1
$\begingroup$

In reading from the book, there's a peculiar definition of the phase, stating:

\begin{align} K[a,b] &= \sum_{\text{All paths from $a$ to $b$}} \phi [x(t)]\\ \phi [x(t)] &= \text{const} \cdot e ^{\frac{i}{\hbar}S[a,b]} \end{align}

Why is the phase defined this way? There is no reasoning in the book apart from the fact that this phase must be proportional to the action of the system for the said path. Can someone provide a more comprehensive reason than stated in the book?

$\endgroup$
1
0
$\begingroup$

There are several ways to see why the path integral's phase is built in such a way.

That phase was first discovered by Dirac, who simply calculated the propagator of a particle and noted that the amplitude was $\propto e^{\frac{i}{\hbar}S_{cl}}$.

You can also show the path integral formalism from the Schrödinger formalism, simply with the identity $\hat{1} = \int dx \vert x \rangle\langle x \vert$ and the Markov property of the time evolution operator $U(t_1, t_2) = U(t_1, t) U(t, t_2)$ :

\begin{eqnarray} K(x,t;x',t') &=& \langle x,t \vert U(t,t') \vert x',t'\rangle = \int dx_1 \langle x,t \vert U(t,t_1) \vert x_1 \rangle\langle x_1 \vert U(t_1,t') \vert x',t'\rangle\\ &=& \int \prod dx_i \langle x_i,t_i \vert U(t_i,t_{i+1}) \vert x_{i+1} \rangle \end{eqnarray}

with $U\propto e^{\frac{i}{\hbar}H}$

You can also use the stationary phase approximation to show that in the limit $\hbar \rightarrow 0$, the probability of the classical path goes to 1.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.