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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:

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

  2. 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 $A_0=1$?

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Maybe I miss some point, but the propagator on a circle (of radius $1$) is just a sum on all winding numbers : $K_{circle}(\phi'', t'';\phi', t') = \sum_{l=-\infty }^{l=+\infty } K(\phi''+2\pi l, t'';\phi', t')$, where $K$ is the standard one-dimensional propagator (Kleinert $6.19$, $6.20$). –  Trimok Oct 6 '13 at 13:42
    
@Trimok: In general, each of the term in the sum, can have a phase factor $e^{in\delta}$ according to Schulman. I am still trying to understand why this is so. An example of such a phase factor occurring in multiply connected spaces if for e.g. the Aharonov-Bohm effect. Would be able to shed more light on why $K_n$ $n$ being the winding number could have a factor $A_n$ multiplying it? –  user7757 Oct 6 '13 at 14:35
1  
Maybe this is not useful, but note that : $\sum_{n=-\infty}^{n=+\infty} e^{2i \pi n\delta} \delta(\phi''+2\pi n - \phi') = e^{i\delta(\phi'-\phi'')} \sum_{n=-\infty}^{n=+\infty} \delta(\phi''+2\pi n - \phi')$ –  Trimok Oct 6 '13 at 15:13
    
@Trimok: The exact words the author uses to justify the claim is that. "As the proof of path integral satisfies the schrodinger equation is a local one, each term in the sum (over winding number) individually satisfies the SE equation. It follows that $\sum_n A_n K_n$ is as good a candidate as $\sum_n K_n$. $A_n$ turns out to be $e^{in \delta}$ (By increasing $\phi^{\prime \prime}$) by a full cycle. –  user7757 Oct 6 '13 at 15:33

1 Answer 1

up vote 1 down vote accepted

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:

  1. 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.

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Thanks a lot. Please could you also comment on why exactly $\sum_n A_n K_n$ rather than $\sum_n K_n$ is taken? This could account for phenomena such as the Aharanov-Bohm effect, but apriori how can I just change the terms of the sum? Schulman says it is because Schrodinger equation is satisfied locally, but I am not being able to see how it is relevant. Isn't the breaking of the sum, to paths with same winding number,just a mathematical convenience? I don't really understand, how you can multiply $K_n$ by factors. Doesn't this contradict Feynmans phase contribution of $e^{iS}$? –  user7757 Oct 6 '13 at 18:46

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