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In Coleman's paper Fate of the false vacuum: Semiclassical theory while working out the exponential coefficient for tunneling probability through a potential barrier, he studies the problem with Wick's rotation $tau=it$, getting to the Euclidean Lagrangian

$$L_E = frac12left(fracdqdtauright)^2+V(q),tag2.14$$

where clearly the potential is inverted. The potential as given in the paper is this

He then states that from conservation of energy formula

$$frac12left(fracdqdtauright)^2-V=0.$$

i quote

"By eq. (2.12)"-the conservation of energy-"the classical equilibrium point, $q_0$, can only be reached asymptotically, as $tau$ goes to minus infinity"
$$lim_taurightarrow-inftyq = q_0.tag2.15$$

• Q1. Why is this true? How you define infinity for a complex number?

Then, by translation invariance, he sets the time at which the particle reaches $sigma$ as $tau=0$ and that

$$left.fracdqdtauright|_0=0.$$

He goes on by saying that this condition

"[...] also tells us that the motion of the particle for positive $tau$ is just the time reversal of its motion for negative $tau$; the particle simply bounces off $sigma$ at $tau=0$ and returns to $q_0$ at $tau=+infty$."

• Q2. Even this isn't very clear for me. Why should the condition for zero velocity at $sigma$ imply that?

Is there something really basic that I'm missing? I'm not very competent in Wick's rotations and such and I have to understand every little bit of this paper for my bachelor's thesis.

1. Here we will work in the Euclidean$^1$ formulation, where Euclidean time $tauinmathbbR$ is real. The bounce solution satisfies the following boundary conditions (BCs)
   $$ dotq(tau_i)~=~0quad wedgequad q(tau_i)~=~q_0quad wedgequad q(0)~=~sigma,tagA $$
   where $qequiv |vecq|$ and dot means differentiation wrt. $tau$. The potential is assumed to satisfy
   $$ V(q_0)~=~0~=~V(sigma) ,tagB $$
   cf. OP's figure.




2. The potential is minus $V$. Energy is conserved (and equal to zero, since that's what it was in the beginning $tau!=!tau_i$). Therefore
   $$ frac12 dotq^2-V(q)~=~0
   qquadLeftrightarrowqquad
   dotq~=~pm sqrt2V(q) .tagC $$
   Therefore we find that (minus) the initial time is
   $$ -tau_i~=~ int_q_0^sigmafracdqsqrt2V(q).tagD $$




3. Now since we assume that the potential is approximately quadratic
   $$V(q) ~propto ~(q-q_0)^2 quadtextforquad q~approx~ q_0tagD,$$
   cf. OP's figure, we see that the integrand (D) has a simple pole at the lower integration limit $q!=! q_0$, implying that the integral $tau_i=-infty$ cannot be finite. This answers OP's first question$^1$.




4. From energy conservation we see that
   $$ dotq(0)~=~0.tagE $$
   By time reversal symmetry and uniqueness of solutions to the first-order ODE (C), it follows that the final time
   $$tau_f~=~ int_q_0^sigmafracdqsqrt2V(q)tagF $$
   is given by the very same formula (D). This answers OP's second question.




5. For more details, see this related Phys.SE post, and links therein.

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$^1$ Concerning Wick rotation, see e.g. this Phys.SE post and links therein.