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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.

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  • $\begingroup$ So your question is about the coefficient in front of the action? $\endgroup$
    – Qmechanic
    Jul 2 '20 at 11:51
  • $\begingroup$ @Qmechanic my question is about the choice $e^{iS/\hbar}$, why not for example $e^{-S/\hbar}$? It satisfies the composition rule, it is a function of $S/\hbar$, i.e. $f(S/\hbar)$, it is also dominated by the classical trajectory. As I see it, it fits as good as the other choice. $\endgroup$
    – Pinkman98
    Jul 2 '20 at 11:56
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    $\begingroup$ are you familiar with the derivation of Eq. (1)? $\endgroup$
    – fqq
    Jul 2 '20 at 11:56
  • $\begingroup$ @fqq yes I am familiar with the one that starts with the Schrodinger amplitude $\left\langle x_f|e^{-i\frac{H}{\hbar}t}|x_i\right\rangle$. But Feynman's book starts directly by saying, the amplitude has a phase proportional to the classical action. $\endgroup$
    – Pinkman98
    Jul 2 '20 at 12:02
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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 $-\frac{1}{\hbar}$.

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  • $\begingroup$ Right. But when proving it this way, we get some left over coefficients in the second order of the time slice due to the Campbell-Baker-Haussdorf formula for commutation. In the proof this coefficient is neglected as the time slice goes to 0. Can we say in general, if this isn't the case, we just regroup it with the measure $\mathcal{D}x$? $\endgroup$
    – Pinkman98
    Jul 2 '20 at 12:46
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    $\begingroup$ That is a much more subtle question, which is partially addressed in the last link. $\endgroup$
    – Qmechanic
    Jul 2 '20 at 12:58

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