The Spectral Theorem for Self-Adjoint Operators
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You should be comfortable with the definition of a Hermitian (self-adjoint) operator, the concept of an orthonormal basis, and the properties of unitary matrices from the previous lecture on Hermitian operators.
Why this matters
The spectral theorem is the mathematical bedrock that allows us to measure physical quantities in quantum mechanics.
In classical mechanics, we often diagonalize matrices to find normal modes, but in quantum mechanics, the act of measurement is fundamentally tied to the eigenvalues and eigenvectors of operators. Without the spectral theorem, we could not guarantee that an observable has a complete set of orthogonal states in which it is diagonal. This theorem ensures that any self-adjoint operator can be decomposed into a sum of projection operators weighted by real eigenvalues, providing the rigorous justification for the probabilistic interpretation of quantum states and the compatibility of simultaneous measurements.
The Core Idea: Diagonalization as Decomposition
Imagine you have a complex machine that transforms vectors in a space. If the machine is "self-adjoint," it has a special symmetry: it treats inputs and outputs in a way that preserves the inner product structure. The spectral theorem tells us that such a machine is not arbitrary. It can always be broken down into a simple sequence of steps: first, project the vector onto a specific axis (eigenvector), then scale it by a real number (eigenvalue), and finally sum these scaled projections. In plain language, a self-adjoint operator is just a weighted sum of "directions" in the space. Because these directions are mutually orthogonal, we can analyze the operator by looking at each direction independently. This is why we can always find a basis where the operator looks like a simple diagonal list of numbers, rather than a complicated matrix of mixed terms.
Formal Statement and Consequences
Let \( \hat{A} \) be a self-adjoint operator on a finite-dimensional complex inner-product space \( V \) with \( \dim V = n \). The spectral theorem guarantees three critical properties: the eigenvalues are real, the eigenvectors corresponding to distinct eigenvalues are orthogonal, and there exists an orthonormal basis of eigenvectors. This last point is the most powerful: it means the operator is diagonalizable by a unitary transformation.
Mathematically, if \( \{|e_1\rangle,\dots,|e_n\rangle\} \) is the orthonormal eigenbasis, we can construct a unitary matrix \( \hat{U} \) whose columns are these eigenvectors. The diagonalization is expressed as:
where \( \lambda_k \in \mathbb{R} \). The inverse relation gives the spectral decomposition of the operator itself:
Here, \( |e_k\rangle\langle e_k| \) acts as a projection operator onto the subspace spanned by \( |e_k\rangle \). This form is crucial because it shows that \( \hat{A} \) is a linear combination of orthogonal projectors, which are idempotent and Hermitian.
Handling Degeneracies
In many physical systems, eigenvalues are not unique; multiple linearly independent eigenvectors may share the same eigenvalue \( \lambda \). This is known as degeneracy. The spectral theorem still holds, but the choice of basis within the degenerate subspace is not unique. Any orthonormal basis spanning the eigenspace associated with \( \lambda \) is valid.
Suppose \( \lambda_1 = \lambda_2 = \lambda \). The subspace \( V_\lambda = \mathrm{span}(|e_1\rangle, |e_2\rangle) \) is invariant under \( \hat{A} \). Within this subspace, \( \hat{A} \) acts as \( \lambda \hat{I} \). The projection operator for this degenerate level is \( \hat{P}_\lambda = |e_1\rangle\langle e_1| + |e_2\rangle\langle e_2| \). The spectral sum becomes:
This grouping is physically significant because it allows us to define the probability of measuring the value \( \lambda \) as the norm squared of the projection of the state onto \( V_\lambda \), regardless of the specific basis chosen within that subspace.
Unitary Symmetry and Change of Basis
The unitary matrix \( \hat{U} \) that diagonalizes \( \hat{A} \) represents a change of basis from the standard computational basis to the eigenbasis. Since \( \hat{U} \) is unitary, it preserves inner products and norms, ensuring that the physical content of the state vector is unchanged, only its representation is rotated.
If we have an arbitrary state \( |\psi\rangle \), its components in the eigenbasis are \( c_k = \langle e_k|\psi\rangle \). The expectation value of \( \hat{A} \) in state \( |\psi\rangle \) is:
This equation reveals the statistical nature of quantum measurements: the expectation value is the weighted average of the eigenvalues, where the weights are the probabilities \( |\langle e_k|\psi\rangle|^2 \). The spectral theorem ensures that these probabilities sum to one, as \( \sum |\langle e_k|\psi\rangle|^2 = \langle \psi|\psi\rangle = 1 \).
The derivation
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 \( \hat{A}=\sum_{k}\lambda_k\,|e_k\rangle\langle e_k| \).
A common misconception is that the eigenvectors of a self-adjoint operator are unique. While the eigenspaces are unique, the basis vectors within a degenerate eigenspace are arbitrary. If you assume a specific basis for a degenerate level without justification, you may incorrectly predict correlations with other observables that do not commute with \( \hat{A} \). Always check if the eigenvalue is degenerate before assuming a specific eigenvector direction. C
Worked example
Problem. Consider the self-adjoint operator \( \hat{A} \) represented by the matrix \( \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \) in the standard basis. Find the eigenvalues, the normalized eigenvectors, and the spectral decomposition of \( \hat{A} \).
- Find the eigenvalues by solving \( \det(\hat{A} - \lambda \hat{I}) = 0 \). The characteristic equation is \( (2-\lambda)^2 - 1 = 0 \), which gives \( (2-\lambda) = \pm 1 \). Thus, \( \lambda_1 = 3 \) and \( \lambda_2 = 1 \).
- For \( \lambda_1 = 3 \), solve \( (\hat{A} - 3\hat{I})|e_1\rangle = 0 \). This yields \( \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = 0 \), so \( a = b \). The normalized eigenvector is \( |e_1\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \).
- For \( \lambda_2 = 1 \), solve \( (\hat{A} - \hat{I})|e_2\rangle = 0 \). This yields \( \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = 0 \), so \( a = -b \). The normalized eigenvector is \( |e_2\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \).
- Construct the projection operators. \( |e_1\rangle\langle e_1| = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \) and \( |e_2\rangle\langle e_2| = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \).
- Compute the spectral sum: \( \hat{A} = 3 \left( \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \right) + 1 \left( \frac{1}{2} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \right) \).
- Sum the matrices: \( \frac{3}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \), which matches the original operator.
Units check. The operator \( \hat{A} \) is dimensionless in this abstract example. If \( \hat{A} \) represented energy, the eigenvalues would have units of Joules, and the projection operators would be dimensionless, ensuring the sum has units of Joules.
Three quick tests
- Why must the eigenvalues of a self-adjoint operator be real?
Answer
Because \( \langle \psi|\hat{A}|\psi\rangle \) must be real for any eigenvector \( |\psi\rangle \), and \( \hat{A}|\psi\rangle = \lambda|\psi\rangle \) implies \( \lambda = \langle \psi|\hat{A}|\psi\rangle / \langle \psi|\psi\rangle \). - What is the role of the unitary matrix \( \hat{U} \) in the spectral theorem?
Answer
It transforms the operator into its diagonal form \( \hat{\Lambda} \) by changing the basis to the eigenbasis, preserving the inner product structure. - If an eigenvalue is degenerate, is the eigenvector unique?
Answer
No, any orthonormal basis spanning the degenerate eigenspace is valid; the specific vectors are not unique.
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.