Path Integral on a circle (calculation of phase and linear independance) I am reading Schulman's "Techniques and applications of path integration" chapter on Path integrals on multiply-connected spaces. In the first section he calculates the path integral of a free particle on a ring. The total path integral can be divided into equivalence classes, each containing paths with the same winding number n, i.e. no. of times we cross a fixes point on the circle. 
$$K(\phi^{\prime \prime}, t^{\prime \prime};\phi^{\prime}, t^{\prime})=\sum_n A_n \sum_{\phi \in g_n} e^{iS(\phi)}=\sum_{n={-\infty}}^{n=\infty} A_n K_n$$
Here $g_n$ is the set (equivalence class) of all paths with winding number n. There is a factor $A_n$ in general, which is obtained if we make the assumption that each term in the sum over winding numbers satisfies the Schrodinger wave eqn locally. Later, it can be proved that $A_{n+1}=e^{i\delta}A_n$
My questions are:


*

*How do I intuitively or rigorously prove that $K_n's$ are linearly independent? Are they also orthogonal?

*Schulman says 'magnitude of $A_0$ is fixed by unitarity to 1'. I cannot see this. I am guessing unitarity means $\int |K(\phi^{\prime \prime}, t^{\prime \prime};\phi^{\prime}, t^{\prime})|^2 d\phi^{\prime \prime}d\phi^{\prime}=1 $, but then there are a lot of cross terms and each of the integrals $\int|K_n|^2$ should be one, so that I am getting the integral diverges.
How do I show $A_0=1$?
 A: Let us suppress the initial and final time, $t_i$ and $t_f$, respectively, from the notation, as they play (almost) no role here.
The full Feynman propagator on a circle $S^1\cong \mathbb{R}/\mathbb{Z}$ is a sum over winding modes$^1$ $n\in\mathbb{Z}$
$$ \tag{23.5}K(\varphi_f;\varphi_i)
~=~ \sum_{n\in\mathbb{Z}}A_n K_n(\varphi_f;\varphi_i), $$
where $K_n(\varphi_f;\varphi_i)=K_0(\Delta\varphi-2\pi n;0)$ is the Feynman propagator for a single winding/instanton sector $n\in\mathbb{Z}$. Here we have defined $\Delta\varphi~:=~\varphi_f-\varphi_i$. The idea is now that if we go around the circle $\varphi_f \to \varphi_f+2\pi$, then the physical result should be the same, i.e. the full Feynman propagator cannot change by more than a phase factor
$$ \tag{A} K(\varphi_f+2\pi;\varphi_i)~=~e^{i\delta}K(\varphi_f;\varphi_i), \qquad \delta\in \mathbb{R}, $$
while by definition the individual sectors get shifted
$$ \tag{B} K_n(\varphi_f+2\pi;\varphi_i)~=~K_{n-1}(\varphi_f;\varphi_i). $$
Thus, if the $K_n$'s are linearly independent, then eqs. (A-B) and (23.5) imply that
$$ \tag{23.6} A_{n+1}~=~e^{i\delta}A_n, $$
so that
$$ \tag{C} K(\varphi_f;\varphi_i)
~=~ A_0 \sum_{n\in\mathbb{Z}}e^{in\delta} K_n(\varphi_f;\varphi_i). $$
It remains to show that $A_0$ is just a phase factor, $|A_0|=1$.  
OP is correct that the modulus $|A_0|$ can in principle be determined from the normalization condition 
$$ \tag{D} \iint_{[0,2\pi]^2}\! \mathrm{d} \varphi_f ~\mathrm{d}\varphi_i ~|K(\varphi_f;\varphi_i)|^2~=~1,$$ 
see also e.g. this Phys.SE post.
Alternatively, if we assume that $A_0$ does not depend on time, then we can prove $|A_0|=1$ by going to the short time limit $|t_f-t_i|\to 0$. In that limit the classical action blows up, so that
$$ \frac{K(\varphi_f;\varphi_i)}{A_0\exp\left[\frac{i\delta}{2\pi}\Delta\varphi\right]}
~\stackrel{(C)}{=}~\sum_{n\in\mathbb{Z}}\exp\left[\frac{i\delta}{2\pi}(2\pi n-\Delta\varphi)\right] K_0(\Delta\varphi-2\pi n;0)$$
$$ ~\quad\longrightarrow\quad~  
\sum_{n\in\mathbb{Z}} \exp\left[\frac{i\delta}{2\pi}(2\pi n-\Delta\varphi)\right] \delta(\Delta\varphi-2\pi n)
~=~\sum_{n\in\mathbb{Z}}  \delta(\Delta\varphi-2\pi n)$$
$$ \tag{E}~=~ \delta(\Delta\varphi-2\pi \mathbb{Z})
~=~\frac{1}{2\pi}\sum_{m\in\mathbb{Z}} e^{im\Delta\varphi}~=:~III_{2\pi}(\Delta\varphi) \quad\text{for} \quad |t_f-t_i|~\to ~0, $$
where we have used the Poisson resummation formula that Trimok wrote in a comment above. See also the Dirac comb function. The rhs. limit of eq. (E) is the Dirac comb function $III_{2\pi}(\Delta\varphi)$, which is the correct limit for the full Feynman propagator $K$ on a circle geometry (up to a phase factor). So $|A_0|=1$. 
The various phase factors appearing between different instanton/winding sectors in the path integral can in principle be accounted for by carefully tracing (i) the physical effects, such as, e.g. the Bohm-Aharonov effect, etc.; and (ii) phase factors implicit in the definition of eigenkets $|\varphi, t\rangle$ localized in angle space.
References:


*

*L.S. Schulman, Techniques and applications of path integration, 1981, chap. 23. 


--
$^1$ It is possible to deduce Schulman's sign convention for the winding mode $n$ from his eq. (23.8) for $K_n$. Incidentally, the explicitly form (23.8) is probably also the simplest and most convincing way to see that the $K_n$'s are indeed linearly independent.
