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The spectrum: where eigenvalues become a continuum

The payoff of the whole track. In infinite dimensions the right notion is not the list of eigenvalues but the spectrum — and it splits into point, continuous, and residual parts. This is the mathematics the observables of quantum mechanics actually live in.

Why eigenvalues are not enough

In Vol I, lambda was an eigenvalue of A exactly when A - lambda I failed to be invertible — and in finite dimensions that is the same as having a nonzero eigenvector, because singular and non-invertible coincide. In infinite dimensions these two ideas split apart. An operator can fail to be invertible *without* any eigenvector at all.

So we widen the definition. The spectrum of a bounded operator T is the set of all complex lambda for which (T - lambda I) has no bounded inverse. The eigenvalues are still in there, but the spectrum can contain values that are not eigenvalues at all. The spectrum, not the eigenvalue list, is the correct infinite-dimensional invariant.

Three ways to be in the spectrum

Why might T - lambda I lack a bounded inverse? There are exactly three failure modes, and they split the spectrum into the point, continuous, and residual spectrum. The classification is just an honest bookkeeping of *how* invertibility breaks: is the map non-injective, or injective-but-not-surjective, and if its range misses the whole space, does it at least come dense?

  1. Point spectrum: T - lambda I is not injective. Then there is a nonzero x with T x = lambda x — lambda is a genuine eigenvalue with an eigenvector, exactly the finite-dimensional case.
  2. Continuous spectrum: T - lambda I is injective with dense range, but the inverse is unbounded. No eigenvector exists, yet there are "almost-eigenvectors" with ||(T - lambda I) x_n|| -> 0 while ||x_n|| = 1.
  3. Residual spectrum: T - lambda I is injective but its range is not even dense. A purely infinite-dimensional possibility, tied by the adjoint: lambda is in the residual spectrum of T iff conj(lambda) is an eigenvalue of T*.
The right shift S on ell^2 -- all three flavors on display
   S(x_1, x_2, ...) = (0, x_1, x_2, ...),   ||S|| = 1

spectrum(S) = closed unit disk { |lambda| <= 1 }.

  POINT spectrum:  EMPTY.
     S x = lambda x forces x = 0  (no eigenvectors at all!).

  RESIDUAL spectrum:  { |lambda| < 1 }  (the open disk).
     range of (S - lambda I) is not dense; and indeed
     conj(lambda) IS an eigenvalue of S* = left shift.

  CONTINUOUS spectrum:  { |lambda| = 1 }  (the boundary circle).
     injective, dense range, but unbounded inverse.

Contrast S* (left shift): every |lambda| < 1 is a true
  EIGENVALUE -- eigenvector (1, lambda, lambda^2, ...) in ell^2.
So S and S* have the SAME spectrum, sorted into different bins.
The shift operator realizes empty point spectrum, a residual disk, and a continuous boundary circle at once.

The payoff: a spectral theory that runs the quantum world

For a compact self-adjoint operator the wildness collapses: by the spectral theorem the spectrum is just its eigenvalues plus the limit point 0 — no continuous or residual part survives. For self-adjoint operators in general the spectrum is always *real*, and that single fact is the mathematical reason physical observables take real values.

This is the destination of the whole track. Quantum mechanics is built on self-adjoint operators on a Hilbert space: position, momentum, and energy are unbounded operators whose spectrum is the set of values a measurement can return. Discrete energy levels are point spectrum; a free particle's continuum of momenta is continuous spectrum. The classification you just learned is literally the table of possible measurement outcomes.