Euclidean functional Integrals: Existence of zero eigenvalue due to time translation symmetry

Question

In the chapter "Uses of Instantons" from the book "aspects of symmetry" by Sidney Coleman I have come across the euclidean version of the path integral in semi-classical approximation. To evaluate the sum over paths, each path is written as the sum of the classical path (for which action is an extremum) and the sum of a set of orthonormal functions obeying the boundary conditions with respective co-efficients. Now action is invariant under time translation. My confusion is that, it is written that due to time translation symmetry, the operator appearing in the path integral formula (whose determinant has to be evaluated) must have a zero eigenvalue, where the eigenfunctions are those orthonormal functions obeying the boundary condition. I can not understand this argument.

Answer 1

Well, besides time translation invariance, it is important that the action $S$ admits instanton solutions $\bar{x}$ in the first place. (This is the case for the double-well potential that Coleman is analyzing in Section 2.2 p.270.) In the $T=t_f-t_i\to\infty$ limit, again because of no explicit time dependence, these instanton solutions must be parametrized by a "collective" time coordinate (which yields the instant of the instanton, so to speak). This signals the existence of a zero-mode.

For more information, see also Remark (3) on the bottom of p. 276 and Appendix 2 p. 341.