Instantons in Altland/Simons I have a question about a statement in Condensed Matter Field Theory (2nd edition) by Altland/Simons on p.124. In short, when we consider a motion in a double well we obtain a classical solution in Euclidean time - instantons. For example, for tunneling amplitude $$G(a,\pm a; t) = <a|e^{iHt}|\pm a>$$  we get a summation over amplituds $A_{n}$, associated with $n$ instantons and we have taken into account the fact that, in order to connect $a$ with $\pm a$, the number of instantons must be  even/odd. We get an amplitude $G(a, a; t)\sim \cosh(..)$  and $G(a, -a; t)\sim \sinh(..)$.

When we consider a problem of quantum mechanical tunneling from a metastable state:

to obtain the survival probability $G(q_{m},q_{m};t)$ we summarize over all $n$:

Summing over all bounce trajectories (note that in this case we
have an exponential series – no even/odd parity effect),

But I didn't understand it, because to get an amplitude to stay at $q_{m}$, as we can see from figure, we have to summarize over even $n$ (we go down the hill and come back).
 A: Let's first investigate again the case of the double-well potential $V_{\text{dw}}(x)$, and why a sum over even/odd number of instantons appear. We will normalize the potential such that $V_\text{dw}(\pm a) = 0$. Denote by $s \mapsto x(s;t)$ the solution
$$
m\frac{d^2x(s;t)}{ds^2} = V_{\text{dw}}(x(s;t)) \ ; \\
x(0;t) = -a \ , \ \ 
x(t;t) = a \ .
$$
Note that there is just one solution satisfying this, it passes through the origin once. This can be called the "instanton" solution.
Observe that starting from $x(s;t)$ we can immediately build an "anti-instanton" solution $s \mapsto y(s;t)$ satisfying
$$
m\frac{d^2y(s;t)}{ds^2} = V_{\text{dw}}(y(s;t)) \ ; \\
y(0;t) = a \ , \ \ 
y(t;t) = -a \ .
$$
simply as $y(s;t) = -x(s;t)$.
Furthermore, for large $t$ the derivatives of $x,y$ close to the endpoints are small. Therefore, we can get an almost smooth function $\psi_n(s;nt)$ as
$$
\psi_n(s;nt) = \begin{cases}
x(s;t) \ , \ s \in [kt,(k+1)t] \ \text{for $k < n$ even}  \ ,\\
y(s;t) \ , \ s \in [kt,(k+1)t] \ \text{for $k < n$ odd}  \ .
\end{cases}
$$
Quite clearly, if $s \mapsto x(s;t)$ contributes an amplitude $A(t)$ to the propagator, then $s \mapsto \psi_n(s;t)$ contributes an amplitude $A(t)^n$.
Importantly for $t\rightarrow \infty$, the positions of these instantons can be moved around. Hence one should include a phases space factor
$$
\int_{0}^t ds_1 \int_{s_1}^t ds_2 \cdots \int_{s_{n-1}}^t ds_n = \frac{t^n}{n!} \ .
$$
Hence in $-a \rightarrow -a$ we will have contributions $\sum_n (A(t)t)^{2n}/(2n!)$ etc.
Note that $\psi_n$ is NOT a solution to the equation of motion. The reason why they are still important is that there is a class of solutions which start at a position very close to (but not exactly at) $-a$ and then oscillate $n$ times before ending up very close (but not exactly at) $\pm a$. In the limit where $t\rightarrow \infty$, they converge to $s\mapsto \psi_n(s;t)$.
Note that this possibility of combining instantons and anti-instantons to these $n$-oscillating solutions rests on the fact that the derivatives of the solutions $x,y$ are small close to $\pm a$.
Now let's investigate the second example, with a potential describing a metastable vacuum $V_\text{ms}(x)$. The classical trajectory is the one where we start at $q_m$ and return to $q_m$ in time $t$. You suggest to break that up into two motions; one where you go from $q_m$ over the barrier, and another one where we go back. The problem isn't that the potential $V_\text{ms}$ is not symmetric, but that an instanton solution has to start and to end at a minimum of the potential. In the double-well, this allowed for gluing together of solutions to get new solutions. In this metastable case, we too glue together solutions; but the solutions we glue are those that start and end at $q_m$, and go over the barrier $n$-times. In particular, the time point where the solution turns around, can, even for large $t$, not be shifted around.
