metastable decay

Question

I am currently dealing with metastable potential in a form of \begin{equation*} V(x) = \begin{cases} \alpha x^2 &\text{ $x\in (-\infty,a] $}\\ -\gamma x &\text{$x>a $} \end{cases} \end{equation*} It is tnteresting to calculate the metastable lifetime in false minimum of parabolic potential. According to general theory we can obtain the probability of a particle to remain in the well in terms of path integral (in imaginary time).
$$G_{E}(0,0,\beta)=\int_{x(-\frac{\beta}{2})=0}^{x(\frac{\beta}{2})=0} Dxe^{-S_{E}},$$ where $$S_{E}=\int_{-\frac{\beta}{2}}^{\frac{\beta}{2}}d\tau \left( (\partial_{\tau}x)^2+ V(x)\right).$$ We can calculate this integral by using saddle-point method. First of all it is necessary to find classical solution of motion equation (instantonic solution). If we think qualitatively then the classical trajectory in the region of the linear potential is a parabola, and for a parabolic region the decaying exponential. But i dont know about point a, of couse we have the derivative of module. P. S. i know how calculate the metastable lifetime in classical smooth cubic parabolic potential.

Answer 1

You need to consider the classical particle in the inverted potential, so the harmonic potential is now inverted and the downward slope of the potential for $x>a$ now runs upward. So, you write down the equation of motion for the potential $V(x) = -\alpha x^2$, and also $\gamma x$ and integrate the action from an initial point to a final point (chosen to be such that $x>a$), taking into account the jump in velocity as the particle crosses the jump in the potential. Note that to get the full decay amplitude, you need to expand around this classical solution, which involves evaluating a functional determinant.

Technical details can be found in the first two chapters of these lecture notes.