#  Lecture notes on

OPERATOR ALGEBRAS AND THEIR APPLICATION IN PHYSICS

Lecturer Dr. A. Kegeles

TU Bergakademie Freiberg
Institute for Geophysics
Geomathematics and Geoinformatics Group

Lecture notes on

OPERATOR ALGEBRAS AND THEIR APPLICATION IN PHYSICS

Lecturer Dr. A. Kegeles

TU Bergakademie Freiberg
Institute for Geophysics
Geomathematics and Geoinformatics Group

## Preamble

Around the year 1927 physicist, lead by Heisenberg and Schrödinger, discovered
a new theory for subatomic particles, called quantum mechanics. This theory
radically differed from classical mechanics in its physical assumptions as
well as in its mathematical formulation. The formal developments of quantum
mechanics rapidly made it one of the most successful physical theories.
Nevertheless, its physical foundations were very obscure and unjustifiable,
and its mathematical basis were often shaky. To provide a clear physical
justification and a rigorous mathematical foundation for the theory, several
mathematicians, lead by von Neumann, Weyl, and Jordan, set out to define an
algebraic approach for quantum mechanics. In this lecture we follow their path
and see how the physical formulation of the standard Schrödinger quantum
mechanics appears from general physical arguments, paired with the brilliant
observations of Heisenberg, and impeccable mathematics of von Neumann and
friends…

In the first part of the lecture, we will develop the basic theory of operator
algebras. Here we will not make any connections to physics and treat the
subject in its pure form. In particular, we discuss the function calculus and
the Gelfand-Naimark-Segal construction of representations for
$C^{\star}$-algebras. These tools will give us the necessary mathematical
background to deal with $C^{\star}$-algebras when they appear in physics.

In the second part of the lecture we establish a connection between pure
mathematics and physics. We will see how general physical principles lead us
to an algebraic description of physical theories. Combining this description
with the Heisenberg’s uncertainty principle and the canonical commutation
relations developed by Born and Jordan we end up with the $C^{\star}$-algebra
that describes a quantum mechanical particle. The final piece of the puzzle
follows from the Stone-von Neumann theorem that justifies the almost exclusive
use of Schrödinger’s formulation for quantum mechanics.

###### Contents

  1. 1 Operator algebras: Banach and $C^{\star}$-algebras
    1. 1.1 Basic properties of operator algebras
    2. 1.2 Function calculus for $C^{\star}$-algebras
    3. 1.3 The GNS construction
      1. 1.3.1 Order structure on the $C^{\star}$-algebra
      2. 1.3.2 States of the $C^{\star}$-algebra
      3. 1.3.3 Representations of a $C^{\star}$-algebra
  2. 2 Operator algebras in physics
    1. 2.1 Classical mechanics
      1. 2.1.1 Kinematics
      2. 2.1.2 Dynamics
    2. 2.2 An algebraic construction for physical theories
  3. 3 A quantum mechanical particle
    1. 3.1 Semi-historical note
    2. 3.2 The Weyl algebra
    3. 3.3 Stone-von Neumann theorem
    4. 3.4 The Schrödinger representation of the Weyl algebra

##  1 Operator algebras: Banach and $C^{\star}$-algebras

###  1.1 Basic properties of operator algebras

Section remarks: This section is based on the volume 1 and 3 of the series by
Kadison and Ringrose [1, 2]. Some of the reminders and definitions that
concern more standard statements of functional analysis are mostly from [3].
Further reading: for the abstract approach to $C^{\star}$-algebras and von
Neumann algebras the book by Sakai [4] is great but is a little more advanced;
another very satisfying read is the book by Pedersen [5]; the four volumes of
the aforementioned Kadison and Ringrose bible are anyway a treat for the soul:
the books cover most of the topics of old fashioned $C^{\star}$-algebras and
develop the theory very slowly and self-consistently.

In a nutshell, an algebra is a vector space in which we can multiply vectors.
More formally we define the algebra as follows.

######  Definition 1.1 (Algebra).

An algebra $\mathcal{A}$ over a field $\mathbb{K}=(\mathbb{R},\mathbb{C})$ is
a vector space over $\mathbb{K}$, equipped with multiplication that

  1. i )

distributes over addition (from left and right):

$\displaystyle f\cdot(g+h)=f\cdot g+f\cdot h,$ $\displaystyle(f+g)\cdot
h=f\cdot h+g\cdot h;$ (1)

  2. ii )

is associative:

$\displaystyle(f\cdot g)\cdot h=f\cdot(g\cdot h);$ (2)

  3. iii )

is compatible with scalar multiplication:

$\displaystyle\lambda(AB)=(\lambda A)B=A(\lambda
B)\qquad(\lambda\in\mathbb{K},\ A,B\in\mathcal{A}).$ (3)

A unit of an algebra $\mathcal{A}$ is an element $I\in\mathcal{A}$ such that
$I\cdot A=A\cdot I=A$ for all $A\in\mathcal{A}$. It is clear that if such an
element exists it must be unique. However, not every algebra contains a unite;
if it does we call the algebra unital. We will always use unital algebras in
the following, even if we do not mention it explicitly. In one of the
exercises, we will see that one can always adjoint a unite to an algebra to
make it unital. For unital algebras we can shorten the definition and state.

######  Definition 1.2.

An algebra $\mathcal{A}$ over a field $\mathbb{K}$ is a vector space, whose
underlying additive group has a ring structure compatible with multiplication
with scalars.

To recall the definitions of the vector spaces and rings consult the following
remarks.  Reminder: Vector space. A vector space over a field $\mathbb{K}$ is
a set of points $V$ equipped with vector addition $+\colon V\times V\to V$ and
scalar multiplication $\cdot\colon F\times V\to V$. These operations have to
satisfy the following requirements. For $f,g\in V$ and
$\alpha,\beta\in\mathbb{K}$ the scalar multiplication has to i ) distribute
over the vector addition: $\alpha(f+g)=\alpha f+\alpha g$, ii ) distribute
over the field addition: $(\alpha+\beta)f=\alpha f+\beta f$, iii ) be
compatible with the field multiplication: $(\alpha\beta)f=\alpha(\beta f)$, iv
) has the neutral element: $1f=f$. The vector addition has to be commutative
and $V$ with $+\colon V\to V$ must be a commutative group. That is for
$f,g,h\in V$ and $\alpha,\beta\in\mathbb{K}$ the addition i ) is commutative
(aka. Abelian): $f+g=g+f$, ii ) is associative: $(f+g)+h=f+(g+h)$, iii ) has a
neutral element $e\in V$: $e+f=f$ iv ) is such that for every $f\in V$ there
is an inverse $f^{-1}\in V$: $f+f^{-1}=e$.

Reminder: Ring. A ring is a commutative group $(R,+)$ equipped with
multiplication $\cdot\colon R\times R\to R$, such that with $f,g,h\in R$ the
multiplication i ) distributes from the left over addition:
$f\cdot(g+h)=f\cdot g+f\cdot h$ ii ) distributes from the right over addition:
$(f+g)\cdot h=f\cdot h+g\cdot h$ iii ) is associative: $(f\cdot g)\cdot
h=f\cdot(g\cdot h)$ iv ) has a neutral element $I\in R$: $I\cdot f=f\cdot
I=f$.

######  Example 1.3.

To visualize the concept let us recall some algebras that we know and love.

  1. i )

The simplest example of an algebra is the algebraic field $\mathbb{K}$ itself.

  2. ii )

A more elaborate example is the space $\mathcal{M}(n,\mathbb{C})$ of $n\times
n$ matrices with complex coefficients. This algebra is unital. The unit being
the identity matrix.

  3. iii )

Yet another example is the space $\mathbb{R}\times\mathbb{R}$ with the
Hadamard product defined by

$\displaystyle(1,0)\cdot(1,0)=(1,0),$ $\displaystyle(0,1)\cdot(0,1)=(0,1),$
$\displaystyle(1,0)\cdot(0,1)=(0,0),$ (4)

and required to be compatible with multiplication by scalars. Notice, however,
that this is not quite natural on the vector space $\mathbb{R}^{2}$. This is
because the definition of the product depends on the choice of the basis. In
particular, this product changes under rotations, which is why it rarely
appears in applications to physics.

  4. iv )

Finally, consider the space of all polynomials in a single variable. We will
call this space $\mathpzc{P}l$. It is a vector space, and equipped with the
pointwise product it becomes a unital algebra, whose unit is the constant $1$.

######  Definition 1.4.

A norm $\|\cdot\|$ on an algebra $\mathcal{A}$ is said to be continuous if for
$A,B\in\mathcal{A}$ it satisfies $\|A\cdot B\|\leq\|A\|\ \|B\|$.

  1. i )

An algebra with a continuous norm is called a normed algebra.

  2. ii )

A normed algebra that is complete (in the metric induced by the norm) is
called a Banach algebra.

As we will see, Banach algebras can be quite small (that is contain few
elements). Nevertheless, they are very general, as they have only little
structure compared to the algebras we will primarily use. We define this
additional structure as follows.

######  Definition 1.5.

Let $\mathcal{A}$ be a Banach algebra. A mapping $x\mapsto x^{\star}$ of
$\mathcal{A}$ into itself is called an involution if it satisfies the
following conditions:

  1. i )

$(x^{\star})^{\star}=x$,

  2. ii )

$(x+y)^{\star}=x^{\star}+y^{\star}$,

  3. iii )

$(xy)^{\star}=y^{\star}x^{\star}$,

  4. iv )

$(\lambda x)^{\star}=\bar{\lambda}x^{\star}$, for $\lambda\in\mathbb{C}$ and
$\bar{\lambda}$ being its complex conjugate.

Reminder: Adjoint. Let $B(H)$ be the space of bounded operators on the Hilbert
space $H$, and let $\langle\cdot,\cdot\rangle$ denote the inner product on
$H$. Then, for $A\in B(H)$ we define its adjoint as the operator
$A^{\dagger}\in B(H)$ such that $\displaystyle\langle x,Ay\rangle=\langle
A^{\dagger}x,y\rangle\qquad(x,y\in H).$ (5)

It is not difficult to see that the adjoint on the space of bounded linear
operators on a Hilbert space satisfies all the requirements of Definition 1.5.
Indeed, the adjoint is the reason why we made these requirements in the first
place. However, in the case of bounded linear operators, we define an adjoint
using the scalar product and then check that it satisfies the properties in
the Definition. For algebras, the logic is reversed. We don’t have the scalar
product at our disposal, and so we require the properties as the defining
features of the mapping. In this sense the involution, sometimes simply called
the star (operation), generalizes the concept of the adjoint for bounded
linear operators.

With this additional structure we can now define the algebras, that will
accompany us for the rest of the lecture.

######  Definition 1.6.

A Banach $\star$-algebra $\mathfrak{B}$ is a Banach algebra with an
(isometric) involution, that is with the property that $\|A^{\star}\|=\|A\|$
for every $A\in\mathfrak{B}$.

######  Definition 1.7.

A $C^{\star}$-algebra $\mathfrak{A}$ is a complex Banach $\star$-algebra that
has the $C^{\star}$-property:

$\displaystyle\|A^{\star}A\|=\|A\|^{2}\qquad(A\in\mathfrak{A}).$ (6)

In the definition of Banach $\star$-algebras some authors do not require the
involution to be isometric. In our case, this difference will not play a big
role, because we will only use either Banach algebras (without the star), or
$C^{\star}$-algebras. For the latter, however, the involution is always
isometric: since if $A$ is an element in the $C^{\star}$-algebra
$\mathfrak{A}$ then the $C^{\star}$-property implies

$\displaystyle\|A\|^{2}=\|A^{\star}A\|\leq\|A^{\star}\|\|A\|,$ (7)

and thus $\|A\|\leq\|A^{\star}\|$. Upon replacing $A\to A^{\star}$ we get the
inverse inequality and thus $\|A\|=\|A^{\star}\|$. Notice, however, that the
converse is not true: an isometric involution is not enough to ensure the
$C^{\star}$-property. Thus, in general $C^{\star}$-algebras have more
structure than Banach $\star$-algebras.

The name giving for the elements of the Banach $\star$-algebra follows the
usual convention. We introduce the standard notation.

Common notation. Let $\mathfrak{A}$ be a unital algebra with an involution.
For $A\in\mathfrak{A}$ we call the element $A^{\star}$ the adjoint of $A$. We
say that $A$ is i ) self-adjoint if $A^{\star}=A$; ii ) normal if
$AA^{\star}=A^{\star}A$; iii ) unitary if $AA^{\star}=A^{\star}A=I$, with the
unit $I\in\mathfrak{A}$; iv ) invertible if there exists a $B\in\mathfrak{A}$
with $AB=BA=I$. If such $B$ exists it is unique and we call it the two-sided
inverse of $A$ and denote it by $A^{-1}$; v ) self-inverse if $A=A^{-1}$; It
follows that the unit $I$ is self-adjoint, self-inverse, and unitary since
$\displaystyle(I^{\star}I)^{\star}=(I^{\star})^{\star}=I=I^{\star}I=I^{\star}.$
(8) Further, we can state the following: i ) The set of all self-adjoint
elements of $\mathfrak{A}$ is a real vector space. ii ) The set of all unitary
elements of $\mathfrak{A}$ forms a multiplicative group. iii ) Each element
$A\in\mathfrak{A}$ can be uniquely expressed as $H+\imath K$, where
$H=\frac{1}{2}(A+A^{\star})$ and $K=\frac{\imath}{2}(A^{\star}-A)$ are self-
adjoint elements of $\mathfrak{A}$, called the real and the imaginary part of
$A$, respectively. Moreover, $A$ is normal if and only if $H$ and $K$ commute.

######  Proposition 1.8.

Let $A$ be an element in a $C^{\star}$-algebra. Then $A$ has a (two-sided)
inverse if and only if $A^{\star}$ is invertible and then
$(A^{\star})^{-1}=(A^{-1})^{\star}$.

###### Proof.

Calculate

$\displaystyle(A^{-1})^{\star}A^{\star}=(AA^{-1})^{\star}=\mathds{1},$ (9)

and

$\displaystyle A^{\star}(A^{-1})^{\star}=(A^{-1}A)^{\star}=\mathds{1}.$ (10)

It follows that $(A^{-1})^{\star}$ is a two-sided inverse of $A^{\star}$. By
uniqueness of the two-sided inverse we get
$(A^{-1})^{\star}=(A^{\star})^{-1}$. ∎

Before we look at some examples of $C^{\star}$-algebras, let us quickly
refresh some basic definitions in topology.  Reminder: Hausdorff space. A
topological space $X$ is called Hausdorff, if for every pair of distinct
points $x,y\in X$ there exists two open sets $U,V\subset X$ with $U\cap
V=\emptyset$ and $x\in U$ and $y\in V$. Reminder: Compact set. A subset $K$ of
a topological space $X$ is compact, if each open cover of $K$, contains a
finite sub-cover of $K$.

######  Example 1.9 (Commutative $C^{\star}$-algebra).

Let $X$ be a compact Hausdorff space. Let $C_{b}(X)$ denote the space of
(bounded) complex valued continuous functions on $X$. Define a ring structure
on $C_{b}(X)$ by the pointwise product: for $f,g\in C_{b}(X)$ define
$(fg)(x)\coloneqq f(x)g(x)$, $(x\in X)$. Next, topologize $C_{b}(X)$ by the
supremum norm: for $f\in C_{b}(X)$ set $\|f\|=\sup_{x\in X}|f(x)|$. Define an
involution on $C_{b}(X)$ by complex conjugation: for $f\in C_{b}(X)$ define
$f^{\star}=\bar{f}$. Then $C_{b}(X)$ is a commutative unital
$C^{\star}$-algebra and the unit is a constant function $I=1$.

###### Proof.

Clearly, $C_{b}(X)$ is a vector space. Since a pointwise product of two
continuous functions is again a continuous function, the space $C_{b}(X)$ is
algebraically closed. Since the uniform limit of continuous functions is
continuous, the space $C_{b}(X)$ is complete. Hence, $C_{b}(X)$ with the
supremum norm and the pointwise product is a Banach algebra. It is straight
forward to check that complex conjugation defines an involution. In fact, to a
great extent complex conjugation inspired our definition of the involution.
Hence, $C_{b}(X)$ is a Banach $\star$-algebra. (Here we do not assume the
involution to be continuous. The continuity will follow, after we prove the
$C^{\star}$-property.) For the $C^{\star}$-property we observe for $f\in
C_{b}(X)$: $|f(x)\bar{f}(x)|=|f(x)||\bar{f}(x)|=|f(x)||f(x)|=|f(x)|^{2}$, at
any $x\in X$. Hence,

$\displaystyle\|f\bar{f}\|=\sup_{x\in X}|f(x)\bar{f}(x)|=\sup_{x\in
X}|f(x)|^{2}=\big{(}\sup_{x\in X}|f(x)|\big{)}^{2}=\|f\|^{2}.$ (11)

This completes the proof. ∎

Comment: Indeed, by the Gelfand transform (that we meet in a few lectures) we
will see that every unital commutative $C^{\star}$-algebra is isometrically
isomorphic to a $C^{\star}$-algebra of continuous functions on some compact
Hausdorff space.

Now we consider a non-commutative example of a $C^{\star}$-algebra.

######  Example 1.10.

Consider our previous example: the algebra $\mathcal{M}(n,\mathbb{C})$ of
$n\times n$ matrices with complex coefficients. Equipped with the operator
norm and adjoint as an involution, $\mathcal{M}(n,\mathbb{C})$ is a unital
$C^{\star}$-algebra.

###### Proof.

This is a special case of the following example. ∎

######  Example 1.11.

Let $H$ be a separable Hilbert space. Denote by $B(H)$ the space of linear
bounded operators on $H$. Then with the operator product, the operator norm,
and the adjoint as involution, the space $B(H)$ is a unital
$C^{\star}$-algebra.

###### Proof.

We only show the $C^{\star}$-property as the rest is trivial. Let $A$ be in
$B(H)$ then

$\displaystyle\|A\|^{2}=\sup_{x\in H;\ \|x\|=1}\|Ax\|=\sup_{x\in H;\
\|x\|=1}\langle Ax,Ax\rangle=\sup_{x\in H;\ \|x\|=1}\langle
x,A^{\dagger}Ax\rangle=\|A^{\dagger}A\|.$ (12)

The last equality holds because $A^{\dagger}A$ is a self-adjoint operator. ∎

Whenever we speak about $C^{\star}$-algebras it is enlightening to consider,
in the back of your head, the space of linear bounded operators on a Hilbert
space. However, you should keep in mind that an abstract $C^{\star}$-algebra
is more general than $B(H)$ for a fixed Hilbert space $H$; it is more general
because a $C^{\star}$-algebra is not defined by the action of operators on a
Hilbert space. Instead, it is only a collection of elements that satisfy some
relations under multiplication. In particular, let $A$ be an element of a
$B(H)$. To characterize it we often define its action on $H$. For example, if
$A$ maps a vector $v$ in $H$ to a rotated vector $\tilde{v}$, we say that $A$
is a rotation. But how do we characterize elements in an abstract algebra if
we don’t define their action… we use their spectrum.

######  Definition 1.12.

If $A$ is an element of a Banach algebra $\mathfrak{A}$ we say that
$\lambda\in\mathbb{C}$ is a spectral value for $A$ (relative to
$\mathfrak{A}$) when $A-\lambda\mathds{1}$ does not have a two-sided inverse
in $\mathfrak{A}$. The set of spectral values of $A$ is called the spectrum of
$A$ and is denoted $\operatorname{sp}_{\mathfrak{A}}(A)$.

######  Example 1.13.

Let $\mathfrak{A}=C_{b}(X)$ be the $C^{\star}$-algebra of continuous real-
valued functions on a compact Hausdorff space $X$. For $f\in\mathfrak{A}$ what
is its spectrum?

For $\lambda\notin\operatorname{Img}(f)$ (the image of $f$) consider
$f-\lambda 1$. The function $f-\lambda 1$ is continuous and defined on a
compact domain. Hence, it attains all its minima. If $f-\lambda 1$ never
vanishes then $\inf|f-\lambda 1|>0$ and the function $g=\frac{1}{f-\lambda}$
is continuous. Clearly, $g$ is the inverse of $f-\lambda 1$. Hence
$\operatorname{sp}_{\mathfrak{A}}(f)\subset\operatorname{Img}(f)$ .

Take $\lambda\in\operatorname{Img}(f)$, so for some $x\in X$ we have
$(f-\lambda 1)(x)=0$. Since $f-\lambda 1$ is continuous there is a sequence
$(x_{n})$ in $X$ that converges to $x$ and satisfies $(f-\lambda
1)(x_{n})<1/n$. Let $g$ be the two-sided inverse such that $(f-\lambda
1)g=g(f-\lambda 1)=1$. Then

$\displaystyle 1=|\big{(}(f-\lambda 1)g\big{)}(x_{n})|=|g(x_{n})||(f-\lambda
1)(x_{n})|\leq|g(x)|\frac{1}{n},$ (13)

for every $x_{n}$. Thus, $|g(x_{n})|\geq n$, and $g$ is not continuous at $x$.
Consequently $g\notin C_{b}(X)$ and
$\operatorname{Img}(f)\subset\operatorname{sp}_{\mathfrak{A}}(f)$. Together
with the opposite inclusion we get
$\operatorname{sp}_{\mathfrak{A}}(f)=\operatorname{Img}(f)$.

######  Example 1.14.

Let $\mathfrak{M}$ be the $C^{\star}$-algebra of complex $n\times n$ matrices.
For $M\in\mathfrak{M}$ consider $M-\lambda\mathds{1}$. This element is
invertible if and only if $\det(M-\lambda\mathds{1})\neq 0$. Hence, if $M$ is
diagonalizable its spectrum is the set of eigenvalues of $M$.

Warning. For a Banach algebra the spectrum of an element does indeed depend on
the algebra. Therefore, we use the subscript to specify the ambient algebra.
The following example shows that changing the algebra can change the spectrum.

######  Example 1.15.

Let $H$ be a separable Hilbert space with the basis $\\{e_{j}\\}$ for
$j\in\mathbb{Z}$. Let $B(H)$ be the Banach algebra of linear bounded operators
on $H$. (Above we have shown that $B(H)$ is a $C^{\star}$-algebra, so in
particular it is a Banach algebra). Consider the left shift $L$ in $B(H)$
defined on the basis of $H$ by $L(e_{j})=e_{j+1}$ and extended by linearity
and continuity to the whole $H$. The adjoint of $L$ is the operator
$L^{\dagger}(e_{j})=e_{j-1}$. One can easily verify that
$LL^{\dagger}=L^{\dagger}L=\mathds{1}$. Thus, $L^{\dagger}$ is the two-sided
inverse of $L$. In other words, $0\notin\operatorname{sp}_{B(H)}(L)$.

Now consider a Banach algebra $\mathfrak{B}$ generated by $\mathds{1}$ and
$L$. It is a subalgebra of $B(H)$. Is it still true that $0$ is not in the
spectrum of $L$ relative to $\mathfrak{B}$? No. To see this consider a closed
subspace $S$ of $H$ spanned by $\\{e_{j}\ \colon\ j=1,2,\dots\\}$. Then $L$
maps this subspace into itself, and thus, each polynomial of $L$ leaves this
subspace invariant. It follows that every element in $\mathfrak{B}$ (the norm
closure of such polynomials) maps $S$ into itself. Assume that $\tilde{L}$ is
a two-sided inverse of $L$ in $\mathfrak{B}$. Then $\tilde{L}$ is also in
$B(H)$ since $\mathfrak{B}\subset B(H)$. By the uniqueness of the two-sided
inverse we thus have $L^{\dagger}=\tilde{L}$. However, $L^{\dagger}$ does not
preserve $S$ as it maps the element $e_{1}$ to $e_{0}\notin S$. Thus,
$L^{\dagger}$ is not in the algebra $\mathfrak{B}$. We can thus conclude that
$L$ does not have a two-sided inverse in $\mathfrak{B}$ and
$0\in\operatorname{sp}_{\mathfrak{B}}(L)$.

The above example shows that the spectrum of an element depends on the ambient
Banach algebra. One of the main results that we will obtain in the first part
of the lecture will be that for $C^{\star}$-algebras this cannot happen.

######  Lemma 1.16.

For a Banach $\star$-algebra $\mathfrak{A}$, let $A$ be in $\mathfrak{A}$ and
let $\operatorname{sp}_{\mathfrak{A}}(A)$ be its spectrum. Then the spectrum
of $A^{\star}$ is

$\displaystyle\operatorname{sp}_{\mathfrak{A}}(A^{\star})=\big{\\{}\
\bar{\lambda}\ \colon\
\lambda\in\operatorname{sp}_{\mathfrak{A}}(A)\big{\\}}\Big{(}=\overline{\operatorname{sp}_{\mathfrak{A}}(A)}\Big{)}.$
(14)

###### Proof.

If $\lambda\notin\operatorname{sp}_{\mathfrak{A}}(A)$ then $(A-\lambda
I)^{-1}$ exists in $\mathfrak{A}$. Hence

$\displaystyle\big{(}(A-\lambda
I)^{\star}\big{)}^{-1}=(A^{\star}-\bar{\lambda}I)^{-1}=\big{(}(A-\lambda
I)^{-1}\big{)}^{\star},$ (15)

and $(A-\lambda I)^{\star}\big{)}^{-1}$ exists. Thus,
$\bar{\lambda}\notin\operatorname{sp}_{\mathfrak{A}}(A^{\star})$.
Interchanging $A$ and $A^{\star}$ in the above argument yields the desired
conclusion. ∎

######  Lemma 1.17.

For a Banach algebra $\mathfrak{A}$, let $A$ be an invertible element in
$\mathfrak{A}$. Then the spectrum of $A^{-1}$ is

$\displaystyle\operatorname{sp}_{\mathfrak{A}}(A^{-1})=\\{\lambda^{-1}\colon\lambda\in\operatorname{sp}_{\mathfrak{A}}(A)\\}\Big{(}=\operatorname{sp}_{\mathfrak{A}}(A)^{-1}\Big{)}$
(16)

###### Proof.

If $A$ is invertible then $0\notin\operatorname{sp}_{\mathfrak{A}}(A)$. Thus,
if $\lambda$ is in $\operatorname{sp}_{\mathfrak{A}}(A)$ then $\lambda\neq 0$
and

$\displaystyle A^{-1}-\lambda^{-1}\mathds{1}=(A\lambda)^{-1}(\lambda-A).$ (17)

Hence, $A^{-1}-\lambda^{-1}\mathds{1}$ is invertible if and only if
$A-\lambda\mathds{1}$ is invertible, meaning

$\displaystyle\operatorname{sp}_{\mathfrak{A}}(A^{-1})=\\{\lambda^{-1}\colon\
\lambda\in\operatorname{sp}_{\mathfrak{A}}(A)\\}.$ (18)

This shows the assertion. ∎

In order to study the spectrum of an element it is good to have better
understanding of invertible elements. The next few lemmas provide us with this
understanding.

######  Lemma 1.18 (Neumann series).

If $A$ is an element of a Banach algebra $\mathfrak{A}$ and $\|A\|<1$ then

$\displaystyle\sum_{k=0}^{n}A^{k}\qquad(\text{with }A^{0}=I),$ (19)

has a limit $B\in\mathfrak{A}$ as $n$ goes to infinity. Moreover, it holds
that

$\displaystyle B(I-A)=(I-A)B=I$ (20)

and $B$ is the (two-sided) inverse of $I-A$.

###### Proof.

For $n<m$ we can estimate

$\displaystyle\|\sum_{k=0}^{n}A^{k}-\sum_{k=0}^{m}A^{k}\|=\|\sum_{k=n+1}^{m}A^{k}\|\leq\sum_{k=n+1}^{m}\|A\|^{k}.$
(21)

Since $\|A\|<1$ the right-hand side of the above estimate goes to zero. Thus,
$\sum_{k=0}^{n}A^{k}$ is a Cauchy sequence. Since $\mathfrak{A}$ is complete
this sequence converges to an element $B$ in $\mathfrak{A}$ as $n$ goes to
infinity. Then

$\displaystyle\big{(}\sum_{k=0}^{n}A^{k}\big{)}(I-A)=\sum_{k=0}^{n}A^{k}-\sum_{k=0}^{n}A^{k+1}=(I-A)\big{(}\sum_{k=0}^{n}A^{k}\big{)}=I-A^{n+1}.$
(22)

By continuity of the product we have
$\|A^{n+1}\|\leq\|A\|^{n+1}\overset{n\to\infty}{\longrightarrow}0$. Thus
$A^{n+1}$ goes to zero as $n$ goes to infinity. Thence the right-hand side of
(22) converges to $I$ and we obtain (again by continuity of multiplication)

$\displaystyle B(I-A)=(I-A)B=I.$ (23)

This completes the proof. ∎

######  Proposition 1.19.

If $\mathfrak{A}$ is a Banach algebra, the set
$\mathpzc{N}\subset\mathfrak{A}$ of invertible elements in $\mathfrak{A}$ is
an open subset of $\mathfrak{A}$ and the operation

$\displaystyle\operatorname{\text{inv}}\colon A\to A^{-1}$ (24)

on $\mathpzc{N}$ is continuous.

The following definition will be convenient for the proof of the proposition.

######  Definition 1.20.

Let $A$ be an element of a Banach algebra $\mathfrak{A}$. We define the left
multiplication by $A$ as the mapping $L_{A}\colon\mathfrak{A}\to\mathfrak{A}$
such that $L_{A}(B)=AB$. In the similar way we define the right multiplication
by $A$ as the mapping $R_{A}\colon\mathfrak{A}\to\mathfrak{A}$ such that
$R_{A}(B)=BA$.

Info: From the (joint) continuity of the product in a Banach algebra it
follows directly that $L_{A}$ and $R_{A}$ are continuous for every $A$ in
$\mathfrak{A}$.

###### Proof of proposition 1.19.

Even though the proof is not complicated, it is still helpful to understand
the general idea of the argument.  Idea: To show that $\mathpzc{N}$ is an open
set we need to show that for every element in $\mathpzc{N}$ there exists an
open set containing this element and that is contained in $\mathpzc{N}$. We
know that an open ball of radius 1 around the identity $I$ is in
$\mathpzc{N}$. The idea is to shift this ball appropriately to prove the
proposition. The multiplication operators that we just defined provide the
“shifting”. We now formalize the idea.

$L_{A}$
---
$I$
---
$A$
---
$\mathpzc{N}$
---
$\mathpzc{N}$
---
$\mathfrak{A}$
---
$\mathfrak{A}$
---
$A$
---
Figure 1: Idea of proof.

For $A\in\mathpzc{N}$ let $L_{A}$ be the left multiplication by $A$. Since $A$
is invertible then so is $L_{A}$ with the inverse $L_{A^{-1}}$. As we
mentioned above $L_{A}$ and $L_{A^{-1}}$ are both continuous (thus $L_{A}$ is
a homeomorphism).

From the Neumann series we know that $B_{1}(I)$ — an open ball of radius 1
around $I$ — is a subset of $\mathpzc{N}$. Since $L_{A}$ is a homeomorphism
(in particular open) it maps $B_{1}(I)$ to an open set $\mathpzc{U}$ in
$\mathfrak{A}$. The set $\mathpzc{U}$ contains $A$ since $I\in B_{1}(I)$ and
$L_{A}(I)=A$; also $\mathpzc{U}$ is contained in $\mathpzc{N}$ since if $A$
and $B$ are invertible then so is $AB$ (with the inverse $B^{-1}A^{-1}$).
Thus, for every $A\in\mathpzc{N}$ there is an open set
$L_{A}(B_{1}(I))\subset\mathpzc{N}$ containing $A$ and $\mathpzc{N}$ is open.

Now we prove that the mapping $\operatorname{\text{inv}}$ is continuous on
$\mathpzc{N}$. Assume that $B\in\mathfrak{A}$ is such that $\|I-B\|<1$, then
by the Neumann series

$\displaystyle\sum_{k=0}^{\infty}(I-B)^{k}=\big{(}I-(I-B)\big{)}^{-1}=B^{-1}.$
(25)

Choosing $\|I-B\|<\epsilon<2^{-1}$ and the fact that $I$ is self-inverse (i.e.
$II=I$) we can estimate

$\displaystyle\|\operatorname{\text{inv}}(B)-\operatorname{\text{inv}}(I)\|$
$\displaystyle=\|B^{-1}-I\|\leq\sum_{k=1}^{\infty}\|I-B\|^{k}$ (26)
$\displaystyle=\|I-B\|\Big{(}\sum_{k=0}^{\infty}\|I-B\|^{k}\Big{)}=\frac{\|I-B\|}{1-\|I-B\|}\leq
2\|I-B\|\leq 2\epsilon.$ (27)

Thus $\operatorname{\text{inv}}$ is continuous at the identity. To show that
it is continuous at every element observe that for any $B\in\mathpzc{N}$ the
following identity holds

$\displaystyle\operatorname{\text{inv}}=R_{B^{-1}}\circ\operatorname{\text{inv}}\circ
L_{B^{-1}}.$ (28)

Thus for each fixed element $A\in\mathpzc{N}$ choosing the right-hand side of
the above equality to be $R_{A^{-1}}\circ\operatorname{\text{inv}}\circ
L_{A^{-1}}$ shows that the map $\operatorname{\text{inv}}$ is a combination of
continuous mappings
$A\overset{L_{A^{-1}}}{\to}I\overset{\operatorname{\text{inv}}}{\to}I\overset{R_{A^{-1}}}{\to}A^{-1}$.
Thus for each $A\in\mathpzc{N}$ the mapping $\operatorname{\text{inv}}$ is
continuous. ∎

We will now show that every element in a Banach algebra has a spectrum. The
proof uses some known, but not entirely trivial statements from the complex
and functional analysis. Since, these topics are themselves very extensive, we
only recall the basic needed statements before continuing with the theorem.
Reminder: Holomorphic function. A complex valued function $f$ is called
holomorphic at $z_{0}\in\mathbb{C}$ if on some neighborhood of $z_{0}$ the
limit $\displaystyle\lim_{z\to z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}$ (29)
exists. A function that is holomorphic at every point of some open set
$\mathpzc{O}\subset\mathbb{C}$ is said to be holomorphic on $\mathpzc{O}$. If
$f(z)$ is holomorphic on an entire complex plain we call it an entire
function. By Liouville’s theorem, every bounded entire function must be
constant. In particular, if an entire function vanishes at infinite it must be
zero.

Reminder: Hahn-Banach theorems. The name “Hahn-Banach theorem” refers to
several related results (consult e.g. [1, chap. 1 2]). There are two types of
such theorems: separation and extension results. One version of the Hahn-
Banach extension theorem is the following.

######  Theorem.

Let $X$ be a vector space and $X_{0}$ be a subspace of $X$. Let
$\rho_{0}\colon X_{0}\to\mathbb{C}$ (or $\mathbb{R}$) be a linear functional,
and $p\colon X\to\mathbb{R}$ be a semi-norm such that
$\displaystyle|\rho_{0}(y)|\leq p(y)\qquad(y\in X_{0}).$ (30) Then there
exists a linear functional $\rho\colon X\to\mathbb{C}$ (or $\mathbb{R}$) such
that $\displaystyle|\rho(x)|\leq p(x)\qquad(x\in X),$
$\displaystyle\rho(y)=\rho_{0}(y)\qquad(y\in X_{0}).$ (31) The corollaries of
this very general theorem are often more useful than the theorem itself. For
example the next corollary assures that we can always extend a linear
functional.

######  Corollary.

Let $X_{0}$ be a subspace of a normed space $X$. Let $\rho_{0}$ be a bounded
linear functional on $X_{0}$. There exists a bounded linear functional $\rho$
on the whole $X$ such that $\|\rho\|=\|\rho_{0}\|$ and $\rho(x)=\rho_{0}(x)$
for $x\in X_{0}$. Another useful corollary is the one we use in the proof of
Theorem 1.21.

######  Corollary.

If $x$ is a non-zero vector in a normed space $X$, there is a bounded linear
functional $\rho$ on $X$ such that $\|\rho\|=1$ and $\rho(x)=\|x\|$.

Equipped with all that fundamental knowledge. We can now prove that the
spectrum of an element in a Banach algebra is never empty.

######  Theorem 1.21.

If $A$ is an element of a Banach algebra $\mathfrak{A}$ then
$\operatorname{sp}_{\mathfrak{A}}(A)$ is a non-empty closed subset of the
closed disc in $\mathbb{C}$ with center 0 and radius $\|A\|$.

###### Proof.

Let $\lambda\in\mathbb{C}$ be such that $|\lambda|>\|A\|$ then

$\displaystyle A-\lambda I=\lambda\Big{(}\frac{1}{\lambda}A-I\Big{)}.$ (32)

Since $\|A\lambda^{-1}\|<1$ the element $\frac{1}{\lambda}A-I$ is invertible
(by the Neumann series). Thus, $A-\lambda I$ has the inverse
$\frac{1}{\lambda}(\frac{1}{\lambda}A-I)^{-1}$. It follows that
$\operatorname{sp}_{\mathfrak{A}}(A)\subset D_{\|A\|}(0)$ (a disc with center
0 and radius $\|A\|$).

Now we need to show that $\operatorname{sp}_{\mathfrak{A}}(A)$ is not empty.
Let $\lambda\notin\operatorname{sp}_{\mathfrak{A}}(A)$. Then by the
proposition 1.19, the element $A-\lambda I$ is invertible in a small
neighborhood around $\lambda$. In particular the set
$\mathbb{C}\setminus\operatorname{sp}_{\mathfrak{A}}(A)$ is open.

Reminder: Resolvent equation. Recall the first resolvent equation [3, thm.
VI.5] $\displaystyle(A-\lambda^{\prime}I)^{-1}-(A-\lambda I)^{-1}$
$\displaystyle=(\lambda^{\prime}-\lambda)(A-\lambda^{\prime}I)(A-\lambda
I)^{-1}.$ (33)

Let $\lambda^{\prime}$ be close enough to $\lambda$ so that
$A-\lambda^{\prime}I$ is invertible. Let $\rho\colon\mathfrak{A}\to\mathbb{C}$
be a continuous linear functional on $\mathfrak{A}$. Then by the resolvent
equation

$\displaystyle\frac{\rho\Big{(}(A-\lambda^{\prime}I)^{-1}\Big{)}-\rho\Big{(}(A-\lambda
I)^{-1}\Big{)}}{\lambda^{\prime}-\lambda}=\rho\Big{(}(A-\lambda^{\prime}I)^{-1}(A-\lambda
I)^{-1}\Big{)}\overset{\lambda^{\prime}\to\lambda}{\to}\rho\Big{(}(A-\lambda
I)^{-2}\Big{)}.$

The last limit follows by the continuity of $\rho$, the product, and the
mapping $\operatorname{\text{inv}}$ on $\mathpzc{N}$. Thus the function
$\lambda\mapsto\rho(A-\lambda I)$ is holomorphic on
$\mathbb{C}\setminus\operatorname{sp}_{\mathfrak{A}}(A)$. Also since
$(\frac{A}{\lambda}-I)$ approaches $-I$ as $|\lambda|$ grows, we have

$\displaystyle\rho\Big{(}(A-\lambda
I)^{-1}\Big{)}=\frac{1}{\lambda}\rho\Big{(}(\frac{1}{\lambda}A-I)^{-1}\Big{)}\to
0,\qquad\text{for }|\lambda|\to\infty.$ (34)

Now, if $\operatorname{sp}_{\mathfrak{A}}(A)$ were empty, the function
$\lambda\mapsto\rho(A-\lambda I)$ would be an entire function that vanishes at
infinity. By the Liouville’s theorem we then had $\rho(A-\lambda I)^{-1}=0$
for all $\lambda\in\mathbb{C}$ and in particular for $\lambda=0$. But then
$\rho(A^{-1})=0$ for every continuous linear functional $\rho$, and by the
Hahn-Banach theorem $A^{-1}=0$. This is a contradiction. Hence
$\operatorname{sp}_{\mathfrak{A}}(A)\neq\emptyset$. This completes the proof.
∎

The above theorem states that the spectrum of an element in the Banach algebra
is bounded by its norm. However, the norm is just an upper bound not the
smallest upper bound. To study how different or similar is the norm and the
smallest upper bound of the spectrum, we give the latter quantity a proper
name first.

######  Definition 1.22 (Spectral radius).

The spectral radius $r_{\mathfrak{A}}(A)$ of an element $A$ in a Banach
algebra $\mathfrak{A}$ is

$\displaystyle
r_{\mathfrak{A}}(A)=\sup\\{|\lambda|\colon\lambda\in\operatorname{sp}_{\mathfrak{A}}(A)\\}$
(35)

Info: The spectral radius is the radius of the smallest disc in $\mathbb{C}$
with center 0 containing $\operatorname{sp}_{\mathfrak{A}}(A)$. Also from the
preceding theorem, $\|A\|$ is the radius of the largest disc in $\mathbb{C}$
with center 0 that contains $\operatorname{sp}_{\mathfrak{A}}(A)$. Hence we
always have $r_{\mathfrak{A}}(A)\leq\|A\|$.

$\mathbb{C}$
---
$\|A\|$
---
$r_{\mathfrak{A}}(A)$
---
$\operatorname{sp}_{\mathfrak{A}}(A)$
---
Figure 2: Spectrum of $A$ relative to $\mathfrak{A}$ bounded by the spectral
radius and by $\|A||$.

Surprisingly, the spectral radius — rather an algebraic property — can be
stated in analytical terms, and thus it relates to topology. For the proof
consult e.g. [1, thm. 3.3.3]

######  Theorem 1.23 (Spectral radius formula).

Let $A$ be an element of a Banach algebra $\mathfrak{A}$. The sequence
$(\|A^{n}\|^{1/n})$ converges to $r_{\mathfrak{A}}(A)$ such that

$\displaystyle r_{\mathfrak{A}}(A)=\lim_{n\to\infty}\|A^{n}\|^{1/n}.$ (36)

This relation between the spectrum and topology is as surprising as it is
powerful. We have already seen that the spectrum of an element depends on the
ambient Banach algebra. Nevertheless, from the spectral radius formula if
follows that if $\mathfrak{B}$ is a Banach subalgebra of $\mathfrak{A}$ then
the spectral radii of $A\in\mathfrak{B}$ and $A\in\mathfrak{A}$ coincide, i.e.
$r_{\mathfrak{A}}(A)=r_{\mathfrak{B}}(A)$. In a $C^{\star}$-algebra the
spectral radius formula provides even stronger results.

######  Proposition 1.24.

Suppose that $A$ is an element of a $C^{\star}$-algebra $\mathfrak{A}$. Then

  1. i )

if $A$ is normal then $r_{\mathfrak{A}(A)}=\|A\|$;

  2. ii )

if $A$ is self-adjoint then $\operatorname{sp}_{\mathfrak{A}}(A)$ is a compact
subset of the real line and it contains at least one of the two real numbers
$\pm\|A\|$;

  3. iii )

if $A$ is unitary then $\|A\|=1$ and $\operatorname{sp}_{\mathfrak{A}}(A)$ is
a compact subset of the unite circle $\\{a\in\mathbb{C}\colon|a|=1\\}$.

###### Proof.

i ) Let $H$ be a self-adjoint element in $\mathfrak{A}$ and $n\in\mathbb{N}$.
Then by the $C^{\star}$-property we have

$\displaystyle\|H^{2n}\|=\|(H^{n})^{\star}H^{n}\|=\|H^{n}\|^{2}.$ (37)

Hence, if $q=2^{m}$ for some $m=1,2,\dots$ then by induction we get

$\displaystyle\|H^{q}\|=\|H\|^{q}.$ (38)

Using the spectral radius formula we get

$\displaystyle r_{\mathfrak{A}}(H)=\lim_{q\to\infty}\|H^{q}\|^{1/q}=\|H\|.$
(39)

So the assertion holds for self-adjoint elements. Now, let $A$ be normal and
$H=A^{\star}A$ then

$\displaystyle\|A\|^{2}\overset{(1)}{=}\|A^{\star}A\|\overset{(2)}{=}r_{\mathfrak{A}}(A^{\star}A)\overset{(3)}{\leq}r_{\mathfrak{A}}(A^{\star})r_{\mathfrak{A}}(A)\overset{(4)}{=}r_{\mathfrak{A}}(A)^{2}\overset{(5)}{\leq}\|A\|^{2}.$
(40)

With the following justifications for the estimates: (1) is the
$C^{\star}$-property; (2) follows from what we have just shown above; (3)
follows from the fact that for normal elements
$\|(A^{\star}A)^{n}\|\leq\|(A^{\star})^{n}\|\|A^{n}\|$ and thus
$\lim\|(A^{\star}A)^{n}\|^{1/n}\leq\lim\|(A^{\star})^{n}\|^{1/n}\|A^{n}\|^{1/n}$;
(4) follows from the fact that
$\operatorname{sp}_{\mathfrak{A}}(A^{\star})=\overline{\operatorname{sp}_{\mathfrak{A}}(A)}$;
(5) follows since $\|A\|$ defines the largest radius of a disc that contains
the spectrum.

Since the left and the right hand side of the above estimate are equal all the
"less or equal" signs in the estimate have to be actual equalities. Hence
$r_{\mathfrak{A}}(A)=\|A\|$.

ii ) By Theorem 1.21 the spectrum of $A\in\mathfrak{A}$ is a compact set of
$\mathbb{C}$. Thus the spectral radius attains its maximum and there is a
$\lambda\in\operatorname{sp}_{\mathfrak{A}}(A)$ such that
$|\lambda|=r_{\mathfrak{A}}(A)$. If $A$ is self-adjoint, then from the first
part of the proposition we also know that $r_{\mathfrak{A}}(A)=\|A\|$ and thus
there is an element $\lambda\in\operatorname{sp}_{\mathfrak{A}}(A)$ with
$|\lambda|=\|A\|$. To prove the assertion we only need to show that the
spectrum of a self-adjoint element is a subset of a real line.

Assume that for some $a,b\in\mathbb{R}$ the element $a+\imath b$ is in
$\operatorname{sp}_{\mathfrak{A}}(A)$. For $n\in\mathbb{N}$ consider the
element $B_{n}=A+\imath nbI$. Then, since
$a+\imath(n+1)b\in\operatorname{sp}_{\mathfrak{A}}(B_{n})$, we have

$\displaystyle|a+\imath(n+1)b|^{2}\leq\|B_{n}\|^{2}\leq\|B_{n}^{\star}B_{n}\|=\|A^{2}+n^{2}b^{2}\|\leq\|A\|^{2}+n^{2}b^{2}.$
(41)

On the other hand,

$\displaystyle|a+\imath(n+1)b|^{2}=(a+\imath(n+1)b)(a-\imath(n+1)b)=a^{2}+n^{2}b^{2}+2nb^{2}+b^{2}.$
(42)

So combining (41) and (42), we get $a^{2}+(2n+1)b^{2}\leq\|A\|^{2}$ for every
$n\in\\{0,1,2,\dots\\}$. Since $A$ is, however, bounded $b$ must be zero.

iii ) If $A$ is unitary in $\mathfrak{A}$, then $\|A\|^{2}=\|A^{\star}A\|=1$.
Thus $\|A\|=1$. Therefore, if $\lambda$ is in
$\operatorname{sp}_{\mathfrak{A}}(A)$ then $|\lambda|\leq\|A\|\leq 1$. On the
other hand, since the inverse of $A$ is its adjoint $A^{\star}$ we have from
Lemma 1.17 that
$\lambda^{-1}\in\operatorname{sp}_{\mathfrak{A}}(A^{-1})=\operatorname{sp}_{\mathfrak{A}}(A^{\star})$
and $|\lambda^{-1}|\leq\|A^{\star}\|=\|A\|\leq 1$. Thus, $|\lambda|\leq 1$ and
$|\lambda^{-1}|\leq 1$ meaning that $|\lambda|=1$. ∎

Algebra vs. bounded operators on a Hilbert space. For bounded operators on a
Hilbert space we can use the inner product to prove that a self-adjoint
operator on a Hilbert space has a real spectrum. This makes the proof simpler
(but also less general). Consider a self-adjoint bounded linear operator $A$
on a Hilbert space $H$. Let $\lambda\in\mathbb{C}$ be such that
$|\operatorname{Im}(\lambda)|>0$. Then for every $x\in H$ we have by the
Cauchy-Schwarz inequality,
$\displaystyle\|x\|\|(A-\lambda\mathds{1})x\|\geq|\langle
x,(A-\lambda\mathds{1})x\rangle|\geq|\operatorname{Im}(\lambda)|\|x\|^{2}>0.$
(43) In particular,
$\|(A-\lambda\mathds{1})x\|\geq|\operatorname{Im}(\lambda)|\|x\|$ for every
$x\in H$. Hence $A-\lambda\mathds{1}$ is injective and has closed range. Also,
since $|\operatorname{Im}(\bar{\lambda})|=|\operatorname{Im}(\lambda)|$, with
the same argument we get that $(A-\lambda\mathds{1})^{\dagger}$ is also
injective, and $A-\lambda\mathds{1}$ has dense range in $H$. It follows that
$A-\lambda\mathds{1}$ is invertible and
$\lambda\notin\operatorname{sp}_{B(H)}(A)$.

###  1.2 Function calculus for $C^{\star}$-algebras

Function calculus on $C^{\star}$-algebras generalizes the idea of
diagonalizing matrices. Diagonalizing a matrix allows us to define powers,
square roots, exponential, logarithms of matrices, and what not… For bounded
operators on a Hilbert space the generalization leads to the spectral
calculus. The functional calculus in $C^{\star}$-algebras essentially bears
the same meaning. We can even consider it a generalization of the spectral
calculus, as it is defined for abstract algebra elements. Generally speaking
the function calculus allows us to associate algebra elements to continuous
functions, calculate with those functions, and then associate the result back
to algebra elements. This process, is very powerful. It is hard to
overestimate the use of the function calculus in the abstract theory and
applications…

In the following we will use the following notation. If $p$ is a polynomial,
say $p(t)=\sum_{k=0}^{n}a_{k}t^{k}$ with $a_{k}\in\mathbb{C}$, then we will
write $p(A)$ to denote the algebra element $\sum_{k=0}^{n}a_{k}A^{k}$.

######  Proposition 1.25 (Spectral calculus for polynomials).

If $A$ is an element in a Banach algebra $\mathfrak{A}$ and $p$ is a
polynomial in one variable, then

$\displaystyle\operatorname{sp}_{\mathfrak{A}}(p(A))=\\{p(\lambda)\colon\lambda\in\operatorname{sp}_{\mathfrak{A}}(A)\\}\Big{(}=p(\operatorname{sp}_{\mathfrak{A}}(A))\Big{)}.$
(44)

###### Proof.

If $\lambda\in\operatorname{sp}_{\mathfrak{A}}(A)$, then $A-\lambda I$ has no
two-sided inverse in $\mathfrak{A}$. If $p(x)=a_{n}x^{n}+\cdots+a_{0}$ then

$\displaystyle
p(A)-p(\lambda)I=a_{n}(A^{n}-\lambda^{n}I)+\cdots+a_{1}(A-\lambda I).$ (45)

For $k\geq 2$ we can rewrite $A^{k}-\lambda^{k}I$ as

$\displaystyle A^{k}-\lambda^{k}I$ $\displaystyle=(A-\lambda
I)(A^{k-1}+\lambda A^{k-2}+\cdots+\lambda^{k-1}I)$
$\displaystyle=(A^{k-1}+\lambda A^{k-2}+\cdots+\lambda^{k-1}I)(A-\lambda I).$

Hence, we get

$\displaystyle p(A)-p(\lambda)I=(A-\lambda I)\big{(}a_{n}(A^{n-1}+\lambda
A^{n-1}+\cdots+\lambda^{n-1}I)+\cdots+a_{1}\big{)}=\big{(}\cdots\big{)}(A-\lambda
I),$

and $p(A)-p(\lambda)I$ does not have a two-sided inverse (otherwise $A-\lambda
I$ would have a two-sided inverse). It follows that
$p(\lambda)\in\operatorname{sp}_{\mathfrak{A}}(p(A))$, and thus
$p(\operatorname{sp}_{\mathfrak{A}}(A))\subset\operatorname{sp}_{\mathfrak{A}}(p(A))$.

For the opposite inclusion consider that if $\gamma$ is in
$\operatorname{sp}_{\mathfrak{A}}(p(A))$ and $\lambda_{1},\dots,\lambda_{n}$
are the $n$ roots of $p(x)-\gamma$ then

$\displaystyle p(A)-\gamma I=c(A-\lambda_{1}I)\cdots(A-\lambda_{n}I),$ (46)

and at least one of $A-\lambda_{j}I$ is not invertible. Hence,
$\lambda_{j}\in\operatorname{sp}_{\mathfrak{A}}(A)$ for at least one of
$\lambda_{j}$’s and by previous discussion
$\gamma=p(\lambda_{j})=\operatorname{sp}_{\mathfrak{A}}(p(A))$ (the first
equality follows since $p(\lambda_{j})-\gamma=0$ by definition of roots).
Hence, $\operatorname{sp}_{\mathfrak{A}}(p(A))\subset
p(\operatorname{sp}_{\mathfrak{A}}(A))$. ∎

To unclutter the notation in the proofs we will use the following common
notations. We summarize them in the following paragraph.

Common notation. By $\mathpzc{P}l$ we will denote the space of polynomials
with complex coefficients. If $A$ is an element in a $C^{\star}$-algebra
$\mathfrak{A}$ with the spectrum $\operatorname{sp}_{\mathfrak{A}}(A)$ then, •
by $C(\operatorname{sp}_{\mathfrak{A}}(A))$ we will denote the
$C^{\star}$-algebra of continuous functions on
$\operatorname{sp}_{\mathfrak{A}}(A)$; • by
$\operatorname{\mathit{id}}\colon\operatorname{sp}_{\mathfrak{A}}(A)\to\operatorname{sp}_{\mathfrak{A}}(A)$
we will denote the identity mapping $z\mapsto z$. • when we write
$\mathpzc{P}l\subset C(\operatorname{sp}_{\mathfrak{A}}(A))$ we consider
elements of $\mathpzc{P}l$ as continuous functions on
$\operatorname{sp}_{\mathfrak{A}}(A)$.

It is time to introduce the function calculus for self-adjoint operators.
Almost the same theorem, with minor modifications, holds also for normal
elements in a $C^{\star}$-algebra (see the Remark at the end of the section).
However, we will prove only the following result, as it does not require any
additional technicalities that are needed for normal elements. Before the
proof, we recall the Weierstrass approximation theorem for convenience.

Reminder: Weierstrass approximation theorem. Any continuous function on a
bounded interval can be uniformly approximated by polynomial functions.

######  Theorem 1.26 (Function calculus for self-adjoint elements).

Let $A$ be a self-adjoint element of a $C^{\star}$-algebra $\mathfrak{A}$.
There is a unique isometric $\star$-isomorphism $\varphi$ from
$C(\operatorname{sp}_{\mathfrak{A}}(A))$ into $\mathfrak{A}$ such that
$\varphi(\operatorname{\mathit{id}})=A$. Moreover, for $f\in
C(\operatorname{sp}_{\mathfrak{A}}(A))$,

  1. i )

$\varphi(f)$ is a normal element and it is self-adjoint if and only if $f$ is
real valued;

  2. ii )

the set $\\{\varphi(f)\colon f\in C(\operatorname{sp}_{\mathfrak{A}}(A))\\}$
is a commutative $C^{\star}$-algebra and it is the smallest
$C^{\star}$-subalgebra of $\mathfrak{A}$ that contains $A$;

  3. iii )

$\varphi(f)$ is the limit of a sequence of polynomials in $I$ and $A$;

  4. iv )

if $B$ is in $\mathfrak{A}$ that commutes with $A$, then it also commutes with
$\varphi(f)$.

###### Proof.

Assume a polynomial $p\in\mathpzc{P}l$ of the form,

$\displaystyle p(t)=\sum_{k=0}^{n}a_{k}\ t^{k}.$ (47)

By Proposition 1.24 ii ), the spectrum of $A$ is a compact subset of the real
line. Then we can write $p$ restricted to this subset using the identity
mapping in $C(\operatorname{sp}_{\mathfrak{A}}(A))$ such that

$\displaystyle p=\sum_{k=0}^{n}a_{k}\
\operatorname{\mathit{id}}^{k}\qquad(\text{with
}\operatorname{\mathit{id}}^{0}=1).$ (48)

Define $\varphi_{0}\colon\mathpzc{P}l\to\mathfrak{A}$ by

$\displaystyle\varphi_{0}(p)=\sum_{k=0}^{n}a_{k}A^{k}=p(A)$ and
$\displaystyle\varphi_{0}(p)^{\star}=\sum_{k=0}^{n}\bar{a}_{k}A^{k}=p(A)^{\star}.$
(49)

Then, $\varphi_{0}$ is linear and
$\varphi_{0}(\operatorname{\mathit{id}}^{k})\mapsto A^{k}$ for $k=\mathbb{N}$.
Also, $\varphi_{0}(p)$ and $\varphi_{0}(p)^{\star}$ commute, implying that
$\varphi_{0}(p)$ is normal. Thus, Proposition 1.24 i ) applies and together
with Proposition 1.25 we have

$\displaystyle\|\varphi_{0}(p)\|=r_{\mathfrak{A}}(p(A))=\sup\\{|\lambda|\colon\lambda\in\operatorname{sp}_{\mathfrak{A}}(p(A))\\}=\sup\\{|p(\lambda)|\colon\lambda\in\operatorname{sp}_{\mathfrak{A}}(A)\\}=\|p\|,$
(50)

where the norm on the left-hand side is the $C^{\star}$-norm on $\mathfrak{A}$
and the norm on the right-hand side is the supremum norm on
$C(\operatorname{sp}_{\mathfrak{A}}(A))$. Thus, $\varphi_{0}$ is an isometry.
If $p$ and $q$ are two polynomials in $\mathpzc{P}l$, identically equal on
$\operatorname{sp}_{\mathfrak{A}}(A)$, then

$\displaystyle\|\varphi_{0}(p)-\varphi_{0}(q)\|=\|p-q\|=0,$ (51)

and then $\varphi_{0}(p)=\varphi_{0}(q)$. Thus, the mapping $\varphi_{0}$ is
well defined (unambiguous).

By the Weierstrass approximation theorem, $\mathpzc{P}l\subset
C(\operatorname{sp}_{\mathfrak{A}}(A))$ is everywhere dense in
$C(\operatorname{sp}_{\mathfrak{A}}(A))$. Since $\mathfrak{A}$ is complete,
$\varphi_{0}$ extends to the mapping $\varphi$ defined on
$C(\operatorname{sp}_{\mathfrak{A}}(A))$. This extension remains isometric;
since if $f\in C(\operatorname{sp}_{\mathfrak{A}}(A))$ and $\\{p_{n}\\}$ is a
sequence of polynomials that approximates $f$ then,

$\displaystyle\|\varphi(f)\|=\lim_{n\to\infty}\|\varphi(p_{n})\|=\lim_{n\to\infty}\|p_{n}\|=\|f\|.$
(52)

(This proves existence of the asserted mapping.)

Thence, since $C(\operatorname{sp}_{\mathfrak{A}}(A))$ is complete its image,
$\\{\varphi(f)\colon f\in C(\operatorname{sp}_{\mathfrak{A}}(A))\\}$, is
complete as well. Therefore, it is a commutative $C^{\star}$-subalgebra of
$\mathfrak{A}$ that contains $I$ and $A$. We will denote this subalgebra
$\mathfrak{A}(A)$. Since polynomials are everywhere dense in
$C(\operatorname{sp}_{\mathfrak{A}}(A))$ then every element of
$\mathfrak{A}(A)$ is a limit of a sequence of polynomials in $A$. A closed
$C^{\star}$-subalgebra $\mathfrak{B}$ of $\mathfrak{A}$ that contains $I$ and
$A$ necessarily contains all polynomials in $A$ and therefore it contains
$\mathfrak{A}(A)$. Hence $\mathfrak{A}(A)$ is the smallest
$C^{\star}$-subalgebra that contains $I$ and $A$. (This proves ii ) and iii
).)

By definition, $\varphi_{0}$ is linear, multiplicative (i.e.
$\varphi(pq)=\varphi(p)\varphi(q)$ for $p,q\in\mathpzc{P}l$), preserves the
star (i.e. $\varphi(\bar{p})=\varphi(p)^{\star}$), and carries the unite of
$\mathpzc{P}l$ onto that of $\mathfrak{A}$. These properties extend by
continuity to $\varphi$ and thus, $\varphi$ is a $\star$-homomorphism. Since
it is isometric it is one-to-one and thus, a $\star$-isomorphism. (This proves
the isometry assertion.)

If $\phi$ is another such $\star$-isomorphism that maps
$\operatorname{\mathit{id}}$ to $A$ then $\phi$ and $\varphi$ coincide on the
dense space $\mathpzc{P}l$; thus they are equal. (This proves uniqueness of
the asserted mapping.)

Since we can approximate every element in $\mathfrak{A}(A)$ by polynomials in
$A$, it follows at once that $\varphi(f)$ is a normal element, and that it is
self-adjoint if and only if $f$ is real valued. If $B\in\mathfrak{A}$ commutes
with $A$ then it commutes with all polynomials in $A$, and thus it commutes
with every element in $\mathfrak{A}(A)$. This concludes the proof. ∎

The mapping $\varphi$ is called the function calculus for the self-adjoint
element $A$. To simplify the notation, it is often common to write $f(A)$
instead of $\varphi(f)$. Henceforth, we will use this convention.

We have already seen that a spectrum of a Banach algebra depends on the
ambient Banach algebra. For $C^{\star}$-algebras this is not the case. Here,
the spectrum of an element is independent of the $C^{\star}$-algebra in which
we consider this element to be. Stated sloppy, $C^{\star}$-algebras are always
large enough to contain an inverse of an element if it exists. With the aid of
the function calculus we can now prove this statement.

######  Proposition 1.27.

Let $\mathfrak{A}$ be a $C^{\star}$-algebra, and $\mathfrak{B}$ be a
$C^{\star}$-subalgebra of $\mathfrak{A}$. If $B$ is in $\mathfrak{B}$, then

$\displaystyle\operatorname{sp}_{\mathfrak{A}}(B)=\operatorname{sp}_{\mathfrak{B}}(B).$
(53)

###### Proof.

We have already discussed that
$\operatorname{sp}_{\mathfrak{A}}(B)\subseteq\operatorname{sp}_{\mathfrak{B}}(B)$.
Thus we need only to prove the opposite. For this we will show that if
$A\in\mathfrak{B}$ has an inverse in $\mathfrak{A}$ then this inverse is also
in $\mathfrak{B}$.

We consider first a self-adjoint element $A$. Since
$0\notin\operatorname{sp}_{\mathfrak{A}}(A)$, the function $f(t)=t^{-1}$ is
continuous on $\operatorname{sp}_{\mathfrak{A}}(A)$. Thus, by the function
calculus relative to the algebra $\mathfrak{A}$ we get an element $f(A)$ in
$\mathfrak{A}$. From Theorem 1.26 ii ) this element is contained in
$\mathfrak{B}$. Since $tf(t)=1$ for each $t$ in
$\operatorname{sp}_{\mathfrak{A}}(A)$, it follows by multiplicativity of the
function calculus that

$\displaystyle I=\varphi(1)=\varphi(\operatorname{\mathit{id}}\ \cdot
f)=\varphi(\operatorname{\mathit{id}})\varphi(f)=Af(A).$ (54)

Thus $A^{-1}=f(A)\in\mathfrak{B}$.

Now, consider a generic $A\in\mathfrak{B}$ that has an inverse $A^{-1}$ in
$\mathfrak{A}$. Then $A^{\star}$ is also in $\mathfrak{B}$ and has an inverse
$(A^{-1})^{\star}$ in $\mathfrak{A}$. Now, $A^{\star}A$ is a self-adjoint
element in $\mathfrak{B}$ and it has an inverse $A^{-1}(A^{-1})^{\star}$ in
$\mathfrak{A}$. By the first part of the proof, $A^{-1}(A^{-1})^{\star}$ is in
$\mathfrak{B}$. Thus,
$A^{-1}=(A^{-1}(A^{-1})^{\star})A^{\star}\in\mathfrak{B}$. ∎

The above proposition shows that the spectrum of an element in a
$C^{\star}$-algebra is independent of the $C^{\star}$-algebra this element is
in. Thus, from now on, we can (and will) omit the subscript referring to the
ambient $C^{\star}$-algebra and write $\operatorname{sp}(A)$ instead of
$\operatorname{sp}_{\mathfrak{A}}(A)$.

In the proof of Proposition 1.27 we use the phrase “… by the function calculus
relative to the algebra $\mathfrak{A}$ …”. Implying, that the function
calculus may depend on the original algebra. Nevertheless, using Proposition
1.27 we can see that this cannot happen. Namely, if $\mathfrak{B}$ is a
$C^{\star}$-subalgebra of $\mathfrak{A}$, and $A$ is a self-adjoint element in
$\mathfrak{B}$, then let $\varphi$ be the spectral calculus
$\operatorname{sp}(A)\to\mathfrak{A}$ and $\psi$ be the spectral calculus
$\operatorname{sp}(A)\to\mathfrak{B}$. However, since the spectrum of $A$
relative to either algebra is the same, we can consider $\psi$ as a spectral
calculus into the larger algebra $\mathfrak{A}$. By uniqueness of the spectral
calculus, $\psi$ must then coincide with $\varphi$. Thus, we do not need to
worry about the ambient $C^{\star}$-algebra when talking about the spectral
calculus. In particular, we can always choose a sufficiently small
$C^{\star}$-algebra, $\mathfrak{A}(A)=\\{f(A)\colon f\in
C(\operatorname{sp}(A))\\}$ such that the spectral calculus,
$C(\operatorname{sp}(A))\to\mathfrak{A}(A)$, is onto. Then, we can state the
following about the spectrum of $f(A)$.

######  Corollary 1.28.

If $A$ is a self-adjoint element of a $C^{\star}$-algebra $\mathfrak{A}$, and
$f\in C(\operatorname{sp}(A))$, then

$\displaystyle\operatorname{sp}(f(A))=\\{f(t)\colon\
t\in\operatorname{sp}(A)\\}.$ (55)

###### Proof.

The function calculus $f\to f(A)$ is a $\star$-isomorphism from
$C(\operatorname{sp}(A))$ onto the $C^{\star}$-subalgebra $\mathfrak{A}(A)$ of
$\mathfrak{A}$, and $\star$-isomorphisms preserve the spectrum. ∎

Another useful consequence of the function calculus is the definition of a
square root.

######  Corollary 1.29.

Let $A$ be a self-adjoint element of a $C^{\star}$-algebra $\mathfrak{A}$,
with $\operatorname{sp}(A)\subseteq\mathbb{R}^{+}$. There exists a unique
self-adjoint element $H$ in the commutative $C^{\star}$-subalgebra
$\mathfrak{A}(A)$ of $\mathfrak{A}$ such that $A=HH$. Moreover,
$\operatorname{sp}(H)\subseteq\mathbb{R}^{+}$, and if $B\in\mathfrak{A}$
commutes with $A$ then it also commutes with $H$.

###### Proof.

Since $\operatorname{sp}(A)$ is a subset of positive real numbers, the
function $f(t)=t^{1/2}\big{(}=\sqrt{\operatorname{\mathit{id}}}\big{)}$ is
real-valued, positive, and continuous on $\operatorname{sp}(A)$. Define $H$ as
$f(A)$. If $\varphi$ denotes the corresponding function calculus then

$\displaystyle
HH=f(A)f(A)=\varphi(f)\varphi(f)=\varphi(f^{2})=\varphi(\operatorname{\mathit{id}})=A.$
(56)

Since, $f$ is real valued, the element $H$ is self-adjoint by 1.26 i ). Since
$f$ is positive the spectrum of $H$ is positive by 1.28. If $B\in\mathfrak{A}$
commutes with $A$ then it commutes with $H$ by 1.26 iv ).

To show uniqueness, suppose that $K$ is any element in $\mathfrak{A}$ with
positive spectrum and such that $K^{2}=A$. Let $\\{p_{n}\\}$ be a sequence of
polynomials that converge to $f=\sqrt{id}$. Set $q_{n}(t)=p_{n}(t^{2})$. Then
we have,

$\displaystyle\lim_{n\to\infty}q_{n}(t)=\lim_{n\to\infty}p_{n}(t^{2})=f(t^{2})=t,$
(57)

uniformly for $t\in\operatorname{sp}(K)$. Hence, by the function calculus for
$K$ we get

$\displaystyle
K=\lim_{n\to\infty}q_{n}(K)=\lim_{n\to\infty}p_{n}(K^{2})=\lim_{n\to\infty}p_{n}(A)=f(A)=H,$
(58)

which proves uniqueness. ∎

There are more useful corollaries that we can derive from the function
calculus. For now, however, we close this chapter and proceed to the next
ingredient that will be necessary to understand the construction of quantum
mechanics: the GNS construction.

Info: The function calculus can be extended to normal elements. The methods to
prove it, however, are a bit different. We provide the full theorem without
proof.

######  Theorem 1.30 (Function calculus for normal elements).

Let $A$ be a normal element of a $C^{\star}$-algebra $\mathfrak{A}$. There is
a unique isometric $\star$-isomorphism $\varphi$ from
$C(\operatorname{sp}_{\mathfrak{A}}(A))$ into $\mathfrak{A}$ such that
$\varphi(\operatorname{\mathit{id}})=A$. Moreover, for $f\in
C(\operatorname{sp}_{\mathfrak{A}}(A))$, i ) $\varphi(f)$ is a normal element
of $\mathfrak{A}$; ii ) the set $\\{\varphi(f)\colon f\in
C(\operatorname{sp}_{\mathfrak{A}}(A))\\}$ is a commutative
$C^{\star}$-algebra and it is the smallest $C^{\star}$-subalgebra of
$\mathfrak{A}$ that contains $A$; iii ) $\varphi(f)$ is the limit of a
sequence of polynomials in $I$, $A$, and $A^{\star}$; iv ) moreover, the
element $A$ is

• self-adjoint iff $\operatorname{sp}(A)\subseteq\mathbb{R}$; • positive iff
$\operatorname{sp}(A)\subseteq\mathbb{R}^{+}$; • unitary iff
$\operatorname{sp}(A)\subseteq\mathbb{C}_{1}$; • projection iff
$\operatorname{sp}(A)\subseteq\\{0,1\\}$.

###  1.3 The GNS construction

In this section we will see that any abstract $C^{\star}$-algebra can be
“represented” as an algebra of bounded linear operators on some appropriate
Hilbert space. The construction used to show this was developed by Gelfand,
Naimark, and Segal, hence the name GNS-construction. This construction is
vital for our purpose: to use $C^{\star}$-algebras in applications to quantum
mechanics. To discuss the GNS construction we need additional tools — the
order structure on the $C^{\star}$-algebra, the space of states, and a brief
discussion of representations of a $C^{\star}$-algebras. We begin with the
order structure.

####  1.3.1 Order structure on the $C^{\star}$-algebra

######  Definition 1.31.

Let $V$ be a vector space. A (positive) cone $\mathpzc{C}$ of $V$ is a subset
of $V$ with the following properties:

  1. i )

if $A$ and $-A$ are in $\mathpzc{C}$, then $A=0$;

  2. ii )

if $a$ is a positive scalar and $A$ is in $\mathpzc{C}$, then $aA$ is in
$\mathpzc{C}$;

  3. iii )

if $A,B$ are in $\mathpzc{C}$ then $A+B$ is also in $\mathpzc{C}$.

A cone on a vector space allows us to compare elements in that vector space.
In particular, we can define an element $A\in V$ to be “smaller then” $B\in V$
whenever $B-A$ is in the cone $\mathpzc{C}$. This relation defines a partial
ordering on $V$. An element $I\in V$ is called an order unit when, for a given
$A\in V$ it holds that $-aI\leq A\leq aI$ for some positive scalar $a$ that
may depend on $A$.

Reminder: partial order. A partial order on a set $\Omega$ is a relation over
the set $\Omega$ such that if $a,b,c\in\Omega$: i ) $a\leq a$, that is the
relation is reflexive; ii ) if $a\leq b$ and $b\leq a$ then $a=b$, that is the
relation is anti-symmetric; iii ) if $a\leq b$ and $b\leq c$ then $a\leq c$,
that is the relation is transitive.

$A$
---
$B$
---
$A+B$
---
$aA$
---
(a) A cone in a vector space.

$A$
---
$B$
---
$B-A$
---
(b) Partial order induced by the cone.

Figure 3:  Cone structure and the partial order.

######  Example 1.32 (Cone on $\mathbb{R}^{2}$).

Let $V=\mathbb{R}^{2}$ a two dimensional Euclidean vector space. Define the
cone $\mathpzc{C}$ on $V$ as the set of all those elements that in the
canonical basis of $\mathbb{R}^{2}$ have positive coefficients, that is

$\displaystyle\mathpzc{C}=\\{v\in\mathbb{R}^{2}\colon v=(v_{1},v_{2}),\text{
and }v_{1}\geq 0,\ v_{2}\geq 0\\}.$ (59)

It is easy to check that $\mathpzc{C}$ is indeed a cone, and thus this
provides a partial ordering for the space $\mathbb{R}^{2}$. However, this cone
(and thus the partial ordering) depends on the basis and is not preserved
under rotations.

Since a (positive) cone defines a set of positive elements, it stands to
reason that a set of positive elements — defined in some meaningful way — will
yield a cone. As is common, we can borrow the intuition of positive elements
from bounded linear operators to define positive elements in a
$C^{\star}$-algebra.

######  Definition 1.33 (Positive element).

A positive element of a $C^{\star}$-algebra is a self-adjoint element with a
positive spectrum. By $\mathfrak{A}^{+}$ we denote the set of all positive
elements of $\mathfrak{A}$.

######  Example 1.34.

Consider a $C^{\star}$-algebra $C(X)$ on a compact Hausdorff space $X$. Then
$f\in C(X)$ is self-adjoint if and only if it is real-valued. Since, the
spectrum of $f$ is its image, it follows that $f$ is positive (in the
$C^{\star}$-algebra sense) if and only if $f(x)\geq 0$ for each $x\in X$. Thus
the notion of positivity for elements in the $C^{\star}$-algebra $C(X)$ is
consistent with the notion of positivity for functions.

######  Example 1.35.

If $B$ is a bounded linear operator on some Hilbert space $H$ then we call it
positive, if $\langle x,Bx\rangle\geq 0$ for every $x\in H$. We have seen in
our tutorial class, that $B$ is positive if and only if it is self-adjoint and
its spectrum is positive. Hence, the notion of positivity for elements in the
$C^{\star}$-algebra $B(H)$ is consistent with the notion of positivity for
bounded linear operators.

Even though, positivity is a statement about “signs”, it has a very useful
characterization in terms of the norm. We state this characterization in the
following lemma.

######  Lemma 1.36.

A self-adjoint element $A$ in a $C^{\star}$-algebra is positive if and only if

$\displaystyle\|A-aI\|\leq a,$ (60)

for some $a\in\mathbb{R}$ with $a\geq\|A\|$.

###### Proof.

Fix $a\geq\|A\|$, then $\operatorname{sp}(A)\subseteq[-a,a]$, and

$\displaystyle\|A-aI\|=r(A-aI)=\sup\\{|t-a|\colon
t\in\operatorname{sp}(A)\\}=\sup\\{(a-t)\colon t\in\operatorname{sp}(A)\\}.$
(61)

Where the first equality holds by 1.24 i ) and the second equality holds by
1.25. Hence $\|A-aI\|\leq a$ if and only if
$\operatorname{sp}(A)\subseteq\mathbb{R}^{+}$. ∎

Now, we can show that the set of positive elements in a $C^{\star}$-algebra
defines a closed positive cone.

######  Theorem 1.37.

Let $\mathfrak{A}$ be a $C^{\star}$-algebra. Then,

  1. i )

$\mathfrak{A}^{+}$ is closed in $\mathfrak{A}$;

  2. ii )

for $A\in\mathfrak{A}^{+}$, if $-A$ is also in $\mathfrak{A}^{+}$ then $A=0$.

  3. iii )

for $A\in\mathfrak{A}^{+}$ and $a\in\mathbb{R}^{+}$ we have
$aA\in\mathfrak{A}^{+}$;

  4. iv )

for $A,B\in\mathfrak{A}^{+}$ we have $A+B\in\mathfrak{A}^{+}$;

  5. v )

for $A,B\in\mathfrak{A}^{+}$ and $AB=BA$ we have $AB\in\mathfrak{A}^{+}$;

###### Proof.

i ) From Lemma 1.36 we have

$\displaystyle\mathfrak{A}^{+}=\\{A\in\mathfrak{A}\colon A=A^{\star}\text{ and
}\big{\|}A-\|A\|I\big{\|}\leq\|A\|\\}.$ (62)

By continuity of the norm, $\mathfrak{A}^{+}$ is closed.

ii ) If $A$ and $-A$ are both in $\mathfrak{A}^{+}$, then
$\operatorname{sp}(A)\subseteq\mathbb{R}^{+}\cap(-\mathbb{R}^{+})=\\{0\\}$; so
$\|A\|=r(A)=0$.

iii ) For $A\in\mathfrak{A}^{+}$ and $a\geq 0$, the element $aA$ is self-
adjoint and

$\displaystyle\operatorname{sp}(aA)=\\{a\lambda\colon\lambda\in\operatorname{sp}(A)\\}\subseteq\mathbb{R}^{+}.$
(63)

iv ) For $A,B\in\mathfrak{A}^{+}$ we have by Lemma 1.36 that

$\displaystyle\big{\|}A-\|A\|I\big{\|}\leq\|A\|$
$\displaystyle\big{\|}B-\|B\|I\big{\|}\leq\|B\|.$ (64)

Therefore,

$\displaystyle\big{\|}A+B-(\|A\|+\|B\|)I\big{\|}\leq\|A\|+\|B\|.$ (65)

With $a=\|A\|+\|B\|\geq\|A+B\|$ it follows again from Lemma 1.36 that
$A+B\in\mathfrak{A}^{+}$.

v ) For $A,B\in\mathfrak{A}^{+}$, let $H$ be the square root of $A$ defined as
in Corollary 1.29, and $K$ be the square root of $B$. By Corollary 1.29, if
$A$ commutes with $B$ then also $H$ commutes with $B$, and by the same
corollary, $K$ commutes with $H$. Thus, $AB=HHKK=HKHK=(HK)^{2}$. Since $H$ and
$K$ are both self-adjoint and since they commute, $HK$ is self-adjoint, and
thus its spectrum is a subset of a real line. By Corollary 1.28 we have
$\operatorname{sp}(AB)=\operatorname{sp}\big{(}(HK)^{2}\big{)}=\big{(}\operatorname{sp}(HK)\big{)}^{2}\subseteq\mathbb{R}^{+}$.
∎

Condition, ii ), iii ), and iv ) state that $\mathfrak{A}^{+}$ is a (positive)
cone. Hence, we can use it to define partial ordering on the
$C^{\star}$-algebra $\mathfrak{A}$. Using this ordering, we can say that an
element $A$ is positive if $A\geq 0$. Also notice, that in one of our
exercises we have shown that $\|A\|I-A$ is always positive, if $A$ is self-
adjoint. Hence, for a unital $C^{\star}$-algebra the unit of the algebra is
also the order unit for this partial ordering,

$\displaystyle-\|A\|I\leq A\leq\|A\|I.$ (66)

####  1.3.2 States of the $C^{\star}$-algebra

So far, we have discussed the elements of the algebra. However, a great deal
of structure on the $C^{\star}$-algebra comes from its dual space: the space
of continuous linear functionals on the algebra. We have encountered the power
of linear functionals in the Theorem 1.21. Now we approach the topic of linear
functionals more systematically. With the aid of the positive cone on the
$C^{\star}$-algebra we can define a cone on its dual space.

Reminder. We briefly recall some basic definitions in topology.

######  Definition 1.38 (Convex set.).

A set $Y$ in a vector space $V$ is called convex if $x$ and $y$ in $Y$ imply
that the line $\\{by_{1}+(1-b)y_{2}\colon 0\leq b\leq 1\\}$ is also in $Y$.

######  Definition 1.39 (Locally convex space.).

A locally convex space is a topological vector space, in which the topology
has a base consisting of convex sets. That means: Let $V$ be a topological
vector space with topology $\tau$. The basis of $\tau$ is the family
$\beta\subset\tau$ such that every open set $O\in\tau$ is of the form
$O=\cup_{\alpha}B_{\alpha}$ for some family $\\{B_{\alpha}\\}\subset\beta$. If
each set in $\beta$ is convex, then $V$ is a locally convex space.

######  Definition 1.40 (Extreme points.).

Let $V$ be a locally convex space and $Y\subset V$ a convex subset of $V$.
Then a point $y_{0}\in Y$ is called an extreme point of $Y$ if
$y_{0}=(1-a)y_{1}+ay_{2}$, with $0<a<1$ and $y_{1},y_{2}\in Y$ implies
$y_{1}=y_{2}=y_{0}$.

######  Theorem 1.41 (Krein-Milman).

If $X$ is a non-empty compact convex set in a locally convex space $V$, then
$X$ has an extreme point. Moreover, $X$ is the closure of the convex hull of
all extreme points of $X$.

After this short reminder we are equipped with everything to dive into the
discussion of linear functional on a $C^{\star}$-algebra. We begin with some
general definitions.

######  Definition 1.42.

Let $V$ be a partially ordered vector space with order unit $I$. Then,

  1. i )

a linear functional $\rho$ on $V$ is called positive if $\rho(v)\geq 0$ for
every $v\geq 0$ with $v\in V$;

  2. ii )

a positive linear functional $\rho$ on $V$ is called a state if $\rho(I)=1$;

  3. iii )

the state space $\mathcal{S}(V)$ is the set of all states on $V$.

If $\rho$ is a linear functional on a $C^{\star}$-algebra $\mathfrak{A}$, we
can define another linear functional $\rho^{\star}$, by the equation
$\rho^{\star}(A)=\overline{\rho(A^{\star})}$ for every $A\in\mathfrak{A}$.

######  Definition 1.43.

Let $\mathfrak{A}$ be a $C^{\star}$-algebra and $\rho$ a linear functional on
it. If $\rho(A)=\overline{\rho(A^{\star})}$ for every $A\in\mathfrak{A}$, we
call $\rho$ hermitian.

Recall that every element of a $C^{\star}$-algebra uniquely decomposes into
its real and imaginary part. Using this decomposition we can rephrase the
condition for a state to be hermitian only in terms of self-adjoint operators:
let $A$ be an element in a $C^{\star}$-algebra $\mathfrak{A}$ with the real
part $H$ and imaginary part $K$, then for a linear functional $\rho$ on
$\mathfrak{A}$ we have,

$\displaystyle\rho(A)=\rho(H+\imath K)=\rho(H)+\imath\rho(K),$
$\displaystyle\overline{\rho(A^{\star})}=\overline{\rho(H-\imath
K)}=\overline{\rho(H)}+\imath\overline{\rho(K)}.$ (67)

It follows that $\rho$ is hermitian if and only if
$\rho(H)=\overline{\rho(H)}$ for every self-adjoint $H\in\mathfrak{A}$.

Using this characterization of hermitian functionals we can deduce that a
positive linear functional is always hermitian: Let $A$ be self-adjoint in
$\mathfrak{A}$ and $\rho$ a positive linear functional on $\mathfrak{A}$.
Since $\|A\|I\pm A\geq 0$ (as we already have seen) it follows from positivity
of $\rho$ that $\rho(\|A\|I\pm A)\geq 0$. In particular, $\rho(\|A\|\pm A)$ is
real, and from

$\displaystyle\rho(A)=\frac{1}{2}\Big{(}\rho\big{(}\|A\|I+A\big{)}-\rho\big{(}\|A\|I-A\big{)}\Big{)},$
(68)

it follows that $\rho(A)$ itself is real.

Additionally, we can characterize positive linear functionals by their norm as
follows.

######  Theorem 1.44.

A linear functional $\rho$ on a (unital) $C^{\star}$-algebra $\mathfrak{A}$ is
positive if and only if $\rho$ is bounded and $\|\rho\|=\rho(I)$.

###### Proof.

For the “only if” direction, assume that $\rho$ is positive. Let $A$ be in
$\mathfrak{A}$, choose $a\in\mathbb{C}$ such that $|a|=1$ and $a\rho(A)\geq
0$, and let $H$ be the real part of $aA$. Since $\|H\|\leq\|A\|$ we have
$H\leq\|H\|I\leq\|A\|I$ and thus

$\displaystyle\|A\|\rho(I)-\rho(H)=\rho(\|A\|I-H)\geq 0.$ (69)

Since $\rho$ is hermitian we have
$\rho(\bar{a}A^{\star})=\overline{\rho(aA)}=\rho(aA)$ and consequently,

$\displaystyle|\rho(A)|=\frac{1}{2}\rho(aA+\bar{a}A^{\star})=\rho(H)\leq\rho(I)\|A\|.$
(70)

Thus, $\rho$ is bounded, with $\|\rho\|\leq\rho(I)$. Simultaneously,
$|\rho(I)|\leq\|\rho\|$; implying $\|\rho\|=\rho(I)$.

For the “if” direction, suppose that $\rho$ is bounded and $\|\rho\|=\rho(I)$.
After suitable rescaling, it is enough to consider the case in which
$\|\rho\|=\rho(I)=1$. With $A\in\mathfrak{A}^{+}$ and some $a,b\in\mathbb{R}$
assume that $\rho(A)=a+\imath b$. We want to show that $a\geq 0$ and $b=0$.
For this we observe that since $\operatorname{sp}(A)\subset\mathbb{R}^{+}$,
there is a small positive number $s$ such that,

$\displaystyle\operatorname{sp}(I-sA)=\\{1-st\colon
t\in\operatorname{sp}(A)\\}\subseteq[0,1].$ (71)

So by the spectral radius formula we have $\|I-sA\|=r(I-sA)\leq 1$. From the
continuity of $\rho$ we then have

$\displaystyle 1-sa\leq|1-s(a+\imath b)|=|\rho(I-sA)|\leq\|I-sA\|\leq 1,$ (72)

and $a\geq 0$. To show that $b=0$ we proceed as in the proof of Proposition
1.24 ii ). For $n\in\mathbb{N}$ consider the element $B_{n}=A+\imath nbI$ in
$\mathfrak{A}$. Then,

$\displaystyle\|B_{n}\|^{2}=\|B^{\star}_{n}B_{n}\|=\|A^{2}+n^{2}b^{2}I\|\leq\|A\|^{2}+n^{2}b^{2},$
(73)

and we get for every $n=1,2,\dots$

$\displaystyle a^{2}+(n^{2}+2n+1)b^{2}=|a+\imath b+\imath
nb|^{2}=|\rho(B_{n})|^{2}\leq\|A\|^{2}+n^{2}b^{2},$ (74)

where the last inequality follows from the continuity of $\rho$ and (73).
Since $A$ is bounded, $b$ has to vanish. ∎

With this understanding we can summarize the similarities and differences
between positive elements in the algebra and its dual.

An element $A$ in $\mathfrak{A}$ is positive if

  1. i )

it is self-adjoint,

  2. ii )

$\operatorname{sp}(A)\subset\mathbb{R}^{+}$,

  3. iii )

$\|A-aI\|\leq a$.

An element $\rho$ in $\mathfrak{A}^{\prime}$ is positive if

  1. i )

it is hermitian,

  2. ii )

$\rho(A)\geq 0$ whenever $A\in\mathfrak{A}^{+}$,

  3. iii )

$\rho(I)=\|\rho\|<\infty$.

Before we proceed with the analysis of states, we recall some ideas regarding
the topologies on vector spaces.

Reminder: weak⋆-topology. Let $V$ be a linear space and let $\mathcal{F}$ be a
family of linear functionals on $V$. Assume that $\mathcal{F}$ separates
points of $V$ in the sense that if $x\in V$ is non-zero then there is a
$\rho\in\mathcal{F}$ such that $\rho(x)\neq 0$. We define the topology
$\sigma(V,\mathcal{F})$ as the coarsest (weakest, one with the fewer open
sets) topology on $V$ relative to which each element of $\mathcal{F}$ is a
continuous mapping from $V$ into $\mathbb{C}$. If $V$ is a locally convex
topological vector space, we denote its continuous (topological dual) as
$V^{\prime}$. If $x$ is in $V$ then the equation
$\displaystyle\hat{x}(\rho)=\rho(x)\qquad(\rho\in V^{\prime}),$ (75) defines a
linear functional $\hat{x}$ on the dual $V^{\prime}$. The set
$\hat{V}=\\{\hat{x}\colon x\in V$ } is a linear subspace of $V^{\prime\prime}$
— the algebraic dual space of $V^{\prime}$ (the space of not necessarily
continuous functionals on $V^{\prime}$). The set $\hat{V}$ separates points of
$V^{\prime}$ since, if $\rho\in V^{\prime}$ and $\hat{x}(\rho)=0$ for all
$x\in V$ then $\rho(x)=0$ for all $x\in V$ and thus $\rho=0$. The
weak⋆-topology on $V^{\prime}$ is the topology $\sigma(V^{\prime},\hat{V})$.
We often denote it simply by $\sigma(V^{\prime},V)$. Hence, the weak⋆-topology
is the coarsest topology on $V^{\prime}$ for which each functional $\hat{x}$
is continuous.

A bounded closed set in an infinite dimensional space is not always compact.
However, in the weak⋆-topology this is indeed true (see the useful Lemma
below). This makes the weak⋆-topology an important tool in analysis.

Useful lemma. Let $X$ be a normed space and $X^{\prime}$ be the topological
dual of $X$. If $S$ is a bounded weak⋆-closed subset of $X^{\prime}$, then $S$
is weak⋆-compact.

From Theorem 1.44 we see that a state on $\mathfrak{A}$ is a bounded linear
functional on $\mathfrak{A}$, with $\|\rho\|=1$. Hence,
$\mathcal{S}(\mathfrak{A})$ is contained in the unit sphere of the Banach dual
space $\mathfrak{A}^{\prime}$ and is therefore bounded; it is also convex,
since for every $\rho_{1},\rho_{2}\in\mathcal{S}(\mathfrak{A})$ and $0\leq
a\leq 1$ the functional $\rho=a\rho_{1}+(1-a)\rho_{2}$ is again positive
(since positive elements form a cone) and $\rho(I)=a+(1-a)=1$. Additionally,
$\mathcal{S}(\mathfrak{A})$ is weak⋆-closed. To see this, observe that

$\displaystyle\mathcal{S}(\mathfrak{A})=\\{\rho\in\mathfrak{A}^{\prime}\colon\rho(I)=1,\
\rho(A)\geq 0\text{ when }A\in\mathfrak{A}^{+}\\}.$ (76)

For $A\in\mathfrak{A}$ let $\hat{A}$ be the dual element of
$\mathfrak{A}^{\prime}$ such that $\rho(A)=\hat{A}(\rho)$ for every
$\rho\in\mathfrak{A}^{\prime}$. Since each $\hat{A}$ is by definition
continuous in the weak⋆-topology, the set
$C_{A}=\hat{A}^{-1}\big{(}[0,\infty)\big{)}$ — the pre-image of $[0,\infty)$
under $\hat{A}$ — is closed, and so is the set $C_{0}=\hat{I}^{-1}(\\{1\\})$.
Then
$\mathcal{S}(\mathfrak{A})=\big{(}\cap_{A\in\mathfrak{A}^{+}}C_{A}\big{)}\cap
C_{0}$ is an intersection of weak⋆-closed sets and thus is weak⋆-closed. In
summary: $\mathcal{S}(\mathfrak{A})$ is a bounded, weak⋆-closed, convex subset
of $\mathfrak{A}^{\prime}$, and thus from the “useful lemma” above, it is
weak⋆-compact. By the Krein-Milman theorem $\mathcal{S}(\mathfrak{A})$ is the
closed convex hull of its extreme points.

######  Definition 1.45.

An extreme point of $\mathcal{S}(\mathfrak{A})$ is called a pure state of
$\mathfrak{A}$. The set of all pure states of $\mathfrak{A}$ will be
$\mathcal{P}(\mathfrak{A})$. The weak⋆-closure of $\mathcal{P}(\mathfrak{A})$
is called the pure state space of $\mathfrak{A}$.

$\mathcal{P}(\mathfrak{A})$
---
$\mathcal{S}(\mathfrak{A})$
---
mixed states
---
Figure 4: Intuition for the space of states $\mathcal{S}(\mathfrak{A})$. The
pure states lie on the black solid boundary line. The line, however, is not
closed; it may miss some points, which are indicated in red. The weak⋆-closure
of that line is the whole boundary and it is the space of pure states
$\mathcal{P}(\mathfrak{A})$. In particular, the space of pure states may
contain elements which are not pure states (the red points). The interior of
the set is sometimes called the space of mixed states.

Visually we can picture the state space as in Figure 4. In general, the set of
pure states is not weak⋆-closed, and $\mathcal{P}(\mathfrak{A})$ can contain
states which are not pure. For commutative $C^{\star}$-algebras, however, this
does not happen. So if $\mathfrak{A}$ is commutative then the space of all
pure states is already weak⋆-closed. This observation relates to a deep result
attributed to Gelfand — sometimes called the Gelfand representation of a
$C^{\star}$-algebra. We state the result without proof.

######  Theorem 1.46 (Gelfand representation.).

Let $\mathfrak{A}$ be a commutative $C^{\star}$-algebra, and
$\mathcal{P}(\mathfrak{A})$ be the set of all pure states of $\mathfrak{A}$.
Then $\mathcal{P}(\mathfrak{A})$ is a compact Hausdorff space, relative to the
weak⋆-topology. For $A\in\mathfrak{A}$ let $\hat{A}$ be a complex-valued
function on $\mathcal{P}(\mathfrak{A})$ defined by $\hat{A}(\rho)=\rho(A)$.
The mapping $A\to\hat{A}$ is a $\star$-isomorphism from $\mathfrak{A}$ onto
the $C^{\star}$-algebra $C(\mathcal{P}(\mathfrak{A}))$.

Remark. By this construction we can associate a compact Hausdorff space to
every commutative $C^{\star}$-algebra. This can be used to show that every
Hausdorff space admits a compactification — the Stone-Čeck compactification.

Remark. If $\mathfrak{A}$ is a commutative $C^{\star}$-algebra then by the
Gelfand representation it is $\star$-isomorphic to the algebra
$C(\mathcal{P}(\mathfrak{A}))$ of continuous functions on the pure state space
$\mathcal{P}(\mathfrak{A})$ of $\mathfrak{A}$. Thus, if $\omega$ is a pure
state of $\mathfrak{A}$ and $f$ is in $C(\mathcal{P}(\mathfrak{A}))$ then
there is a functional $\delta_{\omega}(f)=f(\omega)$. We call such functionals
point evaluations. Therefore a pure state of a commutative $C^{\star}$-algebra
$\mathfrak{A}$ corresponds to a point evaluation on the algebra
$C(\mathcal{P}(\mathfrak{A}))$. Indeed it is a general result required to
prove the Gelfand representation, namely: if $C(X)$ is a $C^{\star}$-algebra
of continuous functions on a compact Hausdorff space $X$ then a state
$\omega\in\mathcal{S}(\mathfrak{A})$ is pure if and only if it is a point
evaluation, i.e. there is an $x\in X$ such that $\omega(f)=\delta_{x}(f)=f(x)$
for every $f\in C(X)$.

####  1.3.3 Representations of a $C^{\star}$-algebra

Up to now we discussed abstract $C^{\star}$-algebras. Now, we will see that
every $C^{\star}$-algebra can be represented as a (sub)algebra of bounded
linear operators on some Hilbert space. From the physicist point of view, it
is this realization that allows us to formulate quantum and classical
mechanics in terms of algebras. After all, as a physicist we understand
operators in terms of how they act, not in terms of to what space they belong.

######  Definition 1.47.

By a representation of $C^{\star}$-algebra $\mathfrak{A}$ on a Hilbert space
$H$, we mean a $\star$-homomorphism $\varphi$ from $\mathfrak{A}$ onto $B(H)$.
If in addition, $\varphi$ is a $\star$-isomorphism (that is one-to-one) then
it is called a faithful representation.

The Gelfand-Naimark theorem assures that every $C^{\star}$-algebra has a
faithful representation on a Hilbert space. However, we only will discus the
construction that provides a (possibly not faithful) representation.

The following terminology is common when discussing representations. Let
$\varphi$ be a representation of a $C^{\star}$-algebra $\mathfrak{A}$ on a
Hilbert space $H$. Recall from our assignment 4 i) that every
$\star$-homomorphism between $C^{\star}$-algebras is continuous, and so
$\varphi$ is continuous, i.e. $\|\varphi(A)\|_{H}\leq\|A\|_{\mathfrak{A}}$ for
every $A\in\mathfrak{A}$. Also recall, that a $\star$-isomorphism (an
injective $\star$-homomorphism) is isometric, and so if $\varphi$ is faithful
$\|\varphi(A)\|_{H}=\|A\|_{\mathfrak{A}}$. If you recall the definition of
ideals from assignment sheet 5, you realize that the set
$\big{\\{}A\in\mathfrak{A}\colon\varphi(A)=0\big{\\}}$ is a closed two-sided
ideal in $\mathfrak{A}$. We call this set the kernel of $\varphi$. If there is
a vector $x\in H$ for which the linear subspace

$\displaystyle\varphi(\mathfrak{A})x=\big{\\{}\varphi(A)x\colon
A\in\mathfrak{A}\big{\\}}$ (77)

is everywhere dense in $H$, then $\varphi$ is said to be a cyclic
representation, and $x$ is said to be the cyclic vector for $\varphi$.

######  Example 1.48.

Recall, that we defines a $\star$-homomorphism as a mapping that among other
things maps preserves the unit of the algebra. If for a second, we do not
consider this requirement then the map $\varphi_{0}(A)=0$ for every
$A\in\mathfrak{A}$ is defines a trivial homomorphism and we can include to say
that $\varphi_{0}$ is a trivial representation.

######  Example 1.49.

Let $\mathfrak{A}$ be a subalgebra of bounded linear operators $B(H)$ on the
Hilbert space $H$. Then the inclusion mapping $\varphi(A)=A$ for every
$A\in\mathfrak{A}$ is a representation. This representation is faithful, since
if $A\in\mathfrak{A}$ is not zero, then it is also non-zero in $B(H)$.

Let $H$ be a Hilbert space $K$ be a closed subspace of $H$, that is invariant
under each operator from $\mathfrak{A}$, i.e. the set $\big{\\{}Ax\colon
A\in\mathfrak{A},\ x\in K\big{\\}}$ is a subset of $K$. Let $A|_{K}$ be the
restriction of the operator $A$ to the subspace $K$. It can be checked that
this restriction is a $\star$-homomorphism and thus $A\to
A|_{K}\colon\mathfrak{A}\to B(K)$ is a representation of $\mathfrak{A}$ on the
Hilbert space $K$. In the same way one can see that the orthogonal complement
$K^{\perp}$ is also invariant under $\mathfrak{A}$ and thus the restriction
$A\to A|_{K^{\perp}}$ is a representation of $\mathfrak{A}$ on $K^{\perp}$.

If the inclusion representation $A\in\mathfrak{A}\to A\in B(H)$ is cyclic then
also the restriction representations are cyclic as well. To see this let $P$
be the orthogonal projection from $H$ onto $K$ and $x\in H$ be the cyclic
vector $x\in H$ for the inclusion representation. If $Px=0$ then $x$ is in
$K^{\perp}$. This, violates the fact that the inclusion representation is
cyclic since $K^{\perp}$ is invariant under the action of $\mathfrak{A}$. Thus
$Px\neq 0$ and $Px\in K$ is a cyclic vector for the representation $A\to
A|_{K}$, and $(I-P)x$ is a cyclic vector for the representation $A\to
A|_{K^{\perp}}$.

For the following discussion it will be convenient to define an inner product
as a sesquilinear form which is not necessarily definite. In the following
reminder we quickly recall the properties of the inner product. We also
refresh that the well known Cauchy-Schwarz inequality does not depend on the
definiteness of the inner product.

Reminder: Inner product and the Cauchy-Schwarz inequality. In the following we
use the convention that an inner product on a complex vector space $V$ may not
be definite. That is, we say that a mapping
$(v,w)\to\big{\langle}v,w\big{\rangle}$ from $V\times V$ to $\mathbb{C}$ is an
inner product on $V$ when it is i ) positive:
$\big{\langle}v,v\big{\rangle}\geq 0$; ii ) hermitian:
$\big{\langle}v,w\big{\rangle}=\overline{\big{\langle}w,v\big{\rangle}}$; iii
) sesquilinear: $\big{\langle}v,\alpha w+\beta
z\big{\rangle}=\alpha\big{\langle}A,B\big{\rangle}+\beta\big{\langle}A,C\big{\rangle}$;
with $v,w,z\in V$ and $\alpha,\beta\in\mathbb{C}$. If in addition it is iv )
definite: $\big{\langle}v,v\big{\rangle}=0$ only when $v=0$, then we call
$\big{\langle}\ ,\ \big{\rangle}$ a definite inner product.

######  Proposition 1.50.

Let $V$ be a complex vector space. Suppose that $\big{\langle}\ ,\
\big{\rangle}$ is an inner product on $V$. Then, i ) the Cauchy-Schwarz
inequality,
$|\big{\langle}v,w\big{\rangle}|^{2}\leq\big{\langle}v,v\big{\rangle}\big{\langle}w,w\big{\rangle}$,
holds for $v,w\in V$; ii ) the set $\mathfrak{L}=\\{v\in V\colon\
\big{\langle}v,v\big{\rangle}=0\\}$ is a linear subspace of $V$; iii ) the
equation
$\displaystyle\big{\langle}v+\mathfrak{L},w+\mathfrak{L}\big{\rangle}_{q}=\big{\langle}v,w\big{\rangle}\qquad\big{(}v,w\in
V\big{)}$ (78) defines a definite inner product $\big{\langle}\ ,\
\big{\rangle}_{q}$ on the quotient space $V/\mathfrak{L}$.

###### Proof.

For the proof of the above Proposition consult e.g. [1, prop. 2.1.1] ∎

######  Proposition 1.51.

Let $\mathfrak{A}$ be a $C^{\star}$-algebra, $\rho$ a state on $\mathfrak{A}$,
and
$\mathfrak{L}_{\rho}=\big{\\{}A\in\mathfrak{A}\colon\rho(A^{\star}A)=0\big{\\}}$.
The state $\rho$ provides a definite inner product on the quotient space
$\mathfrak{A}/\mathfrak{L}_{\rho}$ by the relation

$\displaystyle\big{\langle}A+\mathfrak{L}_{\rho},B+\mathfrak{L}_{\rho}\big{\rangle}=\rho(A^{\star}B)\qquad\big{(}A,B\in\mathfrak{A}\big{)},$
(79)

Moreover, $\mathfrak{L}_{\rho}$ is a closed left ideal in $\mathfrak{A}$, and
$\rho(A^{\star}B)=0$ whenever $A\in\mathfrak{L}_{\rho}$ and
$B\in\mathfrak{A}$.

###### Proof.

For $A,B\in\mathfrak{A}$ define
$\big{\langle}A,B\big{\rangle}_{\rho}=\rho(A^{\star}B)$. Then $\big{\langle}\
,\ \big{\rangle}_{\rho}$ is positive since $\rho$ is positive, hermitian since
$\rho$ is hermitian, and sesquilinear by definition. Thus, $\big{\langle}\ ,\
\big{\rangle}$ is an inner product on $\mathfrak{A}$ with
$\mathfrak{L}_{\rho}=\big{\\{}A\in\mathfrak{A}\colon\big{\langle}A,A\big{\rangle}_{\rho}=0\big{\\}}$.
By Proposition 1.50, the set $\mathfrak{L}_{\rho}$ is a linear subspace of
$\mathfrak{A}$, and
$\big{\langle}A+\mathfrak{L}_{\rho},B+\mathfrak{L}_{\rho}\big{\rangle}=\big{\langle}A,B\big{\rangle}_{\rho}=\rho(A^{\star}B)$
is a definite inner product on $\mathfrak{A}/\mathfrak{L}_{\rho}$.

With $A\in\mathfrak{L}_{\rho}$ and $B\in\mathfrak{A}$ we have by the Cauchy-
Schwarz inequality

$\displaystyle|\rho(A^{\star}B)|^{2}\leq\rho(A^{\star}A)\rho(B^{\star}B)=0,$
(80)

and so $\rho(A^{\star}B)=0$. It follows that

$\displaystyle\rho\big{(}(BA)^{\star}BA\big{)}=\rho\big{(}A^{\star}(B^{\star}BA)\big{)}=0,$
(81)

and thus $BA$ is in $\mathfrak{L}_{\rho}$ whenever $A\in\mathfrak{L}_{\rho}$
and $B\in\mathfrak{A}$. Hence, $\mathfrak{L}_{\rho}$ is a left ideal of
$\mathfrak{A}$, which is closed since $\rho$ and the mapping $A\to A^{\star}A$
are continuous. ∎

We will call the space $\mathfrak{L}_{\rho}$ the left kernel of the state
$\rho$. We can now summarize the idea of the GNS-construction in a loosely way
before proving it rigorously. The preceding Proposition provides us with means
to define a (definite) inner product on $\mathfrak{A}$. With the norm induced
by this inner product, the vector space $\mathfrak{A}$ may not be closed, but
after closing it we obtain a Hilbert space. Moreover, since $\mathfrak{A}$ is
an algebra we can define an action of $A\in\mathfrak{A}$ on the vector space
$\mathfrak{A}$ by setting $B\to AB\colon\mathfrak{A}\to\mathfrak{A}$. After
proving that this mapping is continuous we can extend it to the whole $H$.
This will give us the desired representation. As always, however, the devil
lies in the complement of a dense set…

######  Theorem 1.52 (GNS representation).

For every state $\rho$ on a $C^{\star}$-algebra $\mathfrak{A}$ there is a
cyclic representation $\pi_{\rho}$ of $\mathfrak{A}$ on a Hilbert space
$H_{\rho}$ and a unit cyclic vector $x_{\rho}$ for $\pi_{\rho}$ such that

$\displaystyle\rho(A)=\big{\langle}x_{\rho},\pi_{\rho}(A)x_{\rho}\big{\rangle}\qquad\big{(}A\in\mathfrak{A}\big{)}.$
(82)

If $\varphi$ is a cyclic representation of $\mathfrak{A}$ on a Hilbert space
$H$ with the unit cyclic vector $x$ for $\pi$ such that

$\displaystyle\rho(A)=\big{\langle}x,\varphi(A)x\big{\rangle}\qquad\big{(}A\in\mathfrak{A}\big{)},$
(83)

then there is an isomorphism $U\colon H_{\rho}\to H$ from $H_{\rho}$ onto $H$
such that

$\displaystyle x=Ux_{\rho}$ and
$\displaystyle\varphi(A)=U\pi_{\rho}(A)U^{\ast}\qquad\big{(}A\in\mathfrak{A}\big{)}.$
(84)

###### Proof.

Let $\mathfrak{L}_{\rho}$ be the left kernel of $\rho$. The quotient linear
space $\mathfrak{A}/\mathfrak{L}_{\rho}$ is a pre-Hilbert space relative to
the definite inner product defined by

$\displaystyle\big{\langle}A+\mathfrak{L}_{\rho},B+\mathfrak{L}_{\rho}\big{\rangle}=\rho(A^{\star}B)\qquad\big{(}A,B\in\mathfrak{A}\big{)}.$
(85)

Define $H_{\rho}$ as the completion of $\mathfrak{A}/\mathfrak{L}_{\rho}$ in
the norm induced by the above scalar product.

To show that for $A\in\mathfrak{A}$ the mapping $B+\mathfrak{L}_{\rho}\to
AB+\mathfrak{L}_{\rho}$ defines an unambiguous operator on $H_{\rho}$, we need
to show that it does not depend on the “representative” $B$ of the set
$B+\mathfrak{L}_{\rho}$. For this assume $B,C\in\mathfrak{A}$ such that
$B+\mathfrak{L}_{\rho}=C+\mathfrak{L}_{\rho}$. Then $B-C$ is in
$\mathfrak{L}_{\rho}$ and since $\mathfrak{L}_{\rho}$ is a left ideal of
$\mathfrak{A}$ we have for every $A\in\mathfrak{A}$ that $AB-
AC\in\mathfrak{L}_{\rho}$ and consequently
$AB+\mathfrak{L}_{\rho}=AC+\mathfrak{L}_{\rho}$. Hence, the equation
$\pi(A)(B+\mathfrak{L}_{\rho})=AB+\mathfrak{L}_{\rho}$ defines a linear
operator $\pi(A)$ acting on the pre-Hilbert space
$\mathfrak{A}/\mathfrak{L}_{\rho}$.

Next, we want to show that $\pi$ is continuous. For this recall that

$\displaystyle\|A\|^{2}I-A^{\star}A=\|A^{\star}A\|I-A^{\star}A\in\mathfrak{A}^{+},$
(86)

implies that the positive square root
$K=\big{(}\|A\|^{2}I-A^{\star}A\big{)}^{1/2}$ exists, and
$B^{\star}(\|A\|^{2}I-A^{\star}A)B=(KB)^{\star}(KB)\in\mathfrak{A}^{+}$.
Therefore for each $A,B\in\mathfrak{A}$ we have

$\displaystyle\|A\|^{2}\|B+\mathfrak{L}_{\rho}\|^{2}-\|\pi(A)(B+\mathfrak{L}_{\rho})\|^{2}$
$\displaystyle=\|A\|^{2}\|B+\mathfrak{L}_{\rho}\|^{2}-\|AB+\mathfrak{L}_{\rho}\|^{2}$
$\displaystyle=\|A\|^{2}\big{\langle}B+\mathfrak{L}_{\rho},B+\mathfrak{L}_{\rho}\big{\rangle}-\big{\langle}AB+\mathfrak{L}_{\rho},AB+\mathfrak{L}_{\rho}\big{\rangle}$
$\displaystyle=\|A\|^{2}\rho(B^{\star}B)-\rho(B^{\star}A^{\star}AB)$
$\displaystyle=\rho\big{(}B^{\star}(\|A\|^{2}I-A^{\star}A)B\big{)}\geq 0.$

Thus $\pi(A)$ is bounded, with $\|\pi(A)\|\leq\|A\|$; and $\pi(A)$ extends by
continuity to a bounded linear operator $\pi_{\rho}(A)$ acting on $H_{\rho}$.
It remains to show that $\pi_{\rho}$ is a $\star$-homomorphism and that it is
cyclic.

With $A,B,C\in\mathfrak{A}$ and $\alpha,\beta\in\mathbb{C}$ we get:
linearity,

$\displaystyle\pi_{\rho}(\alpha A+\beta B)(C+\mathfrak{L}_{\rho})=(\alpha
A+\beta B)C+\mathfrak{L}_{\rho}$
$\displaystyle=\alpha(AC+\mathfrak{L}_{\rho})+\beta(BC+\mathfrak{L}_{\rho})$
$\displaystyle=\big{(}\alpha\pi_{\rho}(A)+\beta\pi_{\rho}(B)\big{)}(C+\mathfrak{L}_{\rho});$

multiplicativity,

$\displaystyle\pi_{\rho}(AB)\big{(}C+\mathfrak{L}_{\rho}\big{)}=ABC+\mathfrak{L}_{\rho}=\pi_{\rho}(A)\big{(}BC+\mathfrak{L}_{\rho}\big{)}=\pi_{\rho}(A)\pi_{\rho}(B)\big{(}C+\mathfrak{L}_{\rho}\big{)};$

star preservation,

$\displaystyle\big{\langle}B+\mathfrak{L}_{\rho},\pi_{\rho}(A)(C+\mathfrak{L}_{\rho})\big{\rangle}$
$\displaystyle=\big{\langle}B+\mathfrak{L}_{\rho},AC+\mathfrak{L}_{\rho}\big{\rangle}$
$\displaystyle=\rho\big{(}B^{\star}(AC)\big{)}=\rho\big{(}(A^{\star}B)^{\star}C\big{)}$
$\displaystyle=\big{\langle}A^{\star}B+\mathfrak{L}_{\rho},C+\mathfrak{L}_{\rho}\big{\rangle}=\big{\langle}\pi_{\rho}(A^{\star})(B+\mathfrak{L}_{\rho}),C+\mathfrak{L}_{\rho}\big{\rangle}.$

Since $\mathfrak{A}/\mathfrak{L}_{\rho}$ is everywhere dense in $H_{\rho}$
these properties extend to the whole $H_{\rho}$ and thus $\pi_{\rho}$ is a
representation of $\mathfrak{A}$ on $H_{\rho}$.

To see that $\pi_{\rho}$ is cyclic let $x_{\rho}$ be the vector
$I+\mathfrak{L}_{\rho}$ in $\mathfrak{A}/\mathfrak{L}_{\rho}$, then

$\displaystyle\pi_{\rho}(A)x_{\rho}=\pi_{\rho}(A)(I+\mathfrak{L}_{\rho})=A+\mathfrak{L}_{\rho}\qquad\big{(}A\in\mathfrak{A}\big{)},$
(87)

and $\pi_{\rho}(\mathfrak{A})x_{\rho}$ is $\mathfrak{A}/\mathfrak{L}_{\rho}$ —
an everywhere-dense subset of $H_{\rho}$. Hence $x_{\rho}$ is a cyclic vector
for $\pi_{\rho}$. Moreover,
$\big{\langle}x_{\rho},\pi_{\rho}(A)x_{\rho}\big{\rangle}=\big{\langle}I+\mathfrak{L}_{\rho},A+\mathfrak{L}_{\rho}\big{\rangle}=\rho(A)$
for every $A\in\mathfrak{A}$. In particular, $\|x_{\rho}\|^{2}=\rho(I)=1$.
This proves the first part of the theorem.

For the second part, let $\varphi$ be another representation as stated in the
theorem. For each $A\in\mathfrak{A}$,

$\displaystyle\|\varphi(A)x\|^{2}=\big{\langle}\varphi(A)x,\varphi(A)x\big{\rangle}=\big{\langle}x,\varphi(A^{\star}A)x\big{\rangle}=\rho(A^{\star}A)=\big{\langle}x_{\rho},\pi_{\rho}(A^{\star}A)x_{\rho}\big{\rangle}=\|\pi_{\rho}(A)x_{\rho}\|^{2}.$

Hence, if $A,B\in\mathfrak{A}$ are such that
$\pi_{\rho}(A)x_{\rho}=\pi_{\rho}(B)x_{\rho}$ then inserting $A-B$ instead of
$A$ in the above relations, yields $\varphi(A)x=\varphi(B)x$. Therefore, we
can define a map
$U_{0}\colon\pi_{\rho}(\mathfrak{A})x_{\rho}\to\varphi(\mathfrak{A})x$ by the
equation $U_{0}\pi_{\rho}(A)x_{\rho}=\varphi(A)x$ for each $A\in\mathfrak{A}$.
It follows that $U_{0}$ is a norm-preserving linear operator from
$\pi_{\rho}(\mathfrak{A})x_{\rho}$ onto $\varphi(\mathfrak{A})x$. Since the
closure of $\pi_{\rho}(\mathfrak{A})x_{\rho}$ is the entire $H_{\rho}$ and the
closure of $\varphi(\mathfrak{A})x$ is the entire $H$, it follows that $U_{0}$
extends by continuity to an isomorphism $U$ from $H_{\rho}$ onto $H$, and

$\displaystyle Ux_{\rho}=U_{0}\pi_{\rho}(I)x_{\rho}=\varphi(I)x=x.$ (88)

With $A,B\in\mathfrak{A}$,

$\displaystyle
U\pi_{\rho}(A)\pi_{\rho}(B)x_{\rho}=U\pi_{\rho}(AB)x_{\rho}=\varphi(AB)x=\varphi(A)\varphi(B)x=\varphi(A)U\pi_{\rho}(B)x_{\rho}.$
(89)

Since vectors of the form $\pi_{\rho}(B)x_{\rho}$ with $B\in\mathfrak{A}$ form
an everywhere-dense subset of $H_{\rho}$, it follows that
$U\pi_{\rho}(A)=\varphi(A)U$, and thus $\varphi(A)=U\pi_{\rho}(A)U^{\star}$. ∎

The method used in the proof of the above Theorem is called the GNS (Gelfand-
Naimark-Segal) construction. To avoid bloated notation, we will often write
$(\pi,H,x)$ to refer to a cyclic representation $\pi$ of a $C^{\star}$-algebra
$\mathfrak{A}$ on the Hilbert space $H$ with a cyclic vector $x$ for $\pi$.
When its convenient we will also call the triple $(\pi,H,x)$ itself a cyclic
representation of $\mathfrak{A}$. If $\rho$ is a state on $\mathfrak{A}$, and
$(\pi_{\rho},H_{\rho},x_{\rho})$ is the representation set out in the Theorem
1.52, we will call it the GNS representation.

For a given $C^{\star}$-algebra $\mathfrak{A}$, let $\pi$ be a representation
of $\mathfrak{A}$ on the Hilbert space $H$ and $\varphi$ be another
representation of $\mathfrak{A}$ on the Hilbert space $K$. We will call the
two representations (unitarily) equivalent if there exists an isomorphism
$U\colon H\to K$ such that $\varphi(A)=U\pi(A)U^{\star}$ for each
$A\in\mathfrak{A}$. In particular, the GNS Theorem states that a GNS
representation $\pi_{\omega}$ of $\mathfrak{A}$ and the cyclic state
$x_{\omega}$ is uniquely determined (up to unitary equivalence) by the
condition
$\rho(\cdot)=\big{\langle}x_{\omega},\pi_{\omega}(\cdot)x_{\omega}\big{\rangle}$.

We already mentioned that if $\pi$ is a representation of a
$C^{\star}$-algebra on a Hilbert space $H$ then $H$ may have invariant
subspaces $K\subset H$ such that $\pi(\mathfrak{A})K\subseteq K$. If this
happens, we call $\pi$ a reducible representation. If on the other hand, $H$
has no such invariant subspace, and the only invariant subspace is $H$ itself,
then we call $\pi$ an irreducible representation. By restricting a reducible
representation to the invariant subspaces we can get an irreducible
representation from a reducible one. Conversely, by “piling up” (direct sum)
irreducible representations we can construct new representations, which will
be reducible. In this sense, irreducible representations are the building
blocks of representations. Since $C^{\star}$-algebra representations play a
vital role in physics, it is important to characterize them. From the above it
follows that it is enough to characterize only the irreducible
representations. This is a more tractable undertaking.

Using the GNS construction one can draw a connection between irreducible
representations of a $C^{\star}$-algebra and its pure states. However, we are
not going to do this. Instead we will now transition to the mathematical
description of physical systems to find out how $C^{\star}$-algebras enter
physics. We close this section with an example that will be important for the
next section.

######  Example 1.53.

Consider a $C^{\star}$-algebra $\mathfrak{A}$ and a state $\rho$ on
$\mathfrak{A}$.

  1. i )

The GNS representation for the state $\rho$ is the $\star$-homomorphism
$\pi_{\rho}\colon\mathfrak{A}\to B(H_{\rho})$, with the Hilbert space
$H_{\rho}$ and a cyclic (GNS) vector $x_{\rho}$;

  2. ii )

Let $y$ be a unit vector in $H_{\rho}$. Then we can use $y$ instead of the
cyclic vector for $\pi_{\rho}$. If $y$ is cyclic for $\pi_{\rho}$ then
$(\pi_{\rho},H_{\rho},y)$ is a new cyclic representation which is equivalent
to the GNS representation $\pi_{\rho}$. In fact, the $\star$-homomorphism and
the Hilbert spaces are both the same, and thus $U=I$ is the desired
isomorphism. (Notice that we do not need that the cyclic vectors transform
from one to the other).

  3. iii )

For $y$ from the previous example, set
$\omega_{y}(A)=\big{\langle}y,\pi_{\omega}(A)y\big{\rangle}$ for every
$A\in\mathfrak{A}$. Then $\omega_{y}$ is a positive linear functional on
$\mathfrak{A}$ with $\omega_{y}(I)=1$; thus $\omega_{y}$ is a state on
$\mathfrak{A}$. The GNS representation
$(\pi_{\omega_{y}},H_{\omega_{y}},x_{\omega_{y}})$ is thus unitarily
equivalent to the representation from the previous example.

From the above example we see that every cyclic state of the GNS Hilbert space
$H_{\omega}$ defines a new state $\omega_{y}$ which leads to a new GNS
representation that is equivalent to the original one. In fact, if the
original state $\omega$ is pure, than its GNS representation is irreducible
and every vector $y\in H_{\omega}$ is cyclic. Thus, every vector from the GNS
representation of a pure state, defines an equivalent representation.

$\mathfrak{A}$
---
$H$
---
$B(H_{1})$
---
$B(H_{n})$
---
$\mathcal{S}(\mathfrak{A})$
---
$\mathcal{S}(\mathfrak{A})$
---
Figure 5: The state space $\mathcal{S}(\mathfrak{A})$ yields different GNS
representations. Some are equivalent some are not. In particular, each
(cyclic) vector from the GNS Hilbert space defines a state. The GNS
representations of these states are all equivalent to the original
representation. Also other states (not created from vectors from the Hilbert
space) can yield equivalent representations.

##  2 Operator algebras in physics

Section remarks: This section is based on two books. The part on classical
mechanics as well as the formulation of the Dirac-von Neumann axioms is
adopted from [6]. The axioms for a general physical system are from [7].
Nevertheless, axiom 4 is shortened, because otherwise we would need too much
additional notation. Also the final axiom from [7] is missing entirely,
because by now we have a classification of Jordan-Banach algebras that make
the axiom (more or less) obsolete. The operational motivation (viewing
observables as measurement devices) is again from [6]. Further reading: the
book by Strocchi [6] is very commendable for a simple introduction to the
physical side of the discussion. Despite being a book for mathematicians, it
focuses on the general picture to phrase quantum mechanics in the algebraic
language. It is easy to read and is not too heavy on the mathematical details.
As a complementary read the book by Emch [7] is mathematically much more
precise and helps to understand how careful and rigorous the algebraic
formulation is constructed. For the algebraic formulation of statistical
physics Ruelle [8] is a great reference. Alternatively, the two volumes by
Bratteli and Robinson [9, 10] give a self-consistent introduction to the
subject. The first volume covers the $C^{\star}$-algebra theory, the second
volume discusses the algebraic formulation of quantum mechanics and non-
relativistic quantum field theory. Contrary to Strocchi and Emch, however,
these books focus almost exclusively on the mathematics. The deep physical
ideas that historically lied underneath the subject are not a matter of
discussion in these two books.

###  2.1 Classical mechanics

We separate the description of a physical system into two parts: kinematics
and dynamics. Kinematics describes the possible configurations of the system;
dynamics describes their evolution in time.

####  2.1.1 Kinematics

Kinematics is concerned with describing configurations of a physical system —
objects that uniquely determine all experimental measurements. Such
configurations are called states. To describe the states we usually choose
(independent) variables that determine the state uniquely. Each such
independent physical variable is sometimes termed the (physical) degree of
freedom.

One way to define a system and to identify its degrees of freedom is to use
the Lagrangian formalism. Another, is to use the Hamiltonian description. In
the latter, the state is uniquely described in terms of canonical variables,
which one can think of as (generalized) positions and (generalized) momenta of
a particle. The state of a mechanical system is described by a set of
canonical variables, $\\{q,p\\}$ where $q=(q_{1},\dots
q_{n})\in\mathbb{R}^{n}$ and $p=(p_{1},\dots,p_{n})\in\mathbb{R}^{n}$. The
collection of all possible canonical variables for the system is called the
phase space $\Gamma$. For simplicity, we will assume in the following that the
phase space is compact, that is the range of possible positions and momenta of
a particle are bounded. Since the kinetic energy of a particle with the mass
$m$ is given by

$\displaystyle E_{kin}(q,p)=\frac{p^{2}}{2m},$ (90)

this assumption means that the particle under consideration is confined to a
bounded region and has finite energy. Thus this assumption is reasonable.

##### Observables

of the system are the physical quantities that result from the physical
degrees of freedom. Thus in particular the position $q$ and the momentum $p$
are themselves observables. But all its powers and sums are observables too,
and so the observables include the polynomials of the canonical variables $q$
and $p$. Without loss of generality, we can take the closure of this space in
the uniform topology (supremum norm), to obtain that all observables form the
space of real-valued continuous function $C(\Gamma)_{\mathbb{R}}$.

Equipped with a pointwise multiplication $C(\Gamma)_{\mathbb{R}}$ forms a real
algebra. At the same time, $C(\Gamma)_{\mathbb{R}}$ is the subspace of self-
adjoint elements of the $C^{\star}$-algebra $C(\Gamma)$ equipped with the
conjugation as the star operation and topologized with the supremum norm.
Extending $C(\Gamma)_{\mathbb{R}}$ to $C(\Gamma)$ is quite natural from the
algebraic point of view. It is also not harmful from the physics point of view
since a $C^{\star}$-algebra is uniquely determined by its self-adjoint
elements as we have seen. Thus, we see that the space of observables of a
classical system defines a $C^{\star}$-algebra with identity $I$ — the
constant unit function on $\Gamma$.

##### Physical states

define the physical system at hand. From our description, every state is a
point of the phase space $\\{q,p\\}=P\in\Gamma$, and it uniquely determines
the value of all the observables. Conversely, the value of all the observables
uniquely determines the state by the Urysohn’s and the Stone-Weierstrass
theorem. This is sometimes called the duality relation between the states and
observables.

Identifying states of a classical system with points of a phase space is,
however, an idealization. It requires infinitely precise measurements to
resolve single points of $\Gamma$. For realistic measurements, we always have
measurement errors and noise that corrupt the measurement. The general
procedure (and the one that is in fact used in physics) takes into account
these errors. To overcome the noise problem we typically repeat the experiment
several times and take the average of the measurement results. More precisely,
let us refer to the state of a classical system by $\omega$. Then the value of
an observable $f\in C(\Gamma)$ in the state $\omega$ is given by the following
procedure: we take $n$ replicas of $\omega$ and perform measurements of $f$ on
each of the replicas. With this we obtain the measurement results
$m^{(\omega)}_{1}(f),m^{(\omega)}_{2}(f),\dots,m^{(\omega)}_{n}(f)$;
subsequently we compute the average

$\displaystyle\big{\langle}f\big{\rangle}_{n}^{(\omega)}=\frac{1}{n}\big{(}m^{(\omega)}_{1}(f)+m^{(\omega)}_{2}(f)+\cdots+m^{(\omega)}_{n}(f)\big{)}$
(91)

It is one of the basic assumptions of experimental physics that for
$n\to\infty$ the above quantity converges. This limit value is the expectation
value of the observable $f$ in the state $\omega$. It is this value that we
assign to a measurement

$\displaystyle\omega(f)=\lim_{n\to\infty}\big{\langle}f\big{\rangle}_{n}^{(\omega)}.$
(92)

Since the expectation values of all the observables is all there is that we
can measure, we can identify a state of a physical system with the set of
expectation values of its observables. Since $\omega(f)$ is operationally
defined by the averaging process it follows that such expectations are linear.
That is, for $f,g\in C(\Gamma)$ and $\alpha,\beta\in\mathbb{C}$ we have

$\displaystyle\omega(\alpha f+\beta g)=\alpha\omega(f)+\beta\omega(g).$ (93)

Also such expectations are positive

$\displaystyle\omega(f^{\star}f)\geq 0\qquad\big{(}f\in C(\Gamma)\big{)},$
(94)

since each measurement $m^{(\omega)}_{k}(f^{\star}f)$ yields a positive number
for every $k\in\mathbb{N}$. Thus $\omega$ defines a positive linear functional
on the $C^{\star}$-algebra $C(\Gamma)$.

As we have seen such functionals induce an inner product on the
$C^{\star}$-algebra by

$\displaystyle\big{\langle}f,g\big{\rangle}=\omega(f^{\star}g)\qquad\big{(}f,g\in
C(\Gamma)\big{)},$ (95)

for which the Cauchy-Schwarz inequality holds

$\displaystyle|\big{\langle}f,g\big{\rangle}|^{2}\leq\big{\langle}f,f\big{\rangle}\big{\langle}g,g\big{\rangle}\qquad\big{(}f,g\in
C(\Gamma)\big{)}.$ (96)

In particular, this implies that for every $f\in C(\Gamma)$ we have

$\displaystyle|\omega(f)|\leq\omega(I)\omega(f^{\star}f).$ (97)

If $\omega(I)=0$ then $\omega(f)=0$ for every $f\in C(\Gamma)$, meaning that
$\omega$ is a trivial functional. Conversely, if $\omega$ is non-trivial then
$\omega(f)\neq 0$ for at leas one $f\in C(\Gamma)$ and thus $\omega(I)\neq 0$.
In this case, we can normalize $\omega$ to satisfy $\omega(I)=1$. Thus we can
identify physical states of a classical system with (algebraic) states on the
algebra of observables $C(\Gamma)$.

Reminder: Riesz-Markov representation theorem. Let $X$ be a locally compact
Hausdorff space. For any positive linear functional $\omega$ on $C(X)$, there
is a unique Radon measure $\mu$ on $X$ such that
$\displaystyle\omega(f)=\int_{X}f(x)\ d\mu(x)\qquad\big{(}f\in C(X)\big{)}.$
(98)

###### Proof.

For the proof consult e.g. [3, thm. IV. 14]. ∎

From the Riesz-Markov theorem it follows that a state $\omega$ on $C(\Gamma)$
defines a Radon measure (regular Borel measure) $\mu_{\omega}$ on $\Gamma$
such that

$\displaystyle\omega(f)=\int_{\Gamma}f(q,p)\
d\mu_{\omega}(q,p)\qquad\big{(}f\in C(\Gamma)\big{)}.$ (99)

Since $1=\omega(I)=\int_{\Gamma}d\mu_{\omega}=\mu(\Gamma)$ we see that $\mu$
defines a probability measure on $\Gamma$. Thus such an operational
description of a physical system leads to a probabilistic description of a
state, and observables then get the meaning of random variables.

As a special case, when $\mu_{\delta_{P}}$ is a Dirac measure such that, for
$O$ an open subset of $\Gamma$

$\displaystyle\mu_{\delta_{P}}(O)=\begin{cases}1\text{ when }P\in O;\\\
0\text{ when }P\notin O\end{cases}$ (100)

If $P=\\{q_{0},p_{0}\\}$ we have

$\displaystyle\delta_{P}(f)=\int_{\Gamma}f(q,p)\
d\mu_{\delta_{P}}(q,p)=f(P)=f(q_{0},p_{0}).$ (101)

Thus $\delta_{P}$ is an evaluation state, and from the Gelfand representation
Theorem 1.46 and the remark that follows it, we saw that such evaluation
states define pure state of the $C^{\star}$-algebra $C(\Gamma)$. Thus the
standard description of classical mechanics, where points of the phase space
define the states, corresponds to pure states in the algebraic formalism.

To quantify the precision of measurements for an observable $f\in C(\Gamma)$
in a state $\omega$, we use the variance

$\displaystyle(\Delta_{\omega}f)^{2}=\omega\big{(}(f-\omega(f)^{2})\big{)}.$
(102)

The smaller the variance, the sharper is the measurement — the more “reliable”
is the mean value $\omega(f)$. Pure states such as $\delta_{P}$ ($P\in\Gamma$)
minimize the variance, as for them we have
$\delta_{P}(f^{2})=\delta_{P}(f)^{2}=f(P)^{2}$ and thus
$\Delta_{\delta_{P}}f=0$. Thus such states are called dispersion-free, as they
represent measurements of a classical system without any uncertainty.

Since algebraic states capture both, the pure and the mixed states, they
provide a unified framework to define sharp measurements as well as
statistical fluctuations. For that reason we can use the same formalism to
describe statistical mechanics. In fact, in statistical mechanics the
description of the system with pure states is unfeasible because we would need
to know the exact positions and momenta for $10^{23}$ particles. Therefore,
the only reasonable description of such systems is using probability
distributions for states and random variables to describe observables. This is
what we usually encounter in statistical approach to thermodynamics: the
partition function there defines a probability distribution, and observables
are random variables sampled from that distribution.

####  2.1.2 Dynamics

Dynamics of a physical system describes how the system evolves in time. When
we use canonical variables $\\{q,p\\}$ to describe the system, then dynamics
is defined by the time evolution of these variables

$\displaystyle q\to q(t)$ $\displaystyle=q(t,q_{0},p_{0})$ $\displaystyle p\to
p(t)$ $\displaystyle=p(t,q_{0},p_{0}).$ (103)

In general, the evolution of canonical variables is given by the solution of a
system of differential equation called the Hamiltonian equations of motion,

$\displaystyle\frac{\partial q}{\partial t}$ $\displaystyle=\frac{\partial
H(q,p)}{\partial p}$ $\displaystyle\frac{\partial p}{\partial t}$
$\displaystyle=-\frac{\partial H(q,p)}{\partial p},$ (104) $\displaystyle
q(t_{0})$ $\displaystyle=q_{0}$ $\displaystyle p(t_{0})$
$\displaystyle=p_{0}.$ (105)

where $H(q,p)$ is called the Hamiltonian of the system. If $grad(H)$ is
continuous, then the solution to the Hamiltonian equations of motions exists;
and if $grad(H)$ is continuously differentiable then the solution is unique.
In this case, it can be extended to all times $t\in\mathbb{R}$ and the
evolution $\\{q,p\\}\to\\{q(t),p(t)\\}$ is continuous in time
$t\in\mathbb{R}$, and depends continuously on the initial data
$\\{q(t_{0}),p(t_{0})\\}=\\{q_{0},p_{0}\\}$.

The evolution of the canonical variables implies that the results of
measurements evolve in time. Since all observables are constructed from
canonical variables, all observables evolve in time. In particular, this
implies a one-parameter family of continuous invertible mappings
$\alpha_{t_{0},t}\colon C(\Gamma)\to C(\Gamma)$ from the algebra of
observables into itself so that

$\displaystyle\alpha_{t_{0},t}\colon f(q_{0},p_{0})\to f(q(t),p(t)).$ (106)

If $f,g$ are two polynomials of canonical variables then it is easy to see
that for $\beta,\gamma\in\mathbb{C}$ and every $t\in\mathbb{R}$ we have,

$\displaystyle\alpha_{t_{0},t}(\beta f+\gamma g)$
$\displaystyle=\beta\alpha_{t_{0},t}(f)+\gamma\alpha_{t_{0},t}(g)=\beta
f(t,q_{0},p_{0})+\gamma g(t,q_{0},p_{0}),$ (107)
$\displaystyle\alpha_{t_{0},t}(f^{\star})$
$\displaystyle=\alpha_{t_{0},t}(f)^{\star}=f(t,q_{0},p_{0})^{\star},$ (108)
$\displaystyle\alpha_{t_{0},t}(fg)$
$\displaystyle=\alpha_{t_{0},t}(f)\alpha_{t_{0},t}(g)=f(t,q_{0},p_{0})g(t,q_{0},p_{0}).$
(109)

By continuity, these properties extend to the whole algebra $C(\Gamma)$, and
thus $\alpha_{t_{0},t}$ defines a one-parameter family of
$\star$-automorphisms ($\star$-isomorphisms from the algebra into itself).
When the dynamics depends only on the difference $t-t_{0}$, i.e. it is time-
translation invariant, then the one-parameter family is actually a one-
parameter group of automorphisms $\\{\alpha_{t}\colon t\in\mathbb{R}\\}$ on
$C(\Gamma)$. By duality it defines a one-parameter group of transformations
$\alpha_{t}^{\star}$ on the state space as

$\displaystyle\omega_{t}(f)=(\alpha_{t}^{\star}(\omega))(f)=\omega(\alpha_{t}(f))\qquad\big{(}f\in
C(\Gamma)\big{)}.$ (110)

In summary we see, that observables of a classical system define the
commutative $C^{\star}$-algebra of observables $C(\Gamma)$, and physical
states define (algebraic) state on that algebra; dynamics of the canonical
system is given by a one-parameter group of $\star$-automorphisms on
$C(\Gamma)$. In particular, all physically observable features of a physical
system are captured by the algebraic formulation. However, still in the
standard approach we start with the phase space that describes our system, and
then derive the $C^{\star}$-formulation.

The Gelfand representation Theorem 1.46, however, allows us to reverse the
logic. Instead of starting with the phase space, and deriving the algebra from
it, we can start with a commutative $C^{\star}$-algebra $\mathfrak{A}$ that
captures an abstract characterization of observables of our system; define the
physical states as the state space $\mathcal{S}(\mathfrak{A})$; and define the
dynamics as a $\star$-automorphism on $\mathfrak{A}$. By the Gelfand
representation Theorem the abstract algebra $\mathfrak{A}$ can be represented
as an algebra of continuous functions $C(X)$ on a compact Hausdorff space $X$.
Every dispersion-free state of $\mathfrak{A}$ is associated with the point in
$X$ and so the space $X$ gets the meaning of the phase space. In this sense,
the algebra comes first, from which the phase space is then derived. This is
the starting point for the algebraic construction. Using this motivation we
aim in the next section to define a general algebraic framework for physical
theories.

###  2.2 An algebraic construction for physical theories

The discovery of quantum mechanics soon proved to be pivotal for physics. This
inspired a rapid development in the field. This development was of physical
nature, as more and more quantum mechanical models described more and more
experiments with extraordinary precision. However, on the mathematical side
the theory was shaky, and the concepts were hardly justified on physical
grounds. To formalize the theory, Dirac and von Neumann provided 5 axioms to
summarize a common mathematical structure of quantum mechanics as developed by
Heisenberg and Schrödinger — sometimes referred to as the Dirac-von Neumann
axioms of quantum mechanics. Before we delve into the algebraic framework we
summarize a simplified version of these axioms that define the standard
approach to quantum mechanics.

##### States:

The states of a physical system are described by rays (essentially unit
vectors) in a separable Hilbert space $H$.

##### Observables:

The observables of a physical system are described by the set of bounded self-
adjoint operators in $H$.

##### Expectations:

If a state $\omega$ is described by the vector $\psi\in H$, for any observable
$A$ the expectation $\omega(A)$, defined by the average of the outcomes of
replicated measurements of $A$ performed on the system in the state $\omega$,
is given by the Hilbert space matrix element

$\displaystyle\omega(A)=\big{\langle}\psi,A\psi\big{\rangle}$ (111)

##### Dirac canonical quantization:

Canonical coordinates and momenta satisfy the following commutation relations

$\displaystyle[q_{i},q_{j}]=0=[p_{i},p_{j}],$
$\displaystyle[q_{i},p_{j}]=\imath\hbar\delta_{i,j},\qquad\big{(}i,j=1,2,\dots,n\big{)}.$
(112)

##### Schrödinger representation:

The canonical commutation relations are represented by the following operators
in $H=L^{2}(\mathbb{R}^{n},\lambda)$:

$\displaystyle q_{i}\psi(x)=x_{i}\psi(x),$ $\displaystyle
p_{j}\psi(x)=-\imath\hbar\frac{\partial\psi}{\partial
x_{j}}(x)\qquad\big{(}i,j=1,2,\dots,n\big{)}.$ (113)

Even though the physical framework proved to be very successful, the physical
basis for many of its assumptions was unclear and unjustifiable. For that
reason, a group of mathematicians/physicists pioneered by Jordan, von Neumann,
and Wigner, aimed to provide a conceptually clear, and mathematically rigorous
formalism for the above postulates; to justify some of the tools in use of
quantum mechanics; and to disregard some of the assumptions not supported by
mathematics or physical intuition. In this section we discuss the summary of
that approach that took a long time to be developed.

The idea of the algebraic approach is to use reasonable physical assumptions
to define an algebraic structure on the set of observables of a theory. Then
to show that this algebraic structure is rich enough to provide all the
necessary tools used in quantum mechanics as formulated in the above axioms.
Following the literature, we formulate the necessary assumptions on the set of
observables as axioms.

######  Axiom 1.

To each physical system $\Sigma$ we can associate the triple
$(\mathpzc{A},\mathpzc{S},\big{\langle};\big{\rangle})$ formed by the set
$\mathpzc{A}$ of all observables, the set $\mathpzc{S}$ of all its states, and
a mapping
$\big{\langle};\big{\rangle}\colon(\mathpzc{A},\mathpzc{S})\to\mathbb{R}$ that
assigns to each pair $(A,\phi)\in(\mathpzc{A},\mathpzc{S})$ a real number
$\big{\langle}\phi;A\big{\rangle}$ that we interpret as the expectation value
of the observable $A$ in the state $\phi$.

With a state of a physical system, we identify the method used for its
preparation. That is a (reproducible) way to prepare a configuration of the
system. We assume that a state is fully determined by all its expectation
values, that is if $\phi,\varphi\in\mathpzc{S}$ are such that

$\displaystyle\big{\langle}\phi;A\big{\rangle}=\big{\langle}\varphi;A\big{\rangle}\qquad\big{(}A\in\mathpzc{A}\big{)},$
(114)

then this must imply that $\phi=\varphi$. That is, we assume that there are no
hidden properties that we can never measure. We say the observables separate
the states.

With an observable we identify a measurement device. Thus for any
$A\in\mathpzc{A}$ we can imagine a device with a pointer scale. The
expectation value, is then the average over $n$ measurements of $A$ in
identically prepared states $\omega$. Ideally, $n$ shall go to infinity,
practically it has to be large enough to assure that our average is
meaningful. If any two devices produce the same expectation values for all
states $\phi\in\mathpzc{S}$, then we identify these devices. That is we assume
that each measurement device produces a unique set of expectation values, when
evaluated in all states of $\mathpzc{S}$. We say, that $\mathpzc{S}$ separates
the observables or simply that the set $\mathpzc{S}$ is full with respect to
$\mathpzc{A}$ (see Figure 6 for intuition).

$\ddots$
---
$\mathpzc{A}$
---
$A$
---
$B$
---
$C$
---
(a) Observables: measurement devices.

1) switch on the stove
---
2) take 100g sugar
---
3) eat it
---
4) hope for the best
---
$\vdots$
---
$\cdots$
---
$\mathpzc{S}$
---
(b) States: methods of state preparation.

Figure 6:  The set of observables and the set of states of a physical system.
After preparing a state according to a given method we can measure it with all
observables and identify it with a list of all expectation values. If any to
methods lead to the same list of expectation values we identify these methods.
On the other hand, if any two observables provide two identical lists of
expectation values when measuring all states, we identify them either.

Next we realize that we can use the states to compare observables. For
$A,B\in\mathpzc{A}$ we say that $A\leq B$ if
$\big{\langle}\phi;A\big{\rangle}\leq\big{\langle}\phi;B\big{\rangle}$ for all
state $\phi\in\mathpzc{S}$. In particular, we call $A\geq 0$ positive if
$\big{\langle}\phi;A\big{\rangle}\geq 0$ for every state $\phi\in\mathpzc{S}$.
Especially, this implies that the states are positive functionals on
$\mathpzc{A}$.

######  Axiom 2.

The relation $\leq$ defined above is a partial ordering on $\mathpzc{A}$; in
particular $A\leq B$ and $B\leq A$ imply $A=B$.

With our interpretation of observables as measurement devices we see that the
condition $\big{\langle}\phi;A\big{\rangle}\geq 0$ for every
$\phi\in\mathpzc{S}$ implies that every measurement $A$ is positive. Since the
set of state is full with respect to the set of observables this meas that the
scale of the measurement device represented by $A$ is positive. Thus we can
identify the positive element with measurement devices whose pointer scale
ranges in positive real numbers.

Next we notice that for any two observables $A,B\in\mathpzc{A}$ the set
$\\{\big{\langle}\phi;A\big{\rangle}+\big{\langle}\phi;B\big{\rangle}\colon\phi\in\mathpzc{S}\\}$
can be interpreted as a set of expectation values, thus a measurable quantity
on $\Sigma$. We strengthen this observation to the next axiom.

######  Axiom 3.

  1.   2. i )

For each observable $A\in\mathpzc{A}$ and any $\lambda\in\mathbb{R}$ there
exists an element $(\lambda A)\in\mathpzc{A}$ such that

$\displaystyle\big{\langle}\phi;\lambda
A\big{\rangle}=\lambda\big{\langle}\phi;A\big{\rangle}\qquad\big{(}\phi\in\mathpzc{S}\big{)}.$
(115)

  3. ii )

For any pair of observables $A,B\in\mathpzc{A}$ there exists an element
$(A+B)\in\mathpzc{A}$ such that

$\displaystyle\big{\langle}\phi;A+B\big{\rangle}=\big{\langle}\phi;A\big{\rangle}+\big{\langle}\phi;B\big{\rangle}\qquad\big{(}\phi\in\mathpzc{S}\big{)}.$
(116)

If $A\in\mathpzc{A}$ represents a measurement and $\lambda$ is a real number
then we can think of $\lambda A$ as a new device, whose pointer scale is
rescaled by $\lambda$.

This axiom, even though reasonable, is non-trivial when we interpret elements
in $\mathpzc{A}$ as measurement devices. Using purely operational arguments,
it is not clear that for any two measurement devices represented by $A$ and
$B$ there is always one device $C$ that produces the same expectation values
as the sum of expectation values of $A$ and $B$. Strictly speaking, with this
assumption we cannot interpret the set $\mathpzc{A}$ solely as a set of
measurement devices, but we extend this set to all hypothetical measurements
that result from adding expectation values of measurements together. In
particular, this defines a linear structure on $\mathpzc{A}$ making it into a
vector space. Nevertheless, this assumption is still quite reasonable. Being
pragmatic, we can agree that the set $\mathpzc{A}$ contains all theoretically
measurable devices. Hence also such devices that correspond to measurements of
$A+B$.

Recall from our analysis of a classical system, that dispersion-free states
corresponded to idealized measurements with infinite precision, i.e. to points
of the phase space. Because of this, dispersion-free states uniquely defined
all classical observables. It seems reasonable to expect that this should hold
in a general physical system. We use this intuition to state the next axiom.

######  Axiom 4.

For every observable $A$ the set $\mathpzc{S}$ contains the set of dispersion-
free states $\mathpzc{S}_{A}$ for $A$ that uniquely determines the one-
dimensional subspace of $\mathpzc{A}$ generated by $A$; for any two
observables $A,B$ the dispersion-free states satisfy
$\mathpzc{S}_{A}\cap\mathpzc{S}_{B}\subseteq\mathpzc{S}_{A+B}$ and
$\mathpzc{S}_{I}=\mathpzc{S}$.

In particular, this axiom implies that for two observables $A$ and $B$ if
$\mathpzc{S}_{A}\subseteq\mathpzc{S}_{B}$ and
$\big{\langle}\phi;A\big{\rangle}=\big{\langle}\phi;B\big{\rangle}$ for every
$\phi\in\mathpzc{S}_{A}$ then $A$ is equal to $B$. Thus dispersion-free states
in a certain sense, uniquely determine the observables. This reflects our
original motivation to introduce this axiom. Nevertheless, this axiom requires
an assumption of dispersion-free states, which is an idealization of a
physical process of measurement, as we discussed above. To argue that such
states are reasonable, we can take the standpoint that even though every
realistic measurement is noisy, the noise can be arbitrary reduced. Therefore
there is no fundamental (nature given) threshold which we cannot surpass by
making a measurement more precise/less noisy. (In fact, this may be violated
if one takes General Relativity into account. But we are not going into this
discussion here.)

If $A$ is an observable that represents a measurement device then we can
interpret $A^{2}$ as a measurement device whose scale is the square of the
pointer scale of $A$. In particular, $A^{2}$ is therefore always positive.
Similarly, we can define any power $A^{n}$ for a non-negative integer $n$, and
set $I=A^{0}$. We put the existence of such powers as an additional axiom.

######  Axiom 5.

For any observable $A$ and any non-negative integer $n$ there exists at least
one element $A^{n}\in\mathpzc{A}$ such that

  1. i )

$\mathpzc{S}_{A}\subseteq\mathpzc{S}_{A^{n}}$,

  2. ii )

$\big{\langle}\phi;A^{n}\big{\rangle}=\big{\langle}\phi;A\big{\rangle}^{n}$
for all $\phi\in\mathpzc{S}_{A}$.

From the second point of this Axiom and Axiom 4 we see, that the element
$A^{n}$ is unique in $\mathpzc{A}$ and we can unambiguously call it the $n$th
power of $A$.

With this axiom the vector space $\mathpzc{A}$ obtains a power structure.
Moreover, it follows that $\mathpzc{A}$ contains a unit $I=A^{0}$ for every
$A\in\mathpzc{A}$ and an element $O$ such that

$\displaystyle\big{\langle}\phi;I\big{\rangle}=1$ and
$\displaystyle\big{\langle}\phi;O\big{\rangle}=0\qquad\big{(}\phi\in\mathpzc{S}\big{)}.$
(117)

From the second point of this axiom, it follows that
$\mathpzc{S}_{I}=\mathpzc{S}_{A^{0}}=\mathpzc{S}$, and that the usual rules
apply: $A^{0}=I,\ A^{1}=A,\ I^{n}=I$ and $O^{n}=O$ for $n>1$.

Furthermore, since $\big{\langle}\phi;A^{2}\big{\rangle}\geq 0$ for all
$\phi\in\mathpzc{S}_{A}$ and since $\mathpzc{S}_{A}$ fully determines $A^{2}$
it follows that $A^{2}\geq 0$. In particular this implies that
$\big{\langle}\phi;A^{2}\big{\rangle}\geq 0$ for every state
$\phi\in\mathcal{S}$. Since the set $\mathcal{S}$ is full with respect to
$\mathcal{A}$, we can conclude that $A^{2}+B^{2}+C^{2}+\cdots=0$ implies
$A=B=C=\cdots 0$. This conditions is sometimes called reality, in analogy to
the fact that this implication holds for real numbers (as opposed to complex
numbers).

Even though we have argued for a lot of structure on the space $\mathpzc{A}$,
one important property is still missing: the distributivity of powers over
addition of any two elements $A,B$. In general, for $A,B\in\mathpzc{A}$ we
have that

$\displaystyle(A+B)^{n}=(A+B)(A+B)\neq A(A+B)+B(A+B).$ (118)