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  Why isn't the path integral defined for non homotopic paths?

+ 4 like - 0 dislike
3185 views

Context

In the Aharonov Bohm effect, there is a solenoid which creates a magnetic field. Since the electron cannot be inside the solenoid, the configuration space is not simply connected.

Question

I've read in this paper, that the path integral is defined only for paths in the same homotopy class in the configuration space. But I don't see the reason for this. Could someone explain it or give any reference?

It seems that Laidlaw, DeWitt and Schulman have done some work, but I didn't see any proof. And Feynman & Hibbs don't seem to mention it.

Furthermore, does the same problem arise in standard variational calculus when one applies Hamilton's principle?

This post imported from StackExchange Physics at 2014-11-19 20:06 (UTC), posted by SE-user jinawee
asked Nov 12, 2014 in Theoretical Physics by jinawee (120 points) [ no revision ]
Isn't it just down to the definition of path-homotopy? I mean, by definition, the end-points have to be the same and all paths lying in between are the homotpy class. I think this is just a consequence of the homotopy class being end-point preserving.

This post imported from StackExchange Physics at 2014-11-19 20:06 (UTC), posted by SE-user Autolatry
Isn't it just down to the definition of path-homotopy? I mean, by definition, the end-points have to be the same and all paths lying in between are the homotpy class. I think this is just a consequence of the homotopy class being end-point preserving.

This post imported from StackExchange Physics at 2017-01-08 20:03 (UTC), posted by SE-user Autolatry
@Autolatry But if the domain is not simply connected, even if the endpoints are fixed, they won't necessarily belong to the same homotopy class.

This post imported from StackExchange Physics at 2014-11-19 20:06 (UTC), posted by SE-user jinawee
@Autolatry But if the domain is not simply connected, even if the endpoints are fixed, they won't necessarily belong to the same homotopy class.

This post imported from StackExchange Physics at 2017-01-08 20:03 (UTC), posted by SE-user jinawee
@jinawee Hmmn, I may be confusing the situation here but isn't is true that the domains are (formally) extended through an analytic continuation (in definition of the domain); doesn't simply connectedness become paramount in the definition since the application of a monodromy theorem demands it?

This post imported from StackExchange Physics at 2014-11-19 20:06 (UTC), posted by SE-user Autolatry
@jinawee Hmmn, I may be confusing the situation here but isn't is true that the domains are (formally) extended through an analytic continuation (in definition of the domain); doesn't simply connectedness become paramount in the definition since the application of a monodromy theorem demands it?

This post imported from StackExchange Physics at 2017-01-08 20:03 (UTC), posted by SE-user Autolatry
@Autolatry If you mean that the domain is not simply connected because the function has a singularity, this is not the case. An example is the Aharonov Bohm effect where the electron cannot go through the solenoid, but describes loops around it. So I don't see how you could apply analytic continuation. I'll try to clarify the question when I get home.

This post imported from StackExchange Physics at 2014-11-19 20:06 (UTC), posted by SE-user jinawee
@Autolatry If you mean that the domain is not simply connected because the function has a singularity, this is not the case. An example is the Aharonov Bohm effect where the electron cannot go through the solenoid, but describes loops around it. So I don't see how you could apply analytic continuation. I'll try to clarify the question when I get home.

This post imported from StackExchange Physics at 2017-01-08 20:03 (UTC), posted by SE-user jinawee
The path integral is defined as a sum over all possible paths. In order to do such a summation, you have to integrate within each homotopy class and then sum up all homotopy classes. This is also what the paper states in Theorem 3.1.

This post imported from StackExchange Physics at 2014-11-19 20:06 (UTC), posted by SE-user Heidar
The path integral is defined as a sum over all possible paths. In order to do such a summation, you have to integrate within each homotopy class and then sum up all homotopy classes. This is also what the paper states in Theorem 3.1.

This post imported from StackExchange Physics at 2017-01-08 20:03 (UTC), posted by SE-user Heidar
@Heidar What I've read is that one has to calculate the transition amplitude for each homotopy class and sum them up with their respective weights. But I don't see why this is necessary. Why can't the weight function be one?

This post imported from StackExchange Physics at 2014-11-19 20:06 (UTC), posted by SE-user jinawee
@Heidar What I've read is that one has to calculate the transition amplitude for each homotopy class and sum them up with their respective weights. But I don't see why this is necessary. Why can't the weight function be one?

This post imported from StackExchange Physics at 2017-01-08 20:03 (UTC), posted by SE-user jinawee
Maybe I am misunderstanding this, but if you consider a path integral in a covering space, don't you expect weight function be different? [Update] I found a passage that may help you in p219 of The Global Approach to Quantum Field Theory by B. DeWitt. If I understand him correctly, weight is an ambiguity when we define path integral. In a normal derivation, all weights can be equal, but the derivation assumes the existence of a functional Fourier transform that in turn assumes a vector space structure. So, for topologically non-trivial configuration space, this ambiguity remains.

2 Answers

+ 8 like - 0 dislike

The principle of the superposition of quantum states, or, as I shall refer to it, the sum over the alternatives, holds for particles belonging to a multiply-connected space in the same way as it holds for particles belonging to a simply-connected one, since it is one of the fundamental principles of quantum theory. On the other hand, what must be better explained here is why the sum over the alternatives, or in the present case, over the paths, in a simply-connected space can be constructed as a single path integral, differently from multiply-connected spaces.

First, I shall begin with an intuitive argument. Let $X$ be a "nice" topological space (we mean, for instance, that $X$ is arcwise connected or locally simply connected), $a,b\in X$, $\Omega(a,b)$ the set of paths $[t_{a},t_{b}]\longrightarrow X$ from $a$ to $b$ and $t_{b}>t_{a}>0$. To each $x(t)\in\Omega(a,b)$, we associate an amplitude $\phi[x(t)]$. Recall that, heuristically, we write the following proportionally relation for the propagator $K=K(b,t_{b};a,t_{a})$,

$$ K\sim\sum_{x(t)\in\Omega(a,b)}\phi[x(t)]. $$

If $S$ is the action governing the dynamics of our system and if $t_{b}-t_{a}$ is small enough, we know that $\phi[x(t)]\sim e^{iS[x(t)]}$ (where we have assumed $h=2\pi$). But there is no a priori reason, without evoking any property of $X$, to ensure that all the paths should contribute to $K$ with the same phase. For example, if $x(t),y(t)\in\Omega(a,b)$, why we cannot have

$$ K\sim e^{iS[x(t)]}-e^{iS[y(t)]}+...\,\,\,? $$

It turns out that if our topological space $X$ is simply-connected, we can always deform the path $x(t)$ to $y(t)$ continuously, a deformation which, in effect, should make $\phi[x(t)]$ approach $\phi[y(t)]$ continuously too. Formally, $$ \phi[y(t)]=\lim\phi[x(t)]=e^{iS[y(t)]},\,\,\,\mbox{as }x(t)\rightarrow y(t)\mbox{ continuously.} $$ From this, we can conclude two things:

  • Paths in a simply connected space contribute to the total amplitude with the same phase. So if $X$ is simply connected, we can then write the familiar expression $$ K\sim\sum_{x(t)\in\Omega(a,b)}e^{iS[x(t)]}, $$ which, upon introducing the appropriate measure, results in the Feynman path integral $$ K=\int_{\Omega(a,b)}e^{iS[x(t)]}\mathcal{D}x(t). $$
  • Paths in the same homotopy class contribute to the total amplitude with the same phase. So, for the propagator $K^{q}$ restricted to paths constrained in the homotopy class $q$, we can write similarly $$ K^{q}\sim\sum_{x(t)\in q}e^{iS[x(t)]}, $$ that also becomes a path integral $$ K^{q}=\int_{q}e^{iS[x(t)]}\mathcal{D}x(t), $$ but whose domain of functional integration is now $q$. Each such $K^{q}$ is called a partial amplitude.

Since the principle of the sum over the alternatives allows us to write the propagator $K$ as the sum of the amplitudes of each homotopy class $q$ individually (namely, the partial amplitudes), each one contributing with a phase that will be labeled by $\xi_{q}\in\mathbb{C},|\xi_{q}|=1$, we have that $$ K=\sum_{q\in\pi(a,b)}\xi_{q}K^{q}, $$ where $\pi(a,b)$ is the set of all homotopy classes for the paths from $a$ to $b$. This answers the question raised by jinawee, I hope.

But now, it will be instructive if we sketch on the proof of that result discovered first by Schulman and proved a little later by Laidlaw and DeWitt. Namely, that the set of phases $\{\xi_{q}\}$ can be "identified" with a scalar unitary representation of the fundamental group of $X$. The idea is the following. Let $c\in X$ fixed and choose $C(x)$ to be any path connecting $c$ to whatever $x\in X$. Such $C(\alpha)$ is known as a homotopy mesh. To each pair $(a,b)\in X\times X$, we can construct a mapping

$$ f_{ab}:\pi(c)\longrightarrow\pi(a,b) $$

by $f_{ab}(\alpha)=[C^{-1}(a)]\alpha[C(b)]$. This is an injection between the fundamental group $\pi\equiv\pi(c)$ at $c$ and the homotopy class $\pi(a,b)$, allowing us to label the propagator $K^{q}$ and the phase factor $\xi_{q}$ associated to a homotopy class $q$ with the elements of the fundamental group $\pi$, say,

$$ K^{q}\rightarrow K^{\alpha},\,\,\,\xi_{q}\rightarrow\xi(\alpha) $$

iff $f_{ab}(\alpha)=q$. So finally, our propagator assumes the form of a sum over the elements of a group:

$$ K=\sum_{\alpha\in\pi}\xi(\alpha)K^{\alpha}. $$

The result then follows from this consideration: the association $\alpha\mapsto K^{\alpha}$ of a partial amplitude $K^{\alpha}$ to each element $\alpha$ of the group $\pi$ depends on the injection $f_{ab}$, which in turn, depends on the choice of the mesh function $C$$(x)$. The (absolute value) of the propagator $K$, however, must be the same independently of the adopted mesh function.

The best place to find the details of the proof is still the paper "Feynman Functional Integrals for Systems of Indistinguishable Particles" (1971) by Laidlaw and DeWitt.

Additionally, there is another way to motivate the formula of the propagator $K$ as a sum over partial amplitudes associated to homotopy classes, which is based in a covering space of $X$. This was in fact one of the original reasonings employed by Schulman in "A Path Integral for Spin" (1968) in order to discuss the spin of a (quantum) non-relativistic particle using exclusively the method of path integration.

Roughly, it is like this. Let $\mathbf{X}$ be covering space of $X$ and $\mathrm{p}:\mathbf{X}\longrightarrow X$ the covering projection. Moreover, let $\mathcal{L}$ be the Lagrangian of our system in $X$, for which $S=\int\mathcal{L}dt$, $\mathbf{L}$ the lift of $\mathcal{L}$ to our covering space $\mathbf{X}$ induced by the projection $\mathrm{p}$ and $\mathbf{S}=\int\mathbf{L}dt$ the action on $\mathbf{X}$. To each pair $(a,b)\in X\times X$, choose some $\mathbf{a}\in\mathrm{p}^{-1}(a)$ and let $\mathbf{b}_{\alpha}\in\mathrm{p}^{-1}(b)$ define a sequence in $\mathbf{X}$ indexed by the elements of the fundamental group, that is, $\alpha\in\pi$.

Since the covering space is simply-connected, to each $\alpha\in\pi$, the propagator $K^{\alpha}$ associated to the amplitude for going from $\mathbf{a}$ to $\mathbf{b}_{\alpha}$ in the interval $[t_{a},t_{b}]$ on $\mathbf{X}$ is given, as familiar, by the path integral

$$ K^{\alpha}=\int_{\mathbf{a}}^{\mathbf{b}_{\alpha}}e^{i\mathbf{S}[\mathbf{x}(t)]}\mathcal{D}[\mathbf{x}(t)], $$

where in this case, the functional integral runs over the paths $\mathbf{x}(t):[t_{a},t_{b}]\longrightarrow\mathbf{X}$ connecting $\mathbf{a}$ to $\mathbf{b}_{\alpha}$.

Finally, by the principle of the sum over the alternatives, the propagator $K$ for the amplitude of going from $a$ to $b$ in the time interval $[t_{a},t_{b}]$ on our multiply-connected space $X$ is seem to be the sum over the alternatives for going from $\mathbf{a}$ to $\mathbf{b}_{\alpha}$ for all $\alpha\in\pi$ in the covering $\mathbf{X}$. In effect, we shall again obtain $K=\sum_{\alpha\in\pi}\xi(\alpha)K^{\alpha}$ for some phase factors $\xi(\alpha)\in\mathbb{C}$.

A proof of the result found by Schulman, Laidlaw and DeWitt described above, using the latter approach of covering spaces, may be found in the paper "Quantum mechanics and field theory on multiply connected and on homogenous spaces" (1972) by Dowker. I believe that the best sources for learning the subject are still the original papers cited above (and which I may send upon request). Additionally, if you have access to an university library, it is opportune to give a look at the Chapter 8 of "Functional Integration: Action and Symmetries" by Cartier and DeWitt.

This post imported from StackExchange Physics at 2017-01-08 20:03 (UTC), posted by SE-user Igor Mol
answered Nov 20, 2014 by Igor Mol (550 points) [ no revision ]
+1, informative answer. To nitpick on the terminology, by covering space you actually mean universal covering space.

This post imported from StackExchange Physics at 2017-01-08 20:03 (UTC), posted by SE-user Jia Yiyang
+ 0 like - 0 dislike
It's not true. For example in Yang-Mills theory we must sum over configurations with instantons because of the cluster decomposition principle. These instantons have a topological invariant that distinguishes them from configurations connected to the vacuum, so they are not "homotopic paths".

Though maybe I'm confused what you're asking?
answered Nov 22, 2014 by Ryan Thorngren (1,925 points) [ revision history ]
Based on the comments I think OP is actually asking why the phase factors (in addition to $e^{iS}$) are different for different homotopy classes.

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