$k$QSAT$c$ is the promise problem where the input, given in an explicit encoding with finite number of bits, is a set $\{p_{1},p_{2},\ldots p_{m}\}$ of $k$-local projectors over a $n$-qbits register, such that the least eigenvalue $\varepsilon_{\min}$ of $\sum_{i=1}^{m}p_{i}$ is either $\varepsilon_{\min}=0\tag{y}\label{eq:QSATy-1}$ or $\varepsilon_{\min}(n)\geq n^{-c}\tag{n}\label{eq:QSATn-1}$ and the answer is yes (no) in the y (n) case.
$k$LH$c$ is the promise problem where the input, given in an explicit encoding with finite number of bits, is a set $\{h_{1},h_{2},\ldots h_{m}\}$ of $k$-local hamiltonians, with eigenvalues in $[0,1]$, over a $n$-qbits register, such that the least eigenvalue $\varepsilon_{\min}$ of $\sum_{i=1}^{m}p_{i}$ is either $\varepsilon_{\min}\leq\varepsilon\tag{y}\label{eq:LHy-1}$ where $\varepsilon$ is a parameter given in input, or $\varepsilon_{\min}(n)\geq\varepsilon+n^{-c}\tag{n}\label{eq:LHn-1}$ and the answer is yes (no) in the y (n) case.
With same constants, the QSAT problem restricts LH in two things. First, its promise gap is glued to $0$. Second, its hamiltonian terms are projectors. So I can define two new problems.
QSAT+ Same as QSAT but with general local hamiltonians with eigenvalues in $[0,1]$.
LH- Same as LH but with local projectors.
As pointed out in the answer of a previous question, the classical 5local construction clearly works for projectors, so 5LH- is QMA-hard. Then, since 5LH is in QMA, there is a polynomial reduction from 5LH to 5LH- via a quantum circuit verification for 5LH. Of course, I expect in this case way more natural reductions. However, for $k=2$, we know that 2LH is QMA-hard, and 2QSAT is in P. So what is the complexity of 2LH- and of 2QSAT+?