Hydrogen bond

Including the contribution of the external field we have a total electric potential $\phi(x) = V(x) + U(x) = -\frac{a}{2} x^2 + \frac{b}{4} x^4 - E x$ with the external electric field $E = -\frac{\partial}{\partial x} U$ . In equlibrium, the total electric field (gradient of the potential) must be zero: $\begin{aligned} \frac{\partial \phi}{\partial x} &= -a x + b x^3 - E = 0 \\ \Rightarrow \quad E &= b x^3 - a x =: h(x) \end{aligned}$ Therefore, he have a cubic equation for $x$ . Which solutions this equation possesses, one can make oneself clear by the curve discussion of the function $h(x)$ . This function has two extrema at the points $\begin{aligned} h'(x) &= 3 b x^2 - a = 0 \\ \Rightarrow \quad x_\pm &= \pm \sqrt{\frac{a}{3b}} \\ \Rightarrow \quad h(x_\pm) &= \mp \frac{2 a^{3/2}}{3^{3/2} b^{1/2}} = \pm h_0 \end{aligned}$ If $| E | <h_0$ , the equation $E = h (x)$ has three solutions corresponding to two minima and one maximum of the potential $\phi(x)$ . If $| E | > h_0$ , the equation $E = h (x)$ has only one solution, which corresponds to a single minimum. In the limiting case $| E | = h_0$ there is a saddle point and a minimum, so that here the bond is just broken. Thus, the critical field strength results $E = h_0 \approx 3.48 \,\frac{\text{GV}}{\text{m}}$