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PU-104 / L18 · lecture

Functions of Operators and the Spectral Calculus

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the definition of a self-adjoint operator and the statement of the Spectral Theorem, which guarantees that such operators possess a complete orthonormal basis of eigenvectors with real eigenvalues.

Why this matters

The spectral calculus provides the rigorous bridge between abstract operator algebra and the concrete functions used in physics, such as time evolution and energy expectation values.

In quantum mechanics and classical normal-mode analysis, we frequently encounter expressions like \( e^{-i\hat{H}t/\hbar} \) or \( \hat{H}^{-1} \). Defining these functions directly on operators is non-trivial because operators do not commute in general. However, for self-adjoint operators, the spectral theorem allows us to define any function \( f(\hat{A}) \) by applying the scalar function \( f \) to each eigenvalue in the spectral decomposition. This technique, known as the spectral calculus, transforms complex operator problems into simple algebraic manipulations of eigenvalues, ensuring that physical predictions remain consistent with the underlying linear structure.

The Core Idea: Acting on Eigenvalues

Imagine an operator \( \hat{A} \) as a machine that stretches space along specific axes (eigenvectors) by specific factors (eigenvalues). If you want to apply a function \( f \) to this machine, you do not need to understand the complex geometry of the whole space. Instead, you simply apply \( f \) to the stretching factors. For example, if \( \hat{A} \) stretches a vector by a factor of 2 along a specific axis, then \( \hat{A}^2 \) stretches it by \( 2^2 = 4 \), and \( \sqrt{\hat{A}} \) stretches it by \( \sqrt{2} \). The direction of the stretch (the eigenvector) remains unchanged; only the magnitude is modified by the function. This principle holds for any well-behaved function \( f \), provided \( \hat{A} \) is self-adjoint, because the eigenvectors form a stable, orthogonal basis that does not mix under the operation.

Formal Definition via Spectral Resolution

Let \( \hat{A} \) be a self-adjoint operator with distinct eigenvalues \( \lambda_1, \dots, \lambda_n \) and corresponding orthogonal projectors \( P_k \). According to the spectral resolution, we can write \( \hat{A} = \sum_{k=1}^{n} \lambda_k P_k \). For any function \( f: \mathbb{R} \to \mathbb{C} \) defined on the spectrum of \( \hat{A} \), the function of the operator is defined as:

\[ f(\hat{A}) = \sum_{k=1}^{n} f(\lambda_k) P_k \]

This definition is consistent with polynomial functions. If \( f(x) = x^m \), then \( f(\hat{A}) = \hat{A}^m \). The power of this definition lies in its generality; it extends naturally to transcendental functions like exponentials and logarithms. Because the projectors \( P_k \) are mutually orthogonal and idempotent (\( P_j P_k = \delta_{jk} P_k \)), the algebraic properties of \( f(\hat{A}) \) are determined entirely by the scalar values \( f(\lambda_k) \).

Commutativity and Functional Calculus

A crucial consequence of the spectral calculus is that any two functions of the same self-adjoint operator commute. Let \( f(\hat{A}) \) and \( g(\hat{A}) \) be two such functions. Their product is:

\[ f(\hat{A})g(\hat{A}) = \left( \sum_{j} f(\lambda_j) P_j \right) \left( \sum_{k} g(\lambda_k) P_k \right) = \sum_{j,k} f(\lambda_j)g(\lambda_k) P_j P_k \]

Using the orthogonality condition \( P_j P_k = \delta_{jk} P_k \), the double sum collapses to a single sum:

\[ f(\hat{A})g(\hat{A}) = \sum_{k} f(\lambda_k)g(\lambda_k) P_k = g(\hat{A})f(\hat{A}) \]

This result is fundamental in quantum mechanics. It implies that if two observables are functions of the same self-adjoint operator (or commute with it), they share a common eigenbasis. This is the mathematical underpinning of the concept of compatible observables, where measurements of one quantity do not disturb the state defined by the other.

Physical Applications: Time Evolution

The most prominent application of the spectral calculus is in the time evolution of quantum states. The time-evolution operator is given by \( \hat{U}(t) = e^{-i\hat{H}t/\hbar} \), where \( \hat{H} \) is the Hamiltonian. Since \( \hat{H} \) is self-adjoint, we can apply the spectral definition:

\[ \hat{U}(t) = \sum_{k} e^{-i\lambda_k t/\hbar} P_k \]

Here, \( \lambda_k \) are the energy eigenvalues. If a system is in an energy eigenstate \( |e_k\rangle \), the time evolution is simply a phase factor \( e^{-i\lambda_k t/\hbar} \). For a general state \( |\psi\rangle = \sum_{k} c_k |e_k\rangle \), the evolved state is \( |\psi(t)\rangle = \sum_{k} c_k e^{-i\lambda_k t/\hbar} |e_k\rangle \). This shows that the spectral calculus reduces the complex differential equation \( i\hbar \frac{d}{dt}|\psi\rangle = \hat{H}|\psi\rangle \) to a set of independent scalar phase rotations.

The derivations

◆ DERIVATION EMBED · D-082 · not re-derived here

Spectral Theorem for Self-Adjoint Operators

Let \( \hat{A} \) be a self-adjoint operator on a finite-dimensional complex inner-product space \( V \) with \( \dim V = n \), so that \( \hat{A}=\hat{A}^{\dagger} \). Then \( \hat{A} \) has real eigenvalues, its eigenvectors belonging to distinct eigenvalues are orthogonal, and \( V \) possesses an orthonormal basis \( \{|e_1\rangle,\dots,|e_n\rangle\} \) consisting entirely of eigenvectors of \( \hat{A} \). Equivalently, the unitary \( \hat{U} \) whose columns are those eigenvectors diagonalizes \( \hat{A} \): \( \hat{U}^{\dagger}\hat{A}\hat{U}=\hat{\Lambda}=\mathrm{diag}(\lambda_1,\dots,\lambda_n) \), and \( \hat{A}=\sum_{k}\lambda_k\,|e_k\rangle\langle e_k| \).

\[ \hat{U}^{\dagger}\hat{A}\hat{U}=\hat{\Lambda}=\mathrm{diag}(\lambda_1,\dots,\lambda_n),\quad \lambda_k\in\mathbb{R},\quad \hat{U}^{\dagger}\hat{U}=\hat{I},\quad \hat{A}=\sum_{k=1}^{n}\lambda_k\,|e_k\rangle\langle e_k| \]
Open full derivation →
◆ DERIVATION EMBED · D-089 · not re-derived here

Orthogonal Projections and Spectral Resolution

On a finite-dimensional complex inner-product space \(V\), an operator \(P\) is an orthogonal projector onto a subspace exactly when it is both idempotent, \(P^2=P\), and self-adjoint, \(P^\dagger=P\); such a \(P\) projects every vector onto the closed subspace \(M=\operatorname{ran}P\) along its orthogonal complement \(M^\perp=\ker P\). Consequently, for a self-adjoint operator \(A=A^\dagger\) with distinct eigenvalues \(\{\lambda_k\}\), the spectral theorem produces eigenprojectors \(\{P_k\}\) that are mutually orthogonal, \(P_jP_k=\delta_{jk}P_k\), resolve the identity, \(\sum_k P_k=\mathbf 1\), and resolve the operator itself, \(A=\sum_k\lambda_k P_k\).

\[ P^\dagger=P,\ P^2=P\ \Rightarrow\ V=\operatorname{ran}P\overset{\perp}{\oplus}\ker P;\qquad A=\sum_k\lambda_k P_k,\quad \sum_k P_k=\mathbf 1,\quad P_jP_k=\delta_{jk}P_k \]
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Tier 3 · watch out

When defining \( f(\hat{A}) \), the function \( f \) must be defined on the entire spectrum of \( \hat{A} \). If \( f \) has a singularity at an eigenvalue \( \lambda_k \) (e.g., \( f(x) = 1/x \) when \( \lambda_k = 0 \)), the operator \( f(\hat{A}) \) is not defined on the subspace associated with \( P_k \). In such cases, one must restrict the domain of the operator to the orthogonal complement of the null space, or introduce a regularization scheme. Ignoring this domain issue leads to mathematical inconsistencies, particularly when computing expectation values or commutators involving inverse operators. C

Worked example

Problem. Consider a two-level quantum system with Hamiltonian \( \hat{H} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \) (in units where \( \hbar = 1 \)). Calculate the time-evolution operator \( \hat{U}(t) = e^{-i\hat{H}t} \) and determine the state of the system at time \( t = \pi/2 \) if the initial state is \( |\psi(0)\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \).

  1. Identify the eigenvalues and eigenvectors of \( \hat{H} \). The eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \). The corresponding orthonormal eigenvectors are \( |e_1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \) and \( |e_2\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \).
  2. Construct the projectors \( P_1 = |e_1\rangle\langle e_1| = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \) and \( P_2 = |e_2\rangle\langle e_2| = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \).
  3. Apply the spectral definition for the exponential: \( e^{-i\hat{H}t} = e^{-i(1)t} P_1 + e^{-i(-1)t} P_2 = e^{-it} P_1 + e^{it} P_2 \).
  4. Compute the matrix form: \( \hat{U}(t) = \begin{pmatrix} e^{-it} & 0 \\ 0 & e^{it} \end{pmatrix} \).
  5. Evaluate at \( t = \pi/2 \): \( e^{-i\pi/2} = -i \) and \( e^{i\pi/2} = i \). Thus, \( \hat{U}(\pi/2) = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} \).
  6. Apply to the initial state: \( |\psi(\pi/2)\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} -i \\ i \end{pmatrix} \).
\[ |\psi(\pi/2)\rangle = \frac{i}{\sqrt{2}} \begin{pmatrix} -1 \\ 1 \end{pmatrix} \]

Units check. The Hamiltonian has units of energy, and \( t \) has units of time. The argument of the exponential is \( -iEt/\hbar \). Since \( E \) is in Joules and \( \hbar \) is in Joule-seconds, \( Et/\hbar \) is dimensionless, which is required for the exponential function. The resulting state vector is dimensionless, as expected for a normalized quantum state.

CHECK YOURSELF

Three quick tests

  1. Why must the eigenvalues of a self-adjoint operator be real for the spectral calculus to yield a unitary time-evolution operator?
    AnswerUnitarity requires \( \hat{U}^\dagger \hat{U} = \hat{I} \). For \( \hat{U} = e^{-i\hat{H}t/\hbar} \), this holds if and only if \( \hat{H} \) is self-adjoint, which implies real eigenvalues. If eigenvalues were complex, the exponential would include a decaying or growing amplitude, violating probability conservation.
  2. If \( \hat{A} \) and \( \hat{B} \) commute, does \( f(\hat{A}) \) commute with \( g(\hat{B}) \)?
    AnswerYes. If \( \hat{A} \) and \( \hat{B} \) commute and are self-adjoint, they can be simultaneously diagonalized. In the common eigenbasis, \( f(\hat{A}) \) and \( g(\hat{B}) \) act as diagonal matrices with entries \( f(\lambda_i) \) and \( g(\mu_i) \), which always commute.
  3. What is the result of \( \hat{A}^2 \) if \( \hat{A} \) has eigenvalues \( \{1, -1\} \)?
    AnswerUsing the spectral calculus, \( \hat{A}^2 = \sum \lambda_k^2 P_k \). Since \( 1^2 = 1 \) and \( (-1)^2 = 1 \), \( \hat{A}^2 = 1 \cdot P_1 + 1 \cdot P_2 = P_1 + P_2 = \hat{I} \). The operator is its own inverse.
PROBLEMS

Problem set for PU-104 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-082 · D-089 · Threads: symmetry, waves, matter