Observables as matrices in the energy basis

Completeness licenses the whole move. The sine states satisfy Σₙ |n⟩⟨n| = 1, so A = 1 A 1 = Σₘₙ |m⟩ ⟨m|A|n⟩ ⟨n|, and a Hermitian operator is completely specified by the numbers Aₘₙ = ⟨m|A|n⟩, subject to Aₘₙ = Aₙₘ*. The basis was chosen to diagonalise H, so Hₘₙ = Eₙδₘₙ with Eₙ = n²π²ℏ²/(2ML²). That is the free lunch and the reason for the basis: exp(−iHt/ℏ) is diagonal too, so evolution is cₙ(t) = cₙ(0) exp(−iEₙt/ℏ), one phase per coefficient. Everything else costs an integral. The criterion for a second free lunch is exact, and worth stating as a rule: because no two levels of the box share an energy, A is diagonal in the energy basis if and only if [A, H] = 0. Since [x, H] = iℏp/M is not zero, x cannot be diagonal here. The off-diagonal entries you are about to compute are not clutter, they are the statement that position and energy are incompatible.

Move the origin to the centre and set u = x − L/2. Then ψₙ has parity (−1)ⁿ⁺¹ in u: odd n gives an even function, even n an odd one. The operator u is odd, so ⟨m|u|n⟩ vanishes unless ψₘψₙ is odd, which happens only when m − n is odd. That single argument kills half the matrix with no integration at all. ⟨1|x|3⟩, ⟨2|x|4⟩ and ⟨1|x|5⟩ are exactly zero, while the diagonal is ⟨n|x|n⟩ = L/2 by symmetry alone. For the surviving entries the integral gives xₘₙ = −8Lmn/(π²(m² − n²)²). In units of L that is x₁₂ = −0.18013, x₂₃ = −0.19454 and x₃₄ = −0.19851 next to the diagonal, but only x₁₄ = −0.014410 three steps out: the squared denominator makes distant entries fall like 1/m³, so the position matrix is strongly band-dominated. The shared sign is a phase convention rather than physics, since replacing |2⟩ by −|2⟩ flips x₁₂ with it. No single off-diagonal sign is measurable; |xₘₙ|² is, and so is any product around a closed loop of indices, such as x₁₂x₂₃x₃₄x₄₁, which no rephasing can change.

For any H = p²/2M + V(x) the identity p = (iM/ℏ)[H, x] is exact. Sandwich it between energy eigenstates and the commutator collapses to a number, because H acts on its own eigenvectors: pₘₙ = iM(Eₘ − Eₙ)xₘₙ/ℏ. The momentum matrix therefore falls out of the position matrix and the spectrum, with no new integration. Putting in Eₘ − Eₙ = π²ℏ²(m² − n²)/(2ML²) gives pₘₙ = −4iℏmn/(L(m² − n²)) for m − n odd and zero otherwise. Three checks pay for themselves. The entries are purely imaginary and antisymmetric in their indices, so pₘₙ = pₙₘ* and the matrix is Hermitian. Every diagonal entry vanishes, so ⟨p⟩ = 0 in each stationary state, as it must be for a real bound eigenfunction. And the fall-off is only 1/m, because the factor Eₘ − Eₙ grows like m² and cancels one power of the denominator. In units of ℏ/L the near-diagonal entries are p₁₂ = 2.667i, p₂₃ = 4.800i, p₃₄ = 6.857i: they grow, they do not decay.

To compute, you keep N rows and columns, so A becomes Π_N A Π_N. That is harmless for one operator and wrong for a product, because Π_N A B Π_N equals Π_N A Π_N B Π_N when Π_N commutes with A or with B, and in general not otherwise. The resolution of the identity you inserted between the two factors was truncated along with everything else. The box supplies a clean test case. Inside the walls V = 0, so p² = 2MH is exactly diagonal, (p²)ₘₙ = (nπℏ/L)²δₘₙ. Now truncate p to 3 by 3. In units of ℏ/L only p₁₂ = 2.667i and p₂₃ = 4.800i survive, so ((P₃)²)₁₁ = |p₁₂|² = 64/9 = 7.111 against the exact π² = 9.870, just 72% of it. Worse, ((P₃)²)₁₃ = p₁₂p₂₃ = −12.8, where the true p² has a clean zero: the columns m = 4, 6, 8, … whose contributions cancel that term were thrown away. Widening to N = 9 lifts the (1,1) entry only to 8.977 and drags the spurious (1,3) entry to −2.771.

Completeness also hands you a test you can run without knowing the answer: Σₘ |Aₘₙ|² = ⟨n|A†A|n⟩. For position that reads Σₘ |xₘₙ|² = ⟨n|x²|n⟩ = L²(1/3 − 1/(2n²π²)), which is 0.282673 L² in the ground row. Add the terms up. The diagonal alone gives 0.250000 L², or 88.4%; m = 2 lifts it to 0.282446, or 99.92%; m = 3 adds nothing at all because 3 − 1 is even; and m = 4 reaches 0.282653, or 99.993%. Four columns are already better than a part in ten thousand. Run the same test on momentum and the verdict reverses. Σₘ |pₘₙ|² = 2MEₙ = (nπℏ/L)², and in the ground row the partial sums reach 72.1% at N = 2, 92.6% at N = 10, 99.20% at N = 100 and still only 99.92% at N = 1000, because |p₁ₘ|² falls like 16ℏ²/(L²m²) while |x₁ₘ|² falls like 1/m⁶. A truncated basis is a fine place to compute ⟨x⟩ and a poor place to compute ⟨p²⟩. Take that one from H.

[x, p] = iℏ has no finite-dimensional solution, and the box shows why in one line: the trace of a commutator of finite matrices is tr(XP) − tr(PX) = 0, while tr(iℏ I_N) = iNℏ. A truncated pair must therefore fail the canonical commutation relation, and it is worth seeing where. At N = 3 the diagonal of [X₃, P₃] in units of iℏ is 0.9607, 0.9069, −1.8676: the first entry is already within 4% of the exact 1 and the second within 10%, while the whole deficit is dumped into the last kept row so that the trace comes out exactly zero. At N = 9 the (1,1) entry has climbed to 0.9985 and the last entry has fallen to −7.08. The working rule is to trust the top-left corner of a truncated array, never the last few rows, and to quote no number that leans on them. Diagonalising the truncated x is the same story with a happier ending: the eigenvalues of X₃ are 0.235L, 0.500L and 0.765L, a coarse quadrature grid inside the well, and X₈ gives eight nearly evenly spaced nodes from 0.101L to 0.899L.