Let G=(V,E) be a simple, undirected, unweighted, connected graph. A cut defined by a subset A of V is called trivial if either A or is a singleton set. Let μ be the second smallest eigenvalue of the Laplacian matrix of G. The length of the shortest cycle in the graph is called the girth g of the graph. Let the minimum degree of the graph be δ3. We show that if μ is greater than a threshold, namely 8δ/((δ−1)^(g−1)/2−2), then every minimum cut in G is trivial. The proof is based on the observation that in graphs of large girth and minimum degree δ3, there exists a dichotomy of minimum cuts: Either the minimum cut is trivial or there must be a lot of vertices (Ω((δ−1)^(g−1)/2)) on both sides of the cut. We illustrate that, for large enough values of g, the value obtained by us for this threshold is of the correct order by constructing a graph with girth at least g and minimum degree δ and μ=Ω(g⁻¹(δ−1)^−(g−1)/2), but possessing a nontrivial minimum cut, assuming that a well-known conjecture about the existence of certain high girth graphs is true. Our results in this paper have the obvious algorithmic implication that when we have the a priori information that the value of μ for the given graph is greater than the threshold suggested by our theorems, the algorithmic problems of finding a minimum cut or enumerating all the minimum cuts in the graph becomes trivial.