# Equilibrium points of bounce/instanton solution after Wick's rotation

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

$$L_E = \frac{1}{2}\left(\frac{dq}{d\tau}\right)^2+V(q),\tag{2.14}$$

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

He then states that from conservation of energy formula

$$\frac{1}{2}\left(\frac{dq}{d\tau}\right)^2-V=0.$$

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" $$\lim_{\tau\rightarrow-\infty}q = q_0.\tag{2.15}$$

• 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

$$\left.\frac{dq}{d\tau}\right|_{0}=0.$$

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.

• $\tau$ is real. That's why we call it a rotation -- otherwise it would just be a change of symbol, i.e., merely notation. – AccidentalFourierTransform Sep 8 at 12:52

1. Here we will work in the Euclidean$$^1$$ formulation, where Euclidean time $$\tau\in\mathbb{R}$$ is real. The potential is assumed to satisfy $$V(q_0)~=~0~=~V(\sigma) ,\tag{A}$$ cf. OP's figure. The bounce solution satisfies the following boundary conditions (BCs) $$\dot{q}(\tau_i)~=~0\quad \wedge\quad q(\tau_i)~=~q_0\quad \wedge\quad q(0)~=~\sigma,\tag{B}$$ where $$q\equiv |\vec{q}|$$ and dot means differentiation wrt. $$\tau$$.

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 $$\frac{1}{2} \dot{q}^2-V(q)~=~0 \qquad\Leftrightarrow\qquad \dot{q}~=~\pm \sqrt{2V(q)} .\tag{C}$$ Therefore we find that (minus) the initial time is $$-\tau_i~=~ \int_{q_0}^{\sigma}\frac{dq}{\sqrt{2V(q)}}.\tag{D}$$

3. Now since we assume that the potential is approximately quadratic $$V(q) ~\propto ~(q-q_0)^2 \quad\text{near}\quad q~\approx~ q_0\tag{E},$$ 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 $$\dot{q}(0)~=~0.\tag{F}$$ By time reversal symmetry and uniqueness of solutions to the first-order ODE (C), it follows that the final time $$\tau_f~=~ \int_{q_0}^{\sigma}\frac{dq}{\sqrt{2V(q)}}\tag{G}$$ 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.

• Thank you very much! Now everything is clear – Davide Morgante Sep 8 at 14:48