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  1. Let's say I have 2 commuting Hamiltonians that are not degenerate, I know it means that they a have a common energy basis, yet does it mean that they also have the same ground state? Or is there any way to learn something about the ground state of one from knowing the ground state of the other? Assume that they both have similar structure, for example they are both 2 local on a square lattice acting on n particles.

  2. Is there a sufficient condition for two Hamiltonians to have the same groundstate?

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It is instructive to remind oneself that in finite dimensions, Hamiltonians are represented by Hermitian matrices and think in terms of these matrices. If two Hamiltonians $H_1$ and $H_2$ commute, then -- as you pointed out -- there exists a common eigenbasis in which they are both diagonal: $$ H_1 = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n) \qquad H_2 = \operatorname{diag}(\mu_1, \mu_2, \dots, \mu_n) $$ Note that the eigenvalues will not be ordered and that I start enumerating basis states with $|1\rangle$. The groundstate of $H_1$ ($H_2$) in this basis is the basis vector $\boldsymbol{e}_j = |{j}\rangle$ for which $\lambda_j$ ($\mu_j$) is the minimal eigenvalue.

From this, we can already construct a very simple counterexample to your first question, just consider the two Hamiltonians $$ H_1 = \begin{bmatrix} -1 & 0 \\ 0 & 1\end{bmatrix}\qquad H_2 = \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix} $$ The first has ground state $\boldsymbol{e}_1 = |1\rangle$ whereas the second has groundstate $\boldsymbol{e}_2 = |2\rangle$. The only benefit you get from knowing that the Hamiltonians commute is that you know that one of the eigenstates of $H_1$ will also be an eigenstate of $H_2$. In the non-degenerate case, this means that you could just try all eigenstates of $H_1$ on $H_2$ and you find the groundstate, but nothing more than that.

This highlights that you can't make non-trivial statements about the groundstate of $H_2$ when knowing the groundstate of $H_1$ in general. I personally don't know if this changes if the Hamiltonians in question have more structure like 2-locality.

I'm not able to point you to a reference for your second question. Phrasing it in the language of matrices, the question becomes "Do $H_1$ and $H_2$ have the same eigenvector for their minimal eigenvalue?". It seems obvious to me that there is no further sufficient condition for this.

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