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Canonical MPS environments are isometric and the effective Hamiltonian is Hermitian

Statement

Let a finite open-boundary TT have order $d \geq 2$ and be in the following mixed gauge about a one-site or two-site center window $J = [k, k + w - 1]$ , where $w \in {1, 2}$ : every core strictly left of $J$ is left-isometric and every core strictly right of $J$ is right-isometric. The center coefficient array has shape $r_{k - 1} \times n_{k} \times \dots \times n_{k + w - 1} \times r_{k + w - 1}$ , with boundary ranks interpreted as one, and $x$ is its vectorization in the fixed lexicographic order. Hold every core outside $J$ fixed. Contracting the fixed prefix and suffix defines a linear embedding $E$ from center coefficients to the full tensor space. Then $E^{*} E = I$ , so $∥ E x ∥_{F} = ∥ x ∥_{2}$ . For every finite Hermitian Hamiltonian input $H$ , the effective matrix $H_{e f f} : = E^{*} H E$ is Hermitian and satisfies $⟨ E x, H E x ⟩_{F} = x^{*} H_{e f f} x .$

Facts & Assumptions

Given: A finite open-boundary TT, a one-site or two-site center in mixed canonical gauge, and a self-adjoint finite Hamiltonian.

[F1]

Finite matrix-product states and Hermitian Hamiltonians: a finite Hermitian Hamiltonian input has matrix $H^{*} = H$ in the standard lexicographic basis.

[F2]

Finite tensor arrays and the Frobenius inner product: the Frobenius inner product conjugates the first entry and sums over all coefficient indices.

[F3]

Tensor train cores, bond ranks, and canonical gauges: left core matricizations satisfy $L^{*} L = I$ and right core matricizations satisfy $R R^{*} = I$ .

Proof

1.1givenF3inductionalgebra

Let $L_{j}$ be the prefix interface through site $j$ , with rows indexed by $(i_{1}, \dots, i_{j})$ and columns by $α_{j}$ . Put $L_{0} = [1]$ . For $j < k$ , its recurrence is $L_{j} = (L_{j - 1} \otimes I_{n_{j}}) G_{j}^{(L)}$ , where $G_{j}^{(L)}$ is the left matricization from [F3]. If $L_{j - 1}^{*} L_{j - 1} = I$ , then $(L_{j - 1} \otimes I)^{*} (L_{j - 1} \otimes I) = I \otimes I$ and hence $L_{j}^{*} L_{j} = G_{j}^{(L) *} G_{j}^{(L)} = I$ by [F3]. Induction gives $L_{k - 1}^{*} L_{k - 1} = I$ . For the suffix, put $R_{d + 1} = [1]$ and contract from right to left: for $j > k + w - 1$ , $R_{j} = G_{j}^{(R)} (I_{n_{j}} \otimes R_{j + 1})$ , with rows indexed by $α_{j - 1}$ and columns by $(i_{j}, \dots, i_{d})$ . Since $R_{j + 1} R_{j + 1}^{*} = I$ implies $(I \otimes R_{j + 1}) (I \otimes R_{j + 1})^{*} = I \otimes I$ , the right-isometry in [F3] gives $R_{j} R_{j}^{*} = G_{j}^{(R)} G_{j}^{(R) *} = I$ . Induction yields $R_{k + w} R_{k + w}^{*} = I$ .

2.1F1F2step 1.1algebra

By [F1], the full array entries are the finite contractions of the fixed cores with the center coefficients. Expand $⟨ E x, E y ⟩_{F}$ using the conjugate-first finite sum in [F2], and first sum over the physical prefix and suffix indices. Those sums replace the interface factors by $L^{*} L = I$ and $R R^{*} = I$ from step 1.1. The remaining finite sum is exactly $⟨ x, y ⟩_{F}$ . Hence $E^{*} E = I$ and $∥ E x ∥_{F} = ∥ x ∥_{2}$ .

3.1F1algebra∎

By [F1], $H^{*} = H$ . Therefore $H_{e f f}^{*} = (E^{*} H E)^{*} = E^{*} H^{*} E = E^{*} H E = H_{e f f}$ . Associativity of finite matrix multiplication gives $⟨ E x, H E x ⟩_{F} = x^{*} E^{*} H E x = x^{*} H_{e f f} x$ , as claimed.

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Sources

Ulrich Schollwöck, The density-matrix renormalization group in the age of matrix product states, Annals of Physics 326 (2011), §§4.2, 4.4, 6.2–6.3 (standard reference, not scraped)
Grégoire Misguich, Introduction to matrix-product states and tensor networks, complete lecture notes (2026), §§2.4.3, 2.7 (standard reference, not scraped)