Given a translationally invariant Matrix Product State (assuming periodic boundary condition) on $N$ sites of dimension $d$ each, which takes the form

$\sum_{i_1,i_2\ldots i_N=1}^dTr(A_{i_1}A_{i_2}\ldots A_{i_N})|i_1,i_2\ldots i_N\rangle$,

with $A_i$ being $D\times D$ matrices. If $d<D^2$, then matrices $A_i$ do not span the space of $D\times D$ matrices. But by blocking $l$ sites together, so that $d^l\geq D^2 > d^{l-1}$, one notices that in most cases, the matrices $A_{i_1}A_{i_2}\ldots A_{i_l}$ do span the space of $D\times D$ matrices. Then one can write down a blocked Matrix Product representation, in which case, above state takes the form

$\sum_{i_1,i_2\ldots i_N=1}^{d^l}Tr(B_{i_1}B_{i_2}\ldots B_{i_M})|i_1,i_2\ldots i_M\rangle$,

with $M=\frac{N}{l}$.

Given this representation, there is a 2-local parent hamiltonian in the blocked picture, defined as the projector orthogonal to the subspace $\{\sum_{i_1,i_2}Tr(XB_{i_1}B_{i_2})|i_1,i_2\rangle | X\text{ is set of all } D\times D \text{ matrices }\}$. Thus overall locality of the parent hamiltonian is $2l$. Moreover, this Matrix Product State is unique ground state of the parent hamiltonian constructed above.

But now, lets compute these quantities for AKLT model in one dimension (http://en.wikipedia.org/wiki/AKLT_model) . Since particles are spin-1, we have $d=3, D=2$. The matrices for AKLT do not span the space of all $2\times 2$ matrices. So we need to block two neighbouring sites, which implies the parent hamiltonian will have locality 4. But then why does AKLT hamiltonian have locality 2? Is AKLT hamiltonian not the same as parent hamiltonian for AKLT? Note that AKLT hamiltonian has the AKLT state as its unique ground state, assuming periodic boundary condition.

  • $\begingroup$ Since $d^l>D^2$, the reduced density matrix for $l$ sites $\rho_l$ does not have full rank and one can construct a parent Hamiltonian by projecting to the support of $\rho_l$. So the parent Hamiltonian is $l$-local. $\endgroup$
    – Meng Cheng
    Commented Jun 10, 2015 at 14:33
  • $\begingroup$ Please use \rangle, \langle instead of >, < when using MathJax. $\endgroup$
    – Danu
    Commented Jun 10, 2015 at 14:34
  • $\begingroup$ @Danu, i have changed them to \rangle and \langle. But now, a square box appears in place of > . Am i doing something wrong? $\endgroup$ Commented Jun 10, 2015 at 15:43
  • $\begingroup$ @MengCheng , does the parent hamiltonian, that you suggest, have its corresponding MPS as the unique ground state? I have edited the question, requiring parent hamiltonian to have the corresponding mps as unique ground state. $\endgroup$ Commented Jun 10, 2015 at 15:44
  • 1
    $\begingroup$ @MengCheng As a side remark, for me the fact that the 2-body Hamiltonian for the AKLT has a unique ground state still seems a miracle (or rather a lucky coincidence ;-). It would be very interesting if one could identify (general) criteria which allow to break down parent Hamiltonians into smaller ones. $\endgroup$ Commented Jun 10, 2015 at 22:00

1 Answer 1


It is correct that if you proceed the way you describe it, you obtain a 4-local parent Hamiltonian $H=\sum h$, where $h\ge0$, and $h\vert\Psi\rangle=0$, where $\vert\Psi\rangle$ is the AKLT state. For a parent Hamiltonian constructed this way, one can show (for an arbitrary injective MPS) that the ground state is unique with a gap above.

However, if you now take the AKLT Hamiltonian with 2-local terms $k$, it is easy to see that for $\tilde k:=k_{12}+k_{23}+k_{34}$ (acting on four consecutive sites), $\tilde k\ge0$ (since $k\ge0$), and $\tilde k\vert\Psi\rangle=0$, i.e., $K=\sum \tilde k \propto \sum k$ and $H=\sum h$ have the same ground space. Also, there are always constants s.th. $c_1 \tilde k\le h\le c_2\tilde k$, i.e., $K$ is gapped if and only if $H$ is gapped.

Note that in order to build a parent Hamiltonian which has an MPS as its unique ground state, it is sufficent to consider $l+1$ sites, where $l$ is the number of sites where the blocked $A^{i}$ span the space of all matrices. For the AKLT model, this directly yields a 3-body Hamiltonian (which however still has to be broken down to 2-body Hamiltonians).

While (as Meng Cheng comments) one can indeed also obtain a non-trivial parent Hamiltonian from the 2-body reduced density matrix, I am not aware of a general proof technique which allows to show that the latter has a unique ground state. (For the AKLT model, this is of course true by direct inspection.)


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