# Path integral of free particle with time-dependent mass

The path integral propagator for a free non-relativistic particle in one dimension is,

$$K(x_1,T;x_0,0)=\int_{x(0)=x_0}^{x_1(T)=x_1}\mathcal{D}[x]\exp\left[\frac{i}{\hbar}\int_0^T\mathrm{d}t\frac{1}{2}m\dot{x}^2\right]=\sqrt{\frac{m}{2\pi i\hbar T}}\exp\left[\frac{i}{\hbar}S_{\rm classical}\right].$$

However, now I am dealing with a case where the mass is time dependent, $$K(x_1,T;x_0,0)=\int_{x(0)=x_0}^{x_1(T)=x_1}\mathcal{D}[x]\exp\left[\frac{i}{\hbar}\int_0^T\mathrm{d}t\frac{1}{2}M(t)\dot{x}^2\right].$$

So I was wondering what the propagator becomes in this case? Does it retain a form like $$\sqrt{\frac{M(t)}{2\pi i\hbar T}}\exp\left[\frac{i}{\hbar}S_{\rm classical}\right],$$

where the $$S_{\rm classical}$$ changes to

$$S_{\rm classical}=\frac{im}{2\hbar}\frac{\left(x_1-x_0\right)^2}{T}\rightarrow S_{\rm classical}=\int_0^T{\rm d}t\frac{1}{2}M(t)\left(\frac{-M(t)\pm\sqrt{M^2(t)-c\dot{M}(t)}}{\dot{M}(t)}\right)^2,$$

because the classical velocity satisfies the following equation,

$$\frac{{\rm d}}{{\rm d}t}\left(\dot{M}(t)\dot{x}^2+2M(t)\dot{x}\right)=0.$$

Can someone answer? It would be better if one shows the derivation.

1. Make a time reparametrization $$t^{\prime}~:=~m\int_0^t \frac{\mathrm{d}s}{M(s)} \qquad\Rightarrow \qquad \frac{dt^{\prime}}{dt}~=~\frac{m}{M(t)}.\tag{1}$$
2. Then the action corresponds to a free particle with time-independent mass: $$S~=~\int_0^T\mathrm{d}t\frac{M(t)}{2}\left(\frac{dx}{dt}\right)^2 ~\stackrel{(1)}{=}~\int_0^{T^{\prime}}\mathrm{d}t^{\prime}\frac{m}{2}\left(\frac{dx}{dt^{\prime}}\right)^2,\tag{2}$$ where $$T^{\prime}~:=~m\int_0^T \frac{\mathrm{d}s}{M(s)}.\tag{3}$$