In quantum mechanics, we think of the Feynman Path Integral $\int{D[x] e^{\frac{i}{\hbar}S}}$ (where $S$ is the classical action) as a probability amplitude (propagator) for getting from $x_1$ to $x_2$ in some time $T$. We interpret the expression $\int{D[x] e^{\frac{i}{\hbar}S}}$ as a sum over histories, weighted by $e^{\frac{i}{\hbar}S}$.
Is there a physical interpretation for the weight $e^{\frac{i}{\hbar}S}$? It's certainly not a probability amplitude of any sort because it's modulus squared is one. My motivation for asking this question is that I'm trying to physically interpret the expression $\langle T \{ \phi(x_1)...\phi(x_n) \} \rangle = \frac{\int{D[x] e^{\frac{i}{\hbar}S}\phi(x_1)...\phi(x_n)}}{\int{D[x] e^{\frac{i}{\hbar}S}}}$.
Answer
"It's certainly not a probability amplitude of any sort because it's modulus squared is one." This does not follow... Anyway, an (infinite) normalisation factor is hidden away in the measure. The exponential has the interpretation of an unnormalised probability amplitude. Typically you don't have to worry about the normalisation explicitly because you compute ratios of path integrals, as your example shows. The book about the physical interpretation of path integrals is the original, and very readable, one by Feynman and Hibbs, which now has a very inexpensive Dover edition. I heartily recommend it. :) (Though make sure you get the emended edition as the original had numerous typos.)
No comments:
Post a Comment