Projector Monte Carlo states that if we have a trial wavefunction $|\phi\rangle$ which is not orthogonal to true ground-state $|\psi\rangle$ of the system then application of a projector $$P=\exp{(-\tau H)}$$ $m$ times on $|\phi\rangle$ while $m$ goes to infinity will give us true ground state $|\psi\rangle$ i.e. $$|\psi\rangle=\lim_{m\to\infty}P^m|\phi\rangle$$
My question is how is the project $P$ is projecting out true ground-state when it is applied $m\to\infty$ times?
The projector operator is given by,
\begin{equation} \hat{P} = e^{-\tau \hat{H}} \end{equation}
We have to show that $ | \phi^{m} \rangle = (\hat{P})^{m} | \phi \rangle $ in the large $m$ limit gives, $ | \phi^{m} \rangle = | \psi \rangle $.
\begin{align} | \phi^{m} \rangle = (\hat{P})^{m} | \phi \rangle & = \displaystyle\lim_{m \to \infty} \sum_{\alpha} e^{-m \tau \hat{H}} | \alpha \rangle \langle \alpha | \phi \rangle \\ & = \displaystyle\lim_{m \to \infty} \sum_{\alpha} e^{-m \tau E_{\alpha}} | \alpha \rangle \langle \alpha | \phi \rangle \\ & = | \psi \rangle \end{align}
In the limit of large $m$ and fixed $\tau$, the projection will wipe out all the states with high energy and only the ground state (lowest energy i.e $E_{0}$) will survive. We have assumed that the ground state is non-degenerate (unique!). In case of degenerate vacua, there is some I critical slowdown.
