If T − λ is
one-to-one and
onto, i.e.
bijective, then its inverse is bounded; this follows directly from the
open mapping theorem of functional analysis. So, λ is in the spectrum of T if and only if T − λ is not one-to-one or not onto. One distinguishes three separate cases:
T − λ is not
injective. That is, there exist two distinct elements x,y in X such that (T − λ)(x) = (T − λ)(y). Then z = x − y is a non-zero vector such that T(z) = λz. In other words, λ is an eigenvalue of T in the sense of
linear algebra. In this case, λ is said to be in the point spectrum of T, denoted σp(T).
T − λ is injective, and its
range is a
dense subset R of X; but is not the whole of X. In other words, there exists some element x in X such that (T − λ)(y) can be as close to x as desired, with y in X; but is never equal to x. It can be proved that, in this case, T − λ is not bounded below (i.e. it sends far apart elements of X too close together). Equivalently, the inverse linear operator (T − λ)−1, which is defined on the dense subset R, is not a bounded operator, and therefore cannot be extended to the whole of X. Then λ is said to be in the continuous spectrum, σc(T), of T.
T − λ is injective but does not have dense range. That is, there is some element x in X and a neighborhood N of x such that (T − λ)(y) is never in N. In this case, the map (T − λ)−1x → x may be bounded or unbounded, but in any case does not admit a unique extension to a bounded linear map on all of X. Then λ is said to be in the residual spectrum of T, σr(T).
So σ(T) is the disjoint union of these three sets,
The complement of the spectrum is known as
resolvent set that is .
Surjectivity of T − λ
Injectivity of T − λ
Injective and bounded below
Injective but not bounded below
not injective
Surjective
Resolvent set ρ(T)
Nonexistent
Point spectrum σp(T)
Not surjective but has dense range
Nonexistent
Continuous spectrum σc(T)
Does not have dense range
Residual spectrum σr(T)
In addition, when T − λ does not have dense range, whether is injective or not, then λ is said to be in the compression spectrum of T, σcp(T). The compression spectrum consists of the whole residual spectrum and part of point spectrum.
For unbounded operators
The spectrum of an
unbounded operator can be divided into three parts in the same way as in the bounded case, but because the operator is not defined everywhere, the definitions of domain, inverse, etc. are more involved.
Examples
Multiplication operator
Given a σ-finite
measure space (S, Σ, μ), consider the Banach space
Lp(μ). A function h: S → C is called
essentially bounded if h is bounded μ-almost everywhere. An essentially bounded h induces a bounded multiplication operator Th on Lp(μ):
The operator norm of T is the essential supremum of h. The
essential range of h is defined in the following way: a complex number λ is in the essential range of h if for all ε > 0, the preimage of the open ball Bε(λ) under h has strictly positive measure. We will show first that σ(Th) coincides with the essential range of h and then examine its various parts.
If λ is not in the essential range of h, take ε > 0 such that h−1(Bε(λ)) has zero measure. The function g(s) = 1/(h(s) − λ) is bounded almost everywhere by 1/ε. The multiplication operator Tg satisfies Tg · (Th − λ) = (Th − λ) · Tg = I. So λ does not lie in spectrum of Th. On the other hand, if λ lies in the essential range of h, consider the sequence of sets {Sn =
h−1(B1/n(λ))}. Each Sn has positive measure. Let fn be the characteristic function of Sn. We can compute directly
This shows Th − λ is not bounded below, therefore not invertible.
If λ is such that μ( h−1({λ})) > 0, then λ lies in the point spectrum of Th as follows. Let f be the characteristic function of the measurable set h−1(λ), then by considering two cases, we find
so λ is an eigenvalue of Th.
Any λ in the essential range of h that does not have a positive measure preimage is in the continuous spectrum of Th. To show this, we must show that Th − λ has dense range. Given f ∈ Lp(μ), again we consider the sequence of sets {Sn = h−1(B1/n(λ))}. Let gn be the characteristic function of S − Sn. Define
Therefore, multiplication operators have no residual spectrum. In particular, by the
spectral theorem,
normal operators on a Hilbert space have no residual spectrum.
In the special case when S is the set of natural numbers and μ is the counting measure, the corresponding Lp(μ) is denoted by lp. This space consists of complex valued sequences {xn} such that
T is a
partial isometry with operator norm 1. So σ(T) lies in the closed unit disk of the complex plane.
T* is the right shift (or
unilateral shift), which is an isometry on l q, where 1/p + 1/q = 1:
For λ ∈ C with |λ| < 1,
and T x = λ x. Consequently, the point spectrum of T contains the open unit disk. Now, T* has no eigenvalues, i.e. σp(T*) is empty. Thus, invoking reflexivity and the theorem in
Spectrum_(functional_analysis)#Spectrum_of_the_adjoint_operator (that σp(T) ⊂ σr(T*) ∪ σp(T*)), we can deduce that the open unit disk lies in the residual spectrum of T*.
The spectrum of a bounded operator is closed, which implies the unit circle, { |λ| = 1 } ⊂ C, is in σ(T). Again by reflexivity of l p and the theorem given above (this time, that σr(T) ⊂ σp(T*)), we have that σr(T) is also empty. Therefore, for a complex number λ with unit norm, one must have λ ∈ σp(T) or λ ∈ σc(T). Now if |λ| = 1 and
then
which cannot be in l p, a contradiction. This means the unit circle must lie in the continuous spectrum of T.
So for the left shift T, σp(T) is the open unit disk and σc(T) is the unit circle, whereas for the right shift T*, σr(T*) is the open unit disk and σc(T*) is the unit circle.
For p = 1, one can perform a similar analysis. The results will not be exactly the same, since reflexivity no longer holds.
Self-adjoint operators on Hilbert space
Hilbert spaces are Banach spaces, so the above discussion applies to bounded operators on Hilbert spaces as well. A subtle point concerns the spectrum of T*. For a Banach space, T* denotes the transpose and σ(T*) = σ(T). For a Hilbert space, T* normally denotes the
adjoint of an operator T ∈ B(H), not the transpose, and σ(T*) is not σ(T) but rather its image under complex conjugation.
For a self-adjoint T ∈ B(H), the
Borel functional calculus gives additional ways to break up the spectrum naturally.
This subsection briefly sketches the development of this calculus. The idea is to first establish the continuous functional calculus, and then pass to measurable functions via the
Riesz–Markov–Kakutani representation theorem. For the continuous functional calculus, the key ingredients are the following:
If T is self-adjoint, then for any polynomial P, the operator norm satisfies
The
Stone–Weierstrass theorem, which implies that the family of polynomials (with complex coefficients), is dense in C(σ(T)), the continuous functions on σ(T).
The family C(σ(T)) is a
Banach algebra when endowed with the uniform norm. So the mapping
is an isometric homomorphism from a dense subset of C(σ(T)) to B(H). Extending the mapping by continuity gives f(T) for f ∈ C(σ(T)): let Pn be polynomials such that Pn → f uniformly and define f(T) = lim Pn(T). This is the continuous functional calculus.
For a fixed h ∈ H, we notice that
is a positive linear functional on C(σ(T)). According to the Riesz–Markov–Kakutani representation theorem a unique measure μh on σ(T) exists such that
This measure is sometimes called the spectral measure associated to h. The spectral measures can be used to extend the continuous functional calculus to bounded Borel functions. For a bounded function g that is Borel measurable, define, for a proposed g(T)
In the present context, the spectral measures, combined with a result from measure theory, give a decomposition of σ(T).
Decomposition into absolutely continuous, singular continuous, and pure point
Let h ∈ H and μh be its corresponding spectral measure on σ(T) ⊂ R. According to a refinement of
Lebesgue's decomposition theorem, μh can be decomposed into three mutually singular parts:
where μac is absolutely continuous with respect to the Lebesgue measure, μsc is singular with respect to the Lebesgue measure and atomless, and μpp is a pure point measure.[1][2]
All three types of measures are invariant under linear operations. Let Hac be the subspace consisting of vectors whose spectral measures are absolutely continuous with respect to the
Lebesgue measure. Define Hpp and Hsc in analogous fashion. These subspaces are invariant under T. For example, if h ∈ Hac and k = T h. Let χ be the characteristic function of some Borel set in σ(T), then
So
and k ∈ Hac. Furthermore, applying the spectral theorem gives
This leads to the following definitions:
The spectrum of T restricted to Hac is called the absolutely continuous spectrum of T, σac(T).
The spectrum of T restricted to Hsc is called its singular spectrum, σsc(T).
The set of eigenvalues of T is called the pure point spectrum of T, σpp(T).
The closure of the eigenvalues is the spectrum of T restricted to Hpp.[3][nb 1]
So
Comparison
A bounded self-adjoint operator on Hilbert space is, a fortiori, a bounded operator on a Banach space. Therefore, one can also apply to T the decomposition of the spectrum that was achieved above for bounded operators on a Banach space. Unlike the Banach space formulation,[clarification needed] the union
need not be disjoint. It is disjoint when the operator T is of uniform multiplicity, say m, i.e. if T is unitarily equivalent to multiplication by λ on the direct sum
for some Borel measures . When more than one measure appears in the above expression, we see that it is possible for the union of the three types of spectra to not be disjoint. If λ ∈ σac(T) ∩ σpp(T), λ is sometimes called an eigenvalue embedded in the absolutely continuous spectrum.
When T is unitarily equivalent to multiplication by λ on
the decomposition of σ(T) from Borel functional calculus is a refinement of the Banach space case.
Quantum mechanics
The preceding comments can be extended to the unbounded self-adjoint operators since Riesz-Markov holds for
locally compactHausdorff spaces.
A particle is said to be in a bound state if it remains "localized" in a bounded region of space.[6] Intuitively one might therefore think that the "discreteness" of the spectrum is intimately related to the corresponding states being "localized". However, a careful mathematical analysis shows that this is not true in general.[7] For example, consider the function
This function is normalizable (i.e. ) as
Known as the
Basel problem, this series converges to . Yet, increases as , i.e, the state "escapes to infinity". The phenomena of
Anderson localization and
dynamical localization describe when the eigenfunctions are localized in a physical sense. Anderson Localization means that eigenfunctions decay exponentially as . Dynamical localization is more subtle to define.
Sometimes, when performing quantum mechanical measurements, one encounters "
eigenstates" that are not localized, e.g., quantum states that do not lie in L2(R). These are
free states belonging to the absolutely continuous spectrum. In the
spectral theorem for unbounded self-adjoint operators, these states are referred to as "generalized eigenvectors" of an observable with "generalized eigenvalues" that do not necessarily belong to its spectrum. Alternatively, if it is insisted that the notion of eigenvectors and eigenvalues survive the passage to the rigorous, one can consider operators on
rigged Hilbert spaces.[8]
An example of an observable whose spectrum is purely absolutely continuous is the
position operator of a free particle moving on the entire real line. Also, since the
momentum operator is unitarily equivalent to the position operator, via the
Fourier transform, it has a purely absolutely continuous spectrum as well.
The singular spectrum correspond to physically impossible outcomes. It was believed for some time that the singular spectrum was something artificial. However, examples as the
almost Mathieu operator and
random Schrödinger operators have shown, that all types of spectra arise naturally in physics.[9][10]
Decomposition into essential spectrum and discrete spectrum
Let be a closed operator defined on the domain which is dense in X. Then there is a decomposition of the spectrum of A into a
disjoint union,[11]
Dunford, N.; Schwartz, J. T. (1988). Linear Operators, Part 1: General Theory. John Wiley & Sons.
ISBN0-471-60848-3.
Jitomirskaya, S.; Simon, B. (1994). "Operators with singular continuous spectrum: III. Almost periodic Schrödinger operators". Communications in Mathematical Physics. 165 (1): 201–205.
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10.1007/BF02099743.
ISSN0010-3616.