The operator norm of matrix truncations
Blog Post 2026-08-14

Introduction

An interesting question showed up the other day. First, consider a Hamiltonian \(H = H_0 + \epsilon V\) where \(H_0\) is degeneracy-free, and a transformation \(U = \mathrm e^{T}\) diagonalizing \(H\), i.e. such that \(\mathrm e^{-T}H\mathrm e^{T}\) is diagonal in the eigenbasis of \(H_0\). Let’s write \(T\) as a power series: \(T = \sum_{q=1}^{\infty}\epsilon^qT_q\). Let \(\mathcal T\) denote the superoperator \([T, \cdot]\). Then we can expand \(\mathrm e^{-T}H\mathrm e^{T} = \mathrm e^{\mathcal T}H\) in powers of \(\epsilon\): \[\begin{aligned} \mathrm e^{\mathcal T}H = H_0 + \sum_{q=1}^{\infty}\epsilon^q\mathcal T_qH_0 + V^{(q-1)} \end{aligned}\] where \(V^{(q-1)}\) is Hermitian and a function of \(T_{k < q}\) and \(V\) (the specific form is not important). Clearly, if we choose \(T_q\) such that \(\mathcal T_qH_0 = -O(V^{(q-1)})\), where \(O(V^{(q-1)})\) are the off-diagonal elements of \(V^{(q-1)}\), then the transformed Hamiltonian will be diagonal at all orders.

At this point, we define the superoperator \(\mathbb P^+\) which projects onto the lower-left-triangular part of a matrix, so that if \(A\) is an operator, \(\mathbb P^+A\) strictly raises the eigenvalue of \(H_0\). Then we define the integral operator \[\begin{aligned} I(A) = \int_0^\infty \mathrm e^{-H_0s}\mathbb P^+A \mathrm e^{H_0s} \end{aligned}\] Notice that the norm of \(I(A)\) satisfies e\[\begin{aligned} \Vert I(A) \Vert \leq \Vert \mathbb P^+ A\Vert\int_{0}^{\infty}\dd s\mathrm e^{-\Delta s} = \frac{\Vert \mathbb P^+ A\Vert}{\Delta} \end{aligned}\] where \(\Delta\) is the minimum gap in the spectrum. We then notice that \(I(A)\) satisfies the following property: \[\begin{aligned} [H_0, I(A)] = \int_0^\infty \dd s [H_0, \mathrm e^{-H_0s}\mathbb P^+A\mathrm e^{H_0s}] = \sum_{j>i}(E_j-E_i)\bra{j}A\ket{i} \int \dd s \mathrm e^{-(E_j-E_i)s} = \sum_{j>i}\bra{j}A\ket{i} \end{aligned}\] Thus, defining \(L(A) \equiv I(A) - [I(A)]^\dagger\), if \(A\) is Hermitian, then \[\begin{aligned} [H_0, L(A)] = O(A) \end{aligned}\] This is exactly what we wanted, and putting \(T_q = L(V^{(q-1)})\), then if the series converges, it diagonalizes the Hamiltonian. So when does this series converge? In order to answer that, we need to estimate \(\Vert\mathbb P^+ A\Vert\).

Matrix truncation as averaging

In searching for an answer to this question, I came across a great article by Bhatia (the same person who wrote the matrix bible ). Suppose that \(\Vert \cdot \Vert\) is a unitary-invariant norm. The easiest way to estimate the norm of a super-operator is to describe it as an average over some set of unitary matrices. Let \(U_{\theta} = \operatorname{diag}(1, \mathrm e^{-\mathrm i\theta}, \mathrm e^{-2\mathrm i\theta}, \dots, \mathrm e^{-d\mathrm i\theta})\), where \(d\) is the dimension of the space. Then we notice that \((U^\dagger A U)_{ij} = \mathrm e^{(i-j)\theta}A_{ij}\). Therefore the superoperator \[\begin{aligned} A \mapsto \frac{1}{2\pi}\int_{0}^{2\pi}\dd \theta U_{\theta}^\dagger A U_{\theta}e^{\mathrm ik \theta} \end{aligned}\] projects \(A\) onto the \(k^{\text{th}}\) diagonal, where \(i-j = k\). Thus we can write the superoperator \(\mathbb P^+\) as \[\begin{aligned} \mathbb P^+ A = \frac{1}{2\pi}\sum_{k = 1}^d\int_{0}^{2\pi}\dd \theta U_{\theta}^\dagger A U_{\theta}e^{\mathrm ik \theta} \end{aligned}\] Estimating the norm and using unitary invariance gives \[\begin{aligned} \Vert\mathbb P^+ A\Vert = \frac{\Vert A \Vert}{2\pi}\int_{0}^{2\pi}\dd \theta \left|\sum_{k = 1}^d e^{\mathrm ik \theta}\right| \end{aligned}\] This looks very much like the \(L_1\) norm of the Dirichlet kernel, and we can use a very similar strategy to bound its norm. First we have \[\begin{aligned} \left|\sum_{k = 1}^d e^{\mathrm ik \theta}\right| = \left|\frac{\mathrm e^{\mathrm i\theta}-\mathrm e^{d\mathrm i\theta}}{1- \mathrm e^{\mathrm i\theta}}\right| = \left|-\frac{2\mathrm i\mathrm e^{\mathrm i\theta/2}( 1 - \mathrm e^{(d-1)\mathrm i\theta})}{\sin(\theta/2)}\right| = \left|\frac{4\sin[2]([d-1]\theta/2)}{\sin(\theta/2)}\right| \end{aligned}\] On the interval \([0, \pi/(d-1)]\), we have \[\begin{aligned} \frac{|\sin[2]([d-1]\theta/2)|}{|\sin(\theta/2)|} \leq (d-1)^2\theta \end{aligned}\] and on the interval \([\frac{\pi}{d-1}, \pi - \frac{\pi}{(d-1)})\) we have \[\begin{aligned} \frac{|\sin[2]([d-1]\theta/2)|}{|\sin(\theta/2)|} \leq \frac{4}{\theta} \end{aligned}\] Integrating these, we have \[\begin{aligned} \int_0^{\frac{\pi}{d-1}}(d-1)^2\theta \dd \theta + \int_{\frac{\pi}{d-1}}^{\pi - \frac{\pi}{d-1}}\frac{4}{\theta} \leq \frac{\pi}{2} + 4\ln(d-2) \end{aligned}\] If we repeat this argument for the intervals \([\pi - \pi/(d-1), \pi + \pi/(d-1)), [\pi + \pi/(d-1), 2\pi - \pi/(d-1))\), and \([2\pi - \pi/(d-1), 2\pi]\), we obtain \[\begin{aligned} \int_{0}^{2\pi} \dd \theta \frac{\sin[2]([d-1]\theta/2)}{\sin(\theta/2)} \leq 2\pi + 8\ln(d-2) \end{aligned}\] Taken altogether, we have \[\begin{aligned} \Vert\mathbb P^+ A\Vert \leq \frac{\Vert A \Vert}{\pi}\qty[2+32\ln(d-2)] \leq 12\Vert A \Vert\ln(d) \end{aligned}\] where the last inequality holds when \(d > 1\). Therefore we have a bound that grows very slowly with \(d\), but is in fact unbounded as \(d \to \infty\). The prefactor is not tight; it can be made as small as \(\frac{1}{\pi}\) for large \(d\). It turns out that the form of the bound is tight; the Hilbert matrix, with entries \(H_{ij} = (i-j)^{-1}\), has a norm bounded by \(\pi\), but the norm of its lower-triangular truncation grows as \(\frac{4}{5}\ln(d)\).

For some applications, this is already good enough. In particular, if our system consists of a finite number of sites with a finite onsite Hilbert space dimension, then this bound grows linearly in the number of sites.