# Length path integral

Let's consider a 2-dimensional Euclidean plane. The length between two points $a$ and $b$ can be defined in the following way:

$$(ab) := \inf_{\gamma} \,\int_0^1 d\tau \,\sqrt{\delta_{ab} \,\dot{\gamma}^a \dot{\gamma}^b},$$

where the infimum is taken over all paths $\gamma$ joining $a$ and $b$, $\delta_{ab}$ is the Euclidean metric on the plane and the parametrization is chosen so that $$\gamma(\tau=0)=a \quad \& \quad \gamma(\tau=1)=b.$$

My observation is that this definition exactly reproduces the minimal-action principles from classical mechanics, with reparametrization-invariant action taken to be the length of the path. The length $(ab)$ is then just equal to the action evaluated on the classical trajectory.

Indeed, by solving the equations of motion we find that the extremum of the action is given by the straight line, therefore the Pythagorean theorem holds:

$$(ab) = \sqrt{\delta_{ab} \left( b^a - a^a \right) \left( b^b - a^b \right) }.$$

But there is a quantum path integral approach which generalizes the classical minimal-action principle by replacing the infimum over paths with the sum over exponentials:

$$\inf_{\gamma} \,\int_0^1 d\tau \,\sqrt{\delta_{ab} \,\dot{\gamma}^a \dot{\gamma}^b} \quad \longrightarrow \quad \int D\gamma \, \exp \left\{ - \hbar^{-1} \int_0^1 d\tau \,\sqrt{\delta_{ab} \,\dot{\gamma}^a \dot{\gamma}^b} \right\}.$$

The length $(ab)$ is then given by the Euclidean version of the stationary phase approximation: only paths similar to the straight line give significant contributions to the path integral.

So let me define the "quantum length" $[ab]$:

$$\exp\left\{ - \hbar^{-1} [ab] \right\} := \int D\gamma \, \exp \left\{ - \hbar^{-1} \int_0^1 d\tau \,\sqrt{\delta_{ab} \,\dot{\gamma}^a \dot{\gamma}^b} \right\}.$$

Now I want to understand the difference between $(ab)$ and $[ab]$. There are some properties I expect $[ab]$ to have:

1. It is probably ill-defined as it stands: path integral may need a proper regularisation which would introduce a cut-off with dimension of length, or maybe we could switch from Nambu-Goto-type action to Polyakov-type.

2. It should possess some kind of fractal behaviour due to the non-trivial RG-group flow.

3. It should (at least in some region) coincide with $(ab) + O(\hbar)$.

My question is: has anybody ever seen something similar? I would appreciate references to any papers related to this.

P.S. Maybe there are some other tags which could be added to my question. Feel free to fix this.

• This seems closely related, if not identical, to the quantum effective action without source term. Mar 22 '15 at 14:08
• @ACuriousMind you are absolutely right. So what I want is a way to derive the quantum effective length which I denote $[ab]$. Do you have one? Mar 22 '15 at 18:49
• I'm not sure what exactly you want when you say you seek a "way to derive" it, but it seems to me that it's essentially $E[0]$, which is the "connected VEV for no field", i.e. the zero-point energy (or closely related to it). Mar 22 '15 at 18:59
• @ACuriousMind it should depend at least on $(ab)$ (the distance between two points). I want to understand this dependence. The best thing to do is to try to compute it (perturbatively, numerically, I don't care). Mar 23 '15 at 15:09

The usual way to compute such a path integral is by writing the fields (in your case: the paths) as "classical configuration" (the straight line) plus "quantum fluctuations". So if you write your paths as $\gamma(\tau) = a + \tau b + \hat\gamma(\tau)$ (with $\tau\in[0,1]$ a parameter describing the path), then $\hat\gamma$ will be the "quantum fluctuation" around the classical path. Assuming $\hbar$ is small, quantum fluctuations will be subleading (i.e. we can do the semi-classical approximation), meaning we can write the action as $$S(\gamma) = S(\gamma_0) + \int\hat\gamma(\tau_1) \hat\gamma(\tau_2) \frac{\delta^2S}{\delta\hat\gamma(\tau_1)\delta\hat\gamma(\tau_2)}(\gamma_0) d\tau_1d\tau_2 + \cdots$$ with $\gamma_0$ the straight line. (This is just a Taylor expansion around the straight line, assuming the first derivative vanishes, which it should by the minimization equation.) Then you integrate over $\hat\gamma$ and you're done. The "quantum length" will be equal to the "classical length" plus "quantum corrections".