Now, I'm guessing [itex]N = \left(\frac{m}{i\hbar\tau}\right)^{(n+1)/2}[/itex]. Am I right? Also, what does he mean by "we shall always deal with the normalized transition amplitudes"? What is the normalization here?

Thanks in advance.

PS -- How does one get the nice looking caligraphic D for the functional integral? I used \mathcal{D}, but the effect isn't as pronounced as in Ryder's book

The 2 in the denominator inside the squared-velocity should probably be a tau, and the Lagrangian should probably have minus the potential in it, and not positive the potential.

You're right about 'N'. It goes to infinity as tau, the discrete time interval, goes to zero. Normalization means that [tex]
| \langle q_f t_f|q_i t_i \rangle |^2
[/tex]
summed over all final states [tex]|q_f,t_f> [/tex] equals one, since at time [tex]t_f[/tex] the particle has to be in some [tex]q_f[/tex]. So the N doesn't really matter since you multiply every amplitude by the same constant factor to get the total probability of being somewhere add up to one.

Yes, you're right. Typographical mistake, sorry about that.

Ok, but then

[tex]|\langle q_f t_f|q_i t_i\rangle|^2[/tex]

is proportional to [itex]N^2[/itex], and if you sum it over all possible final states [itex]|q_f t_f\rangle[/itex], you will get a sum multiplied by [itex]N^2[/itex]. If this sum is to become one even as [itex]N \rightarrow \infty[/itex] shouldn't the sum it multiplies tend to zero or something?

I am using

[tex]\langle q_f t_f|q_i t_i\rangle = N \int \mathcal{D}q \exp{\left[\frac{i}{\hbar}\int_{t_i}^{t_f}L(q,\dot{q}) dt\right]}[/tex]

I'm not entirely sure I understand how something as simple as [tex]
\langle q_f t_f|q_i t_i\rangle
[/tex]
happens to be infinity. My guess is that in order to derive the path integral, one inserts too many intermediate states in between the initial and final states (using the completeness relations) and that's where the infinity comes from.

Nevertheless, I would view this normalization as similar to playing the lottery. The more times you play, the more times you win. However, this should not affect the probability. So if you win the lottery 10 times, you have to divide by the 100 times you played, for a 10% winning probability. Similarly, I would view what you're calculating as not the probability, but how many times you've won, or how many times the particle is found at [tex]q_f[/tex]. So what you have is relative probabilities - you can tell that one state is 3 times more likely if you win with that state 3 times more. To get the absolute probability, you have to divide by an overall constant to get everything to add up to 1.

as a "short hand notation" for the discretized form. In the case of a free particle ([itex] V=0[/itex]) you can check by explicit calculation of the discretized form that it is finite and indeed gives you the correct form of the propagator using [itex](n+1)\tau=t_f-t_i[/itex]. When evaluating path integrals for [itex]V\neq 0[/itex], one usually uses the free propagator as a "comparison"/normalization to obtain the correct form. For example:

where [itex]\langle q_f t_f|q_i t_i\rangle_0[/itex] denotes the propagator of a free particle. Then inserting in the denominator the path integral expression for the free particle you find that the "normaliziation factor" [itex]N[/itex] cancels.

where [itex]L^0=m\dot{q}^2/2[/itex], i.e. the Lagrangian of a free particle. Then finally you can replace [itex]\langle q_f t_f|q_i t_i\rangle_0[/itex] with the explicit form of the free particle propagator. This way we don't have to deal with the factor [itex]N[/itex].

Just wanted to add something on top of what I said before. Since almost all we can do with path integrals boils down to some combination of perturbation expansion, saddlepoint approximations and gaussian integrals it is useful to look at the case with quadratic potential [itex]V(q)=m\omega_0^2q^2/2[/itex], i.e. Harmonic oscillator. A saddle point approximation of such an action including gaussian fluctuations around the classical path turns out to be exact. You can then expand the fluctuations in as [itex]\delta q=\sum_n c_n \sin(\omega_n x)[/itex] where [itex]\omega_n=\pi n/(t_f-t_i)[/itex]. Eventually you get something like this

Notice that since the infinite product is divergent this path integral goes to zero. I guess what eventually gives you a nonzero answer is that the normalization factor [itex]N[/itex] becomes infinite, but there is no way to compare the two divergencies. This is why its easier to evaluate the quotient of two path integrals such that the normalization factor N cancels and you will need to evaluate a quotient of determinants which is not divergent: