# Path integral over a ring with a magnetic field

My goal is to obtain the propagator, as a function of Energy, in a ring with initial and final angles $\phi_1=0$ and $\phi_2 \in [0,2\pi]$. Note that the full propagator between $\phi_1=0$ and $\phi_2$ will involve sum over paths, which reach $\phi_2$ from $\phi_1=0$ after completing $m$ loops, where $m \in \mathcal{Z}$. What I intend to find here is not the full propagator, which would be a sum over multiple possible paths between $\phi_1=0$ and $\phi_2$, but a restricted version, where I look only into the contribution by the direct path from $\phi_1=0$ and $\phi_2$. This, I think is given by the following expression \begin{align} W(\phi_2,\phi_1)&=\int_0^{\infty}dt \mathrm{exp}(\frac{iEt}{\hbar})K(\phi_1=0, \phi_2)\\ W(\phi_2,\phi_1)&=\int_0^{\infty}dt\mathrm{exp}(\frac{iEt}{\hbar})\int_{\phi_1=0}^{\phi_2}d\phi \underbrace{\langle\phi_2|\mathrm{exp}(\frac{i}{\hbar}\int_0^tdt\mathcal{L})|\phi_1=0 \rangle}_{K(\phi_2,\phi_1=0)} \end{align} with $\phi$ being the angular coordinate in the ring, and $\mathcal{L}$ being the appropriate Lagrangian for a spin-less particle in a ring with a perpendicular magnetic field. Sum of sectors with different winding numbers ($m$) may be brought about by $\sum_{m=-\infty}^{\infty}K(\phi_2+2\pi m, \phi_1=0 )$, with $K(\phi_2, \phi_1=0)$ (neglecting the magnetic part for the sake of brevity) as defined above. Once again, here I am only interested in $m=0$, ie the direct path from $\phi_1=0$ to $\phi_2$. The outer integration over $t$, with the $\mathrm{exp}(\frac{iEt}{\hbar})$ is to resolve in Energy. The Lagrangian is given by, $$$$\mathcal{L} = \frac{m}{2R^2}\dot{\phi}^2-\frac{\Phi}{\Phi_0}\hbar \dot{\phi}$$$$ Here $\Phi$ is the magnetic flux through the ring, and $\Phi_0=\frac{h}{e}$, is the Aharonov-Bohm flux period. This is the same setting as the Aharonov-Bohm experiment. Now the Lagrangian is separated into two pars, the non-magnetic part, $$\mathcal{L_1}=\frac{m}{2R^2}\dot{\phi}^2$$and the magnetic part, $$\mathcal{L_2}=-\frac{\Phi}{\Phi_0}\hbar \dot{\phi}$$The non-magnetic part, $\mathcal{L_1}$, yields the term in the right bracket [ eq 33 in https://arxiv.org/pdf/quant-ph/0208026.pdf . Note that I am interested only in the direct path. Hence I am neglecting the summation over multiple paths as defined in this reference.] and the magnetic part yields the term in the left bracket in the following equation. $$W(\phi_2,\phi_1)=\left[\mathrm{exp}\left(-i\phi_2\frac{\Phi}{\Phi_0} \right)R\sqrt[2]{\frac{m}{2\pi i \hbar}} \right]\left[\int_0^{\infty}dt \frac{1}{\sqrt[2]{t}} \mathrm{exp}\left(\frac{i}{\hbar}\left[-Et+ \frac{mR^2\phi_2}{2t} \right] \right) \right]$$ The term on the right bracket, is the free particle propagator, but in a ring. The term on the left bracket is the contribution of the magnetic flux, which separates out. Now I am stuck with this integral, which diverges. I found this way of resolving the propagator into a propagator with energy E in (eq 2) of [ https://arxiv.org/pdf/cond-mat/0703216.pdf ]. This seems alright as, \begin{align} \Psi(r',t) &= \int dr K(r',t;r,t=0)\Psi(r,t=0)\\ \Psi(r',E) &= \int_{t=0}^{t=\infty}dt\Psi(r',t)\mathrm{exp}(-\frac{iEt}{\hbar})\\ \Psi(r',E) &= \int dr\underbrace{\int_{t=0}^{t=\infty}dt K(r',t;r,t=0)\mathrm{exp}(-\frac{iEt}{\hbar})}_{K(r',r\bigr|E)=W(r',r)}\Psi(r,t=0)\\ \end{align} My application of this seems to be the source of the error. Any help on what has gone wrong here would be appreciated.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Apr 19 '17 at 10:55
• Related question about obtaining the propagator as a function of angle/position: physics.stackexchange.com/q/297278/50583 – ACuriousMind Apr 19 '17 at 10:55
• The answer suggested by ACuriousMind uses the Fourier series of the $K(\phi_2,\phi_1)$ as mentioned in my question. [ eq 32,34,34 in arxiv.org/pdf/quant-ph/0208026.pdf ]. However what is true is that the $K$s themselves should diverge, regardless of the presence of the magnetic field. Could the divergence of the energy resolved propagator, $W$, as mentioned in my question be related to this diverging $K$? – evening silver fox Apr 19 '17 at 14:05
• Yes. If you read EmilioPisanty's excellent answer fully you find that he suggests several ways of "dealing" with this (prima facie impossible-to-get-rid-of) divergence, mainly thinking of the propagator as a distribution, not a function, and adopting a consistent regularization procedure (stay with the Fourier series/introduce imaginary time/etc.) to evaluate integrals involving it. – ACuriousMind Apr 19 '17 at 14:09
• I'm not sure why you got a divergence there. The divergence in my answer comes from the Fourier series, which you're ignoring. (You're, in effect, doing a particle on a line, with some extra terms that can be gauged away.) – Emilio Pisanty Sep 5 '17 at 7:42