Commutation Relations in Historical Context

Commutation relations are now part of the basic grammar of quantum mechanics. Students learn early that position and momentum satisfy

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

Historically, this relation was not introduced as an abstract axiom. It emerged from a collision between three ideas: the classical Poisson-bracket structure of Hamiltonian mechanics, the matrix multiplication of transition quantities, and the empirical need to replace classical orbital pictures by spectral data.

The formal home of commutators is Commutators. This page explains how noncommutativity entered the theory historically and why it mattered.

Classical Poisson Brackets as Background

Classical Hamiltonian mechanics already had a nontrivial bracket operation. For one degree of freedom, the Poisson bracket of two phase-space functions $f(q,p)$ and $g(q,p)$ is

$$\{f,g\} = \frac{\partial f}{\partial q} \frac{\partial g}{\partial p} - \frac{\partial f}{\partial p} \frac{\partial g}{\partial q}.$$

The canonical variables satisfy

$$\{q,p\} = 1, \qquad \{q,q\} = \{p,p\} = 0.$$

This is not yet quantum noncommutativity. Classical functions still commute under ordinary multiplication:

$$fg = gf.$$

The Poisson bracket is instead a separate antisymmetric operation. It encodes Hamiltonian flow, canonical transformations, and the symplectic structure of phase space. Its importance for quantum mechanics was that it supplied a classical algebraic pattern for dynamics:

$$\frac{df}{dt} = \{f,H\} + \frac{\partial f}{\partial t}.$$

The historical puzzle was how to keep this powerful Hamiltonian structure when the basic quantum quantities no longer behaved like ordinary phase-space functions.

Matrix Multiplication

Heisenberg’s transition quantities were indexed by pairs of stationary states. If $X$ and $Y$ are two such arrays, their product is naturally formed by summing over intermediate states:

$$(XY)_{mn} = \sum_k X_{mk}Y_{kn}.$$

Reversing the order gives

$$(YX)_{mn} = \sum_k Y_{mk}X_{kn}.$$

There is no general reason for these expressions to agree. Thus the multiplication law for the new quantum quantities was not commutative:

$$XY \ne YX.$$

This was not a minor notation issue. Classical observables were functions on phase space, and their ordinary product commuted. Matrix mechanics made the product of physical quantities order-dependent.

Born and Jordan recognized the arrays as matrices. That recognition made the new multiplication law explicit and gave a concise object for measuring order-dependence:

$$[X,Y] = XY-YX.$$

The commutator became the algebraic signature of the new mechanics.

Canonical Commutation Relation

The decisive canonical relation involves position-like and momentum-like quantities. In modern notation,

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

Original papers could write equivalent formulas with different sign conventions because the order $QP-PQ$ versus $PQ-QP$ may be reversed. The stable content is that canonical coordinate and momentum quantities do not commute, and the scale of the failure to commute is set by Planck’s constant.

The relation mirrors the classical bracket

$$\{q,p\} = 1,$$

through the correspondence

$$\{f,g\} \quad \longleftrightarrow \quad \frac{1}{i\hbar} [\widehat f,\widehat g].$$

This correspondence is powerful, but it is not a universal machine. A classical expression such as $q^2p$ does not uniquely say whether the quantum operator should be $Q^2P$, $QPQ$, $PQ^2$, or a symmetrized combination. The commutator relation supplies the algebraic backbone; it does not remove all representation, ordering, and domain questions.

There is also a simple finite-dimensional warning. If $A$ and $B$ are finite matrices, then

$$\operatorname{Tr}([A,B]) = 0.$$

Therefore exact finite matrices cannot satisfy

$$[Q,P] = i\hbar I_N,$$

because the trace of the right-hand side is $i\hbar N$. The canonical relation is an infinite-dimensional operator relation, or an exponentiated Weyl relation, not literally a relation among finite tables.

Noncommuting Measurements

The early matrix mechanics papers did not yet contain the full modern measurement theory. Even so, noncommutativity soon acquired physical meaning beyond algebra.

In the standard finite-dimensional projective setting, commuting observables can be simultaneously diagonalized. Their projectors can be refined into joint alternatives, and measurement order does not change the ideal joint probabilities. Noncommuting observables generally cannot be treated this way.

If $A$ has projectors $P_a$ and $B$ has projectors $Q_b$, the ordered probability for measuring $A$ first and then $B$ is

$$p(a\text{ then }b) = \langle\psi\vert P_aQ_bP_a\vert\psi\rangle.$$

Reversing the order gives

$$p(b\text{ then }a) = \langle\psi\vert Q_bP_aQ_b\vert\psi\rangle.$$

When the projectors do not commute, these are different experimental procedures and can give different statistics. The noncommutativity of the algebra becomes order-dependence in measurement sequences.

Spin provides the cleanest later textbook example. Components such as $S_x$ and $S_z$ do not commute, so an ideal measurement of one component changes the statistics for a subsequent measurement of the other. The historical route to spin has its own pages, but the conceptual lesson is the same: quantum observables need not fit into one shared classical list of pre-existing values.

Modern Formalism Link

In modern quantum mechanics, commutators organize several core ideas:

• compatibility and simultaneous eigenstates;
• uncertainty relations;
• conservation laws and Heisenberg-picture dynamics;
• angular momentum and spin algebras;
• canonical quantization and correspondence with Poisson brackets;
• symmetry generators and transformation laws.

For example, if an observable $A$ has no explicit time dependence, its Heisenberg-picture evolution is

$$\frac{dA}{dt} = \frac{i}{\hbar} [H,A].$$

This is the quantum analogue of Hamiltonian evolution by Poisson brackets. When $[A,H]=0$, the observable is conserved in the corresponding closed-system setting. Thus commutators are not only about uncertainty; they also encode dynamics and symmetry.

The formal pages own the general theory. This historical page owns the origin: noncommutativity entered because the transition quantities demanded by spectroscopy formed arrays whose natural multiplication was matrix multiplication, not classical multiplication of functions.

What This Page Does Not Claim

This page does not claim that quantum mechanics is obtained by replacing every Poisson bracket with a commutator. That slogan is useful only as a first guide. Real quantization also requires a Hilbert space, self-adjoint operators, domains, boundary conditions, representation choices, and ordering conventions.

It also does not claim that noncommutativity by itself is the whole uncertainty principle. Historical Origin of Uncertainty separates Heisenberg’s uncertainty discussions from the later Kennard-Robertson inequalities. Here the point is narrower: the algebra of quantum observables became noncommutative before the full modern interpretation of uncertainty had settled.

Common Mistakes

• Thinking the Poisson bracket already means classical observables do not commute. Classical functions commute under ordinary multiplication; the Poisson bracket is a separate operation.
• Treating $[Q,P]=i\hbar I$ as a relation among ordinary numbers.
• Forgetting that original sign conventions depend on whether the commutator is written as $QP-PQ$ or $PQ-QP$.
• Assuming finite matrices can satisfy the exact canonical commutation relation.
• Treating the Poisson-to-commutator correspondence as a complete quantization algorithm.
• Reducing all noncommutativity to measurement disturbance. Noncommutativity also controls dynamics, compatibility, symmetry, and preparation uncertainty.
• Confusing noncommuting observables with observables that can never have sharp values individually. A state may be sharp in one while not being sharp in the other.

Cross-Links

• Matrix Mechanics and Operator Ideas
• Heisenberg’s Matrix Mechanics
• Born and Jordan’s Matrix Formulation
• Observables Before Operators
• Dirac’s Transformation Theory
• Historical Origin of Uncertainty
• Equivalence of Matrix and Wave Mechanics
• Limits of Old Quantum Theory
• Commutators
• Canonical Commutation Relations
• Noncommuting Observables
• Compatible Observables
• General Uncertainty Relations
• Poisson Brackets
• Classical-Quantum Correspondence
• Sequential Measurements

References

• W. Heisenberg, “Über quantentheoretische Umdeutung kinematischer und mechanischer Beziehungen,” Zeitschrift für Physik 33, 879-893, 1925, DOI: 10.1007/BF01328377.
• M. Born and P. Jordan, “Zur Quantenmechanik,” Zeitschrift für Physik 34, 858-888, 1925, DOI: 10.1007/BF01328531.
• M. Born, W. Heisenberg, and P. Jordan, “Zur Quantenmechanik II,” Zeitschrift für Physik 35, 557-615, 1926, DOI: 10.1007/BF01379806.
• P. A. M. Dirac, “The fundamental equations of quantum mechanics,” Proceedings of the Royal Society A 109, 642-653, 1925, DOI: 10.1098/rspa.1925.0150.
• W. Heisenberg, “Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik,” Zeitschrift für Physik 43, 172-198, 1927, DOI: 10.1007/BF01397280.
• B. L. van der Waerden, ed., Sources of Quantum Mechanics, Dover, 1968.
• M. Jammer, The Conceptual Development of Quantum Mechanics, 2nd ed., American Institute of Physics, 1989.
• J. Mehra and H. Rechenberg, The Historical Development of Quantum Theory, Springer, 1982-2001.
• H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.

Exercises

1. For one classical degree of freedom, compute $\{q,p\}$ from the definition of the Poisson bracket.

Solution

Using

$$\{f,g\} = \frac{\partial f}{\partial q} \frac{\partial g}{\partial p} - \frac{\partial f}{\partial p} \frac{\partial g}{\partial q},$$

with $f=q$ and $g=p$, one obtains

$$\{q,p\} = 1\cdot1-0\cdot0 = 1.$$

2. Let

$$X = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}.$$

Compute $[X,Y]$.

Solution

The products are

$$XY = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}, \qquad YX = \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}.$$

Therefore

$$[X,Y] = XY-YX = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}.$$

3. Use $[Q,P]=i\hbar I$ to compute $[Q,P^2]$.

Solution

Use the product rule for commutators:

$$[Q,P^2] = [Q,P]P+P[Q,P].$$

Substituting $[Q,P]=i\hbar I$ gives

$$[Q,P^2] = i\hbar P+P i\hbar = 2i\hbar P.$$

4. Why does the trace argument rule out exact finite-dimensional matrices satisfying $[Q,P]=i\hbar I_N$?

Solution

For finite matrices,

$$\operatorname{Tr}(QP-PQ) = \operatorname{Tr}(QP)-\operatorname{Tr}(PQ) = 0$$

by cyclicity of trace. But

$$\operatorname{Tr}(i\hbar I_N) = i\hbar N,$$

which is nonzero for $N>0$ and $\hbar\ne0$. Therefore the exact canonical commutator cannot hold in finite dimension.