Feynman & Hibbs, Problem 3-1: Path integral for a free particle

[Trajito]

Hello,

Taking the absolute square of the kernel for a free particle in one
dimension, Feynman calculates the probability of finding the particle
at a location b after a time t passes (p. 43, Problem 3-1). The
solution reads (assuming x_0 and t_0 to be 0)

P(b)dx = m dx / 2 pi hbar t

The book informs us that "clearly this is a relative probability,
since the integral over the complete range of x diverges," and asks
that "what does this particular normalization mean?"

I could not understand what this particular normalization does really
mean. If the integral over the complete range of x to be diverge, why
did not we find a normalization constant that makes the integral 1?
Does the probability need a re-normalization?

Thank you

[zao...@gmail.com]

"the particular normalization" is the very divergence of the probability. I think he wanted you to relate this divergence to the fact that the particle starts off at the exact point a, ie the propagator is a plane wave and the uncertainty at t_a in momentum is infinite. Thus the particle is equally likely to be anywhere in space at t_b. That's why it corresponds to the classical picture described in following this sentence. If it starts off at a small neibourhood around a, the propagator would be a sum of all propagators in the neibourhood, which is a wave packet and is normalizable.