Appel W. Mathematics for Physics and Physicists

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Linear operators 387

This formula has a sense if evaluated at a point y; indeed, this yields the correct formula

ϕ( y) =

+∞

−∞

ϕ(x) δ

( y) dy.

The great strength of D irac’s notation is to lead, very intuitively, to many correct formulas. For

instance, writing

|ϕ〉 =

+∞

−∞

|p〉〈p|ϕ〉 dp

amounts to saying that, for any x ∈ R, we have

ϕ(x) = 〈x|ϕ〉 =

+∞

−∞

〈x| p〉〈p|ϕ〉 dp =

2π }h

+∞

−∞

i p x/ }h

eϕ( p) dp,

which is none other than the inverse Fourier transform formula!

14.3

Linear operators

We will only describe here the simplest properties of linear operators acting

on Hilbert spaces. This is a very complex and subtle theory, which requires by

itself a full volume. The interested reader can read, for instance, the books [4,

95] or [71], which deals with quantum mechanics in Hilbert spaces. We will

deﬁne bounded operators, closed operators, eigenvalues and eigenvectors, and

adjoints of operators.

14.3.a Operators

DEFINITION 14.28 (Linear operator) A linear operator on a Hilbert space H

is a pair



A, D



, wh ere

• D

is a (not necessarily closed) subspace of H, called the domain of A;

• A is a linear map deﬁned on D

and with values in H.

The image of D

by A is called the image of A, and denoted Im( A).

Finally the image of an element ϕ ∈ H by A will be denoted either A(ϕ) or,

more often, simply Aϕ.

Remark 14.29 To say that two operators A and B are equal means, in consequence, that:

• their domains coincide: D

= D

;

• they act identically on their domains: Aϕ = Bϕ for any ϕ ∈ D

Very often, the action is only used as a shorthand for the whole operator; for instance, the

position operator in quantum mechanics, which acts on a wave function ψ by sending it to

the function x 7→ x ψ(x). It may be dangerous, however, to leave th e domain unspeciﬁed; a

number of paradoxes arising from such sloppiness in notation are pointed out in a paper by

F. Gieres [41].

388 Bras, kets, and all that sort of thing

DEFINITION 14.30 (Extension) Let A and B be t wo linear operators on H.

If D

⊂ D

and if Aϕ = Bϕ for any ϕ ∈ D

, then we say that B is an

extension A or that A is the restriction of B to D

Remark 14.31 There is often a large choice available for the domain on which to deﬁne a

linear operator. For instance, the maximal domain on which t h e posit ion operator, denoted

X , is deﬁned is the subspace



ψ ∈ L

(R) ; x 7→ x ψ(x) ∈ L

(R)



However, th is space may not be the most convenient to use. It may be si mpler to restrict X to

the Schwartz space D

= S of C

∞

class functions wi th all derivatives rapidly decaying. This

space is dense in H = L

(R), and the operator X is deﬁned on S . Moreover, this space has

the advantage of being stable with respect to X .

We must insist: the operator (X , D

) is not the same as the operator (X , D

), even if they

are both usually denoted by th e same name “operator X .” The ﬁrst is an extension of the

second (to a strictly larger space).

An important example of an operator is the momentum operator of quan-

tum mechanics. Consider the space H = L

[0, a] of square integrable func-

tions on the interval [0, a]. The wave functions of a particle conﬁned to this

interval a ll satisfy ψ(0) = ψ(a) = 0 (this is a physical constraint). We wish to

deﬁne an operator (t he momentum operator P ) by

P ψ = −i }hψ

′

If ψ ∈ H, its derivative ψ

′

must be taken in the sense of distributions. But

we wish to deﬁne an operator P with image contained in H. Hence this

distribution must be a regular distribution,

which is associated to a square

integrable function. This justiﬁes the following deﬁnition:

DEFINITION 14.32 (Momentum operator P in L

[0, a]) Let H = L

[0, a] .

The momentum operator P is deﬁned, on its domain



ψ ∈ H ; ψ

′

∈ H and ψ(0) = ψ(a) = 0



∀ψ ∈ D

P ψ = −i }h ψ

′

Remark 14.33 (Boundary conditions for the operator P ) One may wonder what the boundary

conditions ψ(0) = ψ(a) really mean; as a matter of fact, ψ, like every function of the L

space,

is deﬁned only up to equality almost everywhere, and therefore, it would seem that a boundary

condition is meaningless. But the fact that ψ

′

∈ L

[a, b ] implies that ψ

′

∈ L

[a, b ] and, in

partic u l ar, that ψ is continuous.

This explains why the boundary conditions make sense.

The momentum operator for a particle th at can range over R is deﬁ ned

similarly:

This does not imply that th e function ψ is dif ferentiable; but it must be conti nuous and

its derivative exists almost ever ywhe re.

Even absolutly continuous, but this is not of concern here.

Linear operators 389

DEFINITION 14.34 (Momentum operator P in L

(RR

RR)) Let H = L

(R). The

momentum operator P is deﬁned, on its domain



ψ ∈ H ; ψ

′

∈ H



P ψ = −i }h ψ

′

∀ψ ∈ D

The limit conditions ψ(−∞) = ψ(+∞) = 0 are automatically satisﬁ ed when

ψ, ψ

′

∈ H.

14.3.b Adjoint

DEFINITION 14.35 (Adjoint) Let



A, D



be a linear operator on H. The

domain of A

†

, denoted D

†

, is the set of vectors ϕ ∈ H for which there

exists a unique vector ω ∈ H satisfying

∀ψ ∈ D

〈ω|ψ〉 = 〈ϕ|Aψ〉.

If D

†

is not empty, the operator adjoint to A, denoted A

†

, is deﬁned by

†

ϕ = ω for any ϕ ∈ D

†

, with th e preceding notation. In other words,

∀ϕ ∈ D

†

∀ψ ∈ D



†





= 〈ϕ|Aψ〉.

PROPOSITION 14.36 (Existence of the adjoint) A has an adjoint if and only if

is dense in H.

Proof. Assume ﬁrst that D

is dense. In the case ϕ = 0, the vector ω = 0 satisﬁes

the equation 〈ω|ψ〉 = 〈ϕ|Aψ〉 for all ψ ∈ D

. So we must show that this vector is

unique. More generally, for a given arbitrary ϕ, l et ω and ω

′

satisfying the required

property. Then 〈ω −ω

′

|ψ〉 = 0 for all ψ ∈ D

. By density, we may ﬁnd a sequence

(ψ

)

n∈N

, with values in D

, which converges to ω −ω

′

. Taking the limit in the scalar

product (which is a continuous map), we obtain kω −ω

′

= 〈ω −ω

′

|ω −ω

′

〉 = 0

and hence ω = ω

′

Conversely, assume that D

is not dense. Then the orthogonal of D

in not equal

to {0} (it is a corollary of Theorem 9.21 ). Choose any χ 6= 0 in th is orthogonal space.

Then 〈χ|ψ〉 = 0 for any ψ ∈ D

. Let ϕ ∈ H. If ω is a candidate vector such that the

relation 〈ω|ψ〉 = 〈ϕ|Aψ〉 holds for all ψ ∈ D

, it is obvious that the vector ω + χ

also satisﬁes this property. So there i s no uniqueness and therefore D

†

= ∅.

Remark 14.37 If D

is dense, it may still be the case that D

†

is equal to {0}.

In the rest of this chapter, we only consider linear operators with dense domain

in H. So the adjoint is always deﬁned.

PROPOSITION 14.38 (Properties of the adjoint) Let A and B be linear opera-

tors.

i) A

†

is a linear operator;

ii) (αA)

†

= α A

†

for all α ∈ C;

390 Bras, kets, and all that sort of thing

iii) if A ⊂ B, then B

†

⊂ A

†

;

iv) if D

†

is dense, then A

†

has un adjoint, denoted A

††

, which is an extension

of A: A ⊂ A

††

In Section 14.4, we will determine the adjoints of some important opera-

tors.

14.3.c B ounded operators, closed operators,

closable operators

In the terminology of Hilbert spaces, a “bounded” operator is s imply a contin-

uous operator. This is justiﬁed by the following deﬁnit ion and proposition:

DEFINITION 14.39 An operator A is bounded if and only if it is bounded

on the unit b all of H, or in other words, if the set



kAϕk ; ϕ ∈ D

and kϕk ¶ 1



is bounded. If A is bounded, we denote

|||A||| = sup

ϕ∈D

kϕk¶1

kAϕk = sup

ϕ∈D

ϕ6=0

kAϕk

kϕk

Just as in any other complete normed vector space, the following holds (it

is just a reminder of Theorem A.51 on page 583 and of sidebar 5 on page 294):

PROPOSITION 14.40 A is bounded if and only if A is continuous.

If A is bounded, it extends uniquely to a continuous linear operator deﬁned on the

whole of H.

A propert y weaker than boundedness b ut which is also important is that

of being a closed operator.

DEFINITION 14.41 (Closed operator, g raph) Let A be a linear operator. The

graph of A is the subspace

Γ(A) =



(ϕ, Aϕ) ; ϕ ∈ D



of H ×H.

The operator A is closed if and only if its graph is closed,

that is, if

for any sequence (ϕ

)

n∈N

with values in D

which converges to a limit ϕ, if

moreover the sequence (Aϕ

)

converges to a vector ψ, then we have ϕ ∈ D

and Aϕ = ψ.

For the topology deﬁned on H ×H by the natural scalar product



(ϕ, ψ)



(ϕ

′

, ψ

′

)



= 〈ϕ|ϕ

′

〉+ 〈ψ|ψ

′

〉.

Linear operators 391

The reader will ﬁnd the following elementary properties easy to check:

PROPOSITION 14.42 Any bounded operator is closed.

THEOREM 14.43 For any operator A, its adjoint A

†

is closed.

DEFINITION 14.44 (Closable operator, closure) An operator A is closable if

there exists a closed extension of A; the smallest such closed extension is called

the closure of A and is denoted A.

PROPOSITION 14.45 A is closable if and only if the domain of A

†

is dense. In

this case, we h ave A = (A

†

)

†

and A

†

= A

†

This notion of closure of an operator may seem very abstract. We will see

below how to construct it in cert ain special cases important in physics.

14.3.d D iscrete and continuous spectra

DEFINITION 14.46 Let



A, D



be an operator. An eigenvalue of A is a

complex number λ such that there exists a nonzero vector ϕ ∈ D

satisfying

the relation

Aϕ = λ ϕ.

Such a vector ϕ is an eigenvector. The set of eigenvectors of A is the discrete

spectrum of A, and is denoted σ

( A).

Example 14.47 (Discrete spectrum of the momentum operator P ) Consider the momentum

operator P on the Hilbert space H = L

(R). It is deﬁned on D

= {ϕ ∈ H ; ϕ

′

∈ H} by

P ψ = −i }hψ

′

Let λ ∈ C. We try to solve, in H, the equation

P ϕ = λϕ.

This implies ϕ

′

= (iλ/ }h) ϕ, and hence ϕ(x) = α e

iλx/ }h

, where α is a constant. However,

functions of this type are square integrable only if α = 0; since the zero function is not an

eigenvector, it follows that P has no eigenvalue: σ

(P ) = ∅.

Example 14.48 (Discrete spectrum of the position operator X) Consider now the position op-

erator X on the Hilbert space H = L

(R). It is deﬁned on D

= {ϕ ∈ H ; x 7→ x ϕ(x) ∈ H}

by X ϕ = ψ, where ψ : x 7→ x ϕ(x).

Let λ ∈ C. As before we try to solve, in H, the equation

X ϕ = λϕ.

It follows that for almost all x ∈ R we have (x −λ) ϕ(x) = 0. Hence the function ϕ is zero

almost everywhere, in other words,

ϕ = 0. Hence X has no eigenvalue: σ

(X ) = ∅.

Since, we recall, the space L

is constructed by identifying together functions which are

equal almost everywhere. Any function which is zero almost everywhere is therefore identiﬁed

with the zero function.

392 Bras, kets, and all that sort of thing

DEFINITION 14.49 Let



A, D



be an operator. A generalized eigenvalue

of A is a complex number λ such that

• λ is not an eigenvalue of A;

• there exists a sequence (ϕ

)

n∈N

of vectors of norm 1 satisfying

lim

n→∞

kAϕ

−λϕ

k = 0.

The set of generalized eigenvalues of A is called the pure continuous spec-

trum of A, and is denoted σ

(A). It is disjoint from the discrete spectrum:

(A) ∩σ

(A) = ∅.

Example 14.50 (Continuous spectrum of the operator X) We saw that the operator X of t h e

Hilbert space L

(R) has no eigenvalue. On the other hand, a function ϕ with graph which is a

very narrow “spike” around a real number λ should satisfy “X ϕ ≈ λϕ.” Hence we will check

that any real number is a generalized ei genvalue, the idea being to construct such a sequence

of spikier and spikier functions. For reasons of normalization, this is not quite the same as a

Dirac sequence.

Let x

∈ R. We consider the sequence (ψ

)

n∈N

of functions deﬁned by ψ

(x) = ψ

(x −x

)

and ψ

(x) =

n Π(nx); recall that the “rectangle” function Π satisﬁes Π(x) = 1 for |x| ¶ 1

and Π (x) = 0 otherwise.

It is easy to ch eck that the functions ψ

are of norm 1. For all n ∈ N, we have



(X − x

)ψ



+∞

−∞

(x)

dx =

1/2n

−1/2n

dx =

12 n

Hence x

is a generalized eigenvalue of the operator X .

One can also show that the continuous spectrum is a subset of R, and therefore conclude

that the continuous spectrum of X is the w h o le real li ne: σ

(X ) = R.

Remark 14.51 In the previous e xample, the sequence (ψ

)

n∈N

converges neither in L

(R) nor

in the sense of distributions (because t h e normalization is cho sen for the purpose of the L

norm). However, the sequence (

n ψ

) is a sequence that converges, in the sense of distribu-

tions, to T = δ(x − x

), which satisﬁes the relation

x δ(x − x

) = x

δ(x − x

in othe r words X T = x

T in the sense of distributions. This is the (generalized) eigenvector

associated to the generalized eigenvalue!

Deﬁning properly what is a (tempered) distribution which is an eigenvector T

of an

operator A requires a proper deﬁnition of what is meant by A T

. This is in fact a delic ate

problem, which we will only solve for a certain type of op erators (see Deﬁnition 14.68).

Example 14.52 (Continuous spectrum of the operator P ) Consider the operator P o n the space

(R). We have already shown that it has no eigenvalue. On the other hand, we can now show

that any p ∈ R is a generalized eigenvalue. The idea is, once more, to ﬁnd a function which

“looks like” the function ω

(which suffers itself from the essential defect of not being square

integrable). For this purpose, we wil l simply average the functions ω

, for values of λ close

to p. Such an average is, up to unimportant factors, the same as the Fourier transform of

a function similar to t h e rectangle function: hence it is a function that decays like 1/x at

inﬁnity, and is therefore square integrable.

To be more precise, deﬁne

(ǫ)

(x)

def

2ǫ

p+ǫ

p−ǫ

(x) dλ =

πǫ

sin(ǫx/ }h)

Hermitian operators; self-adjoint operators 393

for x ∈ R and ǫ > 0. It is easily checked that this function is of norm 1 in L

(R), and that



(P − p) ω

(ǫ)



= C

·ǫ

so that taking ǫ = 1/n with n going to inﬁnity shows t hat p is indeed a generalized eigenvalue.

Later on (see Theorem 14.82) we will see that the same operator deﬁned on L

[0, a] has

no generalized e igenvalue.

To ﬁnish this section, we remark t hat, in addition to the continuous spec-

trum and d iscrete spectrum, there exists a third type.

DEFINITION 14.53 Let A be an operator with dense domain D

. Th e resid-

ual spectrum of A, denoted σ

( A), is the set of complex numbers λ such

that

λ /∈ σ

(A) and λ ∈ σ

†

The spectrum of A is the union of the th ree types of spectra:

σ(A) = σ

(A) ∪σ

(A).

The resolvant set of A is the set of complex numbers which are not in the

spectrum of A: ρ(A) = C\σ(A). It is possible to prove that for any λ ∈ ρ(A),

the operator A −λ Id has a continuous inverse.

Since the residual spectrum of a s elf-adjoint operator is empty, we will

not mention it anymore. The interested reader can look at Exercise 14.3 on

page 403 for another deﬁnition of the various spectra, and may consult a

general reference in functional analysis, such as [71, 94].

14.4

Hermitian operators; self-adjoint operators

The goal of this short section is to prove the following result, which is a

generalization of Theorem 14.7: any self-adjoint operator admits an orthogonal basis

of eigenvectors, where the ba sis may be a generalized basis (i.e., its elements may

be distributions). This fact is absolutely fundamental for quantum mechanics,

where an observable is deﬁned, according to various tex tbooks, as being a

hermitian operator w hich has an orthogonal basis of eigenvectors or, better,

as a self-adjoint operator. In contrast with the ca se of ﬁnite dimension, the

notions of hermitia n operator and of self-adjoint operator no longer coincide,

and this ca us es some difﬁculties.

Finally, we assume in this sect ion that H = L

(R).

Example 14.54 (A classical problem of quantum mechanics) Consider a particle in an inﬁnite

potential si nk, conﬁned to the interval [0, a]. The operator P satisﬁes 〈ϕ|P ψ〉 = 〈P ϕ|ψ〉 for

all f u nct ions ϕ, ψ ∈ D

. However, P has no eigenvalue, and no generalized eigenvalue. Hence

P is not an observable.

394 Bras, kets, and all that sort of thing

14.4.a Deﬁnitions

DEFINITION 14.55 Let A be a linear operator, with d ens e domain D

in H.

Then A is hermitian (or symmetric) if and only if we have

∀ϕ, ψ ∈ D

〈Aϕ|ψ〉 = 〈ϕ|Aψ〉.

THEOREM 14.56 (Characterizations of a hermitian endomorphism) Let A be

an operator on H. Then the following statements are equivalent:

i) A is hermitian;

ii) D

⊂ D

†

and A

†

ϕ = Aϕ for any ϕ ∈ D

;

iii) A ⊂ A

†

Proof. First of all, notice that iii) is just a rephrasing of ii).

Assume that A is hermitian. Then, for any ϕ ∈ D

, the element ω = Aϕ satisﬁes

〈ω|ψ〉 = 〈Aϕ|ψ〉 = 〈ϕ|Aψ〉,

which proves that ϕ ∈ D

†

and A

†

ϕ = Aϕ.

The converse is immediate.

DEFINITION 14.57 Let A be an operator, with D

dense in H. Then A is

self-adjoint if and only if A = A

†

, that is, if

†

= D

and ∀ϕ ∈ D

†

ϕ = Aϕ.

In a ﬁnite-dimensional space, it is equivalent for a n operator to be h ermi-

tian or self-adjoint. Here, however, there is a crucial distinction: the domain

of the ad joint of a hermitian operator A

†

may well be strictly larger than that

of A. In other words, A

†

may be a cting in the same way as A on the domain

of A, but it may also a ct on a larger space. Disregarding this distinction

may lead to unwelcome surprises (an apparent paradox of quantum mechan-

ics is explained in Exercise 14.5). It is often easy to check that an operator is

hermitian, but much more delicate to see whether it is, or is not, self-a djoint.

If A is deﬁned on th e whole of H, then things simplify enormously; since

A ⊂ A

†

, it follows that the domain of A

†

is also equal to H and so if A is

hermitian, it is in fact self-adjoint. The situation is e ven simpler than that:

THEOREM 14.58 (Hellinger-Toeplitz) Let A be a hermitian operator deﬁned on H.

Then A is self-adjoint and bounded.

The adjoint of any operator is closed, and therefore we have

THEOREM 14.59 Any self-adjoint operator is closed.

This result can be made more precise:

Hermitian operators; self-adjoint operators 395

PROPOSITION 14.60 Let A be a hermitian operator. Then A ⊂ A

†

; hence D

†

dense and A is closable. Moreover, A ⊂ A ⊂ A

†

, and A is self-adjoint if and only if

A = A = A

†

If A is hermitian and closed, then A is self-adjoint if and only if A

†

is hermitian.

A last result may be useful.

PROPOSITION 14.61 Let A be a hermitian operator. The defect indices of A are

the quantities, either integers or +∞, deﬁned by

= dim Ker(A

†

−i) and n

−

= dim Ker(A

†

+ i).

The following statements are equivalent:

i) A is self-adjoint;

ii) A is closed and n

= n

−

= 0;

iii) Im(A ±i) = H.

PROPOSITION 14.62 The operator X on the space L

[0, a] is self-adjoint and

bounded.

The operator X on the space L

(R) is self-a djoint and unbounded.

Proof

• Consider the operator X on H = L

[0, a]. Its domain is the whole of H.

Moreover, for any ϕ, ψ ∈ H, we have

〈X ϕ|ψ〉 =

+∞

−∞

x ϕ(x) ψ(x) dx =

+∞

−∞

ϕ(x) x ψ(x) dx = 〈ϕ|X ψ〉,

which proves th at X

†

= X and that X is self-adjoint. It is bounded because

kX ϕk



x ϕ(x)



dx ¶ a



ϕ(x)



dx = a

kϕk

from which we derive in fact that |||X ||| ¶ a. (As an exercise, the reader is invited to

prove that |||X ||| = a.)

• C o nsider now the operator X on the space L

(R). The domain of X is



ϕ ∈ H ; x 7→ x ϕ(x) ∈ H



By the same computation as before, it follows easily that X is hermitian. Let’s prove

that it i s even self-adjoint.

Let ψ ∈ D

†

and write ψ

∗

= X

†

ψ. Then, for any ϕ ∈ D

, we have 〈X ϕ|ψ〉 =

〈ϕ|ψ

∗

〉, by deﬁnition of X

†

. Expressing th e two scalar products as integrals, this gives,

for any ϕ ∈ D

+∞

−∞

ϕ(x)



x ψ(x) −ψ

∗

(x)



dx = 0.

In particular, taking ϕ(x) = x ψ(x) − ψ

∗

(x) for x ∈ [−M , M ] and ϕ(x) = 0 for

|x| > M, we have ϕ ∈ D

and

−M



x ψ(x) −ψ

∗

(x)



dx = 0,

396 Bras, kets, and all that sort of thing

hence ψ

∗

(x) = x ψ(x) almost everywhere on [−M, M], and hence almost everywhere

on R since M was arbitrarily large. It follows that x 7→ x ψ(x) ∈ L

(R), which means

ψ ∈ D

, and in addition ψ

∗

= X ψ.

Thus, we have shown th at D

†

⊂ D

and that X

†

is the restriction of X , that is,

†

⊂ X . Since we know that X ⊂ X

†

, we have X = X

†

Remark 14.63 The operator X on L

[0, +∞[ is also self-adjoint, the previous proof being

valid without change.

PROPOSITION 14.64 The operator P on L

[0, a] is hermitian (symmetric), bu t

not self-adjoint.

Proof. A function ϕ ∈ L

[0, a] belongs to the domain of P if and only if it satisﬁes

the following two conditions:

ϕ and ϕ

′

are in L

[0, a], (∗)

ϕ(0) = ϕ(a) = 0. (∗∗)

(See Remark 14 .33 on p age 38 8 for the meaning of these boundary conditions.) The

subspace D

is dense. Mo reover, for any ϕ, ψ ∈ D

, we have

〈P ϕ|ψ〉 =

−i }h ϕ

′

(x) ψ(x) dx

= −i }h

ϕ(x) ψ

′

(x) dx + i }h



ϕ(a) ψ(a) − ψ(0) ψ(0)



= 〈ϕ|P ψ〉,

since the boundary terms cancel out from condition (∗∗). This establishes that P is

symmetric.

However the domain of P

†

is strictly larger than that of P . Indeed, any function ψ

which sati sﬁes only the condition (∗) is such that 〈P ϕ|ψ〉 = 〈ϕ|P ψ〉 for any ϕ ∈ D

Using techniques similar to that in the proof of Theorem 14.62, it is possible to prove

the converse inclusion, namely, that

†



ϕ ; ϕ, ϕ

′

∈ L

[0, a]



On D

†

, the operator P

†

is deﬁned by P

†

ψ = −i }hψ. So the adjoint of P acts in the

same manner as P on the domain of P , but it has a larger domain.

14.4.b Eigenvectors

The ﬁrst important result concerning hermitian operators is in perfect anal-

ogy with one in ﬁnite dimensions: eigenvectors corresponding to distinct

eigenvalues are orthogonal.

THEOREM 14.65 Let A be a hermitian operator. Then:

i) For any ϕ ∈ D

, 〈ϕ|Aϕ〉 is real.

ii) The eigenvalues of A are real.

iii) Eigenvectors of A associated to distinct eigenvalues are orthogonal.

iv) If (ϕ

)

n∈N

is an orthonormal Hilbert basis where each ϕ

is an eigenvector

of A, then the discrete spectrum of A is e xactly {λ

; n ∈ N}; in oth er words,

there is no other eigenvalue of A.