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The path intgral formalism of quantum mechanics states that the amplitude to go from $\left(x_i,t_i\right)$ to $\left(x_f,t_f\right)$ is $$K\left(x_f,t_f,x_i,t_i\right) = \int \mathcal{D}x\quad e^{i\frac{S\left[\gamma(x)\right]}{\hbar}}\tag{1}$$ where $\gamma$ is a possible trajectory and the integral is the sum on all trajectories. The trajectories that dominates are those $|S-S_{classical}|\leq \hbar$. Those that lie ahead of this limit cancel each other.

My question is why, for example, the choise $e^{-\frac{S\left[\gamma\right]}{\hbar}}$ isn't a possible option to represent the amplitude. Using a saddle point approximation you can see that the biggest contribution comes from the classical trajectory for which $\delta S =0$. And the this amplitude agrees with the composition rule as well.

Qmechanic
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devCharaf
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1 Answers1

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It seems OP is asking about the coefficient$^1$ $\frac{i}{\hbar}$ in front of the action $S$ in the Boltzmann factor $e^{\frac{i}{\hbar}S}$ of the path integral, cf. e.g. this related Phys.SE post. The coefficient is fixed by how the path/functional integral formulation is derived from the operator formalism in the first place (using an operator ordering and time slicing prescription). The coefficient $\frac{i}{\hbar}$ essentially follows by comparing the following 2 facts:

  1. The time evolution operator is $\hat{U}=e^{-\frac{i}{\hbar}\hat{H}\Delta t}$.

  2. The Hamiltonian action is $S=\int_{t_i}^{t_f}\! dt (p_j\dot{q}^j-H)$.

For a full proof, consult any good textbook on QM and path integrals. See also this related Phys.SE post.

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$^1$ Here we assume Minkowski signature. In the Euclidean signature the coefficient is $-\frac{1}{\hbar}$.

Qmechanic
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