Spectral Theorem, Practical Version

The spectral theorem is the mathematical reason self-adjoint operators can represent sharp observables. In finite dimensions it says that a Hermitian matrix can be diagonalized by orthogonal projectors. In infinite dimensions it replaces a simple sum over eigenvalues by a projection-valued measure.

The practical lesson is:

A self-adjoint operator is equivalent to a real spectral decomposition, and probabilities are computed by projecting onto spectral sets.

This page gives the working version used throughout quantum mechanics. It does not prove the theorem.

The exact unbounded theorem, including the square-integrability domain, multiplicity-aware multiplication representation, Borel functional calculus, and proof, is Spectral Theorem for Unbounded Self-Adjoint Operators.

Helpful background. Review finite spectral decomposition to expand a matrix using orthogonal eigenprojectors.

Finite-Dimensional Model

For a finite-dimensional Hermitian operator $A$,

$$A=\sum_a aP_a,$$

where $a$ runs over distinct eigenvalues and $P_a$ projects onto the eigenspace for $a$. The projectors satisfy

$$P_aP_b=\delta_{ab}P_a, \qquad \sum_a P_a=I.$$

For a normalized state $\psi$, the probability of obtaining the value $a$ is

$$\Pr(a) = \langle\psi\vert P_a\psi\rangle.$$

The finite-dimensional version is treated in Spectral Decomposition. The general theorem keeps the same projector logic but allows intervals, continuous spectra, and unbounded operators.

Projection-Valued Measures

For a self-adjoint operator $A$, the spectral theorem assigns to each Borel set $\Delta\subseteq\mathbb R$ a projector

$$E_A(\Delta).$$

The map

$$\Delta\mapsto E_A(\Delta)$$

is called a projection-valued measure, or PVM. It has the following practical meaning:

• $E_A(\Delta)$ projects onto the part of the state whose $A$-measurement values lie in $\Delta$;
• $E_A(\mathbb R)=I$ and $E_A(\emptyset)=0$;
• disjoint spectral sets correspond to orthogonal projectors;
• unions of disjoint sets correspond to sums of the corresponding projectors, in the appropriate limiting sense.

For discrete spectra, $E_A(\Delta)$ is just the sum of eigenspace projectors whose eigenvalues lie in $\Delta$:

$$E_A(\Delta) = \sum_{a\in\Delta}P_a.$$

For continuous spectra, $E_A(\Delta)$ is not usually a finite-rank projector. It is still the object that gives probabilities.

Spectral Integral

The spectral theorem says that a self-adjoint operator can be reconstructed from its spectral measure:

$$A = \int_{\mathbb R} \lambda\,dE_A(\lambda).$$

This notation is the infinite-dimensional analogue of

$$A=\sum_a aP_a.$$

The integral is not an ordinary integral of numbers. It is an operator integral against a projector-valued measure. In calculations, it behaves like “multiply each spectral component by its spectral value.”

Measurement Probabilities

For a normalized state $\psi$, define a probability measure

$$\mu_\psi(\Delta) = \langle\psi\vert E_A(\Delta)\psi\rangle.$$

Then

$$\Pr(A\in\Delta) = \mu_\psi(\Delta).$$

This single formula covers discrete and continuous cases:

• if $\Delta$ contains isolated eigenvalues, the probability is a sum of projector weights;
• if $\Delta$ is an interval in continuous spectrum, the probability is an integral of a density when a density representation exists;
• if the spectrum is mixed, both contributions can appear.

The Core Formalism distinction is introduced in Discrete and Continuous Spectra. The Toolkit interpretation of generalized eigenvectors is Continuous Spectra.

Expectation Values

For a normalized state, a finite mean exists when $\int |\lambda|\,d\mu_\psi(\lambda)<\infty$. In that case,

$$\langle A\rangle_\psi = \int_{\mathbb R} \lambda\,d\mu_\psi(\lambda).$$

In the finite-dimensional case, this reduces to

$$\langle A\rangle_\psi = \sum_a a\,\langle\psi\vert P_a\psi\rangle.$$

For unbounded $A$, not every normalized vector has a finite expectation value. The spectral measure also describes the domain:

$$D(A) = \left\{ \psi\in\mathcal H: \int_{\mathbb R} \lambda^2\,d\mu_\psi(\lambda)<\infty \right\}.$$

The second-moment condition is stronger than the first-moment condition for a finite mean. If $\psi\in D(A)$, the variance is $\|A\psi\|^2-\langle A\rangle_\psi^2$; it does not require $\psi\in D(A^2)$. Exercise 3 gives a normalized state with finite mean energy but infinite variance. See Domains of Operators.

Functions of Operators

If $f$ is a suitable function, the spectral theorem defines

$$f(A) = \int_{\mathbb R} f(\lambda)\,dE_A(\lambda).$$

For finite-dimensional operators, this is just

$$f(A)=\sum_a f(a)P_a.$$

This is how one defines square roots, signs, exponentials, and other functions of self-adjoint operators. If $f$ is unbounded, then $f(A)$ may be an unbounded operator with its own domain.

The Core page Functions of Operators uses this idea for powers, inverses, and time evolution.

Position Operator Example

On $L^2(\mathbb R)$, the position operator $X$ acts as

$$(X\psi)(x)=x\psi(x)$$

on its natural domain. Its spectral projectors are especially concrete:

$$(E_X(\Delta)\psi)(x) = \mathbf 1_{\Delta}(x)\psi(x),$$

where $\mathbf 1_{\Delta}$ is the indicator function of the set $\Delta$.

Therefore

$$\Pr(X\in\Delta) = \int_{\Delta} \lvert\psi(x)\rvert^2\,dx.$$

This is the ordinary position-space Born rule, now expressed as a spectral projection.

Hamiltonian and Time Evolution

If $H$ is a self-adjoint Hamiltonian, the spectral theorem defines

$$U(t) = e^{-iHt/\hbar} = \int_{\mathbb R} e^{-i\lambda t/\hbar}\,dE_H(\lambda).$$

Because each spectral value is multiplied by a phase of modulus one, $U(t)$ is unitary on all of $\mathcal H$, including states outside $D(H)$. The norm derivative $i\hbar\,dU(t)\psi/dt=HU(t)\psi$ is guaranteed for $\psi\in D(H)$. This is the operator-theoretic content behind the statement that a self-adjoint Hamiltonian generates closed-system time evolution.

For a Hamiltonian with discrete bound states and continuous scattering states, the spectral measure contains both parts. One should not force the whole problem into a sum over normalizable eigenvectors.

Common Mistakes

• Assuming the spectral theorem always gives an orthonormal basis of normalizable eigenvectors.
• Treating $dE_A(\lambda)$ as an ordinary scalar measure.
• Forgetting that only self-adjoint operators have the standard real spectral theorem used for observables.
• Ignoring the domain of an unbounded operator recovered from its spectral integral.
• Applying $f$ to matrix entries instead of spectral values.
• Confusing probability density with probability for continuous spectra.
• Forgetting mixed spectra, where discrete and continuous parts coexist.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, Academic Press, 1980.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.

Exercises

1. Let $A=\sum_a aP_a$ be a finite-dimensional spectral decomposition. What is $E_A(\Delta)$ for a set $\Delta\subseteq\mathbb R$?

Solution

It is the sum of projectors whose eigenvalues lie in $\Delta$:

$$E_A(\Delta) = \sum_{a\in\Delta}P_a.$$

If no eigenvalue lies in $\Delta$, then $E_A(\Delta)=0$.

2. For position on the line, show that the spectral projector formula gives the usual probability for an interval $[a,b]$.

Solution

For position,

$$(E_X([a,b])\psi)(x) = \mathbf 1_{[a,b]}(x)\psi(x).$$

Thus

$$\langle\psi\vert E_X([a,b])\psi\rangle = \int_a^b \lvert\psi(x)\rvert^2\,dx.$$

3. Why can an unbounded self-adjoint operator fail to be defined on every normalized state? As an explicit example, take an orthonormal basis $\{e_n\}_{n\ge1}$ of $\ell^2(\mathbb N)$ and the self-adjoint Hamiltonian $He_n=E_0n e_n$, with $E_0>0$, on its natural domain. For

$$\psi=\sum_{n=1}^{\infty}\sqrt{p_n}\,e_n, \qquad p_n=\frac{4}{n(n+1)(n+2)},$$

check normalization, mean energy, and the second moment. Does $e^{-iHt/\hbar}\psi$ exist, and does it have a derivative in the Hilbert-space norm?

Solution

The spectral theorem gives

$$D(A) = \left\{ \psi: \int_{\mathbb R} \lambda^2\,d\mu_\psi(\lambda)<\infty \right\}.$$

A normalized state only requires $\mu_\psi(\mathbb R)=1$. It may still have too much spectral weight at large $\lvert\lambda\rvert$ for the second moment to be finite. Then $\psi\notin D(A)$.

For the stated Hamiltonian, $D(H)$ consists of sequences $c$ with $\sum_n n^2|c_n|^2<\infty$. Normalization follows by telescoping:

$$p_n=\frac{2}{n(n+1)}-\frac{2}{(n+1)(n+2)}, \qquad \sum_{n=1}^{\infty}p_n=1.$$

The mean and second moment behave differently:

$$\begin{aligned} \langle H\rangle_\psi &=4E_0\sum_{n=1}^{\infty}\frac{1}{(n+1)(n+2)} =2E_0,\\ \int E^2\,d\mu_\psi(E) &=4E_0^2\sum_{n=1}^{\infty}\frac{n}{(n+1)(n+2)} =\infty. \end{aligned}$$

Thus $\psi\in D(H^{1/2})$ but $\psi\notin D(H)$, and its energy variance is infinite. The finite mean is the spectral integral, equivalently the quadratic form $\|H^{1/2}\psi\|^2$; the ordinary vector pairing $\langle\psi|H\psi\rangle$ is unavailable because $H\psi$ is not a Hilbert-space vector.

Nevertheless,

$$U(t)\psi=\sum_{n=1}^{\infty}\sqrt{p_n}\,e^{-iE_0nt/\hbar}e_n$$

exists and has norm one for every $t$. A norm derivative would have coefficients $-iE_0n\sqrt{p_n}/\hbar$ at $t=0$, which are not square summable. The unitary orbit therefore has no Hilbert-space norm derivative, even though the state has a finite mean energy. This illustrates the distinction between the spectral domain criterion and unitary evolution (Teschl, 2014, §§3.1 and 5.1).

4. Use the spectral theorem to explain why $e^{-iHt/\hbar}$ is unitary when $H$ is self-adjoint.

Solution

The spectral theorem gives

$$e^{-iHt/\hbar} = \int_{\mathbb R} e^{-i\lambda t/\hbar}\,dE_H(\lambda).$$

The multiplier $e^{-i\lambda t/\hbar}$ has modulus one for real $\lambda$. Therefore the functional calculus produces a unitary operator.