# Can bond dimension vary from bond to bond?

Consider a bipartite system composed of subsystems $$A$$ and $$B$$, with corresponding Hilbert spaces $$\mathcal{H}_A$$ and $$\mathcal{H}_B$$, spanned by $$\{\chi_1,...,\chi_n\}$$ and $$\{\phi_1,...,\phi_m\}$$, respectively.

Now consider a state belonging to this bipartite system, $$|\Psi\rangle_{AB} = \sum_{i,j} \psi_{ij} |i\rangle_A \otimes |j\rangle_B$$. We can perform a Schmidt Decomposition to rewrite this as $$|\Psi\rangle_{AB} = \sum_{k=1}^r \sqrt{\lambda_k} |\chi_k\rangle_A \otimes |\phi_k\rangle_B$$. Here, we have $$r = \min(m,n)$$, the rank of $$\psi_{ij}$$.

Now suppose we have a 1D spin-1/2 chain on $$N$$ sites with nearest-neighbour interactions, and we want to write this as a Matrix Product State (MPS). Suppose we start on the right-hand side and let subsystem $$A$$ correspond to lattice sites $$1,...,N-1$$, and assign subsystem $$B$$ to lattice site $$N$$. Because this is a spin-1/2 system, it seems to me that there are $$2^{N-1}$$ degrees of freedom in subsystem $$A$$ and $$2^1$$ in subsystem $$B$$; therefore, $$r=\min(2^{N-1},2^1)=2$$.

Now I want to introduce the "bond dimension" terminology. Would we say that the bond dimension of this MPS is $$2$$? And if we were now to repartition the total system $$AB$$ into $$A$$ = $$\{$$lattice sites $$1,...,N-2\}$$ and $$B$$ = $$\{$$lattice sites $$N-1,N\}$$, would the bond dimension of this "new" bond be $$r=\min(2^{N-2},2^2)=4$$?

This is my very long-winded way of asking: Suppose we have an MPS of the 1D spin-1/2 system given by $$|\Psi\rangle = \sum_{\{s\}} \text{Tr}[A_1^{[s_1]}A_2^{[s_2]}...A_N^{[s_N]}]|s_1,s_2,...,s_N\rangle$$. Then, assuming that we do not truncate any states of the SVD, would we say that the bond dimension between sites $$n$$ and $$n+1$$ is given by $$r = \min(2^{N-n}, 2^n)$$? And does this correspond to the rank of matrix $$A_n^{[s_n]}$$?

I've heard it mentioned that the bond dimension is connected to entanglement. As such, it seems a bit odd to me that the bond dimension (cf., entanglement) should vary as a function of position in the chain, especially if this means that sites just $$t$$ lattice points apart can vary by as much as $$2^t$$ in their bond dimension. Taking a very rough view on things, this would appear to create very strong "edge" effects that are, in a sense, amplified the deeper one goes into the material.

Apologies for the long question! Any insights are much appreciated :)

The "bond dimension of an MPS" usually refers to the maximum bond dimension. So a bond dimension 16 MPS with 12 sites, each of physical dimension 2 might have bonds like $$2,\;\;4,\;\;8,\;\;16,\;\;16,\;\;16,\;\;16,\;\;16,\;\;8,\;\;4,\;\;2$$.
The bond dimensions do not need to be large on the edges because not as much long range entanglement needs to pass through these bonds. This is true even if we choose to write an arbitrary state in the $$2^N$$ dimensional Hilbert space as an MPS with no truncation, as you have done. Clearly there should be no 1D "chain" structure in the entanglement. That is, the entanglement of a spin in this arbitrary state with the rest of the system should not have to depend on the spin's position in the chain. And it doesn't. But the bipartite entanglement does.
You may be confusing the bipartite entanglement with the entanglement between a spin and its environment. To calculate the latter, you must trace over all other sites, then find the entropy of the spectrum of that density matrix. This would be done by making a copy of the MPS, then contracting all physical legs with their counterparts apart from the site of interest. Once fully contracted, you'd have a $$2\times2$$ reduced density matrix, with a maximal entanglement independent of which site you chose. 