Consider a matrix product state (MPS) with a bond dimension $D$. What is the physical intuition behind the bond dimension?

Is it, in any way, related to the spatial geometry?

In this note, it is mentioned that the von Neumann entropy of the MPS scales as $\mathcal{O}(L \log D)$, where $L$ is the size of the boundary.

Now, let us say that the qubits are arranged in a $2 \mathsf{D}$ grid, with only nearest neighbor interactions allowed between qubits. Then, it is known that the von Neumann entropy is upper bounded by $\mathcal{O}(L)$, where $L$ is the size of the boundary. Does it mean that the bond dimension is $2$ for this case?

In general, is the bond dimension always equal to the spatial dimension?

  • $\begingroup$ When they say the entropy is O(L), they are omitting constant factors. The entropy is O(L) for any fixed bond dimension D, not just D=2 $\endgroup$ Commented Oct 16, 2022 at 15:43

2 Answers 2


The bond dimension across any link is precisely the Schmidt rank of the state across that cut. (Or potentially larger, if you chose a not-so-smart parametrization.)

  • $\begingroup$ Could you elaborate your answer further on why this is so? $\endgroup$
    – BlackHat18
    Commented Oct 20, 2022 at 13:40
  • 1
    $\begingroup$ Hm, basically by definition. It is hard to write a to-the-point answer without knowing what you know (and what you don't know). $\endgroup$ Commented Oct 20, 2022 at 15:36

I don't know if the bond dimension is related to the spatial dimension of many-body systems. As I understand it from this note, the bond dimension can be viewed as a 'refinement parameter', ie the set of states that can be represented by a matrix product state.

A matrix product state of 'bond dimension' $D$ (with periodic boundary conditions) is a pure state vector of the form: $$ c_{j_1, \ldots, j_n}=\sum_{\alpha, \beta, \ldots, \omega=1}^D A_{\alpha, \beta ; j_1}^{(1)} A_{\beta, \gamma ; i_2}^{(2)} \ldots A_{\omega, \alpha ; j_n}^{(n)}=\operatorname{tr}\left(A_{j_1}^{(1)} A_{j_2}^{(2)} \ldots A_{j_n}^{(n)}\right) $$

where the trace and the matrix product are taken over the contracted indices. The remaining physical indices $j_1, \dots, j_n$ are open.

For example, if we take $D=1$, the matrices that we have in the above definition become complex numbers and we obtain a product state.

Small $D$ means that the set of states described by the MPS is 'refined'. For larger bond dimensions, the set of states that can be represented becomes bigger and so the quality of the approximation of the natural states also.

Note however that a MPS already is a good approximation of natural states emerging in physical systems for not-so-big bond dimension $D$.

For open boundary conditions, the matrix $A^{(1)}$ is no longer a matrix from $\mathbb{C}^{D \times D}$, but $A^{(1)} \in \mathbb{C}^{1 \times D}$ is a row vector. Similarly, $A^{(n)} \in \mathbb{C}^{D \times 1}$ is a column vector. The expression of the MPS becomes $$ c_{j_1, \ldots, j_n}=\sum^D A_{\alpha ; j_1}^{(1)} A_{\beta, \gamma ; i_2}^{(2)} \ldots A_{\omega ; j_n}^{(n)}=A_{j_1}^{(1)} A_{j_2}^{(2)} \ldots A_{j_n}^{(n)} $$

but the idea remains the same.


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