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-   -   Simple question regarding anyons (http://www.physicsforums.com/showthread.php?t=29227)

ReyChiquito Jun4-04 07:59 PM

Simple question regarding anyons
 
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 [tex]L_{s}=\frac{\hbar\theta}{\pi}\dot{\phi}[/tex] where [tex]\frac{\theta}{\pi}=\alpha[/tex] is called "anyon parameter" (0 for bosons, 1 for fermions), and [tex]\phi[/tex] is the relative angle between particles.

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

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:

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

Then the book goes
Quote:

[tex]\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 }[/tex]
[tex]a_{r}=0\mbox{, }a_{\phi}=\frac{\Phi}{2\pi}[/tex]
I know it seems simple to deduce this step but i dont get it, here is what ive done:
[tex]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}r sin\phi\\-rcos\phi\end{array}\right)=-\frac{\Phi}{2\pi}\frac{1}{r}\bold{\hat{\phi}}[/tex]

What am i doing wrong??

I asked a friend of mine and he mentioned something about the metric. To tell you the truth, i dont know what he is talking about. Can anybody explain this to me please?

Haelfix Jun4-04 11:39 PM

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 Jun5-04 12:37 AM

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 Jun5-04 09:31 AM

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

ReyChiquito Jun5-04 11:38 AM

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

Why this isnt an ordinary change of variables?

I dont see any rates involved so i dont 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
[tex]\vec{T}=\left(\begin{array}{cc}-y\\x\end{array}\right)=r\hat{\phi}[/tex]

if i used
[tex]\vec{T}=r^2\hat{\phi}[/tex]

i would be describing the wrong point in space right?, plus, how to correct the minus sign? isnt supossed to be a right hand system?

I KNOW that what im doing is wrong, and i understand that the r factor must be included and the book is right, but i dont 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 Jun9-04 12:57 AM

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


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