Does a functional integral really need a Lagrangian?
  Path integral formulation of quantum mechanics (and also other equivalent formulations) depends on a Lagrangian and/or Hamiltonian function that is chosen to describe the underlying classical system. The arbitrariness presented in this choice leads to a phenomenon called Quantization ambiguity. For example both $L_1=\dot{q}^2$ and $L_2=e^\dot{q}$ are suitable Lagrangians on a classical level ($\delta L_1=0=\delta L_2$), but quantum mechanically they are diverse. This paper presents a simple rearrangement of the path integral to a surface functional integral. It is shown that the surface functional integral formulation gives transition probability amplitude which is free of any Lagrangian/Hamiltonian and requires just the underlying classical equations of motion. A simple example examining the functionality of the proposed method is considered. 
Denis Kochan