rangle$ such that:

$$\hat{a}|0\rangle = 0 \implies n = 0$$

Repeated application of $\hat{a}^\dagger$ generates the complete discrete spectrum:

$$E_n = \hbar\omega\left(n + \frac{1}{2}\right), \quad n \in \{0, 1, 2, 3, \dots\}$$ $$|n\rangle = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}}|0\rangle, \quad \hat{a}|n\rangle = \sqrt{n}|n-1\rangle, \quad \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$$

3. Exact Matrix Representation of Operators

In the Fock basis $\{|0\rangle, |1\rangle, |2\rangle, \dots\}$, the operators have the infinite-dimensional matrix representations:

$$\hat{a} \doteq \begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \dots \\ 0 & 0 & \sqrt{2} & 0 & \dots \\ 0 & 0 & 0 & \sqrt{3} & \dots \\ 0 & 0 & 0 & 0 & \ddots \end{pmatrix}, \quad \hat{a}^\dagger \doteq \begin{pmatrix} 0 & 0 & 0 & 0 & \dots \\ \sqrt{1} & 0 & 0 & 0 & \dots \\ 0 & \sqrt{2} & 0 & 0 & \dots \\ 0 & 0 & \sqrt{3} & 0 & \ddots \end{pmatrix}$$

Inverting the definitions of $\hat{a}$ and $\hat{a}^\dagger$ expresses the coordinate and momentum operators purely in terms of ladder operators:

$$\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a} + \hat{a}^\dagger), \quad \hat{p} = -i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a} - \hat{a}^\dagger)$$

This yields the off-diagonal Heisenberg matrix mechanics:

$$\langle n' | \hat{x} | n \rangle = \sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n}\delta_{n', n-1} + \sqrt{n+1}\delta_{n', n+1}\right)$$ $$\langle n' | \hat{p} | n \rangle = -i\sqrt{\frac{m\hbar\omega}{2}}\left(\sqrt{n}\delta_{n', n-1} - \sqrt{n+1}\delta_{n', n+1}\right)$$
EXAM SUCCESS WORKSHOP

Solved University Examination Problems

Step-by-step mathematical solutions to classic university honors examination questions.

SOLVED PROBLEM 1.1

Spectral Decomposition and Projector Analysis of a 3-Level Hamiltonian

A three-level quantum system is governed by the Hamiltonian matrix in an orthonormal basis $\{|1\rangle, |2\rangle, |3\rangle\}$ given by:\n$$\hat{H} = \hbar\omega \begin{pmatrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$$\n(a) Find the eigenvalues and normalized eigenvectors of $\hat{H}$.\n(b) Construct the projection operators $\hat{P}_k$ for each energy level and explicitly verify completeness $\sum_k \hat{P}_k = \hat{I}$ and the spectral decomposition $\hat{H} = \sum_k E_k \hat{P}_k$.\n(c) If the system is initially prepared in state $|\psi(0)\rangle = \frac{1}{\sqrt{3}}(|1\rangle + |2\rangle + |3\rangle)$, calculate the probability of measuring the system in energy state $E = 0$ at time $t > 0$.

RIGOROUS DERIVATION & EXAM SOLUTION
Full Rigorous Analytical Solution
**(a) Eigenvalue Spectrum and Eigenvectors:**\nThe characteristic polynomial is:\n$$\det(\hat{H} - \lambda \hat{I}) = \det\begin{pmatrix} 2\hbar\omega - \lambda & 0 & 0 \\ 0 & -\lambda & \hbar\omega \\ 0 & \hbar\omega & -\lambda \end{pmatrix} = (2\hbar\omega - \lambda)(\lambda^2 - (\hbar\omega)^2) = 0$$\nThe eigenvalues are:\n$$E_1 = 2\hbar\omega, \quad E_2 = +\hbar\omega, \quad E_3 = -\hbar\omega$$\nFinding normalized eigenvectors:\n- For $E_1 = 2\hbar\omega$:\n $$\hat{H}|E_1\rangle = 2\hbar\omega|E_1\rangle \implies |E_1\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = |1\rangle$$\n- For $E_2 = +\hbar\omega$:\n $$\begin{pmatrix} 0 & -\hbar\omega & \hbar\omega \\ 0 & \hbar\omega & -\hbar\omega \end{pmatrix} \begin{pmatrix} c_2 \\ c_3 \end{pmatrix} = 0 \implies c_2 = c_3 \implies |E_2\rangle = \frac{1}{\sqrt{2}}(|2\rangle + |3\rangle)$$\n- For $E_3 = -\hbar\omega$:\n $$\begin{pmatrix} 0 & \hbar\omega & \hbar\omega \\ 0 & \hbar\omega & \hbar\omega \end{pmatrix} \begin{pmatrix} c_2 \\ c_3 \end{pmatrix} = 0 \implies c_2 = -c_3 \implies |E_3\rangle = \frac{1}{\sqrt{2}}(|2\rangle - |3\rangle)$$\n\n**(b) Projection Operators and Spectral Decomposition:**\n$$\hat{P}_1 = |E_1\rangle\langle E_1| = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$\n$$\hat{P}_2 = |E_2\rangle\langle E_2| = \frac{1}{2}\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1/2 & 1/2 \\ 0 & 1/2 & 1/2 \end{pmatrix}$$\n$$\hat{P}_3 = |E_3\rangle\langle E_3| = \frac{1}{2}\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} \begin{pmatrix} 0 & 1 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1/2 & -1/2 \\ 0 & -1/2 & 1/2 \end{pmatrix}$$\nSumming the projectors:\n$$\hat{P}_1 + \hat{P}_2 + \hat{P}_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \hat{I}$$\nEvaluating the spectral decomposition:\n$$E_1\hat{P}_1 + E_2\hat{P}_2 + E_3\hat{P}_3 = 2\hbar\omega\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} + \hbar\omega\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1/2 & 1/2 \\ 0 & 1/2 & 1/2 \end{pmatrix} - \hbar\omega\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1/2 & -1/2 \\ 0 & -1/2 & 1/2 \end{pmatrix} = \hbar\omega\begin{pmatrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} = \hat{H}$$\n\n**(c) Measurement Probability:**\nThe allowed energy eigenvalues are $2\hbar\omega, \hbar\omega, -\hbar\omega$. An eigenvalue of $0$ is **not** present in the spectrum of $\hat{H}$. Therefore, the probability of measuring energy $E = 0$ is strictly **zero**: $\mathcal{P}(E=0) = 0$.
Final Answer & Verification

Complete rigorous derivation and proof detailed above.

SOLVED PROBLEM 1.2

Density Matrix Purity, von Neumann Entropy & Spin Measurement Ensemble

A statistical ensemble of spin-1/2 particles consists of a 75% fraction of particles prepared in the spin-up state along the $z$-axis, $|+z\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, and a 25% fraction prepared in the spin-down state along the $x$-axis, $|-x\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$.\n(a) Write down the density matrix $\hat{\rho}$ of the ensemble in the standard $S_z$ basis.\n(b) Compute $\text{Tr}(\hat{\rho}^2)$ and determine whether the state is pure or mixed.\n(c) Calculate the expectation value $\langle S_x \rangle$ and $\langle S_z \rangle$.\n(d) Calculate the von Neumann entropy $S(\hat{\rho})$.

RIGOROUS DERIVATION & EXAM SOLUTION
Full Rigorous Analytical Solution
**(a) Constructing the Density Matrix:**\nThe statistical mixture has probabilities $p_1 = 3/4$ and $p_2 = 1/4$:\n$$\hat{\rho} = \frac{3}{4}|+z\rangle\langle +z| + \frac{1}{4}|-x\rangle\langle -x|$$\n$$|+z\rangle\langle +z| = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$\n$$|-x\rangle\langle -x| = \frac{1}{2}\begin{pmatrix} 1 \\ -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \end{pmatrix} = \begin{pmatrix} 1/2 & -1/2 \\ -1/2 & 1/2 \end{pmatrix}$$\nCombining both contributions:\n$$\hat{\rho} = \frac{3}{4}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \frac{1}{4}\begin{pmatrix} 1/2 & -1/2 \\ -1/2 & 1/2 \end{pmatrix} = \begin{pmatrix} 3/4 + 1/8 & -1/8 \\ -1/8 & 1/8 \end{pmatrix} = \begin{pmatrix} 7/8 & -1/8 \\ -1/8 & 1/8 \end{pmatrix}$$\nNotice that $\text{Tr}(\hat{\rho}) = 7/8 + 1/8 = 1$ and $\hat{\rho}^\dagger = \hat{\rho}$.\n\n**(b) Purity Check:**\n$$\hat{\rho}^2 = \begin{pmatrix} 7/8 & -1/8 \\ -1/8 & 1/8 \end{pmatrix} \begin{pmatrix} 7/8 & -1/8 \\ -1/8 & 1/8 \end{pmatrix} = \begin{pmatrix} 49/64 + 1/64 & -7/64 - 1/64 \\ -7/64 - 1/64 & 1/64 + 1/64 \end{pmatrix} = \begin{pmatrix} 50/64 & -8/64 \\ -8/64 & 2/64 \end{pmatrix}$$\n$$\text{Tr}(\hat{\rho}^2) = \frac{50}{64} + \frac{2}{64} = \frac{52}{64} = \frac{13}{16} = 0.8125$$\nSince $\text{Tr}(\hat{\rho}^2) = 13/16 < 1$, the ensemble is a **mixed state**.\n\n**(c) Expectation Values:**\n$$\hat{S}_x = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \hat{S}_z = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$\n$$\langle S_x \rangle = \text{Tr}(\hat{\rho}\hat{S}_x) = \frac{\hbar}{2}\text{Tr}\begin{pmatrix} -1/8 & 7/8 \\ 1/8 & -1/8 \end{pmatrix} = \frac{\hbar}{2}\left(-\frac{1}{8} - \frac{1}{8}\right) = -\frac{\hbar}{8}$$\n$$\langle S_z \rangle = \text{Tr}(\hat{\rho}\hat{S}_z) = \frac{\hbar}{2}\text{Tr}\begin{pmatrix} 7/8 & 1/8 \\ -1/8 & -1/8 \end{pmatrix} = \frac{\hbar}{2}\left(\frac{7}{8} - \frac{1}{8}\right) = \frac{3\hbar}{8}$$\n\n**(d) von Neumann Entropy:**\nThe eigenvalues of $\hat{\rho}$ satisfy $\det(\hat{\rho} - \lambda\hat{I}) = 0$:\n$$\left(\frac{7}{8}-\lambda\right)\left(\frac{1}{8}-\lambda\right) - \left(-\frac{1}{8}\right)^2 = \lambda^2 - \lambda + \frac{7}{64} - \frac{1}{64} = \lambda^2 - \lambda + \frac{6}{64} = 0$$\n$$\lambda = \frac{1 \pm \sqrt{1 - 24/64}}{2} = \frac{1 \pm \sqrt{40/64}}{2} = \frac{1 \pm \frac{\sqrt{10}}{4}}{2} = \frac{4 \pm \sqrt{10}}{8}$$\n$$\lambda_1 \approx 0.8953, \quad \lambda_2 \approx 0.1047$$\n$$S(\hat{\rho}) = -k_B(\lambda_1 \ln\lambda_1 + \lambda_2 \ln\lambda_2) \approx -k_B(0.8953(-0.1106) + 0.1047(-2.257)) = 0.335\,k_B$$
Final Answer & Verification

Complete rigorous derivation and proof detailed above.

SOLVED PROBLEM 1.3

Matrix Mechanics Evaluation of Higher-Order Coordinate Moments Using Ladder Operators

Using the algebraic ladder operator definitions $\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a} + \hat{a}^\dagger)$ and $[\hat{a}, \hat{a}^\dagger] = 1$:\n(a) Evaluate the expectation value $\langle n | \hat{x}^4 | n \rangle$ for any arbitrary Fock eigenstate $|n\rangle$ of the 1D harmonic oscillator.\n(b) Deduce the quantum ground state ($n=0$) expectation value $\langle 0 | \hat{x}^4 | 0 \rangle$ and verify the quantum uncertainty relationship $\langle x^4 \rangle > (\langle x^2 \rangle)^2$.

RIGOROUS DERIVATION & EXAM SOLUTION
Full Rigorous Analytical Solution
**(a) Expanding the Fourth Power of Position:**\n$$\hat{x}^4 = \left(\frac{\hbar}{2m\omega}\right)^2 (\hat{a} + \hat{a}^\dagger)^4$$\nLet $\hat{B} \equiv (\hat{a} + \hat{a}^\dagger)^2 = \hat{a}^2 + \hat{a}\hat{a}^\dagger + \hat{a}^\dagger\hat{a} + (\hat{a}^\dagger)^2$.\nUsing $\hat{a}\hat{a}^\dagger = \hat{a}^\dagger\hat{a} + 1 = \hat{N} + 1$:\n$$\hat{B} = \hat{a}^2 + (\hat{a}^\dagger)^2 + 2\hat{N} + 1$$\nNow squaring $\hat{B}$ to obtain $(\hat{a} + \hat{a}^\dagger)^4$:\n$$(\hat{a} + \hat{a}^\dagger)^4 = \hat{B}^2 = [\hat{a}^2 + (\hat{a}^\dagger)^2 + (2\hat{N}+1)]^2$$\nWhen evaluating the diagonal expectation value $\langle n | \dots | n \rangle$, only terms with an equal number of creation and annihilation operators have non-vanishing expectation values. The contributing terms are:\n1. $(2\hat{N} + 1)^2$: $\langle n | (2\hat{N}+1)^2 | n \rangle = (2n + 1)^2$\n2. $\hat{a}^2 (\hat{a}^\dagger)^2$: \n $$\hat{a}^2(\hat{a}^\dagger)^2|n\rangle = \hat{a}^2 \sqrt{n+1}\sqrt{n+2}|n+2\rangle = \sqrt{n+1}\sqrt{n+2}\sqrt{n+2}\sqrt{n+1}|n\rangle = (n+1)(n+2)|n\rangle$$\n3. $(\hat{a}^\dagger)^2 \hat{a}^2$:\n $$(\hat{a}^\dagger)^2 \hat{a}^2|n\rangle = (\hat{a}^\dagger)^2 \sqrt{n}\sqrt{n-1}|n-2\rangle = n(n-1)|n\rangle$$\nSumming these three matrix contributions:\n$$\langle n | (\hat{a} + \hat{a}^\dagger)^4 | n \rangle = (2n + 1)^2 + (n+1)(n+2) + n(n-1)$$\nExpanding algebraically:\n$$= (4n^2 + 4n + 1) + (n^2 + 3n + 2) + (n^2 - n) = 6n^2 + 6n + 3 = 3(2n^2 + 2n + 1)$$\nTherefore, the exact coordinate fourth moment is:\n$$\langle n | \hat{x}^4 | n \rangle = \frac{3\hbar^2}{4m^2\omega^2}(2n^2 + 2n + 1)$$\n\n**(b) Ground State Evaluation:**\nSetting $n = 0$:\n$$\langle 0 | \hat{x}^4 | 0 \rangle = \frac{3\hbar^2}{4m^2\omega^2}$$\nComparing with the second moment $\langle 0 | \hat{x}^2 | 0 \rangle = \frac{\hbar}{2m\omega}$:\n$$(\langle 0 | \hat{x}^2 | 0 \rangle)^2 = \left(\frac{\hbar}{2m\omega}\right)^2 = \frac{\hbar^2}{4m^2\omega^2}$$\n$$\frac{\langle 0 | \hat{x}^4 | 0 \rangle}{(\langle 0 | \hat{x}^2 | 0 \rangle)^2} = \frac{\frac{3\hbar^2}{4m^2\omega^2}}{\frac{\hbar^2}{4m^2\omega^2}} = 3 > 1$$\nThis factor of 3 precisely matches the fourth moment of a Gaussian distribution ($\mathbb{E}[x^4] = 3\sigma^4$), confirming that the quantum harmonic oscillator ground state is an exact Gaussian wavepacket.
Final Answer & Verification

Complete rigorous derivation and proof detailed above.