Finite oscillator commutator boundary defect ($[X_N,P_N]=i\hslash (I_N-N|N-1⟩⟨N-1|)$)

Retain the first $N\ge 2$ number states of a quantum harmonic oscillator. Let ${\Pi }_N$ be the orthogonal projection onto their span and compress $X_N={\Pi }_{N}X{\Pi }_N$, $P_N={\Pi }_{N}P{\Pi }_N$, restricted to this span. The compressed annihilation operator $a_N$ satisfiesFor $n<N-1$, the two products give $n+1$ and $n$; on the top state they give $0$ and $N-1$. Thus, with $\ell =\sqrt{\hslash /(m\omega )}$,On every quantum superposition of $|0⟩,\dots ,|N-2⟩$, this commutator acts exactly as $i\hslash I_N$. For a normalized retained state with coefficients $c_n$,Small probability at the cutoff must therefore be judged relative to $1/N$ for this expectation. For $N=8$, the diagonal of $[X_N,P_N]/(i\hslash )$ is $(1,1,1,1,1,1,1,-7)$, with zero trace.

This is unavoidable for finite matrices: trace of a matrix commutator gives $\mathrm{tr}(XP-PX)=0$, whereas $\mathrm{tr}(i\hslash I_N)=i\hslash N\ne 0$. Compression does not preserve products: ${\Pi }_{N}XP{\Pi }_N$ can use intermediate states outside the retained span, while $X_N{P}_N$ cannot. Accordingly, compressing the full oscillator Hamiltonian operator gives ${\Pi }_{N}H{\Pi }_N=\hslash \omega (a_N^{†}{a}_N+I_N/2)$, but building it from the compressed squares givesThe 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.