One more comment:

The quantum PCP conjecture implies that calculating the energy of the Gibbs state of a quantum system in constant temperature is QMA-Complete. This answers the question:

what implications does its proof have for simulation of Hamiltonians?

Given a Hamiltonian $H$, the Gibbs state is defined by:
$$\rho_{\mathrm{gibbs}} = \frac{1}{Z}\exp(-H/T)$$
where the normalization factor $Z=\operatorname{Tr}(\exp(-H/T))$.

It is known that the Gibbs state minimizes the free energy (H - TS):
$$\rho_{\mathrm{gibbs}}=\operatorname{argmin}_{\rho \succeq 0, \operatorname{Tr}(\rho)=1} \operatorname{Tr}(H\rho) + T \cdot \operatorname{Tr}(\rho \ln \rho)$$

Let $\rho_{\min}$ be one of the pure ground states of $H$, with energy $E_0$. Since $\rho_{\min}$ is a pure state, $Tr(\rho \ln \rho)=0$.

Therefore,
$$\operatorname{Tr}(H\rho_{\mathrm{gibbs}}) + T \cdot Tr(\rho_{\mathrm{gibbs}} \ln \rho_{\mathrm{gibbs}}) \leq \operatorname{Tr}(H \rho_{min}) = E_0.$$

Since $\operatorname{Tr}(\rho \ln \rho) \leq n \ln d$, where $n$ is the number of qudits (of dimension $d$ in the systems,

$$\operatorname{Tr}(H\rho_{gibbs}) \leq E_0 + T n \ln d .$$
By choosing $T = O\left(\frac{\epsilon}{\ln d}\right)$ one gets an additive approximation of $\epsilon n \leq \epsilon m$ to the ground energy, which is $QMA$-hard by the quantum PCP conjecture ($m$ is the number of terms in the Hamiltonian, therefore it is at least linear in the number of qubits in the system).

I believe this result is a folklore, and clearly also holds in the classical setting. I learned about it from the following paper:
Brandão, F. G., & Harrow, A. W. (2013, June). Product-state approximations to quantum ground states. In Proceedings of the 45th annual ACM symposium on Symposium on theory of computing (pp. 871-880). ACM.