We know that when two MPS A, B are contracted, we have:
\langle A|B\rangle = \sum_{i, \sigma} A_1^{\sigma_1} B_1^{\sigma_1} \cdots A_n^{\sigma_n} B_n^{\sigma_n}
where the sum runs over all the indices.
Is there a way to obtain the coefficients of a part of an MPS A (of spinless fermions)? By which I mean (if we are interested in site j, j+1):
|C\rangle = \sum_{i\notin \{ j, j +1\}} A_i \rightarrow a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle
One naive way is that I would define an MPS of all ‘1’, up to normalization, and then perform the partial contraction, but I’m not sure how to define such an all one MPS.
(Everything here I’m taking to be completely analogous to ED, where you can simply sum up all the coefficients of an eigenstate with a specific configuration on site j, j+1, but maybe this is different in MPS)
The other way is I make an outer product:
O = |A\rangle\langle A|
Then compute the partial trace up to j, j+1, but I’m not sure how to recollect the coefficients effectively from the remaining MPO block structure.
Thank you