Spectral Theory of Self-Adjoint Operators

Tier: A (Advanced). Assumes the Banach and Hilbert space material, the Lebesgue integral, and comfort with the Riesz representation theorem. The payoff is the infinite-dimensional generalisation of diagonalisation, and it is the foundation of quantum mechanics, PDE theory and harmonic analysis.

Attempt these from memory before reading.

• State the spectral theorem for a self-adjoint matrix. What does the spectrum consist of?
• Give a bounded operator on a Hilbert space that has no eigenvalues at all.
• Why is a symmetric operator on a Hilbert space not necessarily self-adjoint?
• What is the difference between the spectrum of an operator and its point spectrum?

For a self-adjoint matrix $A$ on $\mathbb{C}^n$, the spectral theorem says $\mathbb{C}^n = \bigoplus_i E_{\lambda_i}$, $A$ acts as multiplication by $\lambda_i$ on $E_{\lambda_i}$, and

where $P_i$ is the orthogonal projection onto $E_{\lambda_i}$.

Two things break on an infinite-dimensional Hilbert space $H$.

Eigenvalues may not exist. The multiplication operator $(Mf)(x) = xf(x)$ on $L^2[0,1]$ has no eigenvalues at all: $Mf = \lambda f$ forces $(x-\lambda)f(x) = 0$ almost everywhere, so $f = 0$. Yet $M$ is the model self-adjoint operator, and its spectral theory is perfectly good. Whatever replaces eigenvalues cannot be a countable set of eigenvectors.

The sum must become an integral. Even when eigenvalues exist they may accumulate, and the projections $P_i$ may be of infinite rank. The finite sum is replaced by integration against a projection-valued measure.

The insight that survives is this:

A self-adjoint operator is multiplication by a function. The only question is which function, on which measure space.

Everything below is bookkeeping for that statement.

Definition (F, U). An operator $T$ on $H$ with dense domain $D(T)$ is symmetric if

It is self-adjoint if $T^* = T$, meaning $D(T^*) = D(T)$ and the operators agree there.

The distinction is invisible on $\mathbb{C}^n$, where $D(T) = H$ always, and it is the single most important technical point in the subject.

Worked example (U). Let $H = L^2[0,1]$ and let $T = i\,\frac{d}{dx}$ on the domain $D(T) = C^1[0,1]$, the continuously differentiable functions.

Integration by parts gives, for $f, g \in C^1$ with $f,g$ vanishing near the endpoints,

so $T$ is symmetric. It is not self-adjoint: one can compute $D(T^*)$ and find it strictly larger than $C^1[0,1]$, because the boundary terms that integration by parts produces can be killed by imposing conditions on $g$ that $g \in C^1$ does not impose.

The consequences are drastic. A symmetric operator need not have real spectrum, need not admit a spectral decomposition, and need not generate a unitary group. Adding a boundary condition makes $T$ self-adjoint, and then all of those return.

Exercise. Show that $i\,d/dx$ on $D(T) = \{f \in H^1[0,1] : f(0) = f(1)\}$ (periodic boundary conditions) has eigenfunctions $e^{2\pi i n x}$, $n \in \mathbb{Z}$, with eigenvalues $2\pi n$. Conclude the spectrum is $2\pi\mathbb{Z}$.

Why the hypothesis is load-bearing. Self-adjointness buys reality of the spectrum, the spectral theorem, and the functional calculus. Symmetry buys none of these. The gap between them is measured by the deficiency indices $(n_+, n_-) = \dim\ker(T^* \mp i)$: symmetric $T$ is self-adjoint exactly when both are zero.

Definition (U). For $T$ closed and densely defined, $\lambda \in \sigma(T)$ if $T - \lambda$ is not bijective onto $H$ with bounded inverse. The spectrum partitions into:

• Point spectrum $\sigma_p$: $T - \lambda$ not injective (eigenvalues).
• Continuous spectrum $\sigma_c$: $T - \lambda$ injective, range dense but not all of $H$.
• Residual spectrum $\sigma_r$: injective, range not dense.

For self-adjoint $T$ there is no residual spectrum, and $\sigma(T) \subseteq \mathbb{R}$.

Worked example (U). For $M$ on $L^2[0,1]$: $M - \lambda$ fails to be invertible as soon as $\lambda \in [0,1]$, since $(x - \lambda)$ vanishes on a set of positive measure and $1/(x-\lambda) \notin L^2$. So $\sigma(M) = [0,1]$, and none of it is point spectrum. The whole spectrum is continuous.

Contrast: $M$ on $\ell^2(\mathbb{N})$, $(a_n) \mapsto (na_n)$, has $\sigma_p = \mathbb{N}$; pure point spectrum. The same formula gives an operator with completely different spectral type, and the difference is which measure space it lives on. This is the sense in which the spectral theorem is a theorem about measure spaces.

Theorem (A, spectral theorem; bounded case). Let $T$ be a bounded self-adjoint operator on $H$. Then there is a unique projection-valued measure $E$ on the Borel sets of $\sigma(T)$ such that

meaning $\langle Tx, y\rangle = \int \lambda \, d\langle E(\lambda)x, y\rangle$ for all $x, y$.

The point of a projection-valued measure is that $E(\Delta)$ is an orthogonal projection for each Borel $\Delta$, with $E(\emptyset) = 0$, $E(\sigma(T)) = I$, and $E(\Delta_1 \cap \Delta_2) = E(\Delta_1)E(\Delta_2)$.

The idea, as opposed to the proof. For a matrix, $E(\{\lambda_i\}) = P_i$ is the projection onto the eigenspace, and the integral is the finite sum. For $M$ on $L^2[0,1]$, $E(\Delta)$ is multiplication by $\mathbf{1}_\Delta$, and

is literally $\int_0^1 \lambda \, d(\text{multiplication by } \mathbf{1}_{[0,\lambda]})$. The spectral theorem says these two pictures are the only pictures: every self-adjoint operator is of the second kind, for some measure space.

The functional calculus (A). For any bounded Borel $\varphi$ on $\sigma(T)$,

is a bounded self-adjoint operator, and $\varphi \mapsto \varphi(T)$ is a $*$-homomorphism: it sends products to products, adjoints to adjoints, and uniform limits to uniform limits. Taking $\varphi$ to be a characteristic function recovers projections; taking it to be $1/(t - z)$ recovers the resolvent.

This is what makes spectral theory useful. Spectral questions become measure questions, and measure theory is much better developed.

Worked example (A). Let $T = M$ on $L^2[0,1]$ and $\varphi(t) = t^2$. Then $\varphi(T) = M^2$, i.e. $(f \mapsto x^2 f(x))$; no surprise. But take $\varphi = \mathbf{1}_{[0,1/2]}$. Then

is the projection onto $L^2[0,\tfrac12]$, and it is a spectral projection even though $M$ has no eigenvalues. The spectral projections replace eigenspaces.

Exercise. Show $\|T\| = \sup\{|\lambda| : \lambda \in \sigma(T)\}$ for self-adjoint $T$. Show also that $\sigma(T)$ is non-empty, using the fact that $T$ has a bounded resolvent only off a closed set.

The position operator $M$ on $L^2(\mathbb{R})$ is unbounded: taking $f$ supported near a large $x$ makes $\|Mf\|/\|f\|$ arbitrarily large. Unbounded operators cannot be defined on all of $H$, so the domain becomes part of the data.

Theorem (A, spectral theorem; unbounded case). Let $T$ be self-adjoint (possibly unbounded). There is a unique projection-valued measure $E$ on $\mathbb{R}$ such that

where $\mu_x(\Delta) = \langle E(\Delta)x, x\rangle$.

The domain is exactly the set where the integral converges. That is why the domain is not an annoyance but part of the theorem.

Where this becomes research. In quantum mechanics a self-adjoint operator is an observable, and its spectrum is the set of possible measurement outcomes. The spectral measure $E$ is the projection-valued measure that gives the Born rule. Two questions drive a great deal of current work:

• What is the spectral type? Decompose $\sigma(T)$ into pure point, singular continuous and absolutely continuous parts. In the Schrödinger operator $-\Delta + V$, the question of which parts occur for a given potential $V$ was open in important cases into the 1990s and remains delicate: the Anderson model exhibits pure point spectrum for large disorder and absolutely continuous spectrum for small disorder, and the transitions between them are a live research area.
• Is the spectrum stable? Weyl’s theorem says compact perturbations do not change the essential spectrum. Singular perturbations do, and spectral pollution; spurious eigenvalues appearing under numerical approximation; is a genuine obstacle in computation.

The page so far has given the picture. This section gives the arguments. They are kept separate because the intuition and the proof do different jobs: the picture tells you what the theorem is for, and the proof tells you what it does not claim.

Theorem 1. If $T$ is self-adjoint then $\sigma(T) \subseteq \mathbb{R}$.

Proof. Let $z = u + iv$ with $v \neq 0$. For $x \in D(T)$,

which follows by expanding $\|(T-u)x - ivx\|^2$ and using $\langle (T-u)x, x\rangle \in \mathbb{R}$ — which is exactly where self-adjointness enters, since $\langle Tx, x \rangle$ is real for self-adjoint $T$. So $T - z$ is bounded below by $|v| > 0$, hence injective with closed range.

The range is dense: $\ker(T - z)^* = \ker(T - \bar z) = \{0\}$ by the same estimate applied to $\bar z$, and $\operatorname{ran}(T-z)^\perp = \ker(T - \bar z)$.

It remains to show the range is all of $H$, and here the estimate does the real work. For $y \in H$, define a functional on $\operatorname{ran}(T-z)$ by $f(Tx) = \langle x, y \rangle$. It is well defined (injectivity) and bounded ($|f(Tx)| \leq \|x\|\|y\| \leq \|y\|\,v^{-1}\|(T-z)x\|$). Extending by Hahn-Banach [Theorem 4.3 in the functional analysis track] gives $\phi \in H^*$ with $\phi((T-z)x) = \langle x, y\rangle$; by Riesz representation there is $w \in H$ with $\langle (T-z)x, w \rangle = \langle x, y \rangle$ for all $x$, i.e. $(T - \bar z)w = y$ and $w \in D(T)$. So $y$ is in the range. $\blacksquare$

Three things are worth noticing. The estimate gives injectivity and closedness at once. Density comes from the adjoint, which is why self-adjointness rather than mere symmetry is needed. And surjectivity is not automatic from the estimate — it needs Hahn-Banach plus Riesz, which is where the proof would break for a symmetric operator that is not self-adjoint.

Theorem 2. A self-adjoint $T$ has no residual spectrum.

Proof. Residual spectrum means $\operatorname{ran}(T - \lambda)$ is not dense. Its orthogonal complement is $\ker(T - \lambda)^* = \ker(T - \bar\lambda)$, which is $\{0\}$ by the estimate in Theorem 1 applied to $\bar\lambda$ with $\operatorname{Im}\lambda \neq 0$; and for $\lambda \in \mathbb{R}$, $\ker(T-\lambda) = \operatorname{ran}(T-\lambda)^\perp$ means the range is dense exactly when $\lambda \notin \sigma_p$, and $\lambda \in \sigma_p$ puts it in the point spectrum, not the residual. Either way residual is empty. $\blacksquare$

This is the statement that makes physics work: a self-adjoint operator’s spectrum consists of eigenvalues and continuous spectrum, and both are things one can integrate against. A residual part would carry no such meaning.

Theorem 3 (existence of the functional calculus, bounded case). There is a unique $*$-homomorphism $\Phi$ from bounded Borel functions on $\sigma(T)$ to $\mathcal{B}(H)$ with $\Phi(1) = I$, $\Phi(\iota) = T$, and $\|\Phi(f)\| \leq \|f\|_\infty$.

Proof, construction. Let $C(\sigma(T))$ be the continuous functions. On the algebra generated by $\iota$ and rational functions of the form $p(\lambda)/(q(\lambda))$ with $q$ non-vanishing on $\sigma(T)$, define $\Phi$ by $\Phi(p(T)q(T)^{-1}) = p(T)q(T)^{-1}$ — well defined because the resolvent identity makes the representation of a rational function independent of how it is written. By the Stone-Weierstrass theorem the polynomials in $\lambda$ are dense in $C(\sigma(T))$, and $\Phi$ is isometric on them because self-adjointness gives $\|p(T)\| = \sup_{\sigma(T)}|p|$ (a $C^*$-identity for self-adjoint $T$, itself a consequence of $\|T^2\| = \|T\|^2$). So $\Phi$ extends to an isometry from $C(\sigma(T))$ into $\mathcal{B}(H)$.

Now fix $x, y \in H$. The map $f \mapsto \langle \Phi(f)x, y\rangle$ is a bounded linear functional on $C(\sigma(T))$ with norm at most $\|x\|\|y\|$. By the Riesz-Markov theorem it is a complex regular Borel measure $\mu_{x,y}$ of total mass at most $\|x\|\|y\|$, and

The assignment $\Delta \mapsto \mu_{x,y}(\Delta)$ is countably additive in $\Delta$, which is the point where measure theory enters and where the matrix picture has nothing to say. Define $E(\Delta)$ by $\langle E(\Delta)x, y \rangle = \mu_{x,y}(\Delta)$; $\mu_{x,y}$ depends sesquilinearly on $(x,y)$, so $E(\Delta)$ is a bounded operator, and $E(\Delta)^2 = E(\Delta)$ because $\mathbf{1}_\Delta^2 = \mathbf{1}_\Delta$ and $\Phi$ is multiplicative on continuous $f$, passing to indicators by dominated convergence. So $E(\Delta)$ is a projection. Uniqueness of the measure from Riesz-Markov gives uniqueness of $E$. $\blacksquare$

Proof, uniqueness. Suppose $E$ and $F$ both satisfy $T = \int \lambda\,dE(\lambda) = \int \lambda\,dF(\lambda)$. Polynomials in $\lambda$ then integrate to the same operator, so $\Phi$ agrees on polynomials, hence on all of $C(\sigma(T))$ by density. Taking indicators of intervals and using regularity to approximate them from above and below by continuous functions gives $E(\Delta) = F(\Delta)$ on every Borel $\Delta$. $\blacksquare$

Where the unbounded case genuinely differs. For an unbounded self-adjoint $T$, the integral defining $\varphi(T)$ has domain

and $D(T) = D(\iota)$ is a proper subspace whenever $T$ is unbounded. Every algebraic identity now comes with a domain caveat: $ST = TS$ means $\overline{D(ST)} = \overline{D(TS)}$ and agreement there, not equality on the nose. Most wrong statements about unbounded operators are wrong because a domain was silently assumed, which is the same failure mode as the deficiency-index counterexample above.

Symmetric but not self-adjoint. $i\,d/dx$ on $C^1[0,1]$, above. Deficiency indices $(1,1)$: self-adjoint extensions exist, but they are a one-parameter family indexed by a phase, and each gives different physics.

Symmetric with no self-adjoint extension. $i\,d/dx$ on $C_c^\infty(0,\infty)$ has deficiency indices $(0,1)$: the equation $T^*\psi = \pm i\psi$ is solved on the half-line by one exponential on one side and none on the other, because square-integrability at $+\infty$ excludes one and permits the other. Unequal indices mean the dimensions cannot match, so no self-adjoint extension exists. The physics consequence is severe: a symmetric Hamiltonian that is not self-adjoint generates a non-unitary time evolution, so probability leaks.

By contrast $-d^2/dx^2$ on $C_c^\infty(0,1)$ has indices $(2,2)$ and does admit self-adjoint extensions — a one-parameter family indexed by a phase, the boundary conditions at the two endpoints. Computing the indices from the domain, rather than assuming them, is the standard error in the subject; always compute $D(T^*)$ from scratch.

A spectral projection that is not a Riesz projection. For $\sigma(T) = \{0\} \cup [1,2]$, the Riesz projection for the spectral set $\{0\}$ agrees with $E(\{0\})$. But for a spectral set $\Delta$ meeting the spectrum in a Cantor set, $E(\Delta)$ has no Riesz-projection analogue, and the holomorphic functional calculus cannot see it. The Borel functional calculus is strictly larger, and that is the point.

• State the spectral theorem for a bounded self-adjoint operator. What replaces the eigenspace decomposition?
• Give an operator with purely continuous spectrum.
• State the difference between symmetric and self-adjoint, and give an example where it matters.
• What are deficiency indices, and what do they tell you?
• What does the functional calculus let you do that the polynomial calculus does not?

Mix these with material from the earlier modules; the selection of method is the part blocked practice never trains.

1. (Measure theory.) Let $\mu_x(\Delta) = \langle E(\Delta)x, x\rangle$ for the operator $M$ on $L^2[0,1]$. Show $\mu_x \ll m$ for every $x$, find its density, and deduce $\sigma(M) = \operatorname{supp}(m)$.
2. (Real analysis.) Show the position operator on $L^2(\mathbb{R})$ is not bounded, by computing $\|M f_n\| / \|f_n\|$ for $f_n = \mathbf{1}_{[n, n+1]}$.
3. (Functional analysis.) Show a compact self-adjoint operator has an orthonormal basis of eigenvectors for the closure of its range, and that its spectrum is countable with $0$ as the only possible accumulation point.
4. (Topology.) Show $\sigma(T)$ is compact and non-empty for bounded $T$.
5. (Advanced.) Prove that for self-adjoint $T$, $\|T\| = \sup_{\lambda \in \sigma(T)} |\lambda|$, using the $C^*$-identity in the functional calculus.

• Normed Spaces and Banach Spaces: the background for bounded operators
• Bounded Linear Operators: resolvents and the Neumann series
• The Fundamental Theorems: Riesz representation, used implicitly throughout
• Lebesgue Measurable Sets: the measure-theoretic background
• How to Use These Notes; the method, and why recall prompts come first