mathematical methods in quantum mechanics · 2020. 2. 16. · mathematical methods in quantum...

Mathematical Methods in Quantum Mechanics

Lars Franke, Martin Lang and Laurin Benz

Winter term 2019/2020

Introduction and Disclaimer

This PDF contains the lecture notes for the lecture “Mathematical Methods in Quantum Me-

chanics I” (http://www.math.kit.edu/iana1/lehre/quantummech2019w/en), given by

Dr. Anapolitanos at the Karlsruhe Institute of Technology (KIT) during the winter term

2019/20.

While we try to keep the content clear of mistakes and in a sensible style of presentation, we do

not take any responsibility for its correctness or any violations of copyright. We do not claim

the rights for any of its content ourselves, but emphasize that the contents were put together

by Dr. Anapolitanos.

The document includes an index of used terminology and symbols at the end. Symbols are,

however, not in the same alphabetical order as humans would probably arrange them. If you

find yourself looking for a certain symbol in the index make sure you search the complete list of

symbols for it.1

1Errors, typos, and suggested improvements can be reported to martin at fachschaft.physik.kit.edu and

laurin.benz at student.kit.edu.

http://www.math.kit.edu/iana1/lehre/quantummech2019w/en

Contents

0 Introduction 7

1 Weak derivatives and Sobolev spaces 13

2 Unbounded operators, spectrum, and resolvent 17

3 Symmetric and self-adjoint operators 23

4 The Schrodinger equation and existence of dynamics 33

5 Observables, Uncertainty Principle, Ground state energy 43

6 Some tools of functional analysis 49

7 Decomposition of an operator 53

8 Discrete and essential spectrum 61

8.1 Exponential decay of eigenfunctions . . . . . . . . . . . . . . . . . . . . . . . . . 71

9 Spectral theorem for self-adjoint operators 75

9.1 Proofs using spectral theorem in projection-valued measure form . . . . . . . . . 85

10 Newton’s theorem and Zhislin’s theorem 89

11 Pure point spectrum, continuous spectrum, RAGE theorem 93

12 Exercises 99

Bibliography 105

Index 107

0 Introduction

When light (from some source like a laser) shines through a small slit on a detector placed

behind the slit, one observes an intensity pattern like the one shown in Figure 0.1.

−40 −20 0 20 40

Angle

0.0

0.2

0.4

0.6

0.8

1.0

Inte

nsity

Single slit experiment

Figure 0.1: Intensity pattern of light shining through a single slit.

If instead the wall with the slit has a second slit (double slit experiment), the light waves cross

both slits simultaneously and an interference pattern appears on the screen.

Now let us replace the source of light by a source of slow (nonrelativistic) electrons. Since

electrons are particles, one would expect that the electron has the choice to pass through either

of the slits, but not through both slits simultaneously, unlike the light waves. Thus, we would

expect a pattern like Figure 0.2 in a double-slit experiment with electrons instead of light.

It came quite surprising in 1927 that electrons show the same behaviour as light in a double-slit

experiment: an interference pattern, as shown in Figure 0.3.

Therefore, a beam of light seems to have some wave-like properties; in particular, there is

8 Chapter 0 Introduction

Angle

0.0

0.1

0.2

0.3

0.4

0.5

Inte

nsity

Two single-slit experiments

Figure 0.2: Naive expectation: the intensity pattern of the electron double-slit experiment is the

sum of the intensity patterns of two electron single-slit experiments. The angle scale

is much smaller than in the previous figure, the distance between the peaks is the

distance between the two slits.

interference from both slits.1 Moreover, even after reducing the flux of electrons drastically so

that there was never more than electron between the source and the screen, the interference

pattern appeared on the screen after enough electrons have been sent through the experiment

successively. Hence, even single electrons seem to “pass both slits simultaneously”, as did the

light waves.

This observation had profound consequences:

1. Physicists had to abandon the idea of knowing the location where a single electron will hit the

screen (and knowing through which slit the electron has passed). Instead, it is only possible

to predict the (probability) distribution of locations on the screen.

2. Interference also applies to particles (and it is sizeable for small enough particles like electrons.

This phenomenon was explained with Quantum Mechanics. We describe a particle by a complex-

valued wave function (state vector) ψ(x, t), where x ∈ R3 is the position and t ∈ R is the time

1It was soon realized that both slits are necessary since the interference pattern disappeared as soon as one of

the two slits was closed.

Chapter 0 Introduction 9

−40 −20 0 20 40

Angle

0.0

0.2

0.4

0.6

0.8

1.0

Inte

nsity

Double slit experiment

double slitsingle slit

Figure 0.3: Electrons show the same interference behaviour as light.

and consider |ψ(x, t)|2 as the probability distribution for fixed time t. Since particles show

interference like waves, we want ψ to satisfy some wave equation.

Since we interpret |ψ(x, t)|2 as the probability distribution, we need that

∫|ψ(x, t)|2 dx <∞∀t;

thus we embed ψ(·, t) into the Hilbert space, ψ(·, t) ∈ L2(R3), where

L2(R3)

=

ψ : R3 → C,

∫|ψ|2 <∞

with the inner product 〈ψ,ϕ〉 =

∫R3

ψ(x)ϕ(x) dx.

The equation that ψ satisfies is a partial differential equation called the Schrodinger equation.

We list some physically sensible properties that ψ and the Schrodinger equation should have.

1. ψ(·, t0) should determine ψ(·, t) for all t > t0.

2. If ψ and ϕ are evolution of states, then aψ + bϕ (with a, b ∈ R) should also be an evolution

of a state (superposition principle).

3.

∫|ψ(x, t)|2 dx should be conserved.

4. Quantum mechanics should be close to classical mechanics in some limit.

From properties 1. and 2., it follows that the Schrodinger equation can be at most a first-order

10 Chapter 0 Introduction

differential equation in t and that it has to be linear:

i∂

∂tψ = Aψ (0.1)

where A is a linear operator.

From 3. we obtain∂

∂t〈ψ,ψ〉 = 0⇒

⟨∂ψ

∂t, ψ

⟩+

⟨ψ,∂ψ

∂t

⟩= 0⇒ 〈−iAψ,ψ〉+ 〈ψ,−iAψ〉 = 0⇒

i 〈Aψ,ψ〉−i 〈ψ,Aψ〉 = 0⇒ 〈Aψ,ψ〉 = 〈ψ,Aψ〉 ∀ψ ∈ L2(R3) Exercise 2⇒ 〈Aψ,ϕ〉 = 〈ψ,Aϕ〉 ∀ϕ,ψ ∈

L2(R3). So A has to be a symmetric operator.

To assure that the initial-value problem id

dtϕt = Hϕt

ϕt|t=0 = u

. (0.2)

has exactly one solution for all ψ0 ∈ L2(R3), we actually need a self-adjoint operator A (more

than symmetric).

Definition 0.1 (Banach and Hilbert spaces)

Let X be a vector space with a norm ‖·‖ (respectively inner product 〈·, ·〉). X is called a

Banach space (respectively Hilbert space) if it is complete (every Cauchy sequence converges)

with respect to the norm (respectively inner product). If in addition X is a complex vector

space, then X is called a complex Banach space (respectively complex Hilbert space).

Example 0.1

L2(R3)

:=

ψ : R3 → C,

∫|ψ|2 dx <∞

equipped with the inner product

〈ψ,ϕ〉 :=

∫R3

ψ (x)ϕ (x) dx is a complex Hilbert space and therefore a complex Banach space

(see e.g. [1]).

Example 0.2

Let X :=ϕ ∈ L2 (R) : ϕ ∈ C1 (R) and ϕ′ ∈ L2 (R)

with norm ‖ϕ‖X := ‖ϕ‖L2 +

∥∥ϕ′∥∥L2 .

Then X is not a Banach space. See Exercise 8.

Chapter 0 Introduction 11

Hint: Let un (x) := exp

(−(

1

n+ |x|2

) 12

). Then un is a Cauchy sequence in X and not

convergent in X. Indeed, if un → u in X for some u ∈ X, then un → u in L2. But

un → exp (−|x|) in L2, so u (x) = exp (−|x|). But u /∈ X because u /∈ C1 (R), therefore X is

not complete.

1 Weak derivatives and Sobolev spaces

We will use the notation C∞c (Rn) := φ : Rn → C : φ ∈ C∞ (Rn) and supp (φ) is compact,where supp (φ) = x ∈ Rn : φ (x) 6= 0. A φ ∈ C∞c (Rn) is called a test function.

Example 1.1

Let f : R → R, f (x) =

e−1x2 , x > 0

0 , x ≤ 0. Then f ∈ C∞ (R) (proof by induction) and

g ∈ C∞c (R), where g(x) := f(x)f(1− x).

We write Lp(X) :=

φ : X → C :

∫X|φ|p <∞

and L1

loc (Rn) :=φ : Rn → C : φ ∈ L1(K) for all closed and bounded subsets K ⊆ Rn

.

Example 1.2

(i) Let φ(x) := ex2. Then φ ∈ L1

loc (R) because φ is bounded on bounded closed sets (as a

continuous function) and thus

∫K|φ| <∞ for all closed and bounded K ⊆ R.

(ii) Let f(x) :=1

|x| . Then f /∈ L1loc (R), since f /∈ L1([−1, 1]).

If α = (α1, . . . , αn), we write ∂α := ∂α1x1. . . ∂αnxn and |α| := α1 + . . .+ αn.

Let u ∈ C∞c (Rn). Then we have ∀φ ∈ C∞c (Rn) :

∫u∂αφ dx = (−1)|α|

∫(∂αu)φ dx by integra-

tion by parts (IBP). In fact, if u, v ∈ C∞c (Rn), then ω = ∂αu if and only if

∀φ ∈ C∞c (Rn) :

∫Rnu∂αφ dx = (−1)|α|

∫Rnωφdx . (1.1)

Definition 1.1

Let u, v ∈ L1loc (Rn). We say that ω = ∂αu is the α-th weak partial derivative of u if

equation (1.1) holds.

14 Chapter 1 Weak derivatives and Sobolev spaces

Example 1.3

Let u : R→ R, u = exp(−|x|) = e−|x|. Then in the sense of weak derivatives u′(x) =x

|x|e−|x|.

Indeed, let φ ∈ C∞c (R), then supp (φ) ⊆ [−M,M ] for some M > 0. We have∫Ruφ′ dx =

∫ M

−Muφ′ dx =

∫ 0

−Muφ′ dx+

∫ M

0uφ′ dx

=

∫ 0

−Mexφ′ dx+

∫ M

0e−xφ′ dx

IBP=

∫ 0

−M(−ex)φ dx+

∫ M

0e−xφ dx

=

∫ M

−M

x

|x|e−|x|φ dx =

∫R

x

|x|e−|x|φ dx ,

verifying equation (1.1).

Proposition 1.2

If u ∈ L1loc (Rn) has a weak derivative then it is unique.

Definition 1.3 (Sobolev space)

Let k ∈ N. Hk (Rn) :=u ∈ L2 (Rn) : ∂αu ∈ L2 (Rn) ∀α with |α| ≤ k

equipped with the

norm ‖u‖Hk(Rn) :=

∑|α|≤k

‖∂αu‖2L2

12

is called Sobolev space of order k.

Example 1.4

Let u be as in Example 1.3. Then u(x) = e−|x|, so u ∈ L2 (R), u′ =x

|x|e−|x|, and u′ ∈ L2 (R).

Therefore, u ∈ H1 (R). (Exercise 8: un = exp

(−(

1

n+ |x|2

) 12

)→ u in H1 (R).)

Theorem 1.4

The Hk (Rn) norm is equivalent to the norm ‖u‖ :=

∫Rn

(1 + |ξ|2

)k|u (ξ)|2 dx, where u is the

Fourier transform of u. In particular, the H2 (Rn) norm is equivalent to

‖u‖ :=(‖u‖2L2(Rn) + ‖∆u‖2L2(Rn)

) 12.

Theorem 1.5 (Completeness)

Hk (Rn) equipped with the norm ‖·‖Hk(Rn) is a Banach space (without proof).

Theorem 1.6

If u ∈ Hk (Rn) there exists a sequence of functions (un)n∈N in C∞c (Rn) with un → u in Hk (Rn)

Chapter 1 Weak derivatives and Sobolev spaces 15

(without proof). So Hk (Rn) is the smallest possible complete space containing C∞c (Rn).

2 Unbounded operators, spectrum, and

resolvent

Let (X, ‖·‖) be a complex Banach space, D ⊆ X a subspace of X. Let A : D → X be a linear

operator. If D = X, then A is called densely defined .

The range (or image of A is defined as Ran (A) := Ax : x ∈ D.The kernel of A is the inverse image of 0: Ker (A) := x ∈ D : Ax = 0.A is called bounded if ‖A‖ := sup‖Ax‖ : x ∈ D, ‖x‖ = 1 <∞, otherwise A is called unbounded .

Remark

A is bounded ⇒ ‖A‖ = sup

‖Ax‖‖x‖ : x ∈ D,x 6= 0

< ∞ ⇒ ∀x ∈ X : ‖Ax‖ ≤

‖A‖‖x‖ ⇒ ‖A(x− y)‖ ≤ ‖A‖‖x− y‖ ⇒ A is Lipschitz continuous⇒ A is continuous.

The converse directions are also true for linear operators A.

Example 2.1

Let A : H2(R3)⊆ L2

(R3)→ L2

(R3), A = −∆ + i.

A is linear and densely defined because H2 (R3) = L2(R3).

A is not bounded:

Let ψ ∈ C∞c(R3)

with ‖ψ‖L2 = 1. Consider (ψn)n∈N, ψn(x) = n32ψ(nx). Then

‖ψn‖2L2 =

∫R3

n3|ψ (nx)|2 dxy=nx

=

∫R3

|ψ (y)|2 dy = 1.

For Ax we have

‖Ax‖ = ‖(−∆ + i)ψn‖ ≥ ‖−∆ψn‖ − ‖ψn‖ = ‖∆ψn‖ − 1. (2.1)

Furthermore (∆ψn)(x)chain rule

= n32n2(∆ψ)(nx).

Therefore ‖∆ψn‖2L2 = n4

∫R3

n3|(∆ψ)(nx)|2 dx = n4‖∆ψ‖2 by repetition of the previous trans-

formation argument. Thus, ‖∆ψn‖L2 →∞ and therefore by equation (2.1) and ‖ψn‖ = 1 we

have that ‖Aψn‖ → ∞. So A is unbounded.

18 Chapter 2 Unbounded operators, spectrum, and resolvent

We find that Ker (A) = 0 because if Aψ = 0, then (−∆ + i)ψ = 0F.T.⇒ (|ξ|2 + i)ψ(ξ) = 0⇒

ψ = 0⇒ ψ = 0.

The image/range of A is Ran (A) = L2(R3): Let f ∈ L2

(R3). We will find a g ∈ H2 with

Ag = (−∆ + i)g = f . We have (−∆ + i)g = f ⇔ (|ξ|2 + i)g(ξ) = f(ξ)⇔ g (ξ) =f(ξ)

|ξ|2 + i.

g ∈ H2 because

‖g‖2H2 =

∫R3

(1 + |ξ|2

)2|g(ξ)|2 dx i =

∫R3

(1 + |ξ|2

)2

∣∣∣f(ξ)∣∣∣2

|ξ|4 + 1dx i ≤ C

∫R3

∣∣∣f(ξ)∣∣∣2 dx i <∞

for some C < ∞. Thus, Ran (A) = L2(R3). In particular, A = −∆ + i : H2

(R3)→ L2

(R3)

is a bijection. This will help us prove that ∆ is self-adjoint later.

Definition

Let A : D ⊆ X → X be a linear operator. A is called closed if the graph of A

ΓA := (x, y) : x ∈ D, y = Ax is closed in the graph norm ‖φ‖A = ‖φ‖+ ‖Aφ‖.This is equivalent to the following statement: For all sequences (xn)n∈N in D and x ∈ X, if

‖xn − x‖ → 0 and ‖y −Ax− n‖ → 0 for some y ∈ X, we have that x ∈ D and y = Ax.

For bounded (continuous) linear operators, xn → x implies Axn → Ax.

For closed linear operators, we have: if xn → x and Axn converges to some y, then y = Ax.

Example 2.2

Let A : H2(R3)⊆ L2

(R3)→ L2

(R3), A = −∆.

Then A is closed because if (fn)n∈N ⊆ H2(R3)

and fn → f in L2 and −∆fn → g in L2 for

some g ∈ L2, then f is Cauchy in the graph norm given by ‖u‖A = ‖u‖L2 + ‖∆u‖L2 ≈ ‖u‖H2 ,

where ≈ denotes the equivalence of the two norms.

But H2 is complete, so fn converges to h for some h ∈ H2. Thus, f = h ∈ H2 and fn → f in

H2. Therefore, −∆fn → −∆f , so −∆f = g.

Proposition (Exercise 5)

(a) Let (X, ‖·‖) be a Banach space, A : D ⊆ X → X closed, and g : [0, T ]→ (D, ‖·‖A) for some

T > 0. Then A

∫ T

0g(t) dt =

∫ T

0Ag(t) dt.

(b) If A = −∆ and g : [0, T ]→ H2(R3)

is continuous, then −∆

∫ T

0g(t) dt =

∫ T

0−∆g(t) dt.

Chapter 2 Unbounded operators, spectrum, and resolvent 19

Definition (Resolvent and spectrum)

The resolvent set of A is defined as:

ρ (A) :=z ∈ C : (z −A) : D(A)→ X is a bijection and (z −A)−1 is bounded

.

Let L(X) := T : X → X : T is linear and bounded.As the resolvent of A one defines the mapping RA : ρ (A)→ L(X), RA(z) := (z −A)−1.

The spectrum of A is defined as σ (A) := C \ ρ (A).

Example 2.3

Let A : H2(R3)⊆ L2

(R3)→ L2

(R3), A = ∆. Then i ∈ ρ (A) because (∆ − i) is a

projection (in Example 2.1 we proved that Ker (∆− i) = 0 and Ran (∆− i) = L2(R3)).

Since∥∥(i∆)−1f

∥∥L2 ≤

∥∥(i−∆)−1f∥∥H2 ≤ C‖f‖L2 , it follows that (i−∆)−1 is bounded.

Lemma 2.1

1. If A is not a closed operator then σ (A) = C.

2. If A is a closed operator and D = X, then A is bounded.

(without proof)

Lemma 2.2

Let K ∈ L(X). If ‖K‖ := sup‖Kx‖ : ‖x‖ = 1 < 1, then the operator I + K is invertible

and (I +K)−1 =∞∑k=0

(−K)n (Exercise 7).

Theorem 2.3

Let X be a Banach space, D ⊆ X a subspace, and A : D → X. Then:

1. ρ (A) is open.

2. σ (A) is closed.

3. The resolvent RA is analytic in ρ (A). More precisely, if z0 ∈ ρ (A), then

B

(z0,

1

‖RA(z0)‖

)⊆ ρ (A)1 and

1B (z0, R) = x ∈ X : ‖x− z0‖ < R


∀z ∈ B(z0,

1

‖RA (z0)‖

): RA(z) =

∞∑n=0

(−1)nRA(z0)n+1(z − z0)n.

In particular, ‖RA(z0)‖ ≥ 1

dist (z0, σ (A)), with dist (z0, σ (A)) := inf|z − y| : y ∈ σ (A).

Proof. Let z ∈ C. Then

z −A = z0 −A+ (z − z0) = (1 + (z − z0)RA(z0))(z0 −A). (2.2)

By Lemma 2.2 the operator 1 + (z − z0)RA(z0) is invertible if ‖(z − z0)RA(z0)‖ < 1

⇔ |z − z0| <1

‖RA(z0)‖ . So B

(z0,

1

‖RA(z0)‖

)⊆ ρ (A), and hence ρ (A) is open and σ (A) is

closed.

If equation (2.2) holds we thus have: (z −A)−1 = (z0 −A)−1(1 + (z − z0)RA(z0))−1

= RA(z0)∞∑n=0

(−(z − z0)RA(z0))n =∞∑n=0

(−1)nRA(z0)n+1(z − z0)n.

Now consider z ∈ σ (A). Then z−A is not invertible, and by equation (2.2) 1+(z−z0)RA(z0) is

not invertible. So ‖(z − z0)RA(z0)‖ ≥ 1 ⇒ |z − z0| ≥1

‖RA(z0)‖ ⇒ infz∈σ(A)

|z − z0| ≥1

‖RA(z0)‖ .

Theorem 2.4

Let V : Rd → C be a continuous function and D :=f ∈ L2

(Rd)

: V · f ∈ L2(Rd)

,

TV : D → L2(Rd)

, (TV f)(x) := V (x) · f(x).

Then σ (TV ) = Ran (V ). In particular, if V is real-valued, then σ (TV ) ⊆ R.

Proof. To prove this, it is sufficient to show

1. If z /∈ Ran (V ), then z ∈ ρ (TV ).

2. If z ∈ Ran (V ), then z ∈ σ (TV ).

From 1. it follows that Ran (V )C ⊆ ρ (TV ), i.e. Ran (V ) ⊇ σ (TV ). We will only show 1. (2. was

shown in Exercise 9): Let z /∈ Ran (V ).

(i) z − TV is injective. Indeed (z − TV )(f) = 0⇔ (z − V (x))f(x) = 0, so either z − V (x) = 0

or f(x) = 0∀x. But since z /∈ Ran (V ), we must have f ≡ 0 and thus Ker (z − TV ) = 0.

(ii) Ran (z − TV ) = L2(Rd)

. Consider a function g ∈ L2(Rd)

. We want to find a function

f ∈ D such that (z − TV )f = g ⇔ (z − V (x))f(x) = g(x). Since z − V (x) 6= 0, we find

f(x) =g(x)

z − V (x). (2.3)

Chapter 2 Unbounded operators, spectrum, and resolvent 21

Since z /∈ Ran (V ), there exists an ε > 0 such that ∀x ∈ Rd : |z − V (x)| ≥ ε. From

equation (2.3) it follows that |f(x)| =|g(x)|

|z − V (x)| ≤|g(x)|ε

, so ‖f‖2 ≤|g(x)|ε

< ∞, and

hence f ∈ L2(Rd)

. Thus, we find V (x)f(x) = zf(x) − g(x), so V · f ∈ L2(Rd)

and

f ∈ D. Finally, since (z − TV )−1g = f , we have∥∥(z − TV )−1g

∥∥2

= ‖f‖2 ∀g ∈ L2(Rd)

. So∥∥(z − TV )−1g∥∥

2≤ ‖g‖2

ε, hence (z − TV )−1 : L2

(Rd)→ L2

(Rd)

is bounded.

Remark

For V : Rd → C measurable (not necessarily continuous) the multiplication operator TV is

defined as well and σ (TV ) = ess sup(V ) :=y ∈ C : ∀ε > 0 : µ(V −1(B (y, ε))) > 0

, where µ

is the Lebesgue measure.

Example 2.4

Let V : Rd → R, V (x) := |x|2. Then by Theorem 2.3 σ (TV ) = Ran (V ) = [0,∞).

Theorem 2.5

Let A : H2(Rd)→ L2

(Rd)

be defined as Af = −∆f . Then σ (−∆) = [0,∞).

Proof. Denote by F and F−1 the Fourier transform on Rd and its inverse. Then F(−∆f)(ξ) =

|ξ|2f(ξ) ⇒ −∆ = F−1|ξ|2F . Hence, z − (−∆) = F−1(z − |ξ|)F ⇒ z − (−∆) is invertible

⇔ z − |ξ|2 is invertible for all ξ ⇔ z /∈ [0,∞).

Definition 2.6

Let X be a Banach space, D1, D2 ⊆ X subspaces, and A : D1 → X, B : D2 → X linear

operators. We say that B is an extension of A if D1 ⊆ D2 and ∀x ∈ D1 : Ax = Bx. We write

A ⊂ B in this case.

Definition 2.7

Let X be a Banach space, D ⊆ X a subspace, and A : D → X a linear operator. A is called

closable if there exists an extension B that is closed.

Example 2.5

Let A : C∞c (Rn) → L2 (Rn), B : H2 (Rn) → L2 (Rn) be defined as Af = −∆f , Bf = −∆f .


Then A ⊂ B.

Theorem 2.8

Let X be a Banach space, D ⊆ X a subspace and A : D → X a linear operator. Then the

following are equivalent:

1. A is closable.

2. ΓA is a graph of a linear operator (which we denote by A).

3. If (xn)n∈N is a sequence in D with xn → 0 and Axn → y for some y ∈ X, then y = 0.

Proof. • 1.⇒ 3.: Let A ⊂ B for some closed operator B. Consider a sequence (xn)n in D

with xn → 0 and Axn → y ∈ X. We need to show that y = 0. Since B is an extension of

A, we have Bxn → y, and since B is closed it follows that y = B · 0 = 0.

• 3.⇒ 2.: Consider (x, y1), (x, y2) ∈ ΓA ⇒ (0, y1 − y2) ∈ ΓA. Thus, there exists a sequence

(xn, Axn)n in ΓA with xn → 0 and Axn → y1 − y23.⇒ y1 − y2 = 0⇒ y1 = y2.

We define the domain of B by D (B) := x ∈ X : ∃y ∈ X such that (x, y) ∈ ΓA and

B : D (B) → X, Bx = y for (x, y) ∈ ΓA. B is well-defined by the argument presented

above. B is linear and ΓB = ΓA. Furthermore, D (B) ⊇ D = D (A).

• By assumption, there exists a linear operator B with ΓB = ΓA. Thus, B is closed and

A ⊂ B.

Example 2.6

Let X = L2 (R), g ∈ X with ‖g‖2 = 1. Define A : Cc(R)→ X, (Af)(x) = f(0) · g(x).

Then A is linear, but not closable. Indeed, consider the sequence fn(x) = f(nx), n ∈ N. Then

fn ∈ Cc(R) ∀n and ∀n : fn(0) = f(0), Afn = Af .

We have

∫R|fn(x)|2 dx =

∫R|f(nx)|2 dx

y=nx=

1

n

∫R|f(x)|2 dx. But ‖fn‖2 =

1

n‖f‖2 → 0 as

n→∞, so fn → 0 in L2.

3 Symmetric and self-adjoint operators

In this chapter letH be a complex Hilbert space and A : D (A) ⊆ H → H a densely defined linear

operator. If for some φ ∈ H there exists one and only φ∗ such that 〈φ,Aψ〉 = 〈φ∗, ψ〉 ∀ψ ∈ D (A),

then φ ∈ D (A∗) and A∗φ = φ∗. The operator A∗ is called the adjoint of A.

Remark

If such a φ∗ exists, then it is unique.

Indeed, if 〈φ∗, ψ〉 =⟨φ∗, ψ

⟩∀ψ ∈ D (A), then ∀ψ ∈ D (A) :

⟨φ∗ − φ∗, ψ

⟩= 0

⇒ ∀ψ ∈ D (A) = H :⟨φ∗ − φ∗, ψ

⟩= 0⇒ φ∗ = φ∗. So A∗ is well-defined only if A is densely

defined. Similarly, A∗∗ = (A∗)∗ is well-defined only if A∗ is densely defined, which is not

always the case.

For φ ∈ H let Tφ : D (A)→ C, Tφ(ψ) = 〈φ,Aψ〉.

Lemma 3.1

We have D (A∗) = φ ∈ H : Tφ is continuous in D (A) (here we are considering D (A) with

respect to the topology induced by the inner product of H).

Proof. If φ ∈ D (A∗), then there exists one (and only one) φ∗ ∈ H with ∀ψ ∈ D (A) : Tφ(ψ) =

〈φ∗, ψ〉, so ‖Tφ(ψ)‖ ≤ C‖ψ‖, where C = ‖φ∗‖ =√〈φ∗, φ∗〉. Thus, Tφ is continuous. If Tφ is

continuous on D (A), then since D (A) = H, by the Riesz representation theorem there exists

one and only one φ∗ ∈ H with Tφ(ψ) = 〈φ∗, ψ〉 ∀ψ ∈ D (A), so φ ∈ D (A∗).

Example 3.1

Let H = L2(R), g ∈ H with ‖g‖ = 1. Let A : Cc(R) ⊆ H → H be defined by Af = f(0)g.

Then D (A∗) = ϕ ∈ H : ϕ ⊥ g and A∗ = 0 in D (A∗). Indeed, if 〈ϕ, g〉 = 0, then

∀f ∈ Cc(R) : 〈ϕ,Af〉 = 〈ϕ, f(0)g〉 = f(0) 〈ϕ, g〉 = 0. Thus, ∀f ∈ Cc(R) : 〈ϕ,Af〉 = 0 = 〈0, f〉.Therefore, ϕ ∈ D (A∗) and A∗ϕ = 0.

24 Chapter 3 Symmetric and self-adjoint operators

Exercise 10: If A ⊂ B, then B∗ ⊂ A∗.Exercise 11: If 〈ϕ, g〉 6= 0, then ϕ /∈ D (A∗).

Theorem 3.2

1.) A∗ is closed.

2.) If A is closable, then (A)∗ = A∗.

3.) A is closable if and only if A∗ is densely defined. In this case A = (A∗)∗.

Proof. 1. Suppose that (xn)n∈N in D (A∗) with xn → x and Axn → y for some x, y ∈ H.

Then we have ∀ϕ ∈ D (A) : 〈x,Aϕ〉 = limn→∞

〈xn, Aϕ〉x∈D(A∗)

= limn→∞

〈A∗xn, ϕ〉 = 〈y, ϕ〉.Thus x ∈ D (A∗) and A∗x = y.

2. If A is closable then (since A ⊂ A) we have (A)∗ ⊂ A∗. It therefore suffices to prove

that D (A∗) ⊆ D(A∗)

. Let x ∈ D (A∗) and ϕ ∈ D(A). Then there exists a sequence

(ϕn)n∈N ⊆ D (A) with ϕn → ϕ and Aϕn → Aϕ. Then⟨x,Aϕ

⟩= lim

n→∞〈x,Aϕn〉

x∈D(A∗)=

limn→∞

〈A∗x, ϕn〉 = 〈A∗x, ϕ〉. Thus, ∀ϕ ∈ D(A)

:⟨x,Aϕ

⟩= 〈A∗x, ϕ〉. It follows that

x ∈ D((A)∗

), so D (A∗) ⊆ D

(A∗)

, as desired.

3. “⇐”: Suppose that A∗ is densely defined. Assume that (xn)n∈N in D (A) with xn → 0 and

Axn → y. Then we have

∀ψ ∈ D (A∗) : 〈ψ, y〉 = limn→∞

〈ψ,Axn〉ψ∈D(A∗)

= limn→∞

〈A∗ψ, xn〉 xn→0= 0.

Since D (A∗) is dense it follows that y = 0.

“⇒”: See [2] Theorem VIII.1.

Theorem 3.3

Ker (A∗) = Ran (A)⊥. Thus, Ran (A) ⊆ D (A∗) and kerA∗ = 0 if and only if Ran (A) = H.

Proof. ϕ ∈ kerA∗ ⇔ A∗ϕ = 0⇔ ∀ψ ∈ D (A) : 〈A∗ϕ,ψ〉 = 0⇔ ∀ψ ∈ D (A) : 〈ϕ,Aψ〉 = 0

⇔ ϕ ∈ Ran (A)⊥ .

Definition 3.4

A linear operator A (densely defined) is called symmetric if A ⊂ A∗. It is called self-adjoint

if A = A∗.

Chapter 3 Symmetric and self-adjoint operators 25

Remark

A ⊂ A∗ ⇔ D (A) ⊆ D (A∗) and ∀ϕ ∈ D (A) : A∗ϕ = Aϕ ⇔ ∀ϕ ∈ D (A) : 〈ϕ,Aψ〉 = 〈Aψ,ϕ〉.A is self-adjoint if and only if A is symmetric and D (A) = D (A∗).

Remark

From Definition 3.4 and 1. of Theorem 3.2 it follows that self-adjoint operators are closed.

Example 3.2

Let B : C∞c (Rn) ⊆ L2 (Rn)→ L2 (Rn) be defined as Bϕ = −∆ϕ. B is symmetric because for

all ϕ,ψ ∈ C∞c (Rn) we have

〈ψ,Bϕ〉 =

∫Rnψ(x)(−∆ϕ)(x) dx

IBP=

∫Rn

(−∆ψ)(x)ϕ (x) dx = 〈Bψ,ϕ〉

by integration by parts (IBP). Similarly, A : H2 (Rn) ⊆ L2 (Rn) → L2 (Rn), Aϕ = −∆ϕ is

symmetric because of the same relation and the fact that C∞c (Rn) = H2 (Rn).

We define C+ := z ∈ C, Im(z) > 0 and C− := z ∈ C, Im(z) < 0.

Theorem 3.5

Let A : D (A) ⊆ H → H be a symmetric operator.

1. For all λ, µ ∈ R and φ ∈ D (A) we have

‖(A− λ− iµ)φ‖2 = ‖(A− λ)φ‖2 + µ2‖φ‖2. (3.1)

2. a) If Ran (A− z+) = H for one z+ ∈ C+, then C+ ⊆ ρ (A).

b) If Ran (A− z−) = H for one z− ∈ C−, then C− ⊆ ρ (A).

Proof. 1. ‖(A− λ− iµ)φ‖2 = 〈(A− λ− iµ)φ, (A− λ− iµ)φ〉= ‖(A− λ)φ‖2 + µ2‖φ‖2 + 〈(A− λ)φ,−iµφ〉︸︷︷︸

=0, since A is symmetric

.

2. We will prove (a), (b) can be proven analogously:

a) Equation (3.1) implies that ‖(A− z+)φ‖ ≥ |Im(z+)|‖φ‖. If (A − z+)φ = 0, then

‖φ‖2 = 0 ⇒ φ = 0 ⇒ A − z+ is 1-1 (injective). By assumption, it is onto. If

ψ = (A − z+)φ, then ‖ψ‖ ≥ |Im(z+)|‖φ‖ ⇒ ‖φ‖ ≤ ‖ψ‖|Im(z+)| ⇒

∥∥(A− z+)−1ψ∥∥ ≤


‖ψ‖|Im(z+)| ⇒ (A− z+)−1 is bounded. Thus z+ ∈ ρ (A) and

∥∥(A− z+)−1∥∥ ≤ 1

|Im(z+)| .

By Theorem 2.3 B(z+,∥∥(A− z+)−1

∥∥−1)⊆ ρ (A) ⇒ B (z+, |Im(z+)|) ⊆ ρ (A). It-

erating this argument for some z′+ ∈ B (z+, |Im(z+)|) ∈ ρ (A) we can prove that

C+ ⊆ ρ (A).

Theorem 3.6 (Basic criterion of self-adjointness)

Let A be a symmetric operator. Then the following are equivalent:

1. A = A∗.

2. σ (A) ⊆ R.

3. Ran (A+ z±) = H for one z+ ∈ C+ and one z− ∈ C−.

4. A is closed and Ker (A∗ + z±) = 0 for one z+ ∈ C+ and one z− ∈ C−.

Proof.

• 1. ⇒ 4.: A = A∗ ⇒ A is closed by Theorem 3.2. Also, Ker (A∗ + z±) = Ker (A+ z±) = 0because A+ z± is injective by 2.) of the proof of Theorem 3.5.

• 4. ⇒ 3.: We have that Ran (A− z±)⊥ = Ker (A+ z±) = 0. Thus Ran (A+ z±) = H.

It remains to prove that Ran (A+ z±) is closed. Let y ∈ Ran (A+ z±). Then there

exists a sequence (ϕn)n∈N in D (A) such that (A + z±)ϕn → ϕ. Thus (A + z±)ϕn is a

Cauchy sequence. Since ‖(A+ z±)(ϕn − ϕm)‖ ≥ |Im(z±)|‖ϕn − ϕm‖, it follows that ϕn is

a Cauchy sequence. Thus, there exists a ϕ ∈ H such that ϕn → ϕ. From this and from

(A+ z±)ϕn → y, it follows (since (A+ z±) is closed) that ϕ ∈ D (A) and (A+ z±)ϕ = y,

so y ∈ Ran (A+ z±).

• 3. ⇒ 2.: Ran (A+ z±) = H Thm. 3.5⇒ C+ ⊆ ρ (A) and C− ⊆ ρ (A). Thus σ (A) = C\ρ (A) ⊆R.

• 2. ⇒ 1.: We want to prove that D (A∗) ⊆ D (A). Let ϕ ∈ D (A∗). We have −i ∈ ρ (A)

(by assumption), so Ran (A+ i) = H. Thus, there exists η ∈ D (A) such that (A + i)η =

(A + i)ϕA⊂A∗⇒ (A∗ + i)η = (A∗ + i)ϕ ⇒ (A∗ + i)(ϕ − η) = 0, so ϕ − η ∈ Ker (A∗ + i) =

Ran (A− i)⊥ = H⊥ = 0. Thus ϕ = η ∈ D (A).


Remark 3.7

From σ (A) ⊆ R alone self-adjointness does not necessarily follow.

Example 3.3

A : H2(R3)⊆ L2

(R3)→ L2

(R3), Aφ = −∆φ is self-adjoint. Indeed, by Example 3.2, A

is symmetric. By Example 2.1, Ran (A+ i) = L2(R3)

and similarly Ran (A− i) = L2(R3).

Thus, by Theorem 3.6, A is self-adjoint.

Theorem 3.8 (Kato-Rellich)

Let A : D (A) ⊆ H → H be a self-adjoint operator and B symmetric with D (B) ⊇ D (A). If

∀ϕ ∈ D (A) : ‖Bϕ‖ ≤ a‖Aϕ‖+ b‖ϕ‖, where a, b ∈ R with a < 1, then A + B : D (A) → H is

self-adjoint.

Proof. Clearly A+B is symmetric on D (A). By Theorem 3.6, it is enough to prove

Ran (A+B + µi) = H for µ ∈ R with |µ| large enough. Notice that (A+B + iµ)

= (I + B(A + µi)−1)(A + µi). We will show that∥∥B(A+ µi)−1

∥∥ < 1 for |µ| large enough. If

this holds, then Lemma 2.2 implies that (I +B(A+ iµ)−1) is invertible. Since A is self-adjoint,

by Theorem 3.6 we have that Ran (A+ iµ) = H ⇒ Ran (A+B + iµ) = H, as desired. It

therefore remains to prove that∥∥B(A+ µi)−1

∥∥ < 1. Let ϕ ∈ H and define ψ := (A+ µi)−1ϕ ∈D (A). Then

∥∥B(A+ µi)−1ϕ∥∥ = ‖Bψ‖ ≤ a‖Aψ‖ + b‖ψ‖. But ‖Aψ‖ ≤ ‖(A+ µi)ψ‖ = ‖ϕ‖

and ‖ϕ‖ = ‖(A+ µi)ψ‖ ≥ |µ|‖ψ‖, so ‖ψ‖ ≤ ‖ϕ‖|µ| . Combining both, we find∥∥B(A+ µi)−1ϕ

∥∥ ≤a‖ϕ‖ +

b

|µ|‖ϕ‖ ≤(a+

b

|µ|

)‖ϕ‖ for all ϕ ∈ H. Thus

∥∥B(A+ µi)−1∥∥ ≤ a +

b

|µ| and since by

assumption a < 1, choosing |µ| large enough we obtain∥∥B(A+ µi)−1

∥∥ < 1, as desired.

A detour in convergence theorems

Theorem (Dominated convergence (DCT))

Let (fn)n∈N be a sequence in L1 (Rn) (complex-valued functions). If

1. fn → f almost everywhere

2. ∃g ∈ L1 such that |fn| ≤ g almost everywhere for all n,

then

∫f dµ = lim

n→∞

∫fn dµ.


Theorem (Monotonic convergence (MCT))

If fn : Rn → [0,∞) is a pointwise increasing sequence of measurable functions and

limn→∞

fn(x) = f(x), then

∫f dµ = lim

n→∞

∫fn dµ.

Lemma (Fatou)

If fn : Rn → [0,∞) is a sequence of measurable functions, we have:∫lim infn→∞

fn dµ ≤ lim infn→∞

∫fn dµ .

Example

f : R → R, f(x) =

1

2sinx, x ∈ [0, π]

0, x /∈ [0, π], fn(x) = nf(nx). Then ∀x ∈ R : fn(x) → 0 as

n → ∞. But

∫Rfn(x) dx =

∫Rn(nx) dx =

∫Rf(y) dy = 1. So lim

n→∞

∫Rfn(x) dx 6= 0 =∫

Rlimn→∞

fn(x) dx (we cannot always exchange limits with integrals).

We conclude: If convergence is dominated or monotone, we can exchange the limit and the

integral.

Proof that monotonic convergence implies Fatou’s Lemma:

Let g(x) := lim infn→∞

fn(x) = limm→∞

inf k ≥ mfk(x)︸︷︷︸:=gm(x)

. Then ∀k ≥ m : gm ≤ fk ⇒ ∀k ≥ m :

∫gm dµ ≤

∫fk dµ ⇒

∫gm dµ ≤ lim inf

n→∞

∫fn dµ. But (gm) is an increasing positive sequence

and gm(x)→ g(x) almost everywhere, so

∫g dµ = lim

m→∞

∫gm dµ ≤ lim inf

n→∞

∫fn dµ.

Theorem 3.9 (Hardy inequality)

For all ψ ∈ H1(R3)

we have that∫ |ψ(x)||x|2

dx ≤ 4

∫|∇ψ(x)|2 dx . (3.2)

Proof. We first assume that ψ ∈ C∞c(R3). Then we have:

∫ (∂i

xi

|xi|2)|ψ(x)|2 dx =

∫ |ψ(x)|2

|x|2dx+

∫ xi

(−∂i|x|2

)|x|4

|ψ (x)|2 dx

=

∫ |ψ(x)|2

|x|2dx−

∫2x2

i

|x|4|ψ(x)|2 dx .


Thus,

3∑i=1

∫ (∂ixi

|x|2)|ψ(x)|2 dx = 3

∫ |ψ(x)|2

|x|2dx− 2

∫ |ψ(x)|2

|x|2dx

⇒∫ |ψ(x)|2

|x|2=

3∑i=1

∫ (∂ixi

|x|2)|ψ (x)|2 dx

IBP=

∫ 3∑i=1

xi

|x|2∂i|ψ (x)|2 dx

=

∫ 3∑i=1

xi

|x|22 Re

(ψ(x)∂iψ(x)

)dx ≤ 2

∫ ∣∣∣∣∣ψ (x)

|x|

∑3i=1 xi∂iψ (x)

|x|

∣∣∣∣∣ dx≤ 2

∫ |ψ(x)||x|

∣∣∣∣x ·∇ψ(x)

|x|

∣∣∣∣ dx ≤ 2

∫ |ψ(x)||x| |∇ψ(x)| dx

From the Cauchy-Schwarz inequality it follows that∫ |ψ(x)|2

|x|2dx ≤ 2

(∫ |ψ(x)|2

|x|2dx

) 12 (∫

|∇ψ(x)|2 dx

) 12

,

which implies equation (3.2) for all ψ ∈ C∞c(R3).

Assume now that ψ ∈ H1(R3). Then by Theorem 1.6 there exists a sequence (ψn)n∈N in

C∞c(R3)

with ψn → ψ in H1.

But then

∫ |ψn(x)|2

|x|2dx ≤

∫|∇ψn(x)|2 dx

n→∞→∫|∇ψ(x)|2 dx. Since ψn → ψ in L2, there

exists a subsequence ψnk with ψnk → ψ almost everywhere. Then by Fatou’s Lemma:∫ |ψ(x)|2

|x|2dx ≤ lim inf

nk→∞

∫ |ψnk(x)|2

|x|2dx ≤

∫|∇ψ(x)|2 dx, as desired.

Example 3.4

Let A : H2(R3)⊆ L2

(R3)→ L2

(R3)

with A = −∆ + V , where V (x) = − 1

|x| (up to physical

units, this is the Hamiltonian of the hydrogen atom). Then A is self-adjoint.

Proof. From Example 3.3 we know that −∆ : H2(R3)⊆ L2

(R3)→ L2

(R3)

is self-adjoint.

With help of the Hardy inequality we will prove that ∀ε > 0 there exists a Cε > 0 such that

for all ψ ∈ H2(R3)

we have V ψ ∈ L2(R3)

and ‖V ψ‖ ≤ ε‖−∆ψ‖ + Cε‖ψ‖. Once we have

this, we can apply Theorem 3.8 to conclude that A is self-adjoint. For all ψ ∈ H2(R3)

we have

that ‖V ψ‖2 =

∫ |ψ(x)|2

|x|2dx

Hardy≤ 4

∫|∇ψ(x)|2 dx. But

∫|∇ψ(x)|2 dx

FT=

∫ ∣∣∣ξψ(ξ)∣∣∣2 dx =∫

ξ2∣∣∣ψ(ξ)

∣∣∣2 dxξ2≤ε2ξ4+ 1

4ε2≤∫ε2ξ4

∣∣∣ψ(ξ)∣∣∣2 dx +

1

4ε2

∫ ∣∣∣ψ(ξ)∣∣∣2 dx = ε2‖−∆ψ‖2 +

1

4ε2‖ψ‖2 ≤(

ε‖−∆ψ‖+1

2ε‖ψ‖

)2

, so ‖V ψ‖ ≤ ε‖−∆ψ‖+1

2ε‖ψ‖.


Exercise 13: Let x := (x1, . . . , xn) ∈ (R3)N = R3N . The operator

B : H2(R3N

)⊆ L2

(R3N

)→ L2

(R3N

),

(Bφ)(x) :=

N∑j=1

(−∆xj −

N

|x|

)φ(x) +

∑1≤i<j≤N

1

|xi − xj |φ(x)

is self-adjoint (up to physical units, this is the Hamilton operator for an atom with N electrons)

Definition 3.10

A symmetric operator is called essentially self-adjoint if its closure A is self-adjoint.

Example 3.5

Let B : C∞c (Rn) ⊆ L2 (Rn)→ L2 (Rn), Bϕ = −∆ϕ. Then B is essentially self-adjoint.

Indeed, A : H2 (Rn) ⊆ L2 (Rn) → L2 (Rn), Aϕ = ∆ϕ is self-adjoint by Example 3.3. So it

suffices to show that B = A (B is closable since e.g. A is closed as a self-adjoint operator

and B ⊂ A). Since B ⊂ A and A is closed, we have that B ⊂ A. Let (ϕ,−∆ϕ) ∈ ΓA. Then

ϕ ∈ H2 (Rn). But by Theorem 1.6 there exists a sequence (ϕn)n∈N in C∞c (Rn) with ϕn → ϕ

in H2 (Rn), so ϕn → ϕ and −∆ϕn → −∆ϕ ⇒ (ϕ,−∆ϕ) ∈ ΓB. Thus A ⊂ B. Hence A = B,

as desired.

Theorem 3.11

Let A be a symmetric operator. Then the following statements are equivalent:

1. A is essentially self-adjoint.

2. Ran (A− z±) = H for a z+ ∈ C+ and a z− ∈ C−.

3. Ker (A∗ − z±) = 0 for a z+ ∈ C+ and a z− ∈ C−.

Proof.

• 1. ⇒ 3.: A is self-adjoint by assumption and (A)∗Thm. 3.2

= A∗.

So Ker (A∗ − z±) = Ker((A)∗ − z±

) Thm. 3.6⇒ Ker (A∗ − z±) = 0.

• 3. ⇒ 2.: Ran (A− z±)⊥Thm. 3.3

= Ker (A∗ − z±) = 0.

• 2. ⇒ 1.: Since A is symmetric, A is closable (by Theorem 3.2 3.)). Let A be its closure.

Claim: Ran(A− z±

)⊆ Ran

(A− z±

). Once this is proven it follows that Ran

(A− z±

)=


H, so by Theorem 3.6 A is self-adjoint. Since Ran (A− z±) ⊆ Ran(A− z±

), it remains

to prove that Ran(A− z±

)is closed.

This can be done as in the proof of Theorem 3.6 “4. ⇒ 3.”.

Theorem 3.12 (Kato-Rellich, general form)

Let A be essentially self-adjoint and B symmetric with D (A) ⊆ D (B). If ‖Bφ‖ ≤ a‖Aφ‖ +

b‖φ‖ ∀φ ∈ D (A), where a, b ∈ R with 0 < a < 1, then A+ B : D (A) ⊆ H → H is essentially

self-adjoint and D(A+B

)= D

(A).

Proof. B is symmetric, thus closable by Theorem 3.2, so let B be its closure. We show:

(i) D(A)⊆ D

(B)

and D(A)⊆ D

(A+B

).

(ii) A+B is self-adjoint.

(iii) A+B ⊂ A+B, D(A+B

)= D

(A), and A+B = A+B.

(i) Let ϕ ∈ D(A). Then there exists a sequence (ϕn)n∈N in D (A) with ϕn → ϕ and Aϕn →

Aϕ. But since ‖B(ϕn − ϕm)‖ ≤ a‖A(ϕn − ϕm)‖+ b‖ϕn − ϕm‖, Bϕn is a Cauchy sequence

and thus convergent. Thus ϕ ∈ D(B)

(and Bϕn → Bϕ). Similarly, ϕ ∈ D(A+B

), since

(A+B)ϕn is a Cauchy sequence.

(ii) Since ‖Bϕn‖ ≤ a‖Aϕn‖+ b‖ϕn‖ ∀n ∈ N, we find∥∥Bϕ∥∥ ≤ ∥∥Aϕ∥∥+ b‖ϕ‖ for n→∞. Since

ϕ ∈ D(A)⊆ D

(B)

was arbitrary, by Theorem 3.8 it follows that A+B is self-adjoint in

D(A).

(iii) Clearly, A + B ⊂ A + B, and since A + B is self-adjoint (and therefore closed), we have

A+B ⊂ A+B. Since D(A+B

)= D

(A)⊆ D

(A+B

), it follows that A+B = A+B,

thus A+B is self-adjoint.

Example 3.6

A : C∞c(R3)⊆ L2

(R3)→ L2

(R3), Aϕ = (−∆ +V )ϕ, where V (x) = − 1

|x| , is essentially self-

adjoint in C∞c(R3), since by Example 3.5 the operator −∆ is essentially self-adjoint and by

Example 3.4 ‖V ψ‖ ≤ ε‖−∆ψ‖+1

2ε‖ψ‖. Thus, the general form of Kato-Rellich is applicable.

4 The Schrodinger equation and existence of

dynamics

Let H : D (H) ⊆ H → H be densely defined. We consider the initial-value problem id

dtϕt = Hϕt

ϕt|t=0 = u

. (4.1)

The partial differential equation id

dtϕt = Hϕt is known as the Schrodinger equation. Let I be a

nontrivial interval with 0 ∈ I. A solution of equation (4.1) is a differentiable function ϕt : I → Hwith

1. ∀t ∈ I : ϕt ∈ D (H).

2. id

dtϕt := lim

h→0iϕt+h − ϕt

h= Hϕt.

3. ϕ0 = u.

Theorem 4.1

(i) If equation (4.1) has for all u ∈ D (H) a solution with constant norm, then H is symmetric.

(ii) If H is symmetric then equation (4.1) has for all u ∈ D (H) at most one solution locally

in time.

Proof.

(i)d

dt〈ϕt, ϕt〉 = 0 ⇒

⟨d

dtϕt, ϕt

⟩+

⟨ϕt,

d

dtϕt

⟩= 0 ⇒ 〈−iHϕt, ϕt〉 + 〈ϕt,−iHϕt〉 = 0 ⇒

i 〈Hϕt, ϕt〉 − i 〈ϕt, Hϕt〉 = 0 ⇒ 〈ϕt, Hϕt〉 = 〈Hϕt, ϕt〉. In particular, ∀u ∈ D (H) :

〈u,Hu〉 = 〈Hu, u〉. From Exercise 2 it follows that 〈u,Hv〉 = 〈Hu, v〉 ∀u, v ∈ D (H), so H

is symmetric.

34 Chapter 4 The Schrodinger equation and existence of dynamics

(ii) If for some u ∈ D (H) equation (4.1) has the two solutions ϕt and ϕt, then id

dt(ϕt − ϕt) = H(ϕt − ϕt)

(ϕt − ϕt)|t=0 = 0

.

Arguing similarly as in (i) we obtaind

dt〈ϕt − ϕt, ϕt − ϕt〉 = 0.

Thus ‖ϕ− ϕt‖ = ‖(ϕt − ϕt)|t=0‖ = 0⇒ ϕ = ϕt.

Theorem 4.2

If H is symmetric and equation (4.1) has a solution in I = R ∀u ∈ D (H), then H is essentially

self-adjoint.

Proof. We will prove that Ker (H∗ + i) = 0 (and similarly Ker (H∗ − i) = 0). Let w ∈Ker (H∗ + i). Then H∗w = −iw. Thus, for any u ∈ D (H), if ϕt is a solution of equation (4.1),

thend

dt〈w,ϕt〉 = 〈w,−iHϕt〉 = 〈iH∗w,ϕt〉 = 〈w,ϕt〉. Solving the differential equation, we

find 〈w,ϕt〉 = et 〈w,ϕ0〉 = et 〈w, u〉. But |〈w,ϕt〉|Cauchy-Schwarz

≤ ‖w‖‖ϕt‖Thm. 4.1≤ ‖w‖‖u‖. Thus

∀t ∈ R :∣∣et 〈w, u〉∣∣ ≤ ‖w‖‖u‖ ⇒ ∀u ∈ D (H) : 〈w, u〉 = 0

D(H) dense⇒ w = 0.

Example 4.1

Let P : H10 (R+) → L2(R+), Pf = −i

d

dxf . Then P is symmetric, but not essentially self-

adjoint (shown in Exercise 14). Thus, if H = P , then equation (4.1) does not have a solution

for all u ∈ H10 (R+) and for all t ∈ R in L2(R+) by Theorem 4.2. This is not surprising because

id

dtϕt = Pϕt ⇒

d

dtϕ = − d

dxϕt. So if u is smooth and ϕt is a solution of equation (4.1), then

ϕt(x) = u(x+ t), but supp (ϕt) is not a subset of R+.

Let us denote by L(H) the set of all bounded linear operators from H to H.

Theorem 4.3 (Existence and properties of a solution, bounded case)

Let H : H → H be bounded and self-adjoint. Then equation (4.1) has ∀u ∈ D (H) = H the

unique solution e−iHtu, where e−iHt =∞∑n=0

(−iHt)n

n!. This solution is global in time. The map

U : R→ L(H), U(t) = e−iHt, fulfills

(1) ∀s, t ∈ R U(t) is unitary and U(t+ s) = U(t)U(s),

Chapter 4 The Schrodinger equation and existence of dynamics 35

(2) limt→0

U(t)ψ = ψ ∀ψ ∈ H.

Proof. (1) That U (t+ s) = U (t)U (s) can be proven using the Cauchy product of power se-

ries as in the case of real numbers. U(t) is unitary since U∗(t) =

( ∞∑n=0

(−iHt)n

n!

)∗=

∞∑n=0

(iHt)n

n!= U(−t). Thus U∗(t)U(t) = U(−t)U(t) = U(−t+ t) = U(0) = I by definition.

(2) U(t)ψ − ψ =∞∑n=1

(−iHt)n

n!ψ = −iHt

∞∑n=1

(−iHt)n−1

n!ψ

⇒ ‖U(t)ψ − ψ‖ ≤ t‖H‖∥∥∥∥∥∞∑n=1

(iHt)n−1

n!ψ

∥∥∥∥∥ t→0→ 0,

so limt→0

U(t)ψ − ψ = 0.

We now prove that ϕt = U(t)u = e−iHtu is a solution of equation (4.1) for all t ∈ R and

uniqueness follows from Theorem 4.1. Trivially ϕ0 = e−iH·0u = u. We have

id

dtϕt = lim

h→0iϕt+h − ϕt

h= i lim

h→0

U(t+ h)u− U(t)u

h= i lim

h→0

U(h)U(t)u− U(t)u

h

= i limh→0

(U(h)− I

h

)U(t)u.

But iU(h)− I

h=

i

h

∞∑n=1

(−iHh)n

n!=

i(−iHh)

h

∞∑n=1

(−iHh)n−1

n!

h→0→ H,

so indeed id

dtϕt = Hϕt and ϕt|t=0 = u.

A map U : R→ L(H) satisfying

(1) ∀s, t ∈ R : U(t) is unitary and U(t+ s) = U(t)U(s) = U(s)U(t), (4.2)

(2) limt→0

U(t)ψ = ψ ∀ψ ∈ H. (4.3)

is called a strongly continuous unitary group .

Example 4.2

The map U : R→ L(L2(R)) defined by [U(t)ϕ](x) = ϕ(x+ t) is a strongly continuous unitary

group.

Proof. [U(t+ s)ϕ](x) = ϕ(x+ t+ s) = [U(s)ϕ](x+ t) = [U(t)U(s)ϕ](x)⇒ U(t+ s) = U(t)U(s).

〈U(t)ϕ,U(t)ϕ〉 =

∫ϕ(t+ x)ϕ(x + t) dx

y:x+t=

∫ϕ(y)ϕ(y) dy = 〈y, y〉, so U(t) is unitary for all

t ∈ R.


If ψ ∈ C∞c (R), then |U(t)ψ(x)| ≤ f ∀t ∈ [−1, 1], where f = max |ψ|χ⋃t∈[−1,1] supp(U(t)ψ) ∈ L1.

Thus, since U(t)ψ → ψ pointwise, we can apply the dominated convergence theorem to conclude

that ‖U(t)ψ − ψ‖L1 → 0. But ‖U(t)ψ − ψ‖L2 ≤ ‖U(t)ψ − ψ‖L∞‖U(t)ψ − ψ‖L1 → 0. Let

ϕ ∈ L2, ε > 0. Then there exists a ϕ0 ∈ C∞c (R) with ‖ϕ− ϕ0‖ <ε

3. But then U(t)ϕ − ϕ =

U(t)(ϕ − ϕ0) + U(t)ϕ − ϕ0 + ϕ0 − ϕ → ‖U(t)ϕ− ϕ‖ ≤ 2‖ϕ− ϕ0‖ + ‖U(t)ϕ0 − ϕ0‖. Choosing

t0 > 0 such that ∀t ∈ (−t0, t0) : ‖U(t)ϕ0 − ϕ0‖ <ε

3, we find ‖U(t)ϕ− ϕ‖ < ε∀t ∈ (−t0, t0).

Remark 4.4

(i) If U is a strongly continuous unitary group then t 7→ U(t)ψ is continuous ∀t ∈ R.

(ii) To verify ∀ψ ∈ H : limt→0

U (t)ψ = ψ it is enough to check it for all ψ in a dense subset of H.

Proof.

(i) limh→0

U(t + h)ψ − U(t)ψ = limh→0

U(h)U(t)ψ − U(t)ψ = limh→0

(U(h) − I)U(t)ψ = 0, by equa-

tion (4.3).

(ii) This can be proven by approximation as in Example 4.2.

Theorem 4.5 (Existence of a solution, general case)

(a) Let H : D (H) ⊆ H → H be a self-adjoint operator (not necessarily bounded). Then

∀u ∈ D (H) equation (4.1) has a unique solution. This solution is global in time.

(b) There exists a strongly continuous unitary group U(t) such that ∀u ∈ D (H) ∀t ∈ R : ϕt =

U(t)u.

Proof. Let Hn := BnHB−n, Bn := in(H + in)−1 (intuition: if H were a real number, then

limn→∞

Bn = 1. Thus we hope that limn→∞

Bn = I in a certain sense and that “ limn→∞

Hn = H”). The

uniqueness of the solution (if it exists) has already been proven. We will prove Theorem 4.5 in

five steps:

(1) Hn is bounded and self-adjoint for all n ∈ N.

(2) Hnψ → Hψ ∀ψ ∈ D (H).

(3) limn→∞

e−iHntψ exists for all ψ ∈ D (H). For fixed ψ ∈ D (H) the limit is uniform on compact

sets [−M,M ] ∀M > 0.


(4) limn→∞

e−iHntψ exists for all ψ ∈ H and U : R → L (H) with U(t)ψ = limn→∞

e−iHntψ is a

strongly continuous unitary group.

(5) ϕt := U(t)u is a solution of id

dtϕt = Hϕt

ϕt|t=0 = u

. (4.4)

(1) B±n is bounded because ±n ∈ ρ (H). ‖BnHBm‖ ≤ ‖Bn‖∥∥∥Hin (H − in)−1

∥∥∥≤ n‖Bn‖

∥∥∥H (H − in)−1∥∥∥ ≤ n‖Bn‖

because ‖Hψ‖Thm. 3.9≤ ‖(H − in)ψ‖ ∀ψ ∈ D (H). If ψ ∈ D (H) then ψ = (H − in)−1ϕ for

some ϕ ∈ H, thus ‖Hψ‖ =∥∥H(H − in)−1ϕ

∥∥ ≤ ‖ϕ‖. So Hn is bounded. Let ϕ,ψ ∈ H.

Then 〈ϕ,Hnψ〉 = 〈ϕ,BnHB−nψ〉H=H∗⇒B∗n=B−n

=

⟨B−nϕ︸︷︷︸∈D(H)

, H B−nψ︸︷︷︸∈D(H)

⟩H self-adjoint

= 〈HB−nϕ,B−nψ〉 = 〈BnHB−nϕ,ψ〉 = 〈Hnϕ,ψ〉. Thus Hn is self-adjoint.

(2) For all ψ ∈ D (H) we have limn→∞

B±nψ = ψ.

Bnψ−ψ = in(H+in)−1ψ−(H+in)(H+in)−1ψ = −H(H+in)−1ψ. Thus, by Exercise 4,(3)

and since ψ ∈ D (H) we have that Bnψ − ψ = − (H + in)−1Hψ. Thus ‖Bnψ − ψ‖ ≤∥∥∥(H + in)−1∥∥∥‖Hψ‖ ≤ 1

|n|‖Hψ‖n→∞→ 0. Similarly for B−n. Hence B±nψ

n→∞→ ψ ∀ψ ∈

D (H). So if ψ ∈ H, let ψ0 ∈ D (H) , ε > 0 with ‖ψ − ψ0‖ <ε

3. Then Bnψ − ψ =

Bnψ−Bnψ0 +Bnψ0−ψ0 +ψ0−ψ and ‖Bnψ − ψ‖ ≤ (‖Bn‖+ 1)‖ψ0 − ψ‖+ ‖Bnψ0 − ψ‖ ≤2ε

3+ ‖Bnψ0 − ψ‖, since ‖Bn‖ =

∥∥in(H + in)−1∥∥ ≤ 1. Now choose n large enough. We have

Hnψ = BnHB−nψψ∈D(H)

= BnB−nHψ. Thus Hnψ−Hψ = BnB−nHψ−Hψ = BnB−nHψ−BnHψ + BnHψ −Hψ = Bn(B−nHψ −Hψ) + BnHψ −Hψ. But Bn(B−nHψ −Hψ) → 0

and BnHψ −Hψ → 0 since B±nψ → ψ and ‖Bn‖ = 1 <∞.

(3) By Theorem 4.3d

dte−iHntψ = e−iHnt(−iHn)ψ = (−iHn)e−iHntψ. Let ψ ∈ D (H). Then

∥∥e−iHntψ − e−iHmtψ∥∥ e−iHmt unitary

=∥∥eiHmte−iHntψ − ψ

∥∥.eiHmte−iHntψ − eiHm·0e−iHn·0ψ =

∫ t

0

d

ds

(eiHmse−iHnsψ

)ds

=

∫ t

0eiHms(iHm − iHn)e−iHnsψ ds

Exercise=

∫ t

0eiHmse−iHmsi(Hm −Hn)ψ ds ,


therefore∥∥eiHmte−iHntψ − ψ

∥∥ ≤ |t|‖(Hn −Hm)ψ‖≤M‖(Hn −Hm)ψ‖ ∀t ∈ [−M,M ] ∀M > 0.

Thus e−iHntψ is uniformly Cauchy in t ∈ [−M,M ] and thus uniformly convergent.

(4) The existence follows by approximation similarly as in Example 4.2.

Now we prove ∀ψ ∈ H : U(t + s)ψ = U(t)U(s)ψ. By approximation it suffices to prove it

∀ψ ∈ D (H). Let Un(t) := e−iHnt. Then Un(s+ t)ψThm. 4.3

= Un(t)Un(s).

But Un(t+ s)ψn→∞→ U(t+ s)ψ by (3) and Un(t)Un(s)ψ

n→∞→ U(t)U(s)ψ since

Un(t)Un(s)ψ − U(t)U(s)ψ = Un(t)(Un(s)− Un(s))ψ + (Un(t)− U(t))U(s)ψ → 0 by (3) and

the fact that ‖Un(t)‖ ≤ 1.

We now show that U(t) is unitary: 〈ϕ,U(t)ψ〉 = limn→∞

〈ϕ,Un(t)ψ〉 Un unitary= lim

n→∞〈U∗n(t)ϕ,ψ〉

= limn→∞

〈U−n(t)ϕ,ψ〉 = 〈U(−t)ϕ,ψ〉. Thus (U(t))∗ = U(−t) and it follows that (U(t))∗U(t)

= U(−t)U(t) = U(−t + t) = U(0) = I. Similarly, U(t)(U(t))∗, hence U(t) is unitary

for all t ∈ R. To prove that U(t) is a strongly continuous unitary group it suffices to

prove that limt→0

U(t)ψ = ψ ∀ψ ∈ D (H) (because D (H) = H). Let fn(t) = Un(t)ψ, where

Un(t) = e−iHnt. Then fn(t) is continuous by Theorem 4.3 (solution if the operator is bounded

and self-adjoint). By (3), fn(t) → U(t)ψ uniformly in t ∈ [−M,M ] ∀M > 0. Since fn is

continous for all n it follows that U(t)ψ is continuous.

(5) Let ϕt := U (t) v. We have to show: ϕt is a solution of equation (4.1) ∀t ∈ R,∀v ∈ D (H).

We have id

dtϕt = i lim

h→0

ϕt+h − ϕth

= i limh→0

U(t+ h)v − U(t)v

h= iU(t) lim

h→0

U(h)− Ih

v. For

fixed h 6= 0,U(h)− I

hv = lim

m→∞

Um(h)− Um(0)

hv = lim

m→∞

1

h

∫ h

0

(d

dsUm(s)v

)ds

Thm. 4.3= lim

m→∞

1

h

∫ h

0Um(s)(−iHm)v ds

to show=

1

h

∫ h

0U(s)(−iHm)v ds. Once we have this, we

find limh→0

U(h)− Ih

v = limh→0

1

h

∫ h

0U(s)(−iH)v ds = U(0)(−iH)v = −iHv. Thus, it follows

that id

dtϕt = U(t)Uv = HU(t)v = Hϕt. To prove the last identity indicated above, we


write ∫ h

0(Um(s)(−iHm)v − U(s)(−iHv)) ds

=

∫ h

0Um(s)(−iHm + iH)v ds+

∫ h

0(Um(s)− U(s))Hv ds

⇒∥∥∥∥∫ h

0(Um(s)(−iHm)v − U(s)(−iH)v) ds

∥∥∥∥U(m) unitary

≤ |h|‖(−iHm + iH)v‖︸︷︷︸m→∞→ 0 by (2)

+

∫ h

0‖(U(s)− U(s))Hv‖︸︷︷︸m→∞→ 0∀s∈[0,h] pointwise

ds .

We have ‖(Um(s)− U(s))Hv‖ ≤ ‖Um(s)Hv‖ + ‖U(s)Hv‖ ≤ 2‖Hv‖, so we can apply the

dominated convergence theorem to the last term on the right-hand side in the above in-

equality.

Remark 4.6

The strongly continuous unitary group U(t) of Theorem 4.5 is denoted by e−iHt. With its

help one can make sense of a solution of equation (4.1) ∀v ∈ H (even if v /∈ D (H)).

Definition 4.7

Let U : R → L(H) be a strongly continuous unitary group. The generator of U is the linear

operator A : D (A) ⊆ H → H, defined by

1.) D (A) =

ψ ∈ H : lim

h→0

U(h)− Ih

ψ exists

2.) Aψ = id

dtU(t) ψ|t=0 = i lim

h→0

U(h)− Ih

ψ.

Example 4.3

In Example 4.2 we showed that U : R→ L2(R), [U(t)ψ](x) = ψ(x− t) is a strongly continuous

unitary group. The generator of it is A : H1(R)→ L2(R), Aϕ = −id

dxϕ.

Proof. Let v ∈ C∞c (R) and ψ ∈ L2(R).

Then

⟨U(h)− I

hψ, v

⟩U∗(h)=U(−h)

=

⟨ψ,U(−h)− I

hv

⟩h→0 (Exercise)→

∫ψ(x)v′(x) dx. Therefore


limh→0

U(h)− Ih

ψ = f in L2 for some f ⇔ 〈f, v〉 =⟨ψ, v′

⟩∀v ∈ C∞c (R)⇔ ψ′ = f (in the sense of

weak derivatives) ⇔ ψ ∈ H1 and ψ′ = f . In this case Aψ = −id

dxψ.

Theorem 4.8 (Stone)

Let A : D (H) ⊆ H → H be the generator of a strongly continuous unitary group. Then:

(1) U(t)D (A) ⊆ D (A) and ∀ϕ ∈ D (A) we have id

dtU(t)ϕ = AU(t)ϕ = U(t)Aϕ

(2) A = A∗

(3) U(t) is uniquely determined by A

Proof.

(1) Aϕ = id

dtU(t) ϕ|t=0. If ϕ ∈ D (A) then:

i limh→0

U(t+ h)− U(t)

hϕ = U(t)i lim

h→0

U(h)− Ih

ϕ = U(t)Aϕ.

Thus U(t)ϕ ∈ D (A) and AU(t)ϕ = id

dtU(t)ϕ = AU(t)ϕ.

(2) (i) A is densely defined.

(ii) A is symmetric.

(iii) A is essentially self-adjoint.

(iv) A is closed.

(i) Let ϕ ∈ H. Then ϕ = limt→0

1

t

∫ t

0U(s)ϕds. Thus it suffices to show that ∀t 6= 0 :

1

t

∫ t

0U(s)ϕds ∈ D (A).

For fixed t we have:

U(h)− Ih

1

t

∫ t

0U(s)ϕds =

1

t

∫ t

0

U(s+ h)− U(s)

hϕds

=1

t

(∫ t+h

h

U(s)

hϕds−

∫ t

0

U(s)

hϕds

)=

1

t

∫ t+h

t

U(s)

hϕds−

∫ h

0

U(s)

hϕds

h→0→ 1

t(U(t)ϕ− U(0)ϕ).


(ii) Since A is densely defined it remains to prove that 〈Aϕ,ψ〉 = 〈ϕ,Aψ〉 ∀ϕ,ψ ∈ D (A).

We have 〈Aϕ,ψ〉 = limh→0

⟨iU(h)− I

hϕ, ψ

⟩= lim

h→0

⟨ϕ,−i

U(−h)− Ih

ψ

⟩=

⟨ϕ, i lim

h→0

U(−h)− I−h ψ

⟩= 〈ϕ,Aψ〉.

(iii) A is symmetric and by (1) equation (4.1) has a solution for all times. Thus, by

Theorem 4.2 A is essentially self-adjoint.

(iv) Let (ϕn)n∈N be a sequence in D (A) with ϕn → ϕ and Aϕn → Aψ for some ψ ∈ H.

We have to show: ϕ ∈ D (A) and Aϕ = ψ. For fixed h 6= 0 we have:

iU(h)− I

hϕ = i lim

n→∞

U(h)ϕn − U(0)ϕnh

= limn→∞

1

h

∫ h

0i

d

dsU(s)ϕn ds

= limn→∞

1

h

∫ h

0U(s)Aϕn ds

Aϕn→ψ uniformly=

1

h

∫ h

0U(s)ψ ds

h→0→ ψ.

Thus ϕ ∈ D (A) and Aϕ = ψ.

(3) For all ϕ ∈ D (A) U(t)ϕ is a solution of

id

dtϕt = Hϕt

ϕt|t=0 = ϕ

(4.5) by (1). If W (t) is another

strongly continuous unitary group generated by A, then ∀ϕ ∈ D (A) : U(t)ϕ = W (t)ϕ

because equation (4.5) has at most one solution by the symmetry of A. Since D (A) is

dense, we find U(t) = W (t).

Example 4.4

From Example 4.2 and Example 4.3 it follows that A : H1(R) → L2(R), Aϕ = −id

dxϕ is

self-adjoint.

5 Observables, Uncertainty Principle, Ground

state energy

Observables are physically measurable quantities that correspond to self-adjoint operators acting

on the Hilbert space of the states (also called wave functions).

Let A be a self-adjoint operator acting on a Hilbert space H.

Definition 5.1

1. The expectation of A in the state ψ is given by 〈ψ,Aψ〉.

2. The variance of A in ψ ∈ D (A) is defined by ∆Aψ =⟨ψ, (A− 〈ψ,Aψ〉)2ψ

⟩=⟨ψ,A2ψ

⟩−

〈ψ,Aψ〉2.

Definition 5.2

The commutator of two operators A,B is defined by [A,B] := AB −BA.

Example 5.1

We compute the expectation (value) of

(i) the position operator (j-th component): 〈ψ, xjψ〉 =

∫xj |ψ (x)|2 dx

(ii) the momentum operator (j-th component):⟨ψ,−i∂xjψ

⟩=⟨ψ, ξjψ

⟩=

∫ξj

∣∣∣ψ (ξ)∣∣∣2 dx

after Fourier transformation and using Plancherel’s identity 〈f, g〉 =⟨f , g⟩

(iii) the energy operator: 〈ψ,Hψ〉, where H =−∆

2m+ V . Here

−∆

2mis the kinetic energy

(classically, it is given byp2

2m, p = mv) and V is the potential energy.

44 Chapter 5 Observables, Uncertainty Principle, Ground state energy

Notation: When we write 〈ψ, xψt〉, where x = (x1, x2, x3), we mean 〈ψ, xψt〉 :=

〈ψt, x1ψt〉〈ψt, x2ψt〉〈ψt, x3ψt〉

.

Analogous short-hand notations hold for the momentum operator and other finite-dimensional

vector-like operators.

Example 5.2

We compute the commutator of the Hamilton operator H with xj : [H,xj ] =

[− ∆

2m+ V, xj

]=[

− ∆

2m,xj

], since V is a real number and therefore commutes with xj . For ψ smooth,[

− ∆

2m,xj

]ψ = −−∆

2mxjψ + xj

∆

2mψ =

1

2m

n∑k=1

(−∂2xk

(xjψ) + xj∂2xkψ)

product rule=

n∑k=1

− 1

mδjk(∂xkxj)∂xkψ = − 1

m∂xjψ, where we have used the Kronecker symbol

δij :=

1, i = j

0, i 6= j.

Example 5.3 ( Informal computations for the evolution of observables )

If id

dtψt = Hψt, how does 〈ψt, Aψt〉 (A self-adjoint and t-independent!) evolve in time? We

haved

dt〈ψt, Aψt〉 = 〈−iHψt, Aψt〉+ 〈ψt, A (−iH)ψt〉 H=H∗

= 〈ψt, (iHA− iAH)ψt〉= 〈ψt, i[H,A]ψt〉. Thus

d

dt〈ψt, Aψt〉 = 〈ψt, i[H,A]ψt〉 . (5.1)

Thus, e.g.d

dt〈ψt, xjψt〉 = 〈ψt, i[H,xj ]ψt〉 Ex. 5.2

=1

m

⟨ψt,−i∂xjψt

⟩, and hence

d

dt〈ψt, xψt〉 =

1

m〈ψt,−i∇ψt〉. If we take one more derivative, we obtain

d2

dt2〈ψt, xψt〉 =

1

m

d

dt〈ψt,−i∇ψt〉 =

1

m〈ψt, i[H,−i∇]ψt〉 =

1

m〈ψt, [H,∇]ψt〉

[∆,−i∇]=0=

1

m〈ψt, [V,∇]ψt〉. Thus

d2

dt2〈ψt, xψt〉 =

1

m〈ψ, (−∇V )ψt〉. 1

Remark

We can write ψt = e−iHtψ0, where ψt|t=0 = ψ0 and ψt solves the Schrodinger equation

1This means that while position and momentum operators do not necessarily obey the equations of motion of

classical mechanics (Newton’s laws),d

dt=

1

mp, a =

d2

dt2x =

1

m(−∇V ) =

1

mF , at least their expectation

values do. This result is known as the Ehrenfest theorem[3].

Chapter 5 Observables, Uncertainty Principle, Ground state energy 45

id

dtψt = Hψt. For an observable A we define A(t) := eiHtAe−iHt. Then

〈ψt, Aψt〉 =⟨e−iHtψ0, Ae−iHtψ0

⟩= 〈ψ0, A(t)ψ0〉 . (5.2)

The left-hand side is called the Schrodinger picture of the expectation of A, while the right-

hand side is called the Heisenberg picture. In the Schrodinger picture, only ψ evolves in time,

while A is constant for all t. In the Heisenberg picture, only the operator A(t) evolves in time,

while ψ0 is constant for all t. Note that we have continued to assume that A (Schrodinger

picture) is t-independent.

At least non-rigorously we have A′ (t) = iHeiHtAe−iHt − eiHtAe−iHtiH, so A′(t) = i[H,A(t)].

Theorem 5.3

Let A,B be self-adjoint operators acting on H. Then for ψ ∈ D (A) ∩D (B) we have 2

∆Aψ∆Bψ ≥1

4|〈ψ, [A,B]ψ〉|2. (5.3)

Proof. 〈ψ, [A,B]ψ〉 = 〈Aψ,Bψ〉 − 〈Bψ,Aψ〉 = 〈Aψ,Bψ〉 − 〈Aψ,Bψ〉 = 2i Im 〈Aψ,Bψ〉. So∣∣∣∣12 〈ψ, [A,B]ψ〉∣∣∣∣ = |Im 〈Aψ,Bψ〉| ≤ |〈Aψ,Bψ〉| ≤ ‖Aψ‖‖Bψ‖ by use of the Cauchy-Schwarz

inequality. Squaring, we obtain1

4|〈ψ, [A,B]ψ〉|2 ≤

⟨ψ,A2ψ

⟩ ⟨ψ,B2ψ

⟩. If 〈ψ,Aψ〉 = 〈ψ,Bψ〉 =

0, then the right-hand side equals ∆Aψ∆Bψ. If A or B does not have the expectation value 0, we

use translation invariance of the commutator (∀c, d ∈ R : [A− c,B − d] = [A,B]) to define A =

A− 〈ψ,Aψ〉, B = B − 〈ψ,Bψ〉, where 〈ψ,Aψ〉 , 〈ψ,Bψ〉 ∈ R since 〈ψ,Aψ〉 = 〈Aψ,ψ〉 = 〈ψ,Aψ〉(and analogously for 〈ψ,Bψ〉). Then we apply the Cauchy-Schwarz inequality to A, B.

In particular, for A = xj , B = ∂xk we find ∆Aψ∆Bψ ≥1

4|〈ψ, [xj , i∂xk ]ψ〉|2. The commutator

vanishes if j 6= k because ∂xkxj = 0 for j 6= k. If j = k, [xj , i∂xk ]ψ = xki∂xkψ − i∂xk(xkψ) =

−iψ by application of the product rule of differentiation. Thus, [xj , i∂xk ] = δjk(−i)I. Hence

∆Aψ∆Bψ ≥1

4δjk|〈ψ,ψ〉|2. In particular, if j = k and if the wavefunction ψ is normalized to

‖ψ‖ = 1, then

∆Aψ∆Bψ ≥ 1

4.3 (5.4)

2By⟨ψ,A2ψ

⟩we mean 〈Aψ,Aψ〉, and by 〈ψ,ABψ〉 we mean 〈Aψ,Bψ〉.

3This inequality has profound consequences in physics: suppose that the position of a particle can be measured

to arbitrary precision. In classical physics, there is nothing that prevents the momentum from being measured

to arbitrary precision as well. However, in Quantum Mechanics, the above inequality implies that the variance

46 Chapter 5 Observables, Uncertainty Principle, Ground state energy

Remark

Here we repeat the argument used in the proof above. If A is a self-adjoint operator 〈ψ,Aψ〉 =

〈Aψ,ψ〉 = 〈ψ,Aψ〉, so 〈ψ,Aψ〉 ∈ R. Hence, the expectation values of self-adjoint operators

are real-valued (or vector-valued with real components).

We consider H : H2(R3N

)⊆ L2

(R3N

)→ L2

(R3N

),

H =N∑j=1

(−∆xj −

Z

|xj |

)+

∑1≤i<j≤N

1

|xi − xj |, (5.5)

where x1, . . . , xN ∈ R3. equation (5.5) is the Hamiltonian (the Hamilton operator) of an ion

with N electrons and a nucleus of charge Z. If N = Z, the system is electrically neutral and is

called an atom.

For notational simplicity, we will write H = −∆ +V , where −∆ = −N∑j=1

∆xj is the total kinetic

energy, i.e. the sum of the kinetic energies of all electrons. Then for all ψ ∈ H2(R3N

)we have

〈ψ,Hψ〉 = 〈ψ,−∆ψ〉+ 〈ψ, V ψ〉 = 〈∇ψ,∇ψ〉+ 〈ψ, V ψ〉, so actually 〈ψ,Hψ〉 makes sense for all

ψ ∈ H1(R3N

). It was proven in Exercise 13 that H is bounded from below, namely ∃C ∈ R

such that 〈ψ,Hψ〉 ≥ C‖ψ‖2. 4

Definition

The ground state energy of H is defined by E0 := infψ∈H1(R3N)‖ψ‖L2=1

〈ψ,Hψ〉. The infimum exists

because by the above argument there exists a lower bound on 〈ψ,Hψ〉. If the infimum is

attained for some ψ0 ∈ H1(R3N

), then ψ0 is called a ground state.

Theorem 5.4

If H has a ground state ψ0, then ψ0 ∈ H2(R3N

)and Hψ0 = E0ψ0.

of the momentum becomes arbitrarily large, i.e. it is not possible to constrain the momentum of the particle

in the limit that the uncertainty of the position goes to zero.

4In classical physics, the total energy is given byp2

2m− 1

|x− y| and it is possible to fix p while decreasing

|x− y|. Thus it is possible for the energy to become arbitrarily negative by fixing p and taking x → y. In

Quantum Mechanics, however, this is forbidden by the Heisenberg Uncertainty Principle: if we take |x− y|smaller and smaller, the variance of p (

⟨ψ, p2ψ

⟩−〈ψ, pψ〉2) increases due to Equation 5.4. Since 〈ψ, pψ〉 = 0 if

〈ψ,ψ〉 is symmetric in p→ −p, we obtain that⟨ψ, p2ψ

⟩increases as |x− y| becomes small. Thus, the energy

p2

2m− 1

|x− y| cannot become arbitrarily negative in Quantum Mechanics and H is bounded from below.

Chapter 5 Observables, Uncertainty Principle, Ground state energy 47

Proof. Let v ∈ C∞c(R3N

), f(s) := 〈(ψ0 + sv), H(ψ0 + sv)〉 − E0 〈ψ0 + sv, ψ0 + sv〉. Then by

assumption f(0) = 0, and by definition of E0 we have f(s) ≥ 0 ∀s ∈ R. Therefore

f ′(0) = 0. (5.6)

But f(s) = 〈ψ0, Hψ0〉+ 2sRe 〈v,Hψ0〉+ s2 〈v,Hv〉 −E0 〈ψ0, ψ0〉 − 2sE0 Re 〈v, ψ0〉 −E0s2 〈v, v〉,

so

f ′(0) = 2 Re 〈v,Hψ0〉 − 2E0 〈v, ψ0〉 = 2 Re(〈v,Hψ0〉 − 〈v,E0ψ0〉). (5.7)

Equation (5.6) and equation (5.7) imply that Re(〈v,Hψ0〉 − 〈v,E0ψ0〉) = 0 ∀v ∈ C∞c(R3N

). So

〈v,Hψ0〉 − 〈v,E0ψ0〉 = 0∀v ∈ C∞c(R3N

). We have 〈∇v,∇ψ0〉+ 〈v, V ψ0〉 = v 〈E0, ψ0〉

v∈C∞c (R3N)⇒ 〈−∆v, ψ0〉 = 〈v,−V ψ0 + E0ψ0〉 ∀v ∈ C∞c(R3N

). Since −V ψ0 + E0ψ0 ∈ L2

(R3N

)by Hardy’s inequality, by the definition of weak derivatives we have −∆ψ0 ∈ L2

(R3N

)and

−∆ψ0 = −V ψ0 + E0ψ0, so Hψ0 = E0ψ0.

Theorem 5.5 (Existence and uniqueness of the ground state of the hydrogen atom)

The Hamiltonian H = −∆− 1

|x| , H : H2(R3)⊆ L2

(R3)→ L2

(R3)

has (up to a constant) a

unique ground state.

Proof. We write E(ψ) = 〈ψ,Hψ〉 and E0 = infψ∈H1(R3)‖ψ‖L2=1

E(ψ). We have E(ψ) = ‖∇ψ‖2L2 −

∫R3

|ψ(x)|2|x| dx. By the Coulomb uncertainty principle (Exercise 17) we have

∫R3

|ψ(x)||x| dx ≤

‖∇ψ‖L2‖ψ‖L2 with equality if and only if ψ(x) = De−c|x|. Thus

E(ψ) ≥ ‖∇ψ‖2L2 − ‖∇ψ‖L2‖ψ‖L2 =

(‖∇ψ‖L2 − 1

2‖ψ‖L2

)2

− 1

4‖ψ‖2L2 (5.8)

with equality if and only if ‖∇ψ‖L2 =1

2‖ψ‖L2 . But in E(ψ) ≥ ‖∇ψ‖2L2−‖∇ψ‖L2‖ψ‖L2 equality

holds if and only if

ψ(x) = De−c|x| (5.9)

for some D ∈ C and c > 0. Since ∇e−c|x| = −c x|x|e−c|x|, we obtain∥∥∥∇e−c|x|

∥∥∥L2

= |c|∥∥∥e−c|x|

∥∥∥L2. (5.10)

Thus, equation (5.9) and equation (5.10) hold simultaneously if and only if ψ(x) = De−12|x|.

Thus, equation (5.8) holds with equality if and only if ψ(x) = De−|x|2 for some D ∈ C.

If we require ‖ψ‖L2 = 1, we obtain |D| = 1√8π

, so E0 ∼1√8π

e−|x|2 .

6 Some tools of functional analysis

In finite dimensions, the property that every bounded sequence has a convergent subsequence

is very useful (e.g. for showing that a function has a minimum or maximum).

However, in infinite dimensions we encounter the problem that this property is not true anymore,

e.g. in a Hilbert space an orthonormal basis (ONB) (en)n∈N is bounded, but does not have a

convergent subsequence.

We try to “weaken” the definition of convergence in infinite dimensions hoping that we will

again have a property similar to the existence of convergent subsequences, as is the case for

finite dimensions.

Definition 6.1

Let (X, ‖·‖) be a real (resp. complex) Banach space and y ∈ X. Let (yn)n∈N be a sequence in

X. We say that yn converges weakly to y, yn y if f(yn)→ f(y) for all linear and continuous

functions f : X → R (resp. f : X → C).

Remark 6.2

Let (X, 〈·, ·〉) is a Hilbert space. Then f : X → R (f : X → C) is linear and continuous if

and only if there exists a y ∈ X with ∀v ∈ X : f(v) = 〈y, v〉 (Riesz representation theorem).

Therefore, yn y ⇔ 〈v, yn〉 n→∞→ 〈v, y〉 ∀v ∈ X.

Example 6.1

Let (X, 〈·, ·〉) be a Hilbert space and (xn)n∈N be an orthonormal sequence, i.e. 〈xn, xm〉 =

δmn =

1, n = m

0, n 6= m. Then xn 0.

Proof. Let v ∈ X. Then by Parseval’s identity∑n∈N|〈xn, v〉|2 ≤ ‖v‖2. Thus

∑n∈N|〈x, v〉|2 is

convergent ⇒ 〈xn, v〉 → 0 as n→∞. However, xn 6→ 0 because ‖xn‖ = 1∀n ∈ N.

50 Chapter 6 Some tools of functional analysis

Theorem 6.3

Let X be a Banach space. If xn x for some x ∈ X, then xn is bounded and

‖x‖ ≤ lim infn→∞

‖xn‖ . (6.1)

Proof. While this holds for Banach spaces, we will only prove it for Hilbert spaces: In this case

‖x‖2 = 〈x, x〉 = limn→∞

〈xn, x〉 ≤ lim infn→∞

‖xn‖‖x‖.

Theorem 6.4 (Banach-Alaoglu, special case)

Let (X, 〈·, ·〉) be a Hilbert space. Then every bounded sequence in X has a weakly convergent

subsequence.

Theorem 6.5

Let (X, 〈·, ·〉) be a Hilbert space. If xn x and ‖xn‖ → ‖x‖, then xn → x.

Proof. ‖xn − x‖2 = 〈xn − x, xn − x〉 = 〈xn, xn〉 − 〈xn, x〉 − 〈x, xn〉+ 〈x, x〉. But 〈xn, x〉 → 〈x, x〉as n→∞, so ‖xn − x‖2 → 〈xn, xn〉 − 〈x, x〉 → 0 as n→∞, since ‖xn‖ → ‖x‖.

Definition 6.6 (Compact operators)

Let X,Y be Banach spaces and K : X → Y be a linear operator. K is called compact if for

every bounded sequence (xn)n∈N in X Kxn has a convergent subsequence in Y .

Theorem 6.7

Let X,Y be Hilbert spaces and K : X → Y linear. Then K is compact if and only if xn x

implies that Kxn → Kx for all sequences (xn)n∈N in X and for all x ∈ X.

Proof. • “⇒”: Let f : Y → R (resp C) be linear and continuous. Then (f K) : X → R(resp. C) is continuous as a composition of continuous functions. Since xn x, we find

f(Kxn)→ f(Kx). Since f was arbitrary, it follows that

Kxn Kx . (6.2)

To show that Kxn → Kx we prove that every subsequence Kxnl has a (sub-)subsequence

Kxnlm with Kxnlm → Kx. xnl is bounded, and since K is compact Kxnl has a convergent

subsequence

Kxnlm → y (6.3)

Chapter 6 Some tools of functional analysis 51

for some y in Y , so xnlm y. (6.4)

From equation (6.2) and equation (6.4) it follows that y = Kx. (6.5)

equation (6.3) and equation (6.5) complete the proof.

• “⇐”: Let (xn)n∈N be bounded. By Theorem 6.4 there exists x ∈ X and a subsequence

xnk with xnk x. But Kxnk → Kx, so Kxn has a convergent subsequence.

Theorem 6.8

Let X,Y be Hilbert spaces (actually, Banach spaces are sufficient) and Kn : X → Y a sequence

of compact operators. If K : X → Y is linear and ‖Kn −K‖ n→∞→ 0, then K is also compact.

Sketch of proof. Let (xm)m∈N be a bounded sequence in X. If K1 is compact, then there exists

a subsequence (xm)m∈A1 , A1 ⊆ N, such that (K1xm)m∈A1 converges. If K2 is compact and

(xm)m∈A1 bounded, then there exists a subsequence (xm)m∈A2 , A2 ⊆ N, such that (K2xm)m∈A2

converges. Repeating the argument we find a family of subsequences A1 ⊇ A2 ⊇ A3 ⊇ . . . ⊇. . . An ⊇ . . . such that (Knxm)m∈An is convergent. Choose m1 ∈ A1, m2 ∈ A2, . . . ,mn ∈ An, . . .such that m1 < m2 < . . . < mn < . . .. Then (Knxml)l∈N is convergent for all n ∈ N. Let

yn := limn→∞

Knxml . Then yn → y for some y ∈ Y and Kxml → y.

Theorem 6.9

Let f ∈ C1c (Rn). Then the operator Tf : H1 (Rn)→ L2 (Rn), Tfψ := fψ is compact.

Idea. Let Tf := J Gf , Gf : H1 (Rn)→ B := g ∈ H1 (Rn) : supp (g) ⊆ supp (f),Gf := fψ, J : B → L2 (Rn), Jg := g. Gf is continuous by the Leibniz rule, B is compact by the

Rellich-Kondrachov theorem, so Tf is compact as a composition of a continuous and a compact

operator.

Theorem 6.10

Let f ∈ C(Rn) with lim|x|→∞

f(x) = 0. Then Tf : H1 (Rn)→ L2 (Rn), Tfψ = fψ is compact.

Proof. Exercise 31. Hint: Combine Theorems 6.8 and 6.9.

7 Decomposition of an operator

Theorem 7.1

Let X be a complex-valued Banach space and A : D ⊆ X → X a linear operator. Let γ be

a simply closed 1 positively oriented (counter-clockwise) curve lying in the resolvent set of A.

If inside of γ there is no point of σ (A), then

(i)

∮γ

(z −A)−1 dz = 0 (Cauchy)

(ii) If w ∈ C is inside of γ, then

(w −A)−1 =1

2πi

∮γ

(z −A)−1

z − w dz (Cauchy integral formula). (7.1)

Proof. (i) Let f(z) := (z −A)−1. Then limz→z0

f(z)− f(z0)

z − z0= lim

z→z0

(z −A)−1 − (z0 −A)−1

z − z0

= limz→z0

(z −A)−1[(z0 −A)− (z −A)](z0 −A)−1

z − z0= −(z0−A)−2. Thus f is complex differ-

entiable in γ and inside of γ and therefore

∮f(z) dz = 0 by repeating the arguments of

the proof of the traditional theorem of Cauchy.

(ii) Since1

2πi

∮γ

1

z − w dz = 1, equation (7.1) is equivalent to1

2πi

∮γ

(w −A)−1

z − w dz

=1

2πi

∮γ

(z −A)−1

z − w dz ⇔∮γ

(w −A)−1 − (z −A)−1

z − w dz = 0

⇔∮γ(w −A)−1(z −A)−1 dz = 0, which holds by part (i).

Definition 7.2

Let X be a complex Banach space and A ∈ L(X). The spectral radius of A is defined by

rA := lim supn→∞

‖An‖ 1n .

1By “simply closed” we mean that γ is closed and that the set enclosed by γ is connected.

54 Chapter 7 Decomposition of an operator

Note: ‖An‖ 1n ≤ ‖A‖ ∀n ∈ N.

Theorem 7.3

Let A ∈ L(X). Then

(1) σ (A) 6= ∅ and rA = supc∈σ(A)

|c|

(2) If X is a Hilbert space and A self-adjoint and bounded, then ‖A‖ = rA = supc∈σ(A)

|c|.

Proof. (1) It was shown in Exercise 23 that σ (A) 6= ∅. If |z| > rA, then lim supn→∞

∥∥∥∥Anzn∥∥∥∥ 1n

=

ra|z| < 1. Therefore

∞∑n=0

An

znconverges absolutely and therefore also in the operator norm.

In particular,An

znn→∞→ 0. Thus we can repeat the argument of Exercise 7 to conclude that(

I − A

z

)is invertibel and

(I − A

z

)−1

=

∞∑n=0

An

zn. Thus z−A = z

(1− A

z

)is invertible for

|z| > rA ⇒ rA ≥ supc∈σ(A)

|c|. It remains to prove that rA ≤ supc∈σ(A)

|c|. Let f(s) := (1− sA)−1.

f(s) is well-defined and bounded for s <1

supc∈σ(A) |c|(because then

1

s/∈ σ (A)). It follows

from Exercise 24 that∞∑n=0

snAn is absolutely convergent. In particular, ∀s < 1

supc∈σ(A) |c|:

lim supn→∞

‖snAn‖ 1n ≤ 1. Therefore ∀s < 1

supc∈σ(A) |c|: rA ≤

1

s. Hence rA ≤ sup

c∈σ(A)|c|.

(2) We showed in (1) that rA = supc∈σ(A)

|c|. By definition, rA = lim supn→∞

‖An‖ 1n ≤ ‖A‖ (this also

holds if A is not self-adjoint). If A is self-adjoint and bounded and η ∈ D (A) (= H, since

A is bounded) with ‖η‖ = 1, then ‖Aη‖2 = 〈Aη,Aη〉 A bounded and self-adjoint=

⟨η,A2η

⟩≤

‖η‖∥∥A2η

∥∥ ≤ ∥∥A2∥∥. Thus

∥∥A2∥∥ = sup

‖η‖=1‖Aη‖2 ≤

∥∥A2∥∥‖A‖2. Hence ‖A‖2 =

∥∥A2∥∥.

By induction, we find ‖A‖2n =∥∥A2n

∥∥. Therefore, ∀n ∈ N :∥∥A2n

∥∥ 12n = ‖A‖. Hence

lim supn→∞

‖An‖ 1

n≥ ‖A‖ ⇒ rA ≥ ‖A‖.

Theorem 7.4

Let H be a complex Hilbert space and A : D (A) ⊆ H → H self-adjoint. Assume that

σ (A) = σ′ ∪ σ′′, where σ′, σ′′ ⊆ R with σ′ ∩ σ′′ = ∅ and σ′ is compact. Let γ be a simply

Chapter 7 Decomposition of an operator 55

closed, positively oriented curve such that σ′ is inside of γ and σ′′ is outside of γ. Let

P :=1

2πi

∮γ(z −A)−1 dz. Then

(1) P is independent of the choice of γ (as long as γ satisfies the conditions stated above)

(2) P is an orthogonal projection

(3) PA ⊂ AP (AP is an extension of PA). In particular, A leaves Ran (P ) and Ran (1− P ) =

Ran (P )⊥ invariant

(4) Let A′ = A|Ran(P ) : D (A)∩Ran (P )→ Ran (P ) and A′′ = A|Ran(P )⊥ : D (A)∩(Ran (P ))⊥ →(Ran (P ))⊥. Then σ

(A′)

= σ′, σ(A′′)

= σ′′ and A′, A′′ are self-adjoint

(5) A′ is bounded

(6) If σ′ = λ for some λ ∈ R, then A′ = λP

Proof. (1) We may assume without loss of generality that one curve is inside of the other (see

below; if not, we can find a third curve inside of both curves) and deform it into the contour

shown on the right-hand side of the figure below. Applying Cauchy’s theorem for δ1, δ2 we

find

∮δ1

(z − A)−1 dz +

∮δ2

(z − A)−1 dz = 0. Therefore

∮γ(z − A)−1 dz =

∮γ(z − A)−1 dz.

Thus P is independent of the choice of γ.

σ′

γ

γ

Rez

Imz

δ1

δ2

σ′

Rez

Imz

Figure 7.1: Deformation of the integration contour; created with MGA-TeX 2.4 by K.

Fritzsche [4].


(2) We have to show: P 2 = P , P ∗ = P (this implies that 〈(1− P )ϕ, Pψ〉 = 〈P (1− P )ϕ,ψ〉 =⟨(P − P 2)ϕ,ψ

⟩= 0). Clearly P is bounded because (z − A)−1 is continuous and therefore

bounded in γ. Let γ1 and γ2 be two simply closed, positively oriented curves with σ′ in the

inside of both curves and γ2 inside of γ1. Then

P 2 =1

(2πi)2

∮γ1

(z −A)−1 dz

∮γ2

(s−A)−1 ds =1

(2πi)

∮γ1

∮γ2

(z −A)−1(s−A)−1 ds dz

Exercise 4=

1

(2πi)2

∮γ1

[∮γ2

((z −A)−1

s− z − (s−A)−1

s− z

)ds

]dz .

The first contribution vanishes by Cauchy’s theorem, so we obtain

P 2 =1

(2πi)2

∮γ1

∮γ2

(s−A)−1

z − s ds dzFubini

=1

(2πi)2

∮γ2

∮γ1

(s−A)−1

z − s dz ds =1

2πi

∮γ2

(s −

A)−1 = P , where we have used the Cauchy integral formula in the second-to-last step. Thus

P is a projection. Choosing γ to be a circle, γ(θ) = c + reiθ with θ ∈ [0, 2π], c ∈ R, d > 0,

we find P =1

2πi

∫ 2π

0

(c+ reiθ −A

)−1rieiθ dθ.

Thus P ∗ = − 1

2πi

∫ 2π

0

(c+ re−iθ −A

)−1(−ir) e−iθ dθ

θ=2π−θ=

1

2πi

∫ 2π

0

(c+ reiθ −A

)−1ireiθ dθ = P .

(3) We prove: if ψ ∈ D (A), then APψ = PAψ. We have Pψ =

(1

2πi

∮γ(z −A)−1 dz

)ψ =

1

2πi

∮γ(z −A)−1ψ dz (Left as an exercise).

A(z −A)−1 Exercise 4= −I + z(z −A)−1 (7.2)

is bounded it follows that (z −A)−1ψ : C→ H is continuous in the graph norm of A. Thus

APψA=A∗⇒A=A∗ closed, Exercise 5

=1

2πi

∮γA(z −A)−1ψ dz

ψ∈D(A), Exercise 4=

1

2πi

∮γ(z −A)−1Aψ dz . (7.3)

Hence APψ =1

2πi

∮γ(z − A)−1 dz Aψ = PAψ. Thus PA ⊂ AP . Let ϕ ∈ D (A) ∩ Ran (P ).

Then AϕPϕ=ϕ

= APϕ ∈ Ran (P ). So

A(D (A) ∩ Ran (P )) ⊆ Ran (P ) . (7.4)

Similarly one shows that

A(D (A) ∩ (Ran (P ))⊥

)⊆ (Ran (P ))⊥ . (7.5)


(4) In view of equations (7.4) and (7.5) A′ = A|Ran(P ) and A′′ = A|(Ran(P ))⊥ are well-defined.

A′ is bounded because A′ = AP |Ran(P ), but APequation (7.3)

=1

2πi

∮γA(z −A)−1 dz

equation (7.2)=

1

2πi

∮γ

(−I + z(z −A)−1

)dz, and −I + z(z − A)−1 is uniformly bounded on γ.

Thus AP is bounded. Let z ∈ C. Then z ∈ ρ (A)⇔ (z−A) : D (A) ⊆ H → H is a bijection

with bounded inverse since A leaves Ran (P ) and (Ran (P ))⊥ invariant

⇔

z −A′ : D (A) ∩ Ran (P )→ Ran (P )

z −A′′ : D (A) ∩ (Ran (P ))⊥ → (Ran (P ))⊥are bijections with bounded inverses ⇔

z ∈ ρ(A′)∩ ρ(A′′). Thus

ρ (A) = ρ(A′)∩ ρ(A′′)⇒ σ

(A′)∪ σ

(A′′)

= σ (A) . (7.6)

We will prove σ(A′)⊆ σ′, σ

(A′′)⊆ σ′′. Let z ∈ ρ (A) \ trγ.

(w −A)−1P =1

2πi

∮γ(w −A)−1(z −A)−1 dz

Exercise 4=

1

2πi

∮γ

((w −A)−1

z − w − (z −A)−1

z − w

)dz . (7.7)

Assume that w is outside of γ, then

∮γ

(w −A)−1

z − w dz = 0, thus (w −A)−1P

=1

2πi

∮γ

(z −A)−1

w − z dz. This formula gives an expression for (w − A)−1 for all w ∈ ρ (A) \

trγ, but it can be naturally extended for all w ∈ trγC . Thus

w ∈ ρ (A)⇒ σ(A′)⊆ σ′. (7.8)

If w is inside of γ, then1

2πi

∮γ

1

z − w dz = 1 and thus (w −A)−1P

= (w − A)−1 − 1

2πi

∮γ

(z −A)−1

z − w dz ⇒ (w − A)−1(1 − P ) =1

2πi

∮γ

(z −A)−1

z − w dz. We have

(w − A′′)−1 =

(1

2πi

∮γ

(z −A)−1

z − w dz

)∣∣∣∣(Ran(P ))⊥

for all w ∈ ρ (A) inside of γ. Since this

formula can be naturally extended for all w inside γ, we find that

σ(A′′)⊆ σ′′. (7.9)

Equations (7.6), (7.8) and (7.9) imply that σ(A′)∪σ

(A′′)

= σ (A) = σ′ ∪σ′′ ⇒ σ′ = σ(A′)

and σ′′ = σ(A′′).

(5) That A′ is bounded was already proven in (4).


(6) If σ′ = λ for some λ ∈ R, then σ(A′)

= λ ⇒ σ(A′ − λP

)= 0. Since A′ − λP is

bounded and self-adjoint, it follows from Theorem 7.3 that∥∥A′ − λP∥∥ = 0⇒ A′ − λP = 0,

hence AP = λP .

Theorem 7.5 (Identification of the spectrum with sequences)

Let A : D (A) ⊆ H → H be self-adjoint (then σ (A) ⊆ R) and λ ∈ R. Then λ ∈ σ (A)⇔ there

exists a sequence (ψn)n∈N in D (A) with ‖ψn‖ = 1 and ‖(A− λ)ψn‖ n→∞→ 0 (there exists a

sequence of “approximate eigenfunctions”).

Proof. • “⇐”: If λ were in ρ (A), then (A− λ) : D (A) ⊆ H → H would be a bijection with

a bounded inverse. So ∃c > 0 with∥∥(A− λ)−1ϕ∥∥ ≤ c‖ϕ‖∀ϕ ∈ H. (7.10)

If ψ ∈ D (A) then ∃ϕ ∈ H with ψ = (A − λ)−1ϕ. Thus ‖ψ‖ =∥∥(A− λ)−1ϕ

∥∥ eqn. (7.10)

≤c‖ϕ‖ = c‖(A− λ)ψ‖ ⇒ ‖(A− λ)ψ‖ ≥ c‖ψ‖∀ψ ∈ D (A), contradicting that ‖ψn‖ = 1 and

‖(A− λ)ψ‖ n→∞→ 0. Thus λ ∈ σ (A).

• “⇒”: Assume λ ∈ σ (A). If there is no sequence (ψn)n∈N in D (A) with ‖ψn‖ = 1 and

‖(A− λ)ψn‖ → 0, then c := infψ∈D(A)‖ψ‖=1

‖(A− λ)ψ‖ > 0. Therefore

∀ψ ∈ D (A) : ‖(A− λ)ψ‖ < c‖ψ‖. (7.11)

We will prove

(i) A− λ is injective.

(ii) Ran (A− λ) is closed.

(iii) Ran (A− λ) is dense, thus A− λ is a bijection.

(iv) (A− λ)−1 is bounded, contradicting λ ∈ σ (A).

Proof:

(i) (A− λ)ψ = 0⇒ ‖ψ‖ = 0⇒ ψ = 0.

(ii) Exercise.

(iii) Exercise.


(iv) Since (A − λ)−1 is well-defined and ∀ψ ∈ D (A) : ‖(A− λ)ψ‖ > c‖ψ‖, we find

∀ϕ ∈ H :∥∥(A− λ)−1ϕ

∥∥ ≤ 1

c‖ϕ‖.

8 Discrete and essential spectrum

Definition 8.1

Let H be a complex Hilbert space, A : D (A) ⊆ H → H self-adjoint.

The discrete spectrum σd (A) of A is defined by

σd (A) := λ ∈ σ (A) : λ eigenvalue of finite multiplicity and isolated in the spectrum.The essential spectrum σess (A) of A is defined by σess (A) := σ (A) \ σd (A).

If H is the Hamiltonian of an atom or an ion, we define ΣR := infψ∈H1

‖ψ‖L2=1

supp(ψ)⊆(BR(0))c

〈ψ,Hψ〉,

Σ := limR→∞

ΣR. Then Σ = inf σess (H). Σ is called ionisation threshold (energy).

Let E = infψ∈H1

‖ψ‖L2

〈ψ,Hψ〉.

Case 1: E = Σ. In this case, it does not cost energy to remove an electron to infinite spatial

distance (measured from the location of the nucleus).

Case 2: E < Σ: In this case, it costs energy to remove the electron to infinite spatial distance.

Moreover E ∈ σd (H) and a ground state exists.

Theorem 8.2 (Weyl’s criterion)

Let A : D ⊆ H → H be self-adjoint, H a complex Hilbert space. Let λ ∈ R.

Then λ ∈ σess (A)⇔ there exists a sequence (ψn)n∈N in D with

(1) ‖ψn‖ = 1

(2) ‖(A− λ)ψn‖ → 0

(3) ψn 0.

Intuition: If ‖ψn‖ = 1 and ψn 0 , then the sequence (ψn)n∈N “lives” in an infinite-dimensional

space (if not it would have a strongly convergent subsequence ψnk → ψ for some ψ ∈ H with

‖ψ‖ = 1, contradicting ψn 0.

62 Chapter 8 Discrete and essential spectrum

Remark

A sequence (ψn)n∈N with the properties (1)-(3) from Theorem 8.2 is called a Weyl sequence

for A, λ.

Proof. • “⇐”: If there is a Weyl sequence for A and λ, then by Theorem 7.5 λ ∈ σ (A).

Assume that λ ∈ σd (A). By Theorem 7.4 we can decompose A to A′, A′′ choosing σ′ = λ.Since λ is an eigenvalue of finite multiplicity it follows that dim Ran (P ) < ∞ (because

AP = λP ). But (A− λ)ψn︸︷︷︸→0

= (A−λ)Pψn+(A−λ)P⊥ψn = (A′−λ)Pψn+(A′′−λ)P⊥ψn.

Now Pψn → 0 by Theorem 6.7 because ψn 0 and P is compact, so (A′ − λ)Pψn → 0,

because A′ is bounded by Theorem 7.4. Hence it follows that (A′′ − λ)P⊥ψ0 → 0 (but∥∥∥P⊥ψn∥∥∥→ 1). Thus (A′′− λ)P⊥ψn‖P⊥ψn‖

→ 0, contradicting (by Theorem 7.5) the fact that

λ /∈ σ(A′′).

• “⇒”: Assume λ ∈ σess (A).

(i) If λ is an eigenvalue of infinite multiplicity then there exists a sequence (ψn)n∈N

orthonormal with (A− λ)ψn = 0. But ψn 0 by Example 6.1.

(ii) If λ is not an eigenvalue of infinite multiplicity, then it is not an isolated point of the

spectrum. Thus there exists a sequence (an)n∈N in R with an → 0, an 6= 0 ∀n ∈ Nand ∀n ∈ N : λ + an ∈ σ (A). Without loss of generality we can assume that an > 0

and

an+1 <an2. (8.1)

By Theorem 7.5, for all n ∈ N there exists a ψn ∈ D with

‖(A− λ− an)ψn‖ ≤ e−1an . (8.2)

From this it follows that ‖(A− λ)ψn‖ ≤ an + e−1an

n→∞→ 0. By the self-adjointness

of A we have 〈ψn, (an − am)ψm〉 = 〈ψn, (A− λ− am)ψm〉 − 〈(A− λ− an)ψn, ψm〉.

Thus, for m 6= n we have |〈ψn, ψm〉| ≤e−

1an + e−

1am

|an − am|by use equation (8.1) and the

triangle and Cauchy-Schwarz inequalities. Thus, if m > n, then (an−am) ≥ an2

, and

thus

|〈ψn, ψm〉| ≤4e−

1an

an. (8.3)

Since ‖ψm‖ = 1, by Theorem 6.4 we may assume that ψm V for some V ∈ Hafter passing to a subsequence. Keeping n fixed in the limit m → ∞, we obtain

Chapter 8 Discrete and essential spectrum 63

(using equation (8.3)) |〈ψn, V 〉| ≤4e−

1an

an. But the limit n→∞ gives |〈V, V 〉| = 0⇒

‖V ‖2 = 0. Therefore ψn 0.

Example 8.1

Let H : H2(R3)⊆ L2

(R3)→ L2

(R3), Hψ =

(−∆− 1

|x|

)ψ. Then σess (H) = [0,∞). In

particular, the ground-state energy (which is −1

4by Theorem 5.5) is in the discrete spectrum

of H.

Proof. Let λ ∈ σess (H). Then by Weyl’s criterion Theorem 8.2 there exists a sequence (ψn)n∈N

in H2(R3)

with ‖ψn‖L2 = 1, ‖(H − λ)ψn‖ → 0 and ψn 0. In particular, 〈ψn, (H − λ)ψn〉 ≤

‖ψn‖‖(H − λ)ψn‖ → 0, so 〈ψn, Hψn〉 → λ. Therefore

∫R3

|∇ψn(x)|2 dx −∫R3

|ψn(x)|2|x| dx → λ

as n→∞. We have to show:

(i) ψn is bounded in H1(R3).

(ii)

∫R3

|ψn(x)|2|x| dx→ 0.

It then follows that λ ≥ 0.

(i)

∫R3

|ψn(x)|2|x| dx

Exercise≤ ‖∇ψn‖‖ψn‖ = ‖∇ψn‖ ⇒

∫R3

|∇ψn|2 dx−∫R3

|ψn(x)|2|x| dx

≥ ‖∇ψn‖2−‖∇ψn‖ ≥(‖∇ψn‖ −

1

2

)2

− 1

4. Thus lim sup

n→∞

(‖∇ψn‖ −

1

2

)2

− 1

4≤ ∞. Thus

(ψn)n∈N is a bounded sequence in H1(R3). 1

(ii) Let ε > 0 and choose R >2

ε. Then

∫R3

|ψn(x)|2|x| dx =

∫|x|≤R

|ψn(x)|2|x| dx+

∫|x|≥R

|ψn(x)|2|x| dx

‖ψn‖=1=

∫|x|≤R

|ψn(x)|2|x| dx+

ε

2.

(8.4)

1Alternatively, one can prove this using Hardy’s inequality:1

|x| = 2√ε

1

2√ε|x|

Binomi

≤ ε

|x|2+

1

4ε. Thus∫

R3

(ψn(x)|2

|x| dx ≤ ε∫R3

|ψn(x)|2

|x| dx+|ψn|2

4ε

Hardy

≤ 4ε

∫|∇ψn|2 dx+

1

4ε. For ε =

1

8we find

∫R3

|ψn(x)|2

|x| dx ≤

1

2‖∇ψn‖2 + 2. Thus

∫R3

|∇ψn|2 dx −∫R3

|ψn(x)|2

|x| dx ≥ ‖∇ψn‖2 −‖∇ψn‖2

2− 2 ⇒ lim sup

n→∞

‖∇ψn‖2

2− 2 ≤

λ <∞.


We have

∫|x|≤R

|ψn(x)|2|x| dx =

∫|x|≤R

|ψn||ψn||x| dx

Cauchy-Schwarz≤

(∫|x|≤R

|ψn|2 dx

) 12(∫|x|≤R

|ψn|2

|x|2dx

) 12

Hardy≤

(∫|x|≤R

|ψn|2 dx

) 12

2‖∇ψn‖(i)

≤ C

(∫|x|≤R

|ψn|2 dx

) 12

. (8.5)

We have∣∣χBR(0)ψn

∣∣ ≤ ∣∣χBR(0)ψn∣∣, where χA(x) =

1, x ∈ A0, x /∈ A

is the indicator func-

tion and χ a “smoothed-out” version of the indicator function (an extension of χA that

goes smoothly to zero outside of A). But TχBR(0): H1

(R3)→ L2

(R3)

is compact

by Theorem 6.9, ψn 0 and ψn is bounded in H1. Hence TχBR(0)ψn

L2

→ 0. Thus∫|x|≤R

|ψn(x)|2 dx → 0. By equations (8.4) and (8.5) it follows that

∫R3

|ψn(x)|2|x| dx < ε

for n large enough.

So far we have shown that σess (H) ⊆ [0,∞). Let λ ∈ [0,∞). Then by Theorem 2.5 λ ∈σ (−∆). Let ε > 0. By Theorem 7.5 there exists a ψε ∈ H2

(R3)

with ‖ψε‖L2 = 1 such that

‖(−∆− λ)ψε‖ <ε

2. Let ψε,h(x) := ψε(x− h). ⇒ ‖(H − λ)ψε,h‖ ≤ ‖(−∆− λ)ψε,h‖︸︷︷︸

=‖(−∆−λ)ψε‖

+

∥∥∥∥ψε,h|x|∥∥∥∥L2

.

Exercise 30: lim|h|→∞

∥∥∥∥ψε,h|x|∥∥∥∥L2

= 0. So, choosing h large enough, we find ‖(H − λ)ψε,h‖ < ε ⇒λ ∈ σ (H). Thus [0,∞) ⊆ σ (H)⇒ [0,∞) ⊆ σess (H).

Theorem 8.3

Let H := −∆ + V , V ∈ C(Rn), lim|x|→∞

V (x) = 0. Then H is self-adjoint in H2 (Rn) and

σess (H) = [0,∞).

Proof. That [0,∞) ⊆ σess (H) can be shown as in the previous proof. The self-adjointness of

H follows from Theorem 3.8 since ‖V ‖ < ∞, so V is a bounded multiplication operator. Let

λ ∈ σess (H). Then there exists a sequence (ψn)n∈N in H2 (Rn) with ‖ψn‖ = 1, ‖(H − λ)ψn‖ → 0

and ψn 0. Thus 〈ψm, (H − λ)ψn〉 → 0 ⇒∫Rn|∇ψn|2 dx +

∫RnV |ψn|2 dx → λ as in the

previous proof (in fact, more easily, ψn is bounded in the H1 norm). Moreover,

∫Rn|ψn|2 dx ≤(∫

Rn|V ψn|2 dx

) 12(∫

Rn|ψn(x)|2 dx

) 12

. By Theorem 6.10 TV : H1 (Rn)→ L2 (Rn), TV ψ = V ψ


is compact. It similarly follows that

∫Rn|V ψn|2 dx→ 0 and thus λ ≥ 0.

Remark

This theorem can be quite easily generalized for V bounded with lim|x|→∞

V (x) = 0.

Theorem 8.4

Assume that V ∈ C (Rn) with lim|x|→∞

V (x) = ∞. Then H = −∆ + V with D (H) =ψ ∈ L2 (Rn) : (−∆ + V )ψ ∈ L2 (Rn)

is self-adjoint on L2 (Rn) and σess (H) = ∅.

Proof. For now, we omit the proof of self-adjointness and that D (H) ⊆ H2 (Rn) (maybe later).

Assume λ ∈ σess (H). Then there exists a sequence (ψn)n∈N in D (H) such that ‖(H − λ)ψn‖ →0, ‖ψn‖ = 1, ψn 0 by Theorem 8.2. Since lim

|x|→∞V (x) = ∞, there exists a D > 0 with

V (x) ≥ −D ∀x ∈ Rn; by adding D we assume V ≥ 0. Since ‖(H − λ)ψn‖ → 0, we have

|〈ψn, (H − λ)ψn〉| ≤ ‖ψn‖‖(H − λ)ψn‖ → 0. Hence

‖∇ψn‖2L2 +

∫V |ψn|2 → λ . (8.6)

Thus lim sup ‖∇ψn‖2 ≤ λ (V ≥ 0) and ψn is bounded in H1 (Rn) because ‖ψn‖ = 1. By Theo-

rem 6.9 we find that χBR(0)ψn → 0 in L2 because ψn 0 in L2 and TχBR(0): H1 (Rn)→ L2 (Rn)

is compact. Thus, if R > 0, there exists n0 ∈ N such that∥∥χBR(0)ψn

∥∥2

L2 <1

3∀n ≥ n0. Hence∫

V |ψn|2 dx ≥∫|x|≥R

V |ψn|2 dx ≥ inf|x|≥R

V (x)

∫|ψn|dx ≥

2

3inf|x|≥R

V (x). Thus, choosing R arbi-

trarily large,

∫V |ψn|dx can become arbitrarily large and lim

n→∞

∫V |ψn|2 dx =∞, contradicting

equation (8.6). Thus σess (H) = ∅.To show that χBR(0)ψn → 0, we show: every subsequence has a further (sub)subsequence con-

verging to zero. Denote a subsequence by χBR(0)ψnk . Since ψnk is bounded in H1, it has a

further subsequence ψnkl with ψnkl v in H1 for some v ∈ H1 (Rn) by Theorem 6.4. We also

have ψnkl 0 in L2 (Rn), so v = 0 since ψnkl v in H1. Thus ψnkl 0 in H1 (Rn) and

TχBR(0): H1 (Rn)→ L2 (Rn) is compact. Hence χBR(0)ψ0 → 0 in L2 (Rn).

Example 8.2

The Hamilton operator of the quantum harmonic oscillator is given by2 H = −∆ + |x|2 and

lim|x|→∞

|x|2 =∞, so the spectrum of H consists only of discrete eigenvalues.

2We obtain this operator from the physical one H = − ~2

2m∆ +

mω2

2|x|2 by choosing ~ = 1, m =

1

2, and ω = 2.


Theorem 8.5 (IMS localization formula)

Let H : H2 (Rn) ⊆ L2 (Rn)→ L2 (Rn), H = −∆ + V be self-adjoint.

(a) If f : Rn → R is a C2-function with ∂αf ∈ L∞(Rn) for 0 ≤ |α| ≤ 2, then 3

fHf =1

2

(f2H +Hf2

)+ |∇f |2. (8.7)

(b) If (Ja)a∈A : Rn → R is a family of C2-functions with ∂βJa ∈ L∞(Rn) for 0 ≤ |β| ≤ 2 ∀a ∈ A,

and∑a∈A

J2a = 1, then 4

H =∑a∈A

JaHJa −∑a∈A|∇Ja|2. (8.8)

Proof. (a) By the properties of f , Tf : H2 (Rn)→ H2 (Rn), Tfψ = fψ is bounded (‖Tfψ‖L2 =

‖fψ‖L2 ≤ ‖f‖L∞‖ψ‖L2). We have ‖∇(fψ)‖L2 ≤ ‖(∇f)ψ‖L2 + ‖f∇ψ‖L2

≤ (‖∇f‖L∞ + ‖f‖L∞) ‖ψ‖H2 . Similarly one can control ‖∇(fψ)‖ ≤ C‖ψ‖H2 for some

C ≥ 0; in particular, fH2 (Rn) ⊆ H2 (Rn). fHf − 1

2(f2H + Hf2) =

1

2(fHf − f2H +

fHf −H2f) =1

2(f(fH −Hf) + (fH −Hf)f) =

1

2f [H, f ]− 1

2[H, f ]f , so

fHf − 1

2(f2H −Hf2) =

1

2[f, [H, f ]] . (8.9)

This holds (at least formally) independently of H. So it suffices to prove [f, [H, f ]] = 2|∇f |2.

We have [H, f ] = [(−∆ + V ), f ] = [−∆, f ] = −(∆f) − 2(∇f) ·∇ − f∆ + f∆ = −(∆f) −2(∇f) · ∇. Thus [f, [H, f ]] = [f,−(∆f)− 2(∇f) ·∇] = [f,−2(∇f) ·∇] = 2|∇f |2, so

equation (8.9) holds true.

(b) By equation (8.7), JaHJa =1

2(J2aH +HJ2

a ) + |∇Ja|2,

so∑a∈A

JaHJa =1

2

(∑a∈A

J2aH +H

∑a∈A

J2a

)+∑a∈A|∇Ja|2. Since

∑a∈A

J2a = 1, this implies∑

a∈AJaHJa = H +

∑a∈A|∇Ja|2.

Lemma

There exists a family (Jk)Nk=0 of functions Jk : R3N → R with ∀ 0 ≤ |α| ≤ 2, k ∈ 0, . . . , N :

∂αJk ∈ L∞(R3N

)such that

3For example, if ψ(x) = e−|x|2

, then (fHψ)(x) = f(x)(−∆ + V (x))e−|x|2

.4The idea behind this formula is to decompose the entire operator H into a sum of localized operators JaHJa

which are easier to analyze, plus a (hopefully small) error.


1. supp (J0) is compact.

2. ∀k ≥ 1 : supp (Jk) ⊆ (x1, . . . , xN ) ∈ R3N : |xk| ≥ 1.

3.N∑k=0

J2k = 1.

Interpretation:

1. Either all electrons are close to nucleus.

2. The k-th electron is far from the nucleus.

3. Possible because either all electrons are close to nucleus or at least one of them is far.

Proof. Let j ∈ C∞ (R; [0, 1]) with j(r) = 0 for r ≤ 1, j(r) = 1 for r ≥ 2. Let f0(x) =

1− j( |x|

2N

), fk(x) = j(|xk|), k = 1, . . . , N , where x = (x1, . . . , xN ). Then supp (f0) is compact

and supp (fk) ⊆ x : |xk| ≥ 1. Moreover, for all x ∈ R3N there exists a k such that fk(x) = 1.

Thus the functions Jk :=fk(∑N

n=0 f2n

) 12

satisfy the desired properties.

Theorem 8.6 (HVZ-Theorem (Hunziker - van Winter - Zhislin))

Let HN : H2(R3N

)⊆ L2

(R3N

)→ L2

(R3N

),

(HNΦ)(x) =N∑j=1

(−∆xj −

Z

|xj |

)Φ(x) +

∑1≤i<j≤N

1

|xi − xj |Φ(x), x = (x1, . . . , xN ) ∈ R3N .

(Hamiltonian of an ion with N electrons and Z protons).

Then σess (HN ) = [inf σ (HN−1) ,∞).

Proof. Let EN−1 = inf σ (HN−1) = inf‖ψ‖L2=1

ψ∈H1(R3(N−1))

〈ψ,HN−1ψ〉

We will prove:

(i) [EN−1,∞) ⊆ σ (HN ) (Easy direction)

(ii) inf σess (HN ) ≥ EN−1 (Hard direction, need Theorem 8.5)

As (i) implies [EN−1,∞) ⊆ σess (HN ) directly, Theorem 8.6 follows directly.


(i)

HN = HN−1 −∆xN −Z

|xN |+

N−1∑i=1

1

|xi − xN |. (8.10)

Let λ ≥ EN−1. To show: λ ∈ σ (HN ).

λ = EN−1 + δ, where EN−1 ∈ σ (HN−1) and δ ∈ [0,∞) = σ (−∆xN ).

Thus by Theorem 7.5 and σ (∆xN ) = [0,∞):

∀ε > 0 ∃ψN−1 ∈ C∞c(R3(N−1)

)and ϕ ∈ C∞c

(R3), ‖ϕ‖ = ‖ψN−1‖ = 1 such that

‖(HN−1 − EN−1)ψN−1‖ <ε

3, (8.11)

‖(∆xN − δ)ϕ‖ <ε

3. (8.12)

Let ϕh(x) := ϕ(x− h).

Then with (ψN−1 ⊗ ϕh)(x1, . . . , xN ) = ψN−1(x1, . . . , xN−1) · ϕh(xN ):

‖(HN − λ)ψN−1 ⊗ ϕh‖ ≤ ‖(HN−1 − EN−1)ψN−1 ⊗ ϕh‖︸︷︷︸<ε

3·1 by eqn. (8.11)

+ ‖(−∆xN − δ)ψN−1 ⊗ ϕh‖︸︷︷︸<1·

ε

3by eqn. (8.12)

+

∥∥∥∥∥(− Z

|xN |+

N−1∑i=1

1

|xi − xN |

)ψN−1 ⊗ ϕh

∥∥∥∥∥︸︷︷︸→0,|h|→∞ as ψN−1,ϕh compact support

.

Therefore for |h| large enough we have ‖(HN − λ)ψN−1 ⊗ ϕh‖ < ε. Thus by Theorem 7.5

we have λ ∈ σ (HN ) as ‖ψN−1 ⊗ ϕh‖ = 1 and ‖(HN − λ)ψN−1 ⊗ ϕh‖ → 0.

(ii) Let Jk,R = Jk

( xR

)for Jk as in the Lemma before Theorem 8.6. Then:

supp (J0,R) compact,

∀k ≥ 1 : supp (Jk,R) ⊆ (x1, . . . , xN ) ∈ R3N : |xk| ≥ R,N∑k=0

J2k,R = 1.

(8.13)

By the IMS localization formula (Theorem 8.5) we have

HN =

N∑k=0

Jk,RHNJk,R −N∑k=0

|∇Jk,R|2. (8.14)


As HN ≥ EN :

J0,RHNJ0,R ≥ ENJ20,R (8.15)

JN,RHNJN,Reqn. (8.10)

= JN,R

HN−1 −∆xN︸︷︷︸≥0

− Z

|xN |−N−1∑i=1

1

|xi − xN |︸︷︷︸≥0

JN,R

≥ JN,R(HN−1 −

Z

|xN |

)JN,R

|xN |≥R in supp(JN,R)≥ JN,R

(HN−1 −

Z

R

)JN,R

≥(EN−1 −

Z

R

)J2N,R

Similarly we get for all k ≥ 1:

Jk,RHNJk,R ≥(EN−1 −

Z

R

)J2k,R (8.16)

From equation (8.14), equation (8.15) and equation (8.16) it follows that

HN ≥ ENJ20,R +

N∑k=1

(EN−1 −

Z

R

)J2k,R −

N∑k=0

|∇Jk,R|2︸︷︷︸≤D

R2for some D>0

as ∇J∈L∞(R3N)

eqn. (8.13)

≥ (EN − EN−1)J20,R +

N∑k=0

EN−1J2k,R −

C

R

Where −CR

comes from −ZR− D

R2. And so with equation (8.13) we have

HN ≥ (EN − EN−1)J20,R + EN−1 −

C

R(8.17)

Now let λ ∈ σess (HN ). By Weyl’s criterion (Theorem 8.2) ∃(ψn)n∈N ⊆ D(HN ) with

‖ψn‖L2 = 1, ‖(HN − λ)ψn‖ → 0 and ψn 0. Thus

|〈ψn, (HN − λ)ψn〉| ≤ ‖ψn‖‖(HN − λ)ψn‖ → 0 (8.18)

But with equation (8.17) we have 〈ψn, HNψn〉 ≥ (EN − EN−1)⟨ψn, J

20,Rψn

⟩+ EN−1 −

C

Rand

⟨ψn, J

20,Rψn

⟩= ‖J0,Rψn‖2 → 0 in L2 by Theorem 6.9 for any R > 0.

Let ε > 0 and choose R withC

R<ε

2. Then for n large enough we have

〈ψn, HNψn〉 ≥ (EN − EN−1)‖J0,Rψn‖2 + EN−1 −C

R≥ EN−1 − ε (8.19)


Together with equation (8.18) we get λ = limn→∞

〈ψn, HNψn〉 ≥ EN−1 − ε for all ε > 0 and

therefore λ ≥ EN−1

Remark

This can be easily generalized for molecules.

The interval from EN = inf σ (HN ) to EN−1 = inf σess (HN ) is called the ionization energy.

This is the energy needed to push one electron to infinity.

For EN−1 the same happens again, this time for the second electron.

Theorem 8.7

We consider two self-adjoint operators A,B with D (A) = D (B). If there exists a z ∈ C \ Rsuch that (z −A)−1 − (z −B)−1 is compact, then σess (A) = σess (B).

Proof. We have to show that σess (A) ⊆ σess (B) (the other inclusion follows from exchang-

ing A and B). Let λ ∈ σess (A). By Weyl’s criterion (Theorem 8.2) there exists a sequence

(ψn)n∈N in D (A) = D (B) such that ‖ψn‖ = 1, ‖(A− λ)ψn‖ → 0, ψn 0. But then[(z −A)−1 − (z − λ)−1

]ψn = (z − A)−1(z − λ)−1 (A− λ)︸︷︷︸

→0

ψn → 0, since z − λ 6= 0 (λ ∈ R,

z ∈ C\R) and because (z−A)−1 is bounded (z ∈ ρ (A)). Similarly,[(z −B)−1 − (z − λ)−1

]ψn =

(z − λ)−1(B − λ)(z −B)−1ψn. But[(z −B)−1 − (z − λ)−1

]ψn =

[(z −B)−1 − (z −A)−1

]ψn +

[(z −A)−1 − (z − λ)−1

]ψn → 0 .

(8.20)

The second term goes to zero by the above argument, the first term goes to zero because ψn 0

and (z − B)−1 − (z − A)−1 is compact. So (B − λ)(z − B)−1ψn → 0. But by Equation (8.20)

limn→∞

∥∥(z −B)−1ψn∥∥ =

∣∣(z − λ)−1∣∣ 6= 0. Thus ϕn =

(z −B)−1ψn‖(z −B)−1ψn‖

is normalized. But ϕn 0

because (z−B)−1ψn 0 and because limn→∞

∥∥(z −B)−1ψn∥∥ 6= 0. So (B−λ)(z−B)−1ψn → 0⇒

(B − λ)(z −B)−1ψn‖(z −B)−1ψn‖

→ 0⇒ (B − λ)ϕn → 0.

Corollary 8.8

Let A,B be two self-adjoint operators with D (A) = D (B). If there is a z ∈ C \ R such that

(B −A)(z −A)−1 is compact, then σess (A) = σess (B).

Proof. (z − B)−1 − (z − A)−1 = −(z − B)−1(B − A)(z − A)−1. But (z − B)−1 is bounded


and (B − A)(z − A)−1 is compact, so (z − B)−1 − (z − A)−1 is compact, so we can apply

Theorem 8.7.

Let EH , EH+ , EH− be the ground-state energy of the hydrogen atom (resp. +1 ion, resp. −1

ion). Then EH + EH < EH+ + EH− (this can be shown using the HVZ theorem).

Definition 8.9

Let H = H∗ be a self-adjoint operator acting on L2 (Rn) that is bounded from below

( infϕ∈D(H)

〈ϕ,Hϕ〉 > −∞) which satisfies the IMS localization formula (Theorem 8.5). For

R > 0 let ΣR = infψ∈D(H)‖ψ‖=1

supp(ψ)⊆(B(R,0))c

〈ψ,Hψ〉 (the IMS localization formula guarantees that the

infimum is assumed over a non-empty set). ΣR is the least possible energy for ψ supported

outside the ball of radius R and center at 0. The ionization threshold of H is defined by

Σ := limR→∞

ΣR.

Interpretation: Σ is the energy that the system should reach so that at least one particle goes

to infinity.

Remark

For HN,Z , Σ = inf σ (HN−1,Z)HVZ= inf σess (HN,Z). So in this case Σ−inf σ (HN,Z)︸︷︷︸

=:EN,Z

= EN−1,Z−

EN,Z is the ionization energy for HN,Z . For an atom with atomic number Z, EZ−1,Z−EZ,Z is

called the first ionization energy . rk := EZ−k,Z−EZ−(k−1),Z is called the k-th ionization energy

of the atom. In mathematics, it is still an open problem whether r1 < r2 < . . . < rk < rk+1,

whereas this ordering is trivial from the physical point of view since the electron-electron

repulsion forces increase with the number of electrons in the atom/ion.

Exercise: Prove that the last ionization energy is larger than the second-last ionization energy.

8.1 Exponential decay of eigenfunctions

Let ϕ ∈ L2(R) be a C2 solution of the one-dimensional (stationary) Schrodinger equation

− ϕ′′ + V ϕ = Eϕ , (8.21)

where V (x) > E for |x| > R. In this case ϕ′′ and ϕ have the same sign (ϕ′′ = aϕ for some

a > 0). A solution to Equation (8.21) is ϕ(x) = C1e√ax+C2e−

√ax. So we expect that ϕ is either


exponentially growing or decaying with |x|. Since ϕ ∈ L2(R), in particular

∫R|ψ(x)|2 dx < ∞,

it can only exponentially decay.

Note: If lim|x|→∞

V (x) exists then lim|x|→∞

V (x) = Σ.

Theorem 8.10 (Exponential decay of eigenfunctions to eigenvalues below Σ)

Let H be as in Definition 8.9 (H = H∗, inf‖ψ‖=1ψ∈D(H)

〈ψ,Hψ〉 > −∞ and if f ∈ C2 with ∂αf ∈ L∞,

0 ≤ |α| ≤ 2, then fD (H) ⊆ D (H) and fHf =1

2

(f2H +Hf2

)+ |∇f |2), e.g. H = −∆ + V .

If

Hψ = Eψ , (8.22)

where E < Σ, then eβxψ ∈ L2 (Rn) ∀β > 0 with β2 > Σ− E.

Proof. Let χ : Rn → [0, 1], χ ∈ C∞ with χ(x) :=

1, |x| ≥ 2

0, |x| ≤ 1and χR(x) := χ

( xR

). For

0 < ε < 1 we define fε(x) :=β|x|

1 + ε|x| and Gε := χR efε . Our goal is to show that

‖Gεψ‖ ≤ DR‖ψ‖ (8.23)

for some suitable R > 0, where DR is independent of ε. Once we have equation (8.23) it follows

that

∫|x|≥2R

e2β|x||ψ(x)|2 dx =

∫|x|≥2R

limε→0

e2fε(x)|ψ(x)|2 dxmonotone

conv.= limε→0

∫|x|≥2R

e2fε(x)|ψ(x)|2 dx.

Since ∫|x|≥2R

e2fε(x)|ψ(x)|2 dx ≤∫χ2R(x)e2fε(x)|ψ(x)|2 dx =

∫Gε(x)|ψ(x)|2 dx

= ‖Gεψ‖2eqn. (8.23)

≤ D2R <∞ (8.24)

and ∫|x|≤2R

e2β|x||ψ(x)|2 dx ≤ e4βR

∫|x|≤2R

|ψ(x)|2 dx <∞ , (8.25)

it follows that eβ|x| ∈ L2 (Rn). We now show equation (8.23): we have

|∇fε(x)| =∣∣∣∣∣ β x

|x|

1 + ε|x| −β|x|ε x|x|

(1 + ε|x|)2

∣∣∣∣∣ =

∣∣∣∣∣βx|x|(1 + ε|x|)− βεx

(1 + ε|x|)2

∣∣∣∣∣ =

∣∣∣∣∣ β x|x|

(1 + ε|x|)2

∣∣∣∣∣ ≤ β . (8.26)

Now ∇Gε = (∇χR) efε + χR efε (∇fε) = (∇χR) efε +Gε (∇fε), so

|∇Gε|2 = |(∇χr)|2e2fε + 2(∇χR)efεGε(∇fε)︸︷︷︸≤CR (ε−independent because supp(∇χR)⊆|x|≤2R

+G2ε|∇fε|2 . (8.27)


By the IMS localization formula (Theorem 8.5) we have GεHGε =1

2

(G2εH +HG2

ε

)+ |∇Gε|2,

so 〈ψ,Gε(H − E)Gεψ〉 =1

2

⟨ψ,G2

ε (H − E)ψ︸︷︷︸=0

⟩+

1

2

⟨(H − E)ψ︸︷︷︸

=0

, G2εψ

⟩+⟨ψ, |∇Gε|2ψ

⟩=⟨ψ, |∇Gε|2ψ

⟩. Hence ⟨

Gεψ,(H − E − |∇fε|2

)Gεψ

⟩≤ CR‖ψ‖2 . (8.28)

But ⟨Gεψ,

(H − E − |∇fε|2

)Gεψ

⟩ eqn. (8.26)

≥⟨Gεψ,

(ΣR − E − β2

)Gεψ

⟩= ‖Gεψ‖2

(ΣR − E − β2

). (8.29)

From equation (8.28) and equation (8.29) if follows that(ΣR − E − β2

)‖Gεψ‖2 ≤ CR‖ψ‖2.

Since Σ − E − β2 > 0, choosing R large enough we find ΣR − E − β2 > 0. Thus ‖Gεψ‖2 ≤CR

ΣR − E − β2‖ψ‖2 <∞ (ε-independent).

Remark

If E is the ground-state energy of a molecule,1

βis interpreted as the radius of the molecule.

9 Spectral theorem for self-adjoint operators

Theorem 9.1 (Spectral theorem in multiplication-operator form)

LetH be a separable Hilbert space and A a self-adjoint operator onH with domain D (A) ⊆ H.

Then there exists a measure space (M,µ) with a finite measure, a unitary operator U : H →L2 (M,dµ) (where dµ denotes the integration with respect to µ), and a real-valued function

f finite almost everywhere such that

(a) ψ ∈ D (A)⇔ f(Uψ) ∈ L2(M, dµ)

(b) If ϕ ∈ U [D (A)] then(UAU−1ϕ

)(m) = f(m)ϕ(m).

Example 9.1

Let A ∈ Cn×n be self-adjoint, ~v1, . . . , ~vn be a basis of Cn with A~vi = λ~vi for all i ∈ 1, . . . , n.Let U : Cn → L2(1, . . . , n, µ) with µ being a counting measure, so µ(B) = #B for all

B ⊆ 1, . . . , n, and [U(c1 ~v1 + · · ·+ cn ~vn)]︸︷︷︸=:ϕ

(j) = cj for all j ∈ 1, . . . , n.

Then UAU−1ϕ = UA(c1 ~v1 + · · ·+ cn ~vn) = U(c1λ1 ~v1 + · · ·+ cnλn ~vn).

So (UAU−1ϕ)(j) = cjλj = λjφ(j) = f(j)ϕ(j) where f : 1, . . . , n → R, f(j) = λ(j).

Example 9.2

• This is the same idea as what we did earlier: If A = −∆ ⇒ A = F−1|ξ|F with F being

the Fourier transformation.

• The conclusion of Theorem 9.1 can also be written as UAU−1 = Tf with Tf being the

multiplication operator of f .

• For matrices Theorem 9.1 ⇔ Matrix is diagonalizable.

Theorem 9.2 (Spectral theorem in functional calculus form)

Let A be a self-adjoint operator on a Hilbert space H. Then there exists a unique map φ from

the bounded Borel functions on R into L(H) with the following properties:

76 Chapter 9 Spectral theorem for self-adjoint operators

(a) φ(fg) = φ(f)φ(g), φ(λf) = λφ(f), φ(1) = I, φ(f) = φ(f)∗. That is φ is an algebraic ∗homomorphism.

(b) ‖φ(h))‖L(H) ≤ ‖h‖L∞ .

(c) Let hn(x) be a sequence of bounded Borel functions with hn(x)→ x pointwise and |hn(x)| ≤|x| for all x and n. Then for all ψ ∈ D(A) : φ(hn)ψ → Aψ.

(d) If hn(x)→ h(x) and ‖hn‖ is bounded, then φ(hn)ψ → φ(h)ψ for all ψ ∈ H.

Those four properties determine φ uniquely. Additionally we have:

(e) If Aψ = λψ then φ(h)ψ = h(λ)ψ.

(f) If h ≥ 0 then φ(h) ≥ 0, namely 〈ψ, φ(h)ψ〉 ≥ 0 for all ψ ∈ H.

Part of idea of proof: By Theorem 9.1 we have that UAU−1 = Tf ⇒ A = U−1TfU . Define

φ(h) = U−1ThfU . Then φ satisfies (a)− (f).

Step 1: Prove functional calculus form for continuous functions if A is bounded (first for polyno-

mials, then by approximation for continuous functions).

Step 2: Multiplication-operator form for bounded self-adjoint operators using step 1 (needs Zorn’s

Lemma).

Step 3: Prove Theorem 9.1 using boundedness of A(A− z)−1 for some z ∈ ρ(A).

Step 4: Theorem 9.1 ⇒ Theorem 9.2 as above.

Example 9.3

Let ht(x) = eitx. Then φ(ht) = eitA. This can be proven using our knowledge that eitA is a

strongly continuous unitary group as well as:

φ(ht)∗φ(ht)

Thm. 9.2= φ(ht)φ(ht)

Thm. 9.2= φ(htht) = φ(|ht|2)

Def of ht= φ(1) = I

Similarly we get φ(ht)φ(ht)∗ = I. Thus φ(ht) is unitary. Now prove with Theorem 9.2 that

φ(ht) is a strongly continuous unitary group. Uniqueness gives φ(ht) = eitA.

Theorem 9.3

Let Ω ⊆ R be Borel and P (Ω) := φ(χΩ). From Theorem 9.2 it follows that the family P (Ω)satisfies:

Chapter 9 Spectral theorem for self-adjoint operators 77

(i) (a) Each P (Ω) is an orthogonal projection.

(b) P (∅) = 0, P ((−∞,∞)) = I.

(c) If Ω =N⋃n=1

Ωn (N ∈ N ∪∞) with ∀n 6= m : Ωn ∩ Ωm = ∅ then P (Ω)ψ =N∑n=1

P (Ωn)ψ.

(d) P (Ω1)P (Ω2) = P (Ω1 ∩ Ω2).

Every such family is called a projection-valued measure.

(ii) For all Borel subsets Ω ⊆ R we have that P (Ω)A ⊆ AP (Ω). If Ω is open then

σ (A) ∩ Ω ⊆ σ(A|Ran(P (Ω))

)⊆ σ (A) ∩ Ω . (9.1)

Proof. (i) Proven in Exercise 40

(ii) As we need Theorem 9.9 this will be proven in Section 9.1.

Remark

If Ω is compact and σ (A)∩Ω is isolated from σ (A)∩Ωc then P (Ω) is just1

2πi

∮γ(z−A)−1 dz

where as in Theorem 7.4 γ is positively oriented and σ (A) ∩ Ω is inside γ while σ (A) ∩ Ωc is

outside γ.

Remark

If A is an observable and ψ is a state then 〈ψ, P (Ω)ψ〉 is the probability that the measurement

of A in the state ψ yields a value in Ω.

Theorem 9.4 (Application: the minimax principle)

Let H : D(H) ⊆ H → H be self-adjoint and bounded from below ( infϕ∈D(H),‖ϕ‖=1

〈ϕ,Hϕ〉 > −∞).

Let µn(H) := infM⊆D(H),dim M=n

( supϕ∈M,‖ϕ‖=1

〈ϕ,Hϕ〉).

Then for all n ∈ N we have either (exclusive):

a) There are at least n eigenvalues (counting multiplicity) below inf σess (H) and µn(H) is the

n-th eigenvalue from below.

b) µn(H) = inf σess (H) = µn+1(H) = µn+2(H) = . . . and there are at most (n− 1) eigenvalues

(counting multiplicity) below µn(H).


Here counting multiplicity means that for example if −3 is a double eigenvalue, −2 is a single

eigenvalue and 0 = inf σess (H), then µ1(H) = −3 = µ2(H), µ3(H) = −2 and 0 = µ4(H) =

µ5(H) = . . . .

Proof. If H has at least n eigenvalues below inf σess (H) let En be the n-th eigenvalue from below

(counting multiplicity), otherwise let En := inf σess (H). We have to show that µn(H) = En for

all n ∈ N.

”≤”: We show that µn(H) ≤ En + ε for all ε > 0.

Let P (Ω) := φ(χΩ) be the functional calculus of H : D(H) ⊆ H → H. If En is the n-th

eigenvalue then

σ(H|Ran(P ((−∞,En+ε)))

) eqn. (9.1)

⊇ σ (H) ∩ (−∞, En + ε) ⊇ E1, . . . , En (9.2)

If En = inf σess (H) then

En /∈ σ (H) ∩ (En + ε,∞) ⊇ σ(H|Ran(P ((En+ε,∞)))

)(9.3)

and so En ∈ σess

(H|Ran(P ((−∞,En+ε)))

)and thus dim Ran (P ((−∞, En + ε))) =∞. So in

any case we have dim Ran (P ((−∞, En + ε))) ≥ n.

If M ⊆ Ran (P ((−∞, En + ε))) has dimension n and is invariant under H then

µn(H)smaller subset

≤ supϕ∈M,‖ϕ‖=1

〈ϕ,Hϕ〉 = max eigenvalues of H|Meqn. (9.1)

≤ En + ε.

”≥”: By equation (9.1) we have that dim Ran (P ((−∞, En))) < n because

σ(H|Ran(P ((−∞,En)))

)⊆ σ (H) ∩ (−∞, En).

So if M has dimension n then there exists a ϕ ∈ M ∩ Ran (P ((−∞, En)))⊥. Therefore

supψ∈M,‖ψ‖=1

〈ψ,Hψ〉 ≥ 〈ϕ,Hϕ〉 ≥ En because ϕ is orthogonal to all eigenspaces to eigenvectors

below En.

An elementary proof in a special case can be constructed rather explicitly:


An elementary proof of Theorem 9.4 in a special case. Simplifying assumptions: We assume

n = 2 and that H has an orthonormal basis of eigenfunctions (vm)m∈N to eigenvalues (Em)m∈N

with E1 ≤ E2 ≤ . . . ≤ Em ≤ . . ., i.e. Hvm = Emvm. Thus we have to prove that

infM⊆D(H)dimM=2

supϕ∈M‖ϕ‖=1

〈ϕ,Hϕ〉 ≤ E2 (9.4)

and

infM⊆D(H)dimM=2

supϕ∈M‖ϕ‖=1

〈ϕ,Hϕ〉 ≥ E2 . (9.5)

To prove equation (9.4) let ϕ ∈ spanv1, v2 with ‖ϕ‖ = 1, i.e. ϕ = c1v1 + c2v2 with |c1|2 +

|c2|2 = 1. Thus 〈ϕ,Hϕ〉 = 〈c1v1 + c2v2, H (c1v1 + c2v2)〉 v1⊥v2= |c1|2 〈v1, Hv1〉+ |c2|2 〈v2, Hv2〉 =

|c1|2E1 + |c2|2E2 ≤ E2 since E1 ≤ E2 and |c1|2 + |c2|2 = 1. Hence supϕ∈M‖ϕ‖=1

〈ϕ,Hϕ〉 ≥ E2 and also

infM⊆D(H)dimM=2

supϕ∈M‖ϕ‖=1

〈ϕ,Hϕ〉 ≤ E2. Thus equation (9.4) holds.

Now we prove equation (9.5). Let M ⊆ D (H) with dimM = 2. Then M = spanu,w with

u,w orthonormal. We can write u as

u =

N1∑j=1

cnjvnj , n1 ≤ n2 ≤ . . . , nj 6= 0∀j,N1∑j=1

∣∣cnj ∣∣2 = 1 (9.6)

and w as

w =

N2∑i=1

dmivmi ,m1 ≤ m2 ≤ . . . ,mi 6= 0∀i,N2∑i=1

|dmi |2 = 1 , (9.7)

where N1, N2 ∈ N ∪ +∞. Thus

〈u,Hu〉 =

N1∑j=1

∣∣cnj ∣∣2Enj , (9.8)

since the vn form a basis of eigenfunctions for H, and similarly 〈w,Hw〉 =

N2∑i=1

|dmi |2Emi . Now

we have to distinguish between two cases:

(1) (i) n1 > 1: In this case 2 ≤ n1 ≤ n2 ≤ . . ., so 〈u,Hu〉 ≥ E2.

(ii) m1 > 1: In this case 2 ≤ m1 ≤ m2 ≤ . . ., so 〈w,Hw〉 ≥ Es.

(2) n1 = m1 = 1: In this case d1u − c1w ∈ M is nonzero (because u ⊥ w) and orthogonal to

v1. Sod1u− c1w

‖d1u− c1w‖can be written in the form of equation (9.6), where all nj > 1 . Thus,

arguing as for equation (9.8), we find

⟨d1u− c1w

‖d1u− c1w‖, H

d1u− c1w

‖d1u− c1w‖

⟩≥ E2.


Thus, in both cases supϕ∈M‖ϕ‖=1

〈ϕ,Hϕ〉 ≥ E2. Since M ⊆ D (H) was arbitrary we find

infM⊆D(H)dimM=2

supϕ∈M‖ϕ‖=1

〈ϕ,Hϕ〉 ≥ E2, so equation (9.5) holds as well.

There is another proof using the spectral theorem in projection-valued measure form in Sec-

tion 9.1.

Remark (for Theorem 9.4)

For example µ1(H) = infϕ∈D(H),‖ϕ‖=1

〈ϕ,Hϕ〉 = ground state energy of H as dim µ = 1 means that

M = span(ϕ).

If A is observable, φ is its functional calculus, ψ is a state and P (Ω) = φ(χΩ), then

〈ψ, P (Ω)ψ〉 = probability that a measurement of A is in Ω.

Corollary 9.5

µ1(H) = infϕ∈D(H),‖ϕ‖=1

〈ϕ,Hϕ〉 (ground state energy of H) is in the spectrum of H.

Proof. From Theorem 9.4 one of

a) µ1(H) < inf σess (H) and it is the lowest eigenvalue of H. Then µ1(H) ∈ σ (H)

b) µ1(H) = inf σess (H) so µ1(H) ∈ σ (H) because σ (H) is closed by Theorem 2.4.

has to hold.

Corollary 9.6 (i) If there exists a function ϕ ∈ D(H) with ‖ϕ‖ = 1 and 〈ϕ,Hϕ〉 < inf σess (H)

then H has a ground state.

(ii) If there exists a subspaceM ⊆ D(H) of dimensionN ∈ N∪∞ with 〈ψ,Hψ〉 < inf σess (H)

for all ψ ∈ M with ‖ψ‖ = 1 then there are at least N eigenvalues below inf σess (H)

(counting multiplicity).

Proof. (i) µ1(H) = inf‖φ‖=1φ∈D(H)

〈φ,Hφ〉 ≤ 〈ϕ,Hϕ〉 < inf σess (H). Thus for µ1(H) case b) of

Theorem 9.4 cannot happen and thus by a) µ1(H) is the lowest eigenvalue of H and it is

below inf σess (H).


(ii) If N ∈ N then µN (H) ≤ supφ∈M,‖φ‖=1

〈φ,Hφ〉 = maxφ∈M,‖φ‖=1

〈φ,Hφ〉 < inf σess (H) (dimM <∞).

So for µN (H) case b) is eliminated again, therefore a) holds and so there are at least N

eigenvalues below inf σess (H).

If N =∞ there are at least n eigenvalues below inf σess (H) for all n ∈ N. Therefore there

are infinitely many eigenvalues below inf σess (H).

Example 9.4

The operatorH : H2(R3)⊆ L2

(R3)→ L2

(R3), H = −∆− 1

|x| (hydrogen atom) has infinitely

many eigenvalues below inf σess (H).

Proof. We have σess (H) = [0,∞) so inf σess (H) = 0. Therefore by Corollary 9.6 it suffices to

find M ⊆ H2(R3)

with dimM =∞ and 〈ψ,Hψ〉 < 0 for all ψ ∈M \ 0.Let ψ ∈ C∞c

(R3)

with supp (ψ) ⊆ x ∈ R3 | 1 < |x| < 2. Define ψn(x) := 2−3n/2ψ(2−nx). To

show:

(i) supp (ψn) ∩ supp (ψm) = ∅ if n 6= m and ‖ψ‖ = 1 for all n ∈ N.

(ii) ∃N0 ∈ N : ∀n ≥ N0 : 〈ψn, Hψn〉 < 0

(iii) If M = span ψn | n ≥ N0 then dimM =∞ and 〈ϕ,Hϕ〉 < 0 for all ϕ ∈M \ 0.

(i) supp (ψm) ⊆ x ∈ R3 | 1 <∣∣2−nx∣∣ < 2 = x ∈ R3 | 2n < |x| < 2n+1.

Thus supp (ψn) ∩ supp (ψm) = ∅ for all m 6= n.

Moreover ψn = U2−nψ where Uλ as in Exercise 33. As Uλ is unitary ‖ψn‖ = 1.

(ii) We have

〈ψn, Hψn〉 = 〈ψn,−∆ψn〉 −⟨ψn,

1

|x|ψn⟩

= 〈U2−nψ,−∆U2−nψ〉 −⟨U2−nψ,

1

|x|U2−nψ

⟩U∗λ=U−1

λ =Uλ−1= 〈ψ,U2n(−∆)U2−nψ〉 −

⟨ψ,U2n

1

|x|U2−nψ

⟩Exercise 33

=1

4n〈ψ,−∆ψ〉 −

⟨ψ,

1

|2nx|ψ⟩

=1

2n

(1

2n〈ψ,−∆ψ〉 −

⟨ψ,

1

|x|ψ⟩)

< 0

for all n ≥ n0 for some n0 ∈ N since the first term vanishes as n→∞.


(iii) Let ϕ ∈M\0. Then ϕ = cn1ψn1+. . .+cnkψnk for some n1, . . . , nk ∈ N with cn1 , . . . , cnk 6=

0. Thus 〈ϕ,Hϕ〉 =

⟨n∑

m=1

cnmψnm , H

k∑l=1

cnlψnl

⟩=

k∑m=1

k∑l=1

cnmcnl 〈ψnm , Hψnl〉. Since

supp (ψi)∩supp (ψj) = ∅ for all i 6= j, the functions ψnm and ψnl are orthogonal if nm 6= nl,

so we have 〈ϕ,Hϕ〉 =k∑l=1

|cnl |2 〈ψnl , Hψnl〉 < 0 = inf σess (H) because of (ii). We have

dimM =∞ since M is the span of infinitely many orthonormal (thus linearly independent)

ψn. Thus, by Corollary 9.6 there are infinitely many eigenvalues below inf σess (H).

Definition 9.7

Let A : D(A) ⊆ H → H be self-adjoint with functional calculus φ from Theorem 9.2. The

map

P : B(R)→ L(H), P (Ω) := PΩ := φ(χΩ) = χΩ(A)

is called the spectral measure of A, where B(R) are the Borel subsets of R. Here

supp (P ) := x ∈ R | P (U) 6= 0 for all neighbourhoods U of x.

Theorem 9.8

Let A be self-adjoint with spectral measure P . Then

1.) S = supp (P ) is closed.

2.) P (S) = I, P (R \ S) = 0.

3.) ∀f ∈ B(R) ∩ C(R) : ‖f(A)‖ = ‖φ(F )‖ = supx∈S|f(x)|.

4.) ∀z ∈ ρ(A) :∥∥(z −A)−1

∥∥ = dist (z, σ (A))−1.

5.) S = σ (A).

Proof. 1.) Let x ∈ R \ S. Then by definition of S there exists a neighborhood Ux of x with

P (Ux) = 0. But then there exists an ε > 0 such that (x − ε, x + ε) ⊆ Ux. For all

y ∈ (x− ε, x+ ε) Ux is a neighborhood of y as well. Thus y ∈ R \ S and R \ S is open.


2.) We will prove that P (R \ S) = 0. Then by Theorem 9.3(i) we have P (S) = P (R) −P (R \ S) = I−0 = I. First we prove P (K) = 0 for all compact subsets K ⊆ R\S: for each

x ∈ K ⊆ R \ S there exists a neighborhood Ux of x such that P (Ux) = 0. But K ⊆⋃x∈K

Ux.

Since K is compact, there exist x1, . . . , xn ∈ K such that K ⊆ Ux1 ∪ Ux2 ∪ . . . ∪ Uxn . But

P(Uxj)

= 0 by 1.). Thereforen∑j=1

P(Uxj)

= 0. Since ∅ ⊆ K ⊆ Ux1 ∪ . . . ∪ Uxn we find

P (∅) ≤ P (K) ≤n∑j=1

P(Uxj)

= 0, so 0 ≤ P (K) ≤ 0. Hence P (K) = 0.

Let now Kn :=

x ∈ R \ S : dist (x, S) ≥ 1

n, |x| ≤ n

. Then Kn ⊆ R \ S is compact. Thus

P (Kn) = 0. But R \S =⋃n∈N

Kn because R \S is open, and since∑n∈N

P (Kn) = 0 we obtain

similarly as before that P (R \ S) = 0.

3.) Since φ(χS) = P (S) = I by Theorem 9.2(a) we have φ(χSf) = φ(χS)φ(f) = φ(f). Thus

‖φ(f)‖ = ‖φ(χSf)‖Thm. 9.2(b)

≤ ‖χSf‖L∞ = supx∈S|f(x)|. Now let x ∈ S. We will prove that

‖φ(f)‖ ≥ |f(x)|: if |f(x)| = 0 there is nothing to be shown since the norm is always non-

negative. So assume |f(x)| > 0. We will prove that ‖φ(f)‖ ≥ f(x)− ε for all ε ∈ (0, |f(x)|).Since f is continuous there exists a neighborhood Ux of x such that |f(y)− ε| > 0 for all

y ∈ Ux. But x ∈ S ⇔ P (Ux) 6= 0. So there exists a ψ ∈ Ran (P (Ux)) with ‖ψ‖ 6= 0.

Thus ‖φ(f)ψ‖ ψ∈Ran(P (Ux))= ‖φ(f)P (Ux)ψ‖‖φ(f)φ(χUx)ψ‖ Thm. 9.2(a)

= ‖φ(fχUx)ψ‖. Hence

‖φ(f)ψ‖2 = ‖φ(fχUx)ψ‖2 = 〈ψ, φ∗(fχUx)φ(fχUx)ψ〉 Thm. 9.2(a)=

⟨ψ, φ(|f |2χUx)ψ

⟩Thm. 9.2(f)

≥⟨ψ, φ

((|f(x)| − ε)2χUx

)ψ⟩ Thm. 9.2(a)

x fixed= (|f(x)| − ε)2 〈ψ, φ(χUx)ψ〉ψ∈Ran(P (Ux))

= (|f(x)| − ε)2 ‖ψ‖2. So ‖φ(f)‖ ≥ f(x)− ε.

4.)∥∥(z −A)−1

∥∥L(H)︸︷︷︸

=‖f(A)‖L(H)

3.)= sup

x∈S

∣∣(z − x)−1∣∣ =

1

infx∈S |z − x|=

1

dist (z, S).

5.) We have λ ∈ σ (A)Thm. 2.4⇔

∥∥∥∥∥(A− λ+

i

n

)−1∥∥∥∥∥ n→∞→ ∞ 4.)⇔ dist

(λ+

i

n, S

)n→∞→ 0

⇔ λ ∈ S S closed= S.

Corollary

(1) λ ∈ σ (A)⇔ λ ∈ S ⇔ P ((λ− ε, λ+ ε)) 6= 0 for all ε > 0.


(2) For all f ∈ B(R) ∩ C(R) we have ‖f(A)‖ = supx∈σ(A)

|f(x)|.

We now tend towards a third version of the spectral theorem. For this we need to introduce

some notations first.

For each ϕ ∈ H the map µ defined by µ(Ω) := 〈ϕ, P (Ω)ψ〉 is a well defined Borel measure,

namely µ(∅) = 0, for all Ω ⊆ R : µ(Ω) ≥ 0 and if Ωn ∩ Ωm = ∅ ∀n 6= m we have µ

(⋃n∈N

Ωn

)=∑

n∈Nµ(Ωn).

Moreover µ(B) = supµ(D) | D ⊆ B,D compact = infµ(O) | B ⊆ O,O open.Thus we can define integration with respect to the measure with the notation d 〈ϕ, P (λ)ϕ〉.

Bounded case: If g is bounded Borel we define

〈ϕ, g(A)ϕ〉 :=

∫ ∞−∞

g(λ) d 〈ϕ, P (λ)ϕ〉 ,

where ∫Ω

d 〈ϕ, P (λ)ϕ〉 = 〈ϕ, P (Ω)ϕ〉 .

Then the functional calculus g(A) satisfies the properties (a) − (d) of Theorem 9.2 and thus

g(A) = φ(g).

With Exercise 2 one can also define 〈ϕ, g(A)ψ〉, 〈ϕ, P (λ)ψ〉 and

〈ϕ, g(A)ψ〉 =

∫ ∞−∞

g(λ) d 〈ϕ, P (λ)ϕ〉 . (9.9)

Unbounded Borel case: If g is an unbounded Borel function and

Dg :=

ϕ ∈ H |

∫ ∞−∞|g(λ)|2 d 〈ϕ, P (λ)ϕ〉 <∞

where

∫ ∞−∞|g(λ)|2 d 〈ϕ, P (λ)ϕ〉 is supposed to be ‖g(A)ϕ‖2.

Then g(A) is defined on Dg as in equation (9.9).

Symbolically g(A) =

∫ ∞−∞

g(λ) dP (λ)supp(P )=σ(A)

=

∫σ(A)

g(λ) dP (λ).


Comparison to measure theory: If µ is a Lebesgue measure on R:

∫RχΩ dµ := µ(Ω),

∫R

n∑k=1

akχΩk dµ :=n∑k=1

akµ(Ωk)

Then we approximate general functions f with simple functions.

Here: ∫RχΩ dP (λ) := P (Ω) ,

∫R

n∑k=1

akχΩk dP (λ) :=n∑k=1

akP (Ωk)

If f is Borel and bounded it can be uniformly approximated by simple functions as well, so∫Rf(λ) dP (λ) can be defined.

Theorem 9.9 (Spectral theorem in projection-valued measure form)

There is a 1− 1 correspondence between self-adjoint operators A and projection-valued mea-

sures P with the correspondence given by

A =

∫ ∞−∞

λdP (λ) =

∫σ(A)

λdP (λ) (9.10)

If g is a real-valued Borel function on R then g(A) :=

∫g(λ) dP (λ) is self-adjoint on Dg.

If g is bounded then g(A) = φ(g) with φ being the functional calculus of A

If P has discrete support P (R) =∞∑n=1

P (λn) then the spectral theorem becomes:

A =

∫σ(A)

λ dP (λ) =∞∑n=1

λnP (λn)

where P (λn) is the projection onto the eigenspace to eigenvalue λn.

9.1 Proofs using spectral theorem in projection-valued measure form

Proof for Theorem 9.3 (ii). Let µ ∈ σ (A) ∩ Ω. Then by Exercise 42 we have for all n ∈ N :

P

((µ− 1

n, µ+

1

n)

)6= 0.


Thus there exists ϕn ∈ Ran

(P

((µ− 1

n, µ+

1

n)

))with ‖ϕn‖ = 1. Then

‖(A− µ)ϕn‖2 =⟨ϕ, (A− µ)2ϕn

⟩=

∫λ∈σ(A)

(λ− µ)2 d 〈ϕn, P (λ)ϕn〉

P ((µ−1/n,µ+1/n)c)ϕn=0=

∫λ∈σ(A)∩(µ−1/n,µ+1/n)

(λ− µ)2︸︷︷︸≤1/n2

d 〈ϕn, P (λ)ϕn〉

≤ 1

n2

∫λ∈σ(A)∩(µ−1/n,µ+1/n)

d 〈ϕn, P (λ)ϕn〉P (λ)≥0

≤ 1

n2

∫R

d 〈ϕn, P (λ)ϕn〉

=1

n2d 〈ϕn, P (R)ϕn〉 =

1

n2‖ϕn‖ =

1

n2

Therefore ‖(A− µ)ϕn‖2 ≤1

n2→ 0 and ‖ϕn‖ = 1. Also ϕn ∈ Ran

(P

((µ− 1

n, µ+

1

n)

))⊆

Ran (P (Ω)) =: HΩ for n large enough because µ ∈ Ω and Ω open. Therefore by Theorem 7.5

µ ∈ σ (A |HΩ).

Suppose that µ ∈ σ (A) ∩ Ω. Then dist(µ, σ (A) ∩ Ω

)=: δ > 0. Thus as before ∀f ∈

P (Ω)D (A) :

‖(A− µ)ϕ‖2 =

∫λ∈σ(A)∩Ω

(λ− µ)2︸︷︷︸≥δ2

d 〈ϕ, P (λ)ϕ〉 ≥ δ2

∫λ∈σ(A)∩Ω

d 〈ϕ, P (λ)ϕ〉

= δ2 〈ϕ, P (Ω)ϕ〉 = δ2‖ϕ‖2.

Therefore @ a set of approximate eigenfunctions to µ. So by Theorem 7.5 µ /∈ σ (A |HΩ).

Alternative proof of Theorem 9.4 using Theorem 9.9.

If H has at least n eigenvalues below inf σess (H) let En be the n-th eigenvalue from below

(counting multiplicity), otherwise let En := inf σess (H). We have to show that µn(H) = En for

all n ∈ N.

”≤”: We show that µn(H) ≤ En + ε for all ε > 0. If En = inf σess (H) then En ∈ σess (H)

(exercise) and thus by Exercise 42:

dim Ran (P ((−∞, En + ε))) =∞.

If En < inf σess (H) then we have E1 ≤ · · · ≤ En < inf σess (H) and let ~vi be orthonormal

with A~vi = Ei~vi, then ~v1, . . . , ~vn ∈ Ran (P ((−∞, En + ε))) because σ(A |(−∞,En+ε)c

)⊆

[En + ε,∞].

So in any case we have dim Ran (P ((−∞, En + ε))) ≥ n.


If M ⊆ Ran (P ((−∞, En + ε))) has dimension n and is invariant under H then

µn(H)smaller subset

≤ supϕ∈M,‖ϕ‖=1

〈ϕ,Hϕ〉 ≤ En + ε.

because

〈ϕ,Hϕ〉 =

∫ En+ε

−∞λ d 〈ϕ, P (λ)ϕ〉 ≤ (En + ε) 〈ϕ,ϕ〉 .

Thus ∀ε > 0 : µn(H) ≤ En + ε and so µn(H) ≤ En.

”≥”: By equation (9.1) we have that dim Ran (P ((−∞, En))) < n because

σ(H|Ran(P ((−∞,En)))

)⊆ σ (H) ∩ (−∞, En).

So if M has dimension n then there exists a ϕ ∈M ∩ Ran (P ((−∞, En)))⊥.

Thus

〈ϕ,Hϕ〉 ϕ∈P ([En,∞])=

∫ ∞En

λ d 〈ϕ, P (λ)ϕ〉 ≥ En∫ ∞En

d 〈ϕ, P (λ)ϕ〉 ‖ϕ‖=1= En.

So supϕ∈M,‖ϕ‖=1

〈ϕ,Hϕ〉 ≥ En and since M was arbirtary we have µn ≥ En.

10 Newton’s theorem and Zhislin’s theorem

Theorem 10.1 (Newton’s theorem)

We consider a spherically symmetric function ρ : R3 → R and assume that for all x ∈ R3

the function g(y) :=ρ(y)

|x− y| is in L1(R3). Then

∫R3

ρ(y)

|x− y| dy =

∫R3

ρ(y)

max |x|, |y| dy =

1

|x|

∫|y|≤|x|

ρ(y) dy +

∫|y|≥|x|

ρ(y)

|y| dy.

Proof. The second equality follows immediately after splitting the integral over R3 into an

integral where |y| ≤ |x| and an integral where |y| ≥ |x|. So it suffices to prove the first equality.

Since ρ(y) =: f(|y|) we have∫R3

ρ(y)

|x− y| dy =

∞∫0

f(r)

[∫∂B(0,r)

1

|x− y| dσ(y)

]dr

Exercise 36=

∞∫0

f(r)4πr2

maxr, |x| dr

=

∞∫0

f(r)

maxr, |x|

(∫∂B(0,r)

1 dσ(y)

)dr

=

∞∫0

∫∂B(0,r)

f(r)

maxr, |x| dσ(y) dr

=

∫R3

ρ(y)

max|x|, |y| dy .

Remark

In particular, if supp (ρ) ⊆ B (0, |x|), then we have

∫R3

ρ(y)

|x− y| dy =1

|x|

∫|y|≤|x|

ρ(y) dy.

Interpretation: Recall that1

|x| is the potential created by a charge +1 located at position

y = 0; mathematically1

|x| is (up to a constant − 1

4π) the fundamental solution (Green’s

function) of the equation ∆φ(x) = δ(x), where δ(x) is the Dirac delta (distribution). We

can use the Poisson equation to interpret φ(x) =q

|x− y| as the potential of a charge q at

90 Chapter 10 Newton’s theorem and Zhislin’s theorem

s0ρ

sx

s0, Q

sx

Figure 10.1: Outside of supp (ρ) (with spherically symmetric ρ) the potential is the same as the

potential of a single charge Q =

∫R3

ρ(y) dy located at 0; created with MGA-TeX

2.4 by K. Fritzsche [4].

position y, measured at position x. By the superposition principle, and taking the charge to

be a (smeared out) distribution, we can interpret

∫R3

ρ(y)

|x− y| dy as the potential of the charge

distribution measured at x.

On the other hand

∫R3

ρ(y) dy just gives the total charge Q and1

|x|

∫R3

ρ(y) dy is therefore

the potential of a charge Q located at 0.

Thus, outside of supp (ρ) it is irrelevant whether the charge is distributed or located at a

single point, as long as ρ is spherically symmetric. This holds not only for electric charges and

their potentials, but e.g. also for gravitational “charge distributions” (i.e. mass distributions)

and the gravitational potential. Assuming that the earth is spherically symmetric it creates a

potentialM

|x| outside of its support, where M is the total mass of the earth.

Theorem 10.2 (Zhislin’s theorem)

We consider the operator HN,Z : H2(R3N

)⊆ L2

(R3N

)→ L2

(R3N

),

(HN,ZΦ)(x) =

N∑j=1

(−∆xj −

Z

|xj |

)Φ(x) +

∑1≤i<j≤N

1

|xi − xj |Φ(x), x = (x1, . . . , xN ) ∈ R3N .

If Z > N − 1 then HN,Z has infinitely many eigenvalues below σess (HN,Z).

Proof. We will prove the statement for N < Z + 1 by induction on N .

N = 1 : proved in Example 9.4 for Z = 1 and for Z > 0 the proof remains the same

Chapter 10 Newton’s theorem and Zhislin’s theorem 91

Induction step: If the statement holds for N − 1 then HN−1,Z has infinitely many eigenvalues

below inf σess (HN−1,Z). In particular it has a ground state

ϕN−1 ∈ L2(R3(N−1)

)with ‖ϕN−1‖ = 1 (10.1)

Let ψ ∈ C∞c(R3), ‖ψ‖ = 1, supp (ψ) ⊆ x ∈ R3 | 1 < |x| < 2 and ψ spherically symmetric.

Let ψn = U2−nψ where Uλψ = λ3/2ψ(λx). Then like in Example 9.4 we have

‖ψn‖ = 1 (10.2)

and supp (ψn) ∩ supp (ψm) = ∅ if m 6= n. Thus by Corollary 9.6 it is enough to prove that for

infinitely many n:

〈ϕN−1 ⊗ ψn, HN,ZϕN−1 ⊗ ψn〉 < 〈ϕN−1, HN−1,ZϕN−1〉︸︷︷︸=:a

= EN−1,Z = inf σ (HN−1,z)

HVZ= inf σess (HN,Z)

(10.3)

where ϕN−1 ⊗ ψn(x) = ϕN−1(x1, . . . , xN−1)ψn(xN ).

Since

HN,Z = HN−1,Z −∆xN −Z

|xN |+N−1∑i=1

1

|xi − xN |(10.4)

we have

〈ϕN−1 ⊗ ψn, HN,ZϕN−1 ⊗ ψn〉

eqn. (10.4)=

⟨ϕN−1 ⊗ ψn,

(HN−1,Z −∆xN −

Z

|xN |+N−1∑i=1

1

|xi − xN |

)(ϕN−1 ⊗ ψn)

⟩

= 〈ϕN−1 ⊗ ψn, HN−1,Z(ϕN−1 ⊗ ψn)〉+

⟨ϕN−1 ⊗ ψn,

(−∆xN −

Z

|xN |

)(ϕN−1 ⊗ ψn)

⟩+N−1∑i=1

⟨ϕN−1 ⊗ ψn,

1

|xi − xN |(ϕN−1 ⊗ ψn)

⟩.

Because HN−1,Z only acts on x1, . . . , xN−1 and equation (10.2) we have:

〈ϕN−1 ⊗ ψn, HN−1,Z(ϕN−1 ⊗ ψn)〉 = 〈ϕN−1, HN−1,ZϕN−1〉 = a.

Because ∆xN −Z

|xN |only acts on xN and equation (10.1) we have

⟨ϕN−1 ⊗ ψn,

(−∆xN −

Z

|xN |

)(ϕN−1 ⊗ ψn)

⟩=

⟨ψn,

(−∆xN −

Z

|xN |

)ψn

⟩.

92 Chapter 10 Newton’s theorem and Zhislin’s theorem

Last but not least we have with Fubini:

N−1∑i=1

⟨ϕN−1 ⊗ ψn,

1

|xi − xN |(ϕN−1 ⊗ ψn)

⟩

=N−1∑i=1

∫|ϕN−1(x1, . . . , xN−1)|2

(∫ |ψn(xN )|2|xi − xN |

dxN

)dx1 . . . dxN−1

Thm. 10.1≤

ψ symmetric

N−1∑i=1

∫|ϕN−1(x1, . . . , xN−1)|2

(∫ |ψn(xN )|2|xN |

dxN

)dx1 . . . dxN−1

So putting those three equations together and using equation (10.1) again we get:

〈ϕN−1 ⊗ ψn, HN,ZϕN−1 ⊗ ψn〉 ≤ a+

⟨ψn,

(−∆xN −

Z − (N − 1)

|xN |

)ψn

⟩= a+ 〈ψn,−∆xNψn〉 − (Z − (N − 1))

⟨ψn,

1

|xN |ψn

⟩< a

for all n ≥ n0 for some n0 for the same reason as in Example 9.4, concluding the proof.

Remark

Theorem 10.2 shows that it costs energy to send an electron at infinity, so the ion with a

nucleus with Z protons and N electrons exists as it is stable.

In the physical case we have U ∈ N and so Z > N − 1⇔ Z ≥ N which means that atoms and

positive ions exist.

Open problems:

1. Do negative ions exist? So is inf σ (HN+1,N ) < inf σ (HN,N )?

2. Does inf σ (HN,Z) < inf σ (HN−1,Z) imply inf σ (HN−1,Z) < inf σ (HN−2,Z)? Or in other

words, if an ion with N electrons exist, does the ion with N − 1 electrons exist? (Only for

negative ions, for positive ions see Theorem 10.2).

11 Pure point spectrum, continuous spectrum,

RAGE theorem

Definition 11.1

Let H : D (H) ⊂ H → H be a self-adjoint operator. We define

Hpp := spanψ ∈ Hψ : ψ eigenfunction of H and Hc := H⊥pp.

Then σpp (H) := λ : λ is eigenvalue of H and we have σpp (H) = σ(H|Hpp

)is called the

pure point spectrum of H and σc (H) := σ(H|Hc

)is called the continuous spectrum of H.

Remark

It can occur that Hpp = ∅, e.g. the Laplace operator −∆ does not have eigenfunctions in L2.

To see this, assume that −∆u = λu for some function u 6= 0 and λ ∈ R, i.e. that there is

an eigenfunction. Then by Fourier transformation(|ξ|2 − λ

)u(ξ) = 0 and hence u(ξ) = 0 if

|ξ|2 6= λ. So u = 0 almost everywhere, and by Fourier transforming again, we have that u = 0

almost everywhere, so u is not an eigenfunction of −∆.

Remark 11.2

We have σ (H) = σpp (H) ∪ σc (H); this union is not necessarily disjoint. If Hϕ = Eϕ then

〈ϕ, P (Ω)ϕ〉 =

1, E ∈ Ω

0, E /∈ Ω, so 〈ϕ, P (Ω)ϕ〉 is a Dirac measure supported in one point. If

ϕ ∈ Hpp then Ω 7→ 〈ϕ, P (Ω)ϕ〉 is a measure with discrete support (pure point measure). If

ϕ ∈ Hc then Ω 7→ 〈ϕ, P (Ω)ϕ〉 is a continuous measure, i.e. 〈ϕ, P (λ)ϕ〉 = 0 for all λ ∈ R.

Theorem 11.3 (RAGE-Theorem: Ruelle, Amrein, Georgescu, Enss)

Let H be a self-adjoint operator on L2(Rd)

. Assume that χB(0,R)(H + i)−1 is compact for

all R > 0. Then

(i)

ϕ ∈ Hpp ⇔ limR→∞

supt≥0

∥∥(1− χB(0,R)

)e−itHϕ

∥∥ = 0 . (11.1)

94 Chapter 11 Pure point spectrum, continuous spectrum, RAGE theorem

(ii)

ϕ ∈ Hc ⇔ limT→∞

1

T

T∫0

∥∥χB(0,R)e−itHϕ

∥∥2dt = 0 for all R > 0 . (11.2)

Remark

If e.g. id

dtϕt = Hϕt

ϕt|t=0 = ϕ

, (11.3)

then ϕt = e−iHtϕ is the solution of the Schrodinger equation. If ϕt describes the evolution

of a particle, then equation (11.1) means that the particle remains localized. In the case of

equation (11.2), the particle escapes the ball B (0, R) for all R > 0 and comes back to it only

rarely in time (on average it is outside the ball).

Before we prove Theorem 11.3, we discuss two prominent examples.

Example

(1) If H = −∆ then χR(H+i)−1 is compact for all R > 0 since (H+i)−1 : L2(Rd)→ H1

(Rd)

is bounded and χR : H1(Rd)→ L2

(Rd)

(acting as a multiplication operator) is compact

by Theorem 6.10. Then H = Hc (there is no pure point spectrum, as explained in the

remark above). Therefore equation (11.2) holds for all ϕ ∈ H = Hc.

(2) Let H = −∆ +V , where V is a real-valued function such that V ψ ∈ L2 and such that there

exist c > 0 and a ∈ (0, 1) with ‖V ψ‖ ≤ a‖−∆ψ‖+ c‖ψ‖ (in some sense, V is considered as a

small perturbation to the free Hamilton operator H = −∆). Then V is called −∆-bounded

with relative bound a. By Kato’s theorem (Theorem 3.8) H is self-adjoint. Moreover

χR(H+i)−1 is compact for all R > 0 since χR(H+i)−1 = χR(−∆ + i)−1︸︷︷︸compact by Thm. 6.10

(−∆+i)(H+i)−1,

so it is enough to show that (−∆ + i)(H + i)−1 is bounded. We have (−∆ + i)(H +

i)−1 = (−∆ + i)(H + µi)−1 (H + µi)(H + i)−1︸︷︷︸=(H+i)(H+i)−1+(µ−1)i(H+i)−1⇒bounded

, so suffices to show that

(−∆ + i)(H +µi)−1 is bounded. Indeed, (−∆ + i)(H +µi)−1 = (−∆ + i)(−∆ +V +µi)−1 =

(−∆ + i)[(I + V (−∆ + µi)−1

)(−∆ + µi)

]−1. Since

∥∥V (−∆ + µi)−1)∥∥ < 1 for |µ| large

enough (see the proof of Theorem 3.8),(I + V (−∆ + µi)−1

)is invertible by Exercise 7

Chapter 11 Pure point spectrum, continuous spectrum, RAGE theorem 95

and we obtain (−∆ + i)(H + µi)−1 = (−∆ + i)(−∆ + µi)−1[I + V + (−∆ + µi)−1

]−1︸︷︷︸bounded

, so

(−∆ + i)(H + µi)−1 is bounded.

Lemma 11.4

Let µ be a complex-valued finite measure on R. We define µ(t) :=

∫R

e−itλ dµ(λ). Then

limT→∞

1

T

∫ T

0

∣∣∣ ˆµ(t)∣∣∣2 dt =

∑y∈R|µ (y)|2. In particular, if µ is continuous (i.e. µ (y) = 0 for all

y ∈ R), then limT→∞

1

T

∫ T

0|µ(t)|2 dt = 0.

Proof. |µ(t)|2 = µ(t)µ(t) =

∫R

e−itx dµ(x)

∫R

eitydµ(y). Applying Fubini’s theorem,

1

T

∫ T

0|µ(t)|2 dt =

∫R

∫R

(1

T

∫ T

0eit(y−x) dt

)dµ(x) dµ(y). But

∣∣∣∣ 1

T

∫ T

0eit(y−x) dt

∣∣∣∣ ≤ 1 for all

T ∈ R and limT→∞

1

T

∫ T

0eit(y−x) dt = χy(x) =

1, x = y

0, x 6= y. Using

∣∣∣∣ 1

T

∫ T

0eit(y−x) dt

∣∣∣∣ ≤ 1 and

the fact that the measure is finite, we can apply the theorem of dominated convergence to obtain

limT→∞

1

T

∫ T

0|µ(t)|2 dt =

∫R

∫Rχy(x) dµ(x)︸︷︷︸

=µ(y)

dµ(y) =

∫Rµ (y) dµ(y) =

∑y∈R|µ (y)|2.

Lemma 11.5

Let H : D (H) ⊆ H → H be a self-adjoint operator. If ϕ ∈ H and ψ ∈ Hc then

limT→∞

1

T

∫ T

0

∣∣⟨ϕ, e−iHtψ⟩∣∣2 dt = 0.

Proof. We may assume without loss of generality that ϕ ∈ Hc as well1. Since ϕ,ψ ∈ Hcthe measure µ : B → C, µ(Ω) := 〈ϕ, P (Ω)ψ〉 is (finite, complex and) continuous (because

of Exercise 2 and since 〈ϕ, P (Ω)ϕ〉 is continuous for all ϕ ∈ Hc). Thus⟨ϕ, e−iHtψ

⟩ Thm. 9.9=⟨

ϕ,

∫e−iλt dP (Ω)ψ

⟩=

∫e−iλt d 〈ϕ, P (Ω)ψ〉 =

∫e−iλt dµ(λ) = µ(t). Hence

limT→∞

1

T

∫ T

0

∣∣⟨ϕ, e−iHtψ⟩∣∣2 dt = lim

T→∞

1

T

∫ T

0|µ(t)|2 dt

Lemma 11.4= 0, since µ is continuous.

We are now finally ready to prove Theorem 11.3.

1If g is an eigenfunction of H (Hg = Eg for some E ∈ R, g 6= 0), then⟨g, e−iHtψ

⟩=⟨

eiHtg, ψ⟩

= eiEt 〈g, ψ〉 = 0,

because g ∈ Hpp ⊥ Hc 3 ψ. Thus if ϕ = ϕpp+ϕc with ϕpp ∈ Hpp, ϕc ∈ Hc, then⟨ϕ, e−iHtψ

⟩=⟨ϕc, e

−iHtψ⟩

.

96 Chapter 11 Pure point spectrum, continuous spectrum, RAGE theorem

Proof of Theorem 11.3. Let Hbound :=

ψ ∈ L2

(Rd)

: limR→∞

supt≥0

∥∥(1− χR) e−iHtψ∥∥ = 0

and Hleaving :=

ψ ∈ L2

(Rd)

: limT→∞

1

T

∫ T

0

∥∥χRe−itHψ∥∥2

dt = 0 for all R > 0

. By the triangle

inequality, Hbound and Hleaving are subspaces of H = L2(Rd)

. The proof will proceed in four

steps:

1.) Hbound is closed

2.) Hpp ⊆ Hbound

3.) Hbound ⊥ Hleaving

4.) Hc ⊆ Hleaving

The other direction is simple: to show thatHbound ⊆ Hpp, assume that ϕ ∈ Hbound, but ϕ /∈ Hpp.

Then ϕ ∈ (Hpp)c = Hc ⊆ Hleaving, so ϕ ∈ Hleaving. But Hleaving ⊥ Hbound, so ϕ ∈ Hbound and

ϕ /∈ Hbound, a contradiction. So if ϕ ∈ Hbound then automatically ϕ ∈ Hpp, i.e. Hbound ⊆ Hpp.

In the same manner one shows that Hleaving ⊆ Hc, i.e. we have Hbound = Hpp and Hleaving = Hcwhich then concludes the proof of Theorem 11.3.

1.) Let ψ ∈ Hbound and ε > 0. Then there exists ψε ∈ Hbound : ‖ψ − ψε‖ <ε

2. As ψε ∈ Hbound

there exists R0 > 0 such that

supt∈R

∥∥(1− χR)e−itHψε∥∥ < ε

2(11.4)

for all R > R0. Therefore we have for all R > R0:

supt≥0

∥∥(1− χR)e−itHψ∥∥ ∆−ineq.≤ sup

t≥0

∥∥(1− χR)e−itHψε∥∥+ sup

t≥0

∥∥(1− χR)e−itH(ψ − ψε)∥∥

eqn. (11.4)

≤ ε

2+ sup

t≥0

∥∥e−itH(ψ − ψε)∥∥ =

ε

2+ sup

t≥0‖(ψ − ψε)‖ <

ε

2+ε

2= ε

2.) Since by step 1 Hbound is a closed subspace it suffices to prove Hψ = λψ ⇒ ψ ∈ Hbound.

Hψ = λψThm. 9.2 f)

= eitHψ = eiλtψ. Thus

limR→∞

supt≥0

∥∥(1− χR)e−itHψ∥∥ = lim

R→∞supt≥0

∥∥∥(1− χR)e−itλψ∥∥∥ ≤ lim

R→∞‖(1− χR)ψ‖ = 0

3.) Let ϕ ∈ Hbound, ψ ∈ Hleaving, ‖ϕ‖ = ‖ψ‖ = 1. To show: ∀ε > 0 : |〈ϕ,ψ〉| < ε.

〈ϕ,ψ〉 unitary=

⟨e−itλϕ, e−itλψ

⟩=⟨

(1− χR)e−itλϕ, e−itλψ⟩

+⟨e−itλϕ, χRe

−itλψ⟩.

Chapter 11 Pure point spectrum, continuous spectrum, RAGE theorem 97

Thus for all R and all t ≥ 0:

|〈ϕ,ψ〉|CS≤∥∥(1− χR)e−itHϕ

∥∥+∥∥χRe−itHψ

∥∥⇒ |〈ϕ,ψ〉|2

CS≤ 2 ·

∥∥(1− χR)e−itHϕ∥∥2

+ 2 ·∥∥χRe−itHψ

∥∥2.

We therefore have

1

T

∫ T

0|〈ϕ,ψ〉|2 dt ≤ sup

t≥02 ·∥∥(1− χR)e−itHϕ

∥∥2+ 2 · 1

T

∫ T

0


dt < ε

because by the definition of Hpp the first addend is less thanε

2for some R = R0 > 0 and

for the same R0 there exists a T0 > 0 such that the second addend is also less thanε

2.

4.) By Lemma 11.5 and the Riesz representation theorem we have for all f : H → C linear and

bounded:

limT→∞

1

T

∫ T

0

∥∥f(e−itHψ)∥∥2

dt = 0 (11.5)

It follows that for all linear, bounded and finite rank g : H → H:

limT→∞

1

T

∫ T

0

∥∥g(e−itHψ)∥∥2

dt = 0 (11.6)

because g(h) = g1(h)h1 + · · ·+ gn(h)hn where each gi : H → C is linear and bounded. Then

we use equation (11.5).

Since compact operators K : H → H can be approximated with finite rank operators

(without proof) it follows from equation (11.6) that for all compact oeprators K:

limT→∞

1

T

∫ T

0

∥∥K(e−itHψ)∥∥2

dt = 0 (11.7)

Let ψ ∈ D (H) ∩ Hc. Then there exists ϕ ∈ H : ψ = (H + i)−1ϕ because i ∈ ρ(H) (H is

self-adjoint).

Then ϕ ∈ Hc because if Hu = Eu we have:

0 = 〈ψ, u〉 =⟨(H + i)−1ϕ, u

⟩=⟨ϕ, (H − i)−1u

⟩= (λ− i)−1 〈ϕ, u〉

⇒ 〈ϕ, u〉 = 0

Thus for all R > 0:

limT→∞

1

T

∫ T

0


dt = limT→∞

1

T

∫ T

0

∥∥χRe−itH(H + i)−1ϕ∥∥2

dt

= limT→∞

1

T

∫ T

0

∥∥χR(H + i)−1e−itHϕ∥∥2

dteqn. (11.7)→ 0

As D (H) ∩Hc is dense in Hc we thus finished our proof.

12 Exercises

Here the statements from the exercise session used in the lecture will be restated.

Exercise 2

Let A : D(A) ⊂ X → X be a linear operator such that 〈ψ,Aψ〉 = 〈Aψ,ψ〉 for all ψ ∈ D(X).

Then

〈ϕ,Aψ〉 = 〈Aϕ,ψ〉

for all ϕ,ψ ∈ D(X).

Hint : Use the polarization identity.

Exercise 4

Let X be a Banach space. Let A : D(A) ⊆ X → X be a closed linear operator and z, w ∈ ρ(A).

Then

1. RA(z)−RA(w) = (w − z)RA(z)RA(w).

2. RA(z)RA(w) = RA(w)RA(z).

3. RA(z)A ⊆ ARA(z) = zRA(z)− I.

Exercise 5

1. Let (X, ‖ · ‖) be a Banach space, let A : D(A) ⊆ X → X be a closed operator and

g : [0, T ]→ (D(A), ‖ · ‖A) be continuous where ‖ · ‖A is a graph norm. Then

A

∫ T

0g(t)dt =

∫ T

0Ag(t)dt .

Hint : Integrals can be understood as Riemannian integrals.

2. Let g : [0, T ]→ H2(R3) be continuous. Then

−∆x

∫ T

0g(t)dt =

∫ T

0−∆xg(t)dt .

100 Chapter 12 Exercises

Exercise 7

Show the following: Let X be a Banach space, K : X → X be linear operator s.t. ‖K‖ < 1.

Then I +K is invertible and (I +K)−1 =

∞∑n=0

(−1)nKn.

Exercise 8

Define X := (ϕ ∈ L2(R)|ϕ ∈ C1(R),∇ϕ ∈ L2(R), ‖ · ‖2 + ‖∇ · ‖2). Show that

fn = exp

[−(

1

n+ |x|2

) 12

].

is a Cauchy sequence in X but fn does not converge in X. Furthermore show that fn → f in

H1(R).

Exercise 9

Let V ∈ C(Rn,C). Then we define MV : D(MV ) ⊆ L2(Rn)→ L2(Rn) as

MV f(x) = V (x)f(x)

where D(MV ) := f ∈ L2(Rn) | V × f ∈ L2(Rn). Show that

Ran(V ) ⊆ σ(MV ) .

Exercise 10

Let A,B be densely defined linear operators on a Hilbert space H. Show the following

A ⊂ B ⇒ B∗ ⊂ A∗ .

Exercise 11

Let H = L2(R) and g ∈ H with ‖g‖ = 1. Let A : Cc(R) ⊂ H → H, Af = f(0)g. Show that

〈ϕ, g〉 6= 0, ϕ ∈ H ⇒ ϕ /∈ D(A∗) .

Exercise 12

Let V and T be densely defined linear operators on a Hilbert space H with D(T ) ⊂ D(V ).

Consider two assumptions:

1. For some a, b ∈ R and all ψ ∈ D(T )

‖V ψ‖ ≤ a‖Tψ‖+ b‖ψ‖ . (12.1)

Chapter 12 Exercises 101

2. For some a, b ∈ R and all ψ ∈ D(T )

‖V ψ‖2 ≤ a2‖Tψ‖2 + b2‖ψ‖2 . (12.2)

Show the following statements:

1. If (12.2) holds, then (12.1) holds with a = a and b = b.

2. If (12.1) holds, then (12.2) holds for a2 = (1 + ε)a2 and b2 = (1 + ε−1)b2 for each ε > 0.

Thus the infimum of all a in (12.1) is the same as the infimum of all a in (12.1).

Exercise 13

Consider a Hamiltonian describing n electron atom in Born-Oppenheimer approximation. Let

H = H2(R3n),

T =

n∑j=1

−∆j , D(T ) = H

where −∆j is a kinetic energy operator of j-th electron and V is a multiplication operator

defined using the function

v(x) = −n∑j=1

Z

|xj |+

n∑k>j=1

1

|xj − xk|

where |xj | is a coordinate of the j-th electron.

Show the following:

1. V is bounded w.r.t. T with a < 1, i.e. there exists 0 ≤ a < 1 and b ∈ R+0 s.t. (12.1) holds

2. T + V is lower semibounded, i.e. there exists a ∈ R s.t. ∀ϕ ∈ H: 〈ϕ, (T + V )ϕ〉 ≥ a‖ϕ‖2,

3. T + V is self-adjoint.

Exercise 14

Let H10 (R+) = C∞c (R+). Then we define “momentum operator” P : H1

0 (R+)→ L2(R+) as

Pf(x) = −i d

dxf(x) .

Show the following:

1. P is symmetric and P ∗ is not symmetric,

2. P is closed and

102 Chapter 12 Exercises

3. P is not essentially self-adjoint.

Exercise 17

Let ψ ∈ H1(R3). Prove that ∫R3

|ψ(x)|2|x| dx ≤ ‖∇ψ‖2‖ψ‖2 .

Equality holds only if

ψ(x) ∼ e−c|x|

for some constant c > 0.

Hint : Use that1

|x| =1

2∇(x

|x|

).

Exercise 23

Let H 6= 0 be a Hilbert space and A ∈ L(H). Show that the spectrum of A is not empty.

Exercise 24

Let H be a Hilbert space and A ∈ L(H). Assume that for all |s| < r the function f(s) :=

(1− sA)−1 exists and is bounded. Prove that f(s) has a series expansion in s and that for all

ε > 0 the series converges uniformly in |s| < r − ε.

Exercise 30

Let ψ ∈ H1(R3). Show that

limh→∞

∥∥∥∥ψ(x− h)

x

∥∥∥∥ = 0 .

Exercise 31

Let f ∈ ψ ∈ C(Rd)| lim|x|→∞

ψ = 0 and Mf : H1(Rd)→ L2(Rd),Mfϕ := fϕ. Show that Mf is

compact.

Exercise 33

Let λ > 0 then the operator Uλ : L2(Rn)→ L2(Rn) is defined as

[Uλφ](x) := λn/2φ(λx) .

Show that:

1. Uλ is unitary, U−1λ = Uλ−1 and Uλ leaves Hs(Rn) invariant,

Chapter 12 Exercises 103

2. On S(Rn) the following holds

Uλ∆U−1λ =

1

λ2∆ , UλV (x)U−1

λ = V (λx) ,

with V being multiplication operator satisfying V S(Rn) ⊂ L2(Rn).

Exercise 36

Prove that the following equality holds∫|x|=r

dσ(x)

|x−R| =4πr2

maxr, |R|

where dσ(x) is a surface measure on a sphere and R ∈ R3.

This is known as Newton’s Shell Theorem.

Exercise 40

Let A be a self-adjoint operator on H. Let PΩ := χΩ(A) where χΩ is a characteristic function

of the measurable set Ω ⊂ R. Show that the family of operators PΩ has the following

properties:

1. Each PΩ is an orthogonal projection.

2. P∅ = 0, PR = I.

3. Let N ∈ N ∪ 0. If Ω =N⋃n=1

Ωn with Ωn ∩ Ωm = ∅ if n 6= m, then PΩψ =N∑n=1

PΩnψ,

∀ψ ∈ H.

4. PΩ1PΩ2 = PΩ1∩Ω2 .

Exercise 42

Let A be a self-adjoint operator and let PΩ := χΩ(A). Prove the following

1. λ ∈ σ(A) if and only if for all ε > 0 : P(λ−ε,λ+ε) 6= 0 .

2. λ ∈ σd(A) if and only if λ ∈ σ(A) and there exists ε > 0 s.t. P(λ−ε,λ+ε)H <∞.

3. λ ∈ σess(A) if and only if dim(P(λ−ε,λ+ε)H) =∞ for all ε > 0.

Bibliography

[1] Gerald B. Folland. Real analysis : modern techniques and their applications. A Wiley-

Interscience publication. Wiley, New York [i.a.], 1984.

[2] Michael Reed and Barry Simon. Methods of modern mathematical physics, volume 1: Func-

tional analysis. Academic Press, San Diego [a.o.], rev. and enlarged ed. edition, 2005.

[3] P. Ehrenfest. Bemerkung uber die angenaherte Gultigkeit der klassischen Mechanik innerhalb

der Quantenmechanik. Zeitschrift fur Physik, 45(7):455–457, Jul 1927.

[4] K. Fritzsche. MGA-TeX. http://www2.math.uni-wuppertal.de/~fritzsch/kf_latx.

html, 2013. Accessed: 2020-01-24.

http://www2.math.uni-wuppertal.de/~fritzsch/kf_latx.html

http://www2.math.uni-wuppertal.de/~fritzsch/kf_latx.html

Index

−∆-bounded function, 94

A∗, 23

Bn, 36

E0, 46

EN , 67

H10 (R+), 34

HN , 67

Hn, 36

Lp(X), 13

P , 55

RA, 19

Tφ(ψ), 23

U(t), 34

C+, 25

C−, 25

C∞c (Rn), 13

C∞ (Rn), 13

∆Aψ, 43

ΓA, graph, 18

Hbound, 96

Hc, 93

Hk (Rn), 14

Hleaving, 96

Hpp, 93

Ker (A), 17

L2(R3), 9

L1loc (Rn), 13

P (Ω), 76

Ran (A), 17

〈ψ,ϕ〉, 9

Σ, 61, 71

ΣR, 61, 71

B (z0, R), 19

χA(x), 64

d 〈ϕ, P (λ)ϕ〉, 84

δij , 44

dist (z0, A), 20

µ(t), 95

(Ja)a∈A : Rn → R, 66

B(R), 82

F , F−1, 21

L(X), 19

L(H), 34

ess sup(V ), 21

µ(Ω), 84

‖·‖Hk(Rn), 14

‖·‖L2(Rn), 14

∂α, 13

ψ0, 46

ρ (A), 19

σ′, 54

σ′′, 54

108 Chapter Index

σc (H), 93

σd (A), 61

σess (A), 61

σpp (H), 93

σ (A), 19

supp (P ), 82

supp (φ), 13

χA(x), 64

k-th ionization energy, 71

rk, 71

H, 23

adjoint, 23

adjoint operator, 23

atom, N electrons, 30

Banach space, 10, 14

Banach space, complex, 10

basic criterion of self-adjointness, 26

bounded linear operator, 17

Cauchy integral formula, 53

Cauchy’s theorem, 53

closable linear operator, 21

closed linear operator, 18

commutator, 43

compact operator, 50

continuous spectrum, 93

densely defined linear operator, 17

discrete spectrum, 61

dominated convergence, 27

double-slit experiment, 7

Ehrenfest theorem, 44

energy operator, 43

essential spectrum, 61

essentially self-adjoint linear operator, 30

evolution (of a state), 9

expectation, 43

expectation value, 43

extension, 21

Fatou, lemma of, 28

first ionization energy, 71

Fourier transformation, 21

graph, 18

graph norm, 18

Green’s function, 89

ground state, 46, 47

ground state energy, 46

Hamilton operator, 46

Hamiltonian, 46

Hardy inequality, 28

Heisenberg picture, 45

Hilbert space, 10

Hilbert space, complex, 10

HVZ theorem, 67

hydrogen atom, 29, 47

image, 17

IMS localization formula, 66

initial-value problem, 10, 33

inner product, 9, 10

interference, 7

ion, N electrons, nucleus of charge Z, 46

ionization energy, 61

ionization threshold, 61, 71

kernel, 17

Kronecker symbol, 44

Lebesgue measure, 21

Chapter Index 109

lemma of Fatou, 28

linear operator, 17

linear operator, image, 17

linear operator, kernel, 17

momentum operator, 43

monotonic convergence, 27

Newton’s theorem, 89

norm, 10

norm, graph, 18

observable, 43

ONB, 49

orthogonal projection, 55

orthonormal basis, 49

Parseval’s identity, 49

partial differential equation, 9

Plancherel’s identity, 43

Poisson equation, 89

position operator, 43

positively oriented curve, 53

probability distribution, 9

projection-valued measure, 77

pure point measure, 93

pure point spectrum, 93

range, 17

range (of a linear operator), 17

resolvent, 19

resolvent set, 19

Riesz representation theorem, 23, 49, 97

Schrodinger equation, 9, 33

Schrodinger picture, 45

self-adjoint, 24

self-adjoint linear operator, 24

simply closed curve, 53

Sobolev space, 14

spectral measure, 82

spectral radius, 53

spectral theorem, 75

spectrum, 19

state vector, 8, 43

strongly continuous unitary group, 35

strongly continuous unitary group, genera-

tor, 39

superposition principle, 9

symmetric, 24

symmetric linear operator, 24

test function, 13

theorem of Banach-Alaoglu, 50

theorem of dominated convergence, 27

theorem of Kato-Rellich, 27, 31

theorem of monotonic convergence, 27

theorem of Rellich-Kondrachov, 51

theorem of Stone, 40

unbounded linear operator, 17

unitary group, 35

variance, 43

wave equation, 9

wave function, 8, 43

weak convergence, 49

weak derivative, 13

weak partial derivative, 13

Weyl sequence, 62

Weyl’s criterion, 61

Zhislin’s theorem, 90

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