# Why the classical configuration always static when applying saddle point (semi-classical) approximation?

For an Green function/partition function: $$\int D[\phi]e^{\frac{i S[\phi]}{\hbar}}$$ We can make saddle point approximation and gives classical configuration: $$\delta \mathcal{S}=0\Longrightarrow \phi_{cl}.$$ I can understand that when $$\hbar \rightarrow 0$$ (or other equivalent control parameters), the configurations witch deviate from such classical configuration will decay rapidly due to the quickly changing phase. As the result, we can begin with the classical configuration and consider fluctuation around it. However, I think all the solutions for $$\delta \mathcal{S}=0$$ can be seemed as start points of such semi-classical approximation. In the other words, generally speaking, all the configurations which satisfy Euler-Lagrangian equation can be the case and they may be time dependent. But, In fact, most textbook only use the static one:

$$\frac{d}{dt}\phi_{cl}=0$$

and I don't know the reason behind it,

## Example

In Ch.10 of Auerbach, Interacting Electrons and quantum magnetism, the spin path integral gives: $$\mathcal{S}[\Omega]=\int_{0}^{\beta} d \tau S\left(-i \mathcal{H} \partial_{\tau} \Omega+H[\Omega]\right)$$ where $$\Omega$$ is the unit vector and spin $$S$$ play the role of $$1/\hbar$$. When we take the large $$S$$ limit, it is the same as the saddle point approximation. And $$\delta \mathcal{S}=0$$ gives the classical configurations which satisfy: $$\partial_{\tau} \Omega_{\mathrm{cl}}(\tau)=\Omega_{\mathrm{cl}}(\tau) \times \frac{\partial H}{\partial \Omega}$$ which means it can be time dependent, and I think the total partition function need to consider all the classical configurations. However, in fact, it argues that at large $$S$$ limit, there only remains static configuration : $$\partial_{\tau} \Omega_{\mathrm{cl}}(\tau)=0$$. And the following expansion (spin wave theory) also only around the static configuration.