# Momentum conservation in path integral formulation

In the path integral formulation of quantum mechanics, the amplitudes of all possible paths from starting point A to end point B are added up.

For a free particle, how can all these paths be physically possible? If the particle is to conserve momentum, it must travel in a straight line. It is not interacting with anything else, so the path cannot be a curve in flat spacetime. Why does the integral include physically impossible paths?

• "Physically" is not synonymous with "classically". Jun 23, 2021 at 2:00
• Jun 23, 2021 at 5:13
• @ConnorBehan: I'm not aware of any mechanism that allows a free particle to "physically" violate momentum conservation. Please illuminate me. Jun 23, 2021 at 16:41
• There doesn't need to be one. Every time a new theory becomes accepted, it's because of a prediction which the old theory couldn't make. In this case, the virtual paths you drew have no way of being predicted by definite phase space trajectories. But experiments have shown that these don't exist. Jun 23, 2021 at 18:55

2. In the semiclassical limit $$\hbar\to 0$$, the path integral is dominated by contributions from on-shell classical paths, i.e. solutions to Euler-Lagrange (EL) equations, cf. e.g. this Phys.SE post.
• 1. Off-sell paths still conserve momentum. It violates only $E^2 - p ^2 = m^2$. Besides, in the non-relativistic formulation, this is not relevant. 2. The size of the contribution is not the problem here unless paths not conserving momentum have zero contribution. 3. The question is about the simplest case - the free particle. 4. The non-relativistic free particle lagrangian is translational invariant. Jun 23, 2021 at 16:33