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The dominant eigenvalue of the transfer matrix of a matrix product state

Physics Asked on August 2, 2021

Consider a translation-invariant matrix product state
begin{equation}
|psi_Lrangle= mathrm{Tr}[A(s_1)A(s_2)ldots A(s_L)]|s_1 s_2ldots s_Lrangle.
end{equation}

I’m interested in the expectation value of a local observable $hat{O}$~(say a single site operator) in this state $langlehat{O}rangle_{psi_L}=langlepsi_L|hat{O}|psi_Lrangle/langlepsi_L|psi_Lrangle$. To calculate this, define the transfer operator
begin{eqnarray}
T^{im}_{jn}&=&sum_s A_{ij}(s)A^*_{mn}(s),
W^{im}_{jn}&=&sum_{s,s’} A_{ij}(s)A^*_{mn}(s’) langle s’|hat{O}_l|srangle.nonumber
end{eqnarray}

Let the eigenvalues of $T$ be
${lambda_j}^{d^2}_{j=1}$, where $d$ is the bond dimension. Let the left, right eigenvectors be $L_j, R_j$, i.e. $T=sum^{d^2}_{j=1}lambda_j R_jL_j^T$. Then
begin{equation}label{eq:initialstateassumption}
langlehat{O}rangle_{psi_L}=frac{mathrm{Tr}[T^{L-1}W]}{mathrm{Tr}[T^L]}=frac{sum^{d^2}_{j=1}lambda_j^{L-1} (L_j^T W R_j)}{sum^{d^2}_{j=1}lambda_j^L}. tag{1}
end{equation}

My question: is the limit $Lto infty$ is guaranteed to exist?

When $Lto infty$, only the largest (in absolute value) eigenvalues of $T$ contribute, all other eigenvalues can be neglected. Let the largest eigenvalues be $lambda_1,lambda_2,ldots,lambda_m$, with $|lambda_1|=|lambda_2|=ldots=|lambda_m|>|lambda_{m+1}|>ldots$. If $lambda_1=lambda_2=ldots=lambda_m>0$, then the limit (1) clearly exists. But if $lambda_1,lambda_2,ldots,lambda_m$ are not all equal, and contains some negative/complex values, then $langlehat{O}rangle_{psi_L}$ in (1) would be oscillating at large $L$. Is this at all possible?

One Answer

You are right: If $|lambda_1|=dots=|lambda_m|$ with different phases, then the convergence is not clear.

However, this is the case even in more generally: Even if $lambda_1=dots=lambda_m>0$, the issue is that local expectation values in the thermodynamic limit are not well-defined, since they depend on the boundary conditions $L_i$ and $R_i$, and you generally do not want properties in the thermodynamic limit to depend on boundary conditions.

The mathematically clean solution in the latter case is to focus on the states which are obtained from the extremal boundary conditions: The possible boundary conditions for a degenerate $lambda$ (i.e. eigenvectors) form a convex set (if you think of them as density operators on virtual ket+bra index), and you choose the extremal points of this convex set. These define the "pure states" (using the terminology of Fannes, Nachtergaele, and Werner -- -- their seminal paper formally introduced MPS (called "Finitely Correlated States" there) as a way to define a class of states in the thermodynamic limit, and is the place to check for the formal mathematical construction of MPS for the thermodynamic limit), and the bulk properties for any boundary condition can be obtained by mixing those states. This is analogous to a scenario with symmetry breaking, where you usually consider the extremal symmetry broken states $vert00cdots00rangle$ and $vert11cdots11rangle$. (In essence, there are no cross-terms between different "pure states" in the thermodynamic limit, so "global" superpositions of those states amount to mixing of the local reduced density matrices, or equivalently averaging of the expectation values.)

Now how to deal with a non-positive leading eigenvalue? It can be shown that in that case, those eigenvalues all must be some integer ($p$'th) root of unity (see https://arxiv.org/abs/quant-ph/0608197, Proof of Theorem 5, quoting a result from Fannes, Nachtergaele, and Werner). In that case, you can block $p$ sites, after which all leading eigenvalues are again positive. So you are back to the aforementioned case -- you just have to take into account that the observable $O$ can now be at either one of the $p$ different positions in the block, and all of those can appear as the expectation value in a possible extremal state, which is indeed what you would expect from a system which has some kind of periodicity (such as an Ising antiferromagnet).

Correct answer by Norbert Schuch on August 2, 2021

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