Heisenberg Commutation Relation

The Heisenberg canonical commutation relation is $[Q,P]=i\hbar I$, interpreted on vectors for which both products exist. It specifies the infinitesimal noncommutativity of a canonical pair. It is not an identity between bounded matrices, nor does an identity on an arbitrary dense domain automatically determine a global representation. The bounded, exponentiated relation and its sign convention are developed in The Weyl Form of the CCR.

Required background. Self-Adjoint Operators explains why operator actions, domains, and closures must be distinguished.

The canonical identity includes its input space

For linear operators $Q,P$ and a linear subspace $\mathcal D$, the statement means

$$\mathcal D\subseteq D(QP)\cap D(PQ), \qquad QP\psi-PQ\psi=i\hbar\psi \quad(\psi\in\mathcal D).$$

The product domains are part of the hypothesis. Even membership in $D(Q)\cap D(P)$ does not by itself guarantee membership in both product domains.

For the usual position and momentum on the line, $\mathcal S(\mathbb R)$ is a dense invariant choice. Multiplication by $x$ and differentiation preserve that space, and the ordinary product rule gives the canonical identity. The Schrödinger Representation gives the maximal self-adjoint domains and the full realization. Here that example supplies a nontrivial representation of the infinitesimal algebra.

For $d$ canonical degrees of freedom the corresponding relations are

$$[Q_j,P_k]=i\hbar\delta_{jk}I, \qquad [Q_j,Q_k]=[P_j,P_k]=0,$$

on a stated common domain. The number $d$ is the number of canonical pairs, not the Hilbert-space dimension. A single canonical pair already requires an infinite-dimensional Hilbert-space realization when the relation is nontrivial.

Why finite matrices cannot satisfy the relation

For finite square matrices, cyclicity gives $\operatorname{Tr}(QP-PQ)=0$. If the canonical relation held in dimension $N>0$, taking traces would instead give

$$0=i\hbar N,$$

which is impossible for $\hbar\ne0$. This argument uses the ordinary finite trace. In infinite dimension neither $QP$ nor $PQ$ need be trace class, and $I$ has infinite trace, so the same formal manipulation is unavailable.

The trace obstruction explains why finite numerical models can approximate selected canonical predictions without realizing the exact canonical algebra. A discretization must contain a defect somewhere, often near a cutoff or boundary.

The bounded-operator obstruction in any dimension

The obstruction is stronger than the finite-matrix trace argument. No two bounded operators on a nonzero Banach space can obey $AB-BA=I$. The following norm proof also explains why domains enter the quantum theory.

Assume otherwise. By induction using the commutator product rule,

$$[A,B^n]=nB^{n-1}.$$

Taking norms gives

$$n\|B^{n-1}\|\leq2\|A\|\,\|B^n\|.$$

Since $\|I\|=1$, iteration implies

$$\|B^n\|\geq\frac{n!}{(2\|A\|)^n}.$$

But boundedness also gives $\|B^n\|\leq\|B\|^n$, so $(n!)^{1/n}\leq2\|A\|\|B\|$ for every $n$. The left side is unbounded, a contradiction. Rescaling $A=Q/(i\hbar)$ and $B=P$ excludes a bounded canonical pair.

If bounded $Q,P$ satisfied the relation merely on a dense domain, continuity would extend it to all vectors and reproduce the contradiction. Thus an exact canonical pair must leave the bounded-operator setting. This does not prove that every formal unbounded pair is admissible: self-adjointness, common domains, and exponentiation remain separate checks.

The defect in a truncated oscillator

Keep the oscillator basis $|0\rangle,\ldots,|N-1\rangle$. Define $a_N|n\rangle=\sqrt n|n-1\rangle$, with $a_N|0\rangle=0$, and let $a_N^*$ be its finite-matrix adjoint. At the top state the raising action vanishes. Direct multiplication yields

$$[a_N,a_N^*]=I_N-N|N-1\rangle\langle N-1|.$$

The diagonal entries are one below the cutoff and $1-N$ at the top, so the trace is zero. Defining truncated position and momentum with a length scale $\ell$,

$$Q_N=\frac{\ell}{\sqrt2}(a_N+a_N^*), \qquad P_N=\frac{\hbar}{i\sqrt2\ell}(a_N-a_N^*),$$

gives the same defect multiplied by $i\hbar$. The discrepancy has operator norm $N\hbar$, so it does not vanish uniformly as the cutoff grows. Yet it vanishes on every fixed finite-support state once the cutoff is above that support. This is a precise example of statewise validity without uniform operator validity.

Exercise: quantify the cutoff error

For a normalized truncated state $\psi=\sum_{n=0}^{N-1}c_n|n\rangle$, compute the expectation of $[Q_N,P_N]$. What tail condition makes its canonical value accurate as $N$ changes?

Solution

The expectation is $i\hbar(1-N|c_{N-1}|^2)$. Thus the necessary and sufficient condition for this particular commutator expectation to approach $i\hbar$ is $N|c_{N-1}|^2\to0$. Small top-state probability alone is insufficient unless it decreases faster than $1/N$. This condition controls this observable diagnostic; other errors need their own estimates.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — canonical operators, oscillator algebra, and their domains.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Section 8.4 — abstract commutation relations and boundedness obstructions.