In this paper, it says that one can "determine the ground state wave function by applying the projection operator $\exp(-\tau H)$ to an arbitrary initial state $|\Psi\rangle$," and that in the limit of $\tau \to \infty$, we have that $\exp(-\tau H)|\Psi\rangle$ converges to the true ground state of the system. It then goes on to state that that this projection cannot be done in a single step since terms in the Hamiltonian $H$ do not commute with each other. My question is two-fold:
- How do we know that $\exp(-\tau H)|\Psi\rangle$ converges to the true ground state of the system in the limit $\tau \to \infty$? I don't recall this coming up in any of my QM courses, but I might be missing something very self-evident here.
- Why does a Hamiltonian with non-commuting terms mean that we cannot do this projection in a single step? I understand how non-commuting terms cause difficulty when we want to split $\tau$ into many small pieces and "build up" to the full $\tau$ by applying the projection operator iteratively. However, the way this sentence is written in the paper makes me think that there is some a priori reason for why $\exp(-\tau H)|\Psi\rangle$ does not give the true ground state, after a single application, when $H$ contains non-commuting terms.
Thanks for the help!