Inverse Wick rotation

 $$\omega\equiv i\frac{\partial}{\partial t}\log\psi$$

$$\beta:=\frac{1}{k_B T}\equiv -it$$

so

$$\beta\omega=t\frac{\partial}{\partial t}\log\psi,$$

thus

$$e^{\frac{\hbar\omega}{k_B T}}=e^{\hbar\beta\omega}=\exp (\hbar t\frac{\dot{\psi}}{\psi}).$$

In operator form 

$$\hat{e}^{\hbar\beta\omega}=\exp(\hbar t\frac{\partial}{\partial t})$$