General Settings

Consider a particle circles on a ring, and the radius of the ring is $d$. Suppose its rotation angle at time $t$ is $\theta(t)$. Obviously, this system has rotational invariance and time transition invariance. Its Lagrangian is

$$ L=\frac{1}2 mR^2\dot{\theta}^2 $$

Set boundary conditions (initial and final position):

$$ \theta_i=0,\ t_i=0\quad\text{meanwhile}\quad \theta_f=\theta,\ t_f=T\ (0\leq\theta<2\pi). $$

Because for any $\theta$, it plus $2\pi$ gives the same final position, so we have $\infty$-many classical paths with the corresponding $\theta$.

$$ \theta_{cl}^{(n)}(t)=\frac{t}{T}(\theta+2n\pi),\quad n=0,\pm1,\pm2,\cdots $$

$n$ is the winding number. To use path integral form, compute classical action first.

$$ \begin{aligned} S_{cl}^{(n)}& =\int_{0}^{T}dt\ L(\theta,\dot{\theta},t)=\int_{0}^{T}dt\ \frac{1}2 mR^2(\dot{\theta}_{cl}^{(n)})^2\\ &=\frac{mR^2}{2T}(\theta+2\pi n)^2, \end{aligned} $$

It's the classical action corresponding to the winding number $n$.

Sum Over Paths

Now we can sum transition amplitude over all classical paths. Recall the path integral form of free particle

$$ K(\theta,T;0,0)=\sum_{n=-\infty}^{\infty}\sqrt{\frac{m}{2\pi i\hbar T}}\exp\Big[\frac{imR^2}{2\hbar T}(\theta+2\pi n)^2\Big]. $$

Here, by intuition, we suppose $F(T)$ (the square root term) is the same as free particle, by regarding $(\theta+2\pi n)$ distribute on a real line. It can be shown strictly, but not here.

It can be more general by introducing a pre-factor $A_n$

$$ K(\theta,T;0,0)=\sum_{n=-\infty}^{\infty}A_n\sqrt{\frac{m}{2\pi i\hbar T}}\exp\Big[\frac{imR^2}{2\hbar T}(\theta+2\pi n)^2\Big] $$

and $A_n$ is mot arbitrary. We can claim

$$ A_n=e^{in\delta},\ \delta\in \mathbb{R}. $$

Obviously, $\delta$ stands for phase, but later we will see $\delta$ has its physical meaning, which is related to energy eigenvalues.

Use the notation

$$ K_n(\theta,T;0,0)=\sqrt{\frac{m}{2\pi i\hbar T}}\exp\Big[\frac{imR^2}{2\hbar T}(\theta+2\pi n)^2\Big]. $$

And it's easy to show

$$ \begin{aligned} K(\theta+2\pi,T;0,0)& =\sum_{n=-\infty}^{\infty}A_nK_{n+1}(\theta,T;0,0)=\sum_{n=-\infty}^{\infty}A_{n-1}K_{n}(\theta,T;0,0)\\ &=e^{-i\delta}\sum_{n=-\infty}^{\infty}A_nK_n=e^{-i\delta}K(\theta,T;0,0). \end{aligned} $$

That is, physically, as a wavefunction, $K$ can change at most by variable $\delta$ (phase). And we can show

$$ A_n $$

To be continued.