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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

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

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

enter image description here

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.

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

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

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