Hilbert Space

The state of a quantum system is described by a vector in a complex vector space called Hilbert space. A Hilbert space is complete (all Cauchy sequences converge) and equipped with an inner product.

For a finite-dimensional system, Hilbert space is just $\mathbb{C}^N$ — the space of $N$-component complex column vectors. For infinite-dimensional systems (like a particle in a box), Hilbert space is the space of square-integrable functions $L^2$.

The inner product of two vectors $|f\rangle$ and $|g\rangle$ is denoted $\langle f|g\rangle$. It satisfies: (1) $\langle f|g\rangle = \langle g|f\rangle^*$, (2) $\langle f|f\rangle \ge 0$ with equality only when $|f\rangle = 0$, and (3) linearity in the second argument: $\langle f|(a|g\rangle + b|h\rangle) = a\langle f|g\rangle + b\langle f|h\rangle$.

Two states are orthogonal if $\langle f|g\rangle = 0$. A state is normalized if $\langle f|f\rangle = 1$. A set of states is orthonormal if $\langle i|j\rangle = \delta_{ij}$.

An operator $\hat{Q}$ is a linear map that takes a vector to another vector: $\hat{Q}|f\rangle = |g\rangle$. In quantum mechanics, physical observables are represented by Hermitian operators.

The Hermitian conjugate (adjoint) $\hat{Q}^\dagger$ is defined by $\langle f|\hat{Q}^\dagger|g\rangle = \langle g|\hat{Q}|f\rangle^*$. An operator is Hermitian if $\hat{Q}^\dagger = \hat{Q}$.

Hermitian operators have three crucial properties: (1) their eigenvalues are real, (2) their eigenvectors corresponding to distinct eigenvalues are orthogonal, and (3) their eigenvectors form a complete set (span the Hilbert space).

A unitary operator $\hat{U}$ satisfies $\hat{U}^\dagger \hat{U} = \hat{U} \hat{U}^\dagger = \hat{I}$. Unitary operators preserve inner products and norms. Time evolution in quantum mechanics is generated by a unitary operator.

Dirac notation introduces two kinds of vectors: kets $|\psi\rangle$ (column vectors) and bras $\langle\psi|$ (row vectors, the Hermitian conjugate). The inner product is written $\langle\phi|\psi\rangle$.

An operator can be written as a sum of outer products: $\hat{Q} = \sum_{ij} Q_{ij} |i\rangle\langle j|$, where $Q_{ij} = \langle i|\hat{Q}|j\rangle$ are the matrix elements in the basis $\{|i\rangle\}$.

The completeness relation for an orthonormal basis $\{|i\rangle\}$ is: $\sum_i |i\rangle\langle i| = \hat{I}$. This is also called the resolution of the identity. Inserting this anywhere in an expression expands it in the basis.

For example, the matrix element $\langle i|\hat{Q}|j\rangle$ can be written as $\langle i|\hat{Q}|j\rangle = \sum_k \langle i|\hat{Q}|k\rangle\langle k|j\rangle = \sum_k Q_{ik}\delta_{kj} = Q_{ij}$. This is just matrix multiplication in Dirac notation.

Suppose we have two orthonormal bases: $\{|i\rangle\}$ (the old basis) and $\{|\alpha\rangle\}$ (the new basis). Any state can be expanded in either basis:

The expansion coefficients are related by:

The numbers $\langle\alpha|i\rangle$ are the elements of the transformation matrix from the old basis to the new basis. Since both bases are orthonormal, the transformation matrix is unitary.

The matrix elements of an operator also transform under a change of basis:

This is just a similarity transformation: $Q' = S Q S^\dagger$, where $S_{\alpha i} = \langle\alpha|i\rangle$.

Consider a two-dimensional Hilbert space with orthonormal basis $\{|1\rangle, |2\rangle\}$. An operator $\hat{A}$ has matrix elements $A_{ij} = \langle i|\hat{A}|j\rangle$ given by the matrix $A = \begin{pmatrix} 3 & 1 \\ 1 & 1 \end{pmatrix}$. Find the eigenvalues, eigenvectors, and verify the spectral theorem.

Step 1 — Find eigenvalues. $\det(A - \lambda I) = (3-\lambda)(1-\lambda) - 1 = \lambda^2 - 4\lambda + 2 = 0$. Thus $\lambda = 2 \pm \sqrt{2}$.

Step 2 — Find eigenvectors. For $\lambda_1 = 2 + \sqrt{2}$: solving $(A - \lambda_1 I)v = 0$ gives $v_1 \propto (1, \sqrt{2}-1)^T$. Normalizing: $|v_1\rangle = \frac{1}{\sqrt{4-2\sqrt{2}}} (1, \sqrt{2}-1)^T$. For $\lambda_2 = 2 - \sqrt{2}$: $|v_2\rangle = \frac{1}{\sqrt{4+2\sqrt{2}}} (1, -\sqrt{2}-1)^T$.

Step 3 — Verify orthonormality. $\langle v_1|v_2\rangle = 0$, as expected for eigenvectors of a Hermitian matrix with distinct eigenvalues. Also $\langle v_1|v_1\rangle = \langle v_2|v_2\rangle = 1$.

Step 4 — Verify completeness. $|v_1\rangle\langle v_1| + |v_2\rangle\langle v_2| = \hat{I}$, the $2\times2$ identity matrix. This confirms the spectral theorem.

Step 5 — Spectral decomposition. $A = \lambda_1 |v_1\rangle\langle v_1| + \lambda_2 |v_2\rangle\langle v_2|$. Reconstructing $A$ from this formula gives the original matrix.