How Does the Vector Potential Transform in the Flux-Tube Model for Anyons?

[ReyChiquito]

I have a simple question regarding the flux-tube model for anyons. It may sound complicated but it isnt. So here we go.

Considering the interaction term $$L_{s}=\frac{\hbar\theta}{\pi}\dot{\phi}$$ where $$\frac{\theta}{\pi}=\alpha$$ is called "anyon parameter" (0 for bosons, 1 for fermions), and $$\phi$$ is the relative angle between particles.

I have proven that the Hamiltonian in relative coordinates for that kind of system can be written as
$$H_{r}=\frac{p_{r}^2}{m}+\frac{(p_{\phi}-\hbar\alpha)^2}{mr^2}.}$$

In order to generalize the Hamiltonian for a N partices system, the book (Fractional Statistics and Quantum Theory by Khare) introduces the next vector potential:

$$a_{i}(\bold{r})=\frac{\Phi}{2\pi}\frac{\epsilon_{ij}r_{j}}{\bold{r^2}}$$ where $$\epsilon_{ij}$$ is the antisimetric tensor (i asume).

Then the book goes

$$\mbox{Thus }a_{x}=\frac{\Phi}{2\pi}\frac{y}{x^2+y^2}\mbox{, }a_{y}=\frac{\Phi}{2\pi}\frac{-x}{x^2+y^2}\mbox{, or in polar coordinates }$$
$$a_{r}=0\mbox{, }a_{\phi}=\frac{\Phi}{2\pi}$$

I know it seems simple to deduce this step but i don't get it, here is what I've done:
$$a_{i}(\bold{r})=\frac{\Phi}{2\pi\bold{r^2}}\left(\begin{array}{cc}y\\-x\end{array}\right)=\frac{\Phi}{2\pi\bold{r^2}}\left(\begin{array}{cc}rsin\phi\\-rcos\phi\end{array}\right)=-\frac{\Phi}{2\pi}\frac{1}{r}\bold{\hat{\phi}}$$

What am i doing wrong??

I asked a friend of mine and he mentioned something about the metric. To tell you the truth, i don't know what he is talking about. Can anybody explain this to me please?

 

[Haelfix]

Yea that looks right up to the last step, you basically have it in front of you *the last equality is wrong tho*.. Keep in mind you pick up a factor of r from the transformation..

You know, ds^2 = dr^2 + r^2 d(theta)^2

 

[ReyChiquito]

Thx.. i knew i was missing something.

Can you explain me from where this factor arrives?

Its directly from calculus (i.e. jacobian) or has to do something with tensors and metric?

 

[selfAdjoint]

It's the radial coordinate of the polar coordinates. The formula Haelfix gave is the length differential in polar coordinates.

 

[ReyChiquito]

I understand that, maybe i need to refrace my question.

Why this isn't an ordinary change of variables?

I don't see any rates involved so i don't understand why do i have to include that factor.

If this where a simple calculus problem the tangent vector can be described in polar coordinates as
$$\vec{T}=\left(\begin{array}{cc}-y\\x\end{array}\right)=r\hat{\phi}$$

if i used
$$\vec{T}=r^2\hat{\phi}$$

i would be describing the wrong point in space right?, plus, how to correct the minus sign? isn't supossed to be a right hand system?

I KNOW that what I am doing is wrong, and i understand that the r factor must be included and the book is right, but i don't see any reason for including that factor.

Am i missing something simple and i need to review my clac notes?

ps. my quantum mechanics course sucked big time, it was like half spetial functions course and half "learn the dirac algebra and the conmutator operator", i harldy saw any of the stuff u should see in this subject (ie Cohen)

 

[ReyChiquito]

Dumping here. Just to see if anybody can explain it to me like i was a 4 year old :P