= Finite oscillator commutator boundary defect
{title2=$[X_N,P_N]=i\hbar(I_N-N|N-1\rangle\langle N-1|)$}
Retain the first $N\ge2$ <number states> of a <quantum harmonic oscillator>. Let $\Pi_N$ be the <orthogonal projection> onto their span and compress $X_N=\Pi_NX\Pi_N$, $P_N=\Pi_NP\Pi_N$, restricted to this span. The compressed <annihilation operator> $a_N$ satisfies
$$
a_N|n\rangle=\sqrt n|n-1\rangle,\qquad a_N^\dagger|N-1\rangle=0,\qquad [a_N,a_N^\dagger]=I_N-N|N-1\rangle\langle N-1|.
$$
For $n<N-1$, the two products give $n+1$ and $n$; on the top state they give $0$ and $N-1$. Thus, with $\ell=\sqrt{\hbar/(m\omega)}$,
$$
X_N=\frac{\ell}{\sqrt2}(a_N+a_N^\dagger),\qquad P_N=\frac{i\hbar}{\sqrt2\ell}(a_N^\dagger-a_N),\qquad [X_N,P_N]=i\hbar(I_N-N|N-1\rangle\langle N-1|).
$$
On every <quantum superposition> of $|0\rangle,\ldots,|N-2\rangle$, this <commutator> acts exactly as $i\hbar I_N$. For a normalized retained state with coefficients $c_n$,
$$
\frac{\langle[X_N,P_N]\rangle}{i\hbar}=1-N|c_{N-1}|^2.
$$
Small probability at the cutoff must therefore be judged relative to $1/N$ for this expectation. For $N=8$, the diagonal of $[X_N,P_N]/(i\hbar)$ is $(1,1,1,1,1,1,1,-7)$, with zero <trace>.
\Image[/oscillator-matrix-commutator-cutoff.png]
{title=Compressed <quantum harmonic oscillator> matrices and their exact <commutator> boundary defect}
{description=Original calculation with eight retained <number states>. Seven diagonal entries reproduce the <canonical commutation relation>; the last compensates their <trace>. The missing raising path through $|8\rangle$ creates the defect.}
This is unavoidable for finite <matrices>: <trace of a matrix commutator> gives $\operatorname{tr}(XP-PX)=0$, whereas $\operatorname{tr}(i\hbar I_N)=i\hbar N\ne0$. Compression does not preserve products: $\Pi_NXP\Pi_N$ can use intermediate states outside the retained span, while $X_NP_N$ cannot. Accordingly, compressing the full oscillator <Hamiltonian operator> gives $\Pi_NH\Pi_N=\hbar\omega(a_N^\dagger a_N+I_N/2)$, but building it from the compressed squares gives
$$
\frac{P_N^2}{2m}+\frac{m\omega^2X_N^2}{2}=\Pi_NH\Pi_N-\frac{N\hbar\omega}{2}|N-1\rangle\langle N-1|.
$$
The two prescriptions agree on lower <number states> and disagree at the boundary. In the infinite oscillator, the <canonical commutation relation> holds on the common invariant <dense subspace> of finite <number state> combinations. Its unbounded <position> and <momentum operators> do not admit the finite <trace> argument; writing the formal identity does not assert that both products act on every vector of the <Hilbert space>.
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