If we assume that \( \lambda_0 \) is the largest eigenvector we see that in the limit \( t\rightarrow \infty \), \( \mathbf{\hat{w}}(t) \) becomes proportional to the corresponding eigenvector \( \mathbf{\hat{v}}_0 \). This is our steady state or final distribution.

We can relate this property to an observable like the mean energy. With the probabilty \( \mathbf{\hat{w}}(t) \) (which in our case is the squared trial wave function) we can write the expectation values as

$$ \langle \mathbf{M}(t) \rangle = \sum_{\mu} \mathbf{\hat{w}}(t)_{\mu}\mathbf{M}_{\mu}, $$

or as the scalar of a vector product

$$ \langle \mathbf{M}(t) \rangle = \mathbf{\hat{w}}(t)\mathbf{m}, $$

with \( \mathbf{m} \) being the vector whose elements are the values of \( \mathbf{M}_{\mu} \) in its various microstates \( \mu \).