In this lecture at around 53:31, the equation for total amplitude in terms of elementary amplitude along a path is written down.

$$\phi_{\text{total} } ( \overline{x_0} , \overline{x_n} , t_N- t_0) = \lim_{N \to \infty} \prod_{i=1}^n \int_{\overline{x_0}}^{\overline{x_n } } dx_i \phi_{path(x_1,x_2...,x_N)}^N \tag{1}$$

The set up is that a particle is shot by a source to a detector screen and the total probably of the particle being at position $\overline{x_n}$ starting at position $x_0$ starting at time $t_0$ is found by consider the integral of the elementary probability amplitude over all possible paths. Later the elementary probability amplitude is shown to be given as

$$ \phi_{path(x_1,x_2...,x_N)}^N = e^{ i \frac{S \left[\text{path} \right]}{\overline{h}}}.\tag{2}$$

Now my confusion happens due to the successive lecture-3 at 15:41. There he introduces two factors $A$ and $\epsilon$ into the equation. $A$ is a normalization factor and $\epsilon$ is the total time and calls the equation the path integral. Here is a picture:

$$ \phi( x_A,x_B; t_A,t_B) = \lim_{N \to \infty} \left( \prod_{i=1}^{N-1} \int dx_i \right) \frac{1}{A(\epsilon)^N} \exp \{ \frac{i}{ \hbar } S\left[ \text{path} (x_0,...,x_N ) \right] \} \tag{3}$$

What is the difference between (1) and (3)? Has prof maybe made typo in (1)?


1 Answer 1


Yes, $$A(\epsilon)~=~\sqrt{\frac{2\pi i\hbar \epsilon}{m}}, \qquad \epsilon~=~\frac{t_N-t_0}{N},$$ in eq. (3) is the famous Feynman fudge factor, which needs to be included in the path integral measure. For details, see e.g. this, this & this Phys.SE posts.


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