Consider a matrix product state on $\mathbb{C}^{d N}$:

$$ \Psi = \sum_{\sigma_1,...\sigma_N} A_1(\sigma_1) ... A_N(\sigma_N) |\sigma_1 ... \sigma_N \rangle \quad \quad (\text{OBC MPS}) $$

with some matrices $ D_{i-1} \times D_{i}$-matrices $A_i(\sigma_i)$ and $D_1 = 1, D_N = 1$. The bond dimension is then $D = \max_{i} D_i$.

Such a state may always be written as a translation invariant matrix product state, that is, in the form

$$ \Psi = \sum_{\sigma_1,...,\sigma_N} \text{tr}(B(\sigma_1) \cdots B(\sigma_N) )|\sigma_1 ... \sigma_N \rangle \quad \quad (\text{TI MPS})$$

where the $B(\sigma)$ are $D'\times D'$-matrices. For general MPS, we have to chose $D' \in \mathcal{O}(N D)$. This is even true for TI MPS representations of translation invariant states. To be more explicit: In the cited article they consider the $W$-state:

$$ W_N := \frac{1}{\sqrt{N}} \left( |10000 \cdots 0 \rangle + |0100 \cdots 0 \rangle + |0010 \cdots \rangle + \cdots |0000 \cdots 1\rangle \right) \ .$$

They then argue that any TI MPS representation of the $W$-state needs bond dimension at least of order $N^\frac{1}{3}$, although there is a OBC MPS representation with bond dimension $2$. They prove this from a conjecture about injectivity of MPS.

I came across a peculiarity of this state which i will describe below; i would be glad for somebody putting this into perspective.

Namely, it seems to me that the thermodynamic limit of the W-state is a mean-field state (a MPS of bond dimension 1). To see this i will use the following facts

  1. There is the following Schmidt decomposition

$$W_{k+1} = \sqrt{1 - \frac{1}{k+1}} |0\rangle \otimes W_k + \frac{1}{\sqrt{k+1}} |1\rangle \otimes \Omega_k \ , \quad \Omega_k := |0\rangle^{\otimes k} \ .$$

  1. It holds that

$$ W_N = ( v_L, B_1 \cdots B_N v_R ) \ ,$$


$$ v_L = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \ , \quad v_R = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \ , \quad B_k = \begin{pmatrix} \sqrt{1-c_k}|0\rangle & \sqrt{c_k}|1\rangle \\ 0 & |0\rangle \end{pmatrix} \ , \quad c_k := \frac{1}{N-k+1} \ . $$

  1. Introduce the linear map on operators on $\mathbb{C}^2$, indexed by a $2\times2$-matrix $a$:

$$ \mathbb{B}_{\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}}[k] \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} = \sum_i B_k(i) x B_k(i)^* a_{ji} = \\ = \begin{pmatrix} a_{11}(1-c_k)x_{11} + a_{22} c_k x_{22} + \sqrt{c_k(1-c_k)}(a_{12} x_{21} + a_{21} x_{12}) & a_{11}\sqrt{1-c_k}x_{12} + a_{12} \sqrt{c_k} x_{22} \\ a_{11}\sqrt{1-c_k}x_{21} + a_{21} \sqrt{c_k} x_{22} & a_{11} x_{22} \end{pmatrix} \ . $$

Those people which are familar with MPS know that this object allows to compute expectation values of observables.

I will use two further facts, with $\mathbb{B}[k] = \mathbb{B}_1[k]$:

  1. $$\mathbb{B}[k] \circ \cdots \circ \mathbb{B}[N](v_R \otimes v_R^*) = 1 \ .$$
  2. $$ (v_L, \mathbb{B}[1] \circ \cdots \circ \mathbb{B}[k](X) v_L) =\text{tr}(qX) + (1-k/N) \text{tr}(\sigma_3 X) \ , \quad q = \frac{1 - \sigma_3}{2} \ .$$

This finally allows to compute the expectation value of an observable $A = a_1 \otimes \cdots a_{2k}$, supported on sites $N/2 -k,...N/2+k$, where we take $N$ even for simplicity:

$$ \langle W_N, A W_N \rangle = \text{tr}\left(\left[q + \left(\frac{1}{2} + \frac{k-1}{N}\right) \sigma_3 \right] \mathbb{B}[N/2-k]_{a_1} \cdots \mathbb{B}[N/2+k]_{a_{2k}}(1)\right) \ . $$

For $N \rightarrow \infty$ with $k$ finite:

$$ \mathbb{B}_a[N/2+k](x) \rightarrow (v_L,a v_L) x \ ; $$

so that

$$ \langle W_N, A W_N \rangle \rightarrow \prod_{l=1}^{2k} (v_L,a_l v_L) \ . $$

Sorry for the long and tedious calculation which also has a lot of steps missing, but i thought the result to be too odd to just state it. The weird thing is that for finite $N$, this $W_N$-state can be represented as a non-manifestly translationally invariant MPS of bond dimension $2$ or as TIMPS of bond dimension growing with $N$. But taking the limit $N\rightarrow \infty$, we can represent it as a mean-field state (with bond dimension $1$).

What is happening here?


1 Answer 1


Maybe you're asking to resolve the tension between the following facts:

(1) For finite length $N$, the $W$-state $|W_N\rangle = \frac{1}{\sqrt{N}}(|10...0\rangle + |010...0\rangle + ... + |00...01\rangle)$ is not a product state. (In particular, it has up to one bit of entanglement on sub-regions, and an MPS representation requires bond dimension at least two.)

(2) In the thermodynamic limit $N \to \infty$, the $W$-state is "like a product state."

There's no direct paradox of course, but maybe it will be comforting to explore fact (2) a bit. First note that for any fixed $k$ while $N \to \infty$, the reduced density matrix $\rho_A=\textrm{Tr}_{\bar{A}} |W_N\rangle\langle W_N|$ on a region of size $k$ approaches $(|0\rangle\langle0|)^{\otimes k}$. So the $W$-state looks $\textit{locally}$ like the product state $|0...0\rangle$ as $N \to \infty$, even though the full state is in fact orthogonal to $|0\rangle^{\otimes N}$ for any finite $N$.

The situation here is actually similar to considering a particle on a line, with wavefunction $|\psi\rangle \in L^2(\mathbb{R})$, in the case that $|\psi\rangle$ is uniformly spread over the interval $[-N,N]$. As $N \to \infty$, the probability of the particle being in any fixed finite region tends to zero, so the state becomes locally indistinguishable from the state with no particles. (Well, the ``state of no particles'' is not actually represented in $L^2(\mathbb{R})$.) Likewise, one can think of the $W$-state as an excitation (spin flip) with uniformly spread wavefunction.

We can also formalize the sense in which the $W$-state becomes equal to the $|0\rangle^{\otimes N }$ in the strict $N=\infty$ limit, even though the states are orthogonal for all finite $N$. The formalization depends on how you choose to define the algebra of observables and the space of states for $N=\infty$. (The naïve definitions involving infinite tensor products are a bit tricky, because then inner products and norms can yield infinities.) One common choice is to define the algebra of observables as the so-called “quasi-local” algebra, which is (the completion of) the space of finite range (i.e. local) observables. See e.g. [1]. Then one usually defines “states” as positive linear functionals on the algebra of observables (mapping observables to expectation values). In that case, a state is fully defined by how it assigns expectations to only $\textit{finite-range}$ observables. Then a sequence of $W$-states for increasing $N$ really does approach the $|0\rangle^{\otimes N }$ state, because the W-states act identically on finite-range observables in the $N \to \infty$ limit . And in fact there is no “analog” of the $W$-state for $N=\infty$, besides the $|0\rangle^{\otimes N }$ state: the former doesn’t really exist except insofar as we choose to define it by the latter.

[1] “Quantum Spin Systems on Infinite Lattices,” Pieter Naaijkens, https://arxiv.org/pdf/1311.2717.pdf


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