But let's say I'm only interested in the eigenvector associated with the lowest eigenvalue (as is the case when using Lanczos to find the ground-state of a quantum mechanical system).
You'd want to use a shift-and-invert spectral transformation with an estimate of the eigenvalue you care about. Lanczos normally finds the largest magnitude eigenvalue and its eigenvector (after some post-processing); inversion transforms the problem to solving the smallest magnitude eigenvalue, and shifting transforms the problem to solving for the eigenvalue closest to an eigenvalue of interest (the shift).
Wouldn't then any start vector that has some non-zero overlap with the ground-state do?
If I remember correctly, if you give Lanczos the exact eigenvector corresponding to a given eigenvalue, it will only generate the Krylov subspace corresponding to that eigenvector-eigenvalue pair, assuming all calculations are performed in exact arithmetic. In double precision arithmetic, with some partial reorthogonalization, it will eventually generate a basis, given enough iterations (with probability 1, since on a set of measure zero, there are pathological inputs, as Kirill points out). So, essentially any starting vector will work.