Observables as an Algebra

The algebraic formulation begins with a simple observation: quantum observables do not form an unstructured list. They can be added, multiplied, conjugated, and compared through commutators. These operations encode coarse-graining, sequential action, compatibility, symmetry, and dynamics.

The appropriate basic object is therefore a complex unital *-algebra $\mathcal A$. Its elements include the self-adjoint quantities interpreted as observables, but $\mathcal A$ must also contain non-self-adjoint elements. Products of observables need not be observables, and useful operators such as ladder operators are not self-adjoint. Restricting the algebra to self-adjoint elements would discard the multiplication that carries much of the theory’s structure.

This page develops that structure. States as Positive Linear Functionals and Dynamics as Automorphisms have separate canonical homes.

Algebraic Structure

A complex associative algebra $\mathcal A$ is a complex vector space with a bilinear multiplication. For $A,B,C\in\mathcal A$ and $\alpha,\beta\in\mathbb C$,

$$\begin{aligned} (AB)C&=A(BC),\\ (\alpha A+\beta B)C &=\alpha AC+\beta BC,\\ C(\alpha A+\beta B) &=\alpha CA+\beta CB. \end{aligned}$$

Associativity says that a string of products can be regrouped without ambiguity. It does not imply commutativity:

$$AB\neq BA$$

in general. This distinction is decisive. Classical observables multiply as ordinary functions on phase space and therefore commute. Quantum observables are represented by elements of a generally noncommutative algebra.

An algebra is unital if it contains an identity element $I$ satisfying

$$IA=AI=A$$

for every $A\in\mathcal A$. A star algebra, conventionally written *-algebra, also carries an involution $A\mapsto A^*$ compatible with its linear and multiplicative structure. In a Hilbert-space representation, this operation becomes the operator adjoint.

The complex algebra $\mathcal A$ is larger than the real vector space of physical observables:

$$\mathcal A_{\mathrm{sa}} = \{A\in\mathcal A:A^*=A\}.$$

The self-adjoint part $\mathcal A_{\mathrm{sa}}$ is closed under real linear combinations, but it is not generally closed under multiplication. That failure is not a defect; it is precisely what records noncommutativity.

Sums and Products

If $A$ and $B$ represent observables with the same physical dimension, then a real linear combination

$$C=aA+bB, \qquad a,b\in\mathbb R,$$

is again self-adjoint. Operationally, $C$ is a new observable whose expectation value is the corresponding linear combination:

$$\langle C\rangle = a\langle A\rangle+b\langle B\rangle.$$

Complex linear combinations are required for the full algebra even though they need not be self-adjoint. For example,

$$A+iB$$

is generally not an observable by itself, but excluding it would prevent $\mathcal A$ from being a complex vector space and would obstruct useful constructions.

Products require more care. For bounded self-adjoint operators,

$$(AB)^*=BA.$$

Consequently,

$$AB=(AB)^* \quad\Longleftrightarrow\quad AB=BA.$$

Thus the product of two self-adjoint elements is self-adjoint exactly when they commute. For unbounded operators, the same formal identity is not enough: the domains of $AB$, $BA$, and their adjoints must also be controlled.

Two combinations separate the symmetric and antisymmetric content of the product:

$$A\circ B = \frac{1}{2}(AB+BA),$$

and

$$A\times B = \frac{1}{2i}(AB-BA).$$

For self-adjoint $A$ and $B$, both $A\circ B$ and $A\times B$ are self-adjoint in the bounded setting. The first is the Jordan product; the second is the self-adjoint version of the commutator product. Together they reconstruct the original multiplication:

$$AB=A\circ B+iA\times B.$$

This decomposition is useful conceptually. The Jordan product captures the symmetric product structure used in moments and correlations, while the antisymmetric product captures incompatibility and infinitesimal transformations. Neither part alone recovers the complete associative algebra.

Finite-Dimensional Example: A Qubit

For a qubit,

$$\mathcal A=M_2(\mathbb C),$$

the algebra of all complex $2\times2$ matrices. Every element can be written

$$A=a_0I+\sum_{i=1}^{3}a_i\sigma_i, \qquad a_\mu\in\mathbb C,$$

where $\sigma_i$ are the Pauli matrices. The element is self-adjoint exactly when all four coefficients are real.

The multiplication law

$$\sigma_i\sigma_j = \delta_{ij}I + i\sum_{k=1}^{3}\epsilon_{ijk}\sigma_k$$

contains both the symmetric and antisymmetric structures:

$$\frac{1}{2} \{\sigma_i,\sigma_j\} = \delta_{ij}I, \qquad \frac{1}{2i} [\sigma_i,\sigma_j] = \sum_k\epsilon_{ijk}\sigma_k.$$

For example,

$$\sigma_x\sigma_y=i\sigma_z.$$

Both $\sigma_x$ and $\sigma_y$ are observables, but their product is not self-adjoint. The algebra still contains it. This elementary example shows why “the algebra of observables” conventionally means the full *-algebra whose self-adjoint elements are observables, not merely the self-adjoint subset.

The qubit algebra is also closed under functions of its elements. If

$$A=a_0I+\mathbf a\cdot\boldsymbol{\sigma},$$

then the spectral theorem determines $f(A)$ from the two eigenvalues $a_0\pm\lVert\mathbf a\rVert$. The general functional calculus is developed in Functions of Operators.

The Adjoint Operation

The involution of a *-algebra obeys

$$\begin{aligned} (A^*)^*&=A,\\ (\alpha A+\beta B)^* &=\overline{\alpha}A^* +\overline{\beta}B^*,\\ (AB)^*&=B^*A^*. \end{aligned}$$

The reversal of order in the last line is essential. On a Hilbert space it follows from

$$\langle\psi\vert AB\phi\rangle = \langle B^*A^*\psi\vert\phi\rangle,$$

provided the operator domains make the expressions meaningful.

The adjoint identifies several physically important classes of elements:

• self-adjoint elements, $A^*=A$, model real-valued observables;
• unitary elements, $U^*U=UU^*=I$, model reversible transformations;
• projections, $P^*=P=P^2$, model sharp yes-or-no events;
• positive elements, $A\geq0$, support probabilities, effects, and state positivity.

Every element $A^*A$ is positive. Conversely, every positive element of a C*-algebra has a positive square root and can be written as $B^*B$. This makes

$$\omega(A^*A)\geq0$$

the natural positivity requirement for a state $\omega$.

An effect is a positive element $E$ satisfying

$$0\leq E\leq I.$$

Projections are special effects. The broader effect language is useful for generalized measurements, but the measurement theory itself belongs to the POVMs and Quantum Instruments pages.

Commutators and Derivations

The commutator

$$[A,B]=AB-BA$$

measures the failure of two elements to commute. It is bilinear and antisymmetric:

$$[A,B]=-[B,A].$$

It also satisfies the Jacobi identity,

$$[A,[B,C]] + [B,[C,A]] + [C,[A,B]] =0.$$

These properties make the algebra into a Lie algebra under the commutator bracket. At the same time, the commutator acts as a derivation with respect to the associative product:

$$[A,BC] = [A,B]C+B[A,C].$$

This Leibniz rule is why commutators generate infinitesimal transformations. If $G=G^*$ generates a unitary family, then formally

$$\delta_G(A) = \frac{i}{\hbar}[G,A]$$

obeys

$$\delta_G(AB) = \delta_G(A)B+A\delta_G(B).$$

The Commutator Dynamics page develops this dynamical role. Here the point is structural: the associative product simultaneously determines the Jordan product and the commutator, so statistical and transformational information live in one algebra.

For self-adjoint $A$ and $B$,

$$[A,B]^*=-[A,B].$$

The commutator itself is anti-self-adjoint, while $[A,B]/i$ is self-adjoint. Forgetting the factor of $i$ is a common source of sign and interpretation errors.

Why the Identity Matters

The identity element has several related roles:

1. It represents the constant observable with value one.

2. It normalizes states through $\omega(I)=1$.

3. It appears in canonical commutation relations such as

   $$[Q,P]=i\hbar I.$$

4. It lets constants be added to observables:

   $$A\longmapsto A+cI.$$

5. It anchors complements of effects and projections:

   $$E^\perp=I-E.$$

A nonunital algebra can often be enlarged by adjoining an identity, and some applications naturally begin with nonunital algebras. For ordinary quantum mechanics, however, a unital observable algebra is the clean default because normalization and constant quantities are intrinsic.

Composite and Local Algebras

For finite systems $A$ and $B$, the composite observable algebra is

$$\mathcal A_{AB} = \mathcal A_A\otimes\mathcal A_B.$$

An observable local to the first subsystem has the form $X\otimes I$, while one local to the second has the form $I\otimes Y$. They commute:

$$[X\otimes I,I\otimes Y]=0.$$

This algebraic fact expresses compatibility of operations on distinct subsystems. The full construction, including entangling observables and basis conventions, is developed in Operators on Composite Systems.

In algebraic quantum field theory, locality is encoded through a net of algebras

$$\mathcal O\longmapsto\mathcal A(\mathcal O).$$

Observables assigned to suitable spacelike separated regions commute. This is a far-reaching extension of the same structural idea, but local QFT algebras require their own treatment and should not be inferred naively from finite tensor products.

Bounded Versus Unbounded Operators

If $\mathcal H$ is a Hilbert space, the bounded operators

$$\mathcal B(\mathcal H)$$

form a unital -algebra under operator addition, composition, and adjoint. Equipped with the operator norm, they satisfy the C-identity

$$\lVert A^*A\rVert=\lVert A\rVert^2.$$

A norm-closed -subalgebra of $\mathcal B(\mathcal H)$ is a concrete C-algebra. Abstract C*-algebras axiomatize this bounded-operator structure without choosing a particular Hilbert-space realization.

Many central observables are unbounded. Position, momentum, and typical Hamiltonians are not elements of $\mathcal B(\mathcal H)$. Their domains are proper dense subspaces, products may have smaller domains, and formal adjoint identities can fail to imply equality of operators. In particular,

$$D(AB) = \{\psi\in D(B):B\psi\in D(A)\}$$

depends on both operators. Even if $A$ and $B$ are self-adjoint, $AB$ need not be densely defined or closed.

There are several standard ways to retain a bounded algebraic framework:

• use resolvents such as $(A-zI)^{-1}$ for nonreal $z$;
• use spectral projections of self-adjoint operators;
• use bounded functions $f(A)$ from the functional calculus;
• exponentiate generators to unitaries, such as $e^{itA}$;
• formulate canonical relations through Weyl unitaries rather than raw $Q$ and $P$.

These choices are not cosmetic. Writing unbounded symbols as though they formed an everywhere-defined matrix algebra can hide false products, invalid commutators, and nonexistent adjoints. See Bounded Operators, Unbounded Operators, and Domains of Operators for the analytic details.

Relation to Hilbert-Space Operators

An abstract observable algebra does not initially come with a preferred Hilbert space. A representation is a *-homomorphism

$$\pi:\mathcal A\to\mathcal B(\mathcal H)$$

that preserves sums, products, adjoints, and the identity:

$$\begin{aligned} \pi(AB)&=\pi(A)\pi(B),\\ \pi(A^*)&=\pi(A)^*,\\ \pi(I)&=I_{\mathcal H}. \end{aligned}$$

If $\pi$ is injective, it is faithful: distinct abstract algebra elements remain distinct as operators. A nonfaithful representation identifies elements in the kernel of $\pi$.

Finite-dimensional textbook quantum mechanics usually starts after a faithful representation has already been chosen. Then $\mathcal A=M_n(\mathbb C)$ acts on $\mathbb C^n$, and the distinction between abstract element and representing matrix is easy to overlook.

In infinite systems, different physically relevant representations can be unitarily inequivalent. The algebra describes common relations, while a state can select a representation through the GNS construction. The Algebraic Formulation Overview explains that reconstruction and its role in many-body theory and QFT.

The bounded algebra should also not be confused with every field symbol used in a calculation. Quantum fields are generally operator-valued distributions and are unbounded after smearing. Bounded local observable algebras can instead be generated from suitable functions or exponentials of those fields. Keeping these levels distinct prevents the finite-dimensional matrix picture from being extended beyond its domain of validity.

Common Mistakes

• Calling every element of $\mathcal A$ an observable without noting that physical observables are normally the self-adjoint elements.
• Assuming the product of self-adjoint operators is self-adjoint; this requires commutativity in the bounded setting.
• Treating the self-adjoint part as a complex associative algebra. It is closed under real linear combinations and Jordan products, not arbitrary products.
• Reversing the adjoint incorrectly: $(AB)^*=B^*A^*$, not $A^*B^*$.
• Forgetting that $[A,B]$ is anti-self-adjoint when $A$ and $B$ are self-adjoint.
• Using the canonical commutator as an identity between bounded operators. Bounded operators cannot satisfy $[Q,P]=i\hbar I$ under the usual hypotheses.
• Manipulating unbounded products without checking their domains.
• Assuming that an abstract algebra has one preferred Hilbert-space representation.
• Importing finite tensor-product intuition directly into continuum QFT.

References

• O. Bratteli and D. W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987.
• R. V. Kadison and J. R. Ringrose, Fundamentals of the Theory of Operator Algebras, Volume I, AMS, 1997.
• G. J. Murphy, C-Algebras and Operator Theory*, Academic Press, 1990.
• F. Strocchi, An Introduction to the Mathematical Structure of Quantum Mechanics, 2nd ed., World Scientific, 2008.
• R. Haag, Local Quantum Physics: Fields, Particles, Algebras, 2nd ed., Springer, 1996.
• E. M. Alfsen and F. W. Shultz, State Spaces of Operator Algebras: Basic Theory, Orientations, and C-Products*, Birkhäuser, 2001.

Exercises

1. Let $A$ and $B$ be bounded self-adjoint operators. Prove that $AB$ is self-adjoint if and only if $A$ and $B$ commute.

Solution

Because $A^*=A$ and $B^*=B$,

$$(AB)^*=B^*A^*=BA.$$

Therefore $AB$ is self-adjoint exactly when

$$AB=(AB)^*=BA,$$

which is precisely $[A,B]=0$. Boundedness ensures that all products and adjoints are defined on the whole Hilbert space. The same one-line argument is not sufficient for unbounded operators because the relevant domains may differ.

2. Decompose the product of two self-adjoint elements into self-adjoint Jordan and commutator parts.

Solution

Define

$$A\circ B=\frac{AB+BA}{2}, \qquad A\times B=\frac{AB-BA}{2i}.$$

Using $(AB)^*=BA$ gives

$$(A\circ B)^*=A\circ B, \qquad (A\times B)^*=A\times B.$$

Adding the two contributions,

$$A\circ B+iA\times B = \frac{AB+BA}{2} + \frac{AB-BA}{2} =AB.$$

Thus the two self-adjoint products jointly determine the associative product.

3. For $A=a_0I+\mathbf a\cdot\boldsymbol{\sigma}$ and $B=b_0I+\mathbf b\cdot\boldsymbol{\sigma}$ with real coefficients, compute their commutator and determine when they commute.

Solution

The identity terms commute with everything. Using

$$\sigma_i\sigma_j = \delta_{ij}I + i\epsilon_{ijk}\sigma_k,$$

one finds

$$(\mathbf a\cdot\boldsymbol{\sigma}) (\mathbf b\cdot\boldsymbol{\sigma}) = (\mathbf a\cdot\mathbf b)I + i(\mathbf a\times\mathbf b)\cdot\boldsymbol{\sigma}.$$

Reversing $\mathbf a$ and $\mathbf b$ changes the sign of the cross product, so

$$[A,B] = 2i(\mathbf a\times\mathbf b) \cdot\boldsymbol{\sigma}.$$

They commute exactly when $\mathbf a\times\mathbf b=0$, meaning that their Bloch vectors are parallel or one vanishes.

4. Verify the derivation identity for the commutator.

Solution

Expand directly:

$$\begin{aligned} [A,BC] &=ABC-BCA\\ &=ABC-BAC+BAC-BCA\\ &=(AB-BA)C+B(AC-CA)\\ &=[A,B]C+B[A,C]. \end{aligned}$$

Thus the map $\delta_A(B)=[A,B]$ obeys a Leibniz rule. Multiplication by a fixed scalar, such as $i/\hbar$, preserves this property.

5. Explain why the canonical commutation relation cannot be realized by bounded operators $Q$ and $P$ on a Hilbert space.

Solution

Suppose bounded operators satisfied

$$[Q,P]=i\hbar I.$$

The commutator identity gives by induction

$$[Q,P^n]=ni\hbar P^{n-1}.$$

Taking operator norms,

$$n\hbar\lVert P\rVert^{n-1} \leq 2\lVert Q\rVert\lVert P\rVert^n.$$

If $P\neq0$, division by $\lVert P\rVert^{n-1}$ yields

$$n\hbar \leq 2\lVert Q\rVert\lVert P\rVert$$

for every positive integer $n$, which is impossible. If $P=0$, the original commutator also fails. Canonical position and momentum must therefore be unbounded, or the relations must be expressed in a bounded form such as the Weyl relations.