The spectral theorem for physicists

A Hamiltonian need not possess a complete list of normalizable eigenvectors. In an infinite-volume field theory its spectrum can contain continuous parts. The spectral theorem is the replacement for the familiar instruction “insert a sum over energy eigenstates.” It lets us discuss a gap without assuming in advance that the spectrum consists of isolated particles.

Begin with a finite matrix

Prerequisites: diagonalizing Hermitian matrices in quantum mechanics.
Outcomes: replace an eigenstate sum by a spectral measure; derive the decay bound from a gap; identify why a single probe can miss lighter states.

If $H$ is a Hermitian matrix, orthogonal projectors $P_n$ onto its eigenspaces satisfy

$$H=\sum_n E_nP_n,\qquad \sum_nP_n=I.$$

For any function $f$ defined on its eigenvalues,

$$f(H)=\sum_n f(E_n)P_n.$$

Thus Euclidean evolution is $e^{-tH}=\sum_n e^{-tE_n}P_n$. For a vector $\psi$,

$$\langle\psi,e^{-tH}\psi\rangle =\sum_n \|P_n\psi\|^2e^{-tE_n}.$$

The coefficients are nonnegative. This positivity is the reason an energy spectrum can be extracted from decay.

Replace the sum by a measure

For a self-adjoint operator on a Hilbert space, the corresponding statement uses a projection-valued measure $P(dE)$. The notation means that an energy interval is assigned an orthogonal projection. A vector $\psi$ then defines an ordinary positive measure

$$d\mu_\psi(E)=\langle\psi,P(dE)\psi\rangle, \qquad \langle\psi,e^{-tH}\psi\rangle =\int e^{-tE}\,d\mu_\psi(E).$$

An isolated eigenvalue contributes an atom to the measure. A continuum contributes a continuous part. Both can coexist. For an unbounded Hamiltonian, self-adjointness includes a specification of its domain; being formally symmetric is not enough. Once a nonnegative self-adjoint Hamiltonian is established, $e^{-tH}$ is a bounded operator for $t\geq0$.

Take a normalized vacuum $\Omega$ with $H\Omega=0$. If $O$ is a suitable smeared physical observable, define

$$\psi=(O-\langle\Omega,O\Omega\rangle)\Omega.$$

Then $\psi$ is orthogonal to the vacuum, and the integral is a connected Euclidean two-point function. Smearing and domain assumptions make this expression meaningful.

What a gap says

A gap $\Delta>0$ means that the Hamiltonian has no spectrum in $(0,\Delta)$. If the vacuum is unique, every vector orthogonal to it obeys

$$0\leq\langle\psi,e^{-tH}\psi\rangle \leq e^{-t\Delta}\|\psi\|^2.$$

This follows simply because $e^{-tE}\leq e^{-t\Delta}$ on the support of the measure. It does not require an isolated eigenvalue at $\Delta$.

The converse needs care. Fast decay for one $\psi$ constrains only $\mu_\psi$. That vector may have zero overlap with lower-energy states. A statement about the whole Hilbert space requires an appropriately rich class of vectors and sufficient control of their correlators.

A three-level example

Let $H=\operatorname{diag}(0,1,3)$ and $\Omega=(1,0,0)$. The vector $\psi=(0,0,1)$ has correlator $e^{-3t}$, even though the Hamiltonian’s gap is $1$. The missing energy is a property of the chosen probe, not of the theory.

Exercise. Take $\psi=(0,\sqrt{w},\sqrt{1-w})$. Find the large-$t$ decay rate for $w>0$ and for $w=0$.

Solution. The correlator is $w e^{-t}+(1-w)e^{-3t}$. Its asymptotic logarithmic decay rate is $1$ for every fixed $w>0$, but $3$ when $w=0$. The limit $w\to0$ is not uniform in the time needed to reveal the first state.

Continue

For the operator statement, see Teschl, Mathematical Methods in Quantum Mechanics, online version 12 February 2009, chapter 3. Use operator domains and quadratic forms when a formally written Hamiltonian has not yet been shown self-adjoint. The free-field model gives a continuum spectral threshold without assuming an isolated normalizable particle state at that threshold.

Use this bridge with Euclidean correlators, operator overlap, and the correlator lab.

For the role of the physical Hamiltonian and gap in the actual problem, see Jaffe and Witten, Quantum Yang–Mills Theory, sections 3–5. The finite-dimensional calculations above supply the elementary bridge; they do not construct that Hamiltonian.