Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,345 answers , 22,720 comments
1,470 users with positive rep
818 active unimported users
More ...

  Sign error in the (Chern-Simon induced) Bose-Fermi transmutation literature?

+ 4 like - 0 dislike
537 views


In Bose-Fermi transmutation [(started with Polyakov 1988)][1] one considers a Chern-Simon action
\begin{align}
S_{CS}= \int d^3 x \frac{\epsilon^{\mu\nu\rho}}{4\pi} a_\mu \partial_\nu a_\rho
\end{align}
(my $\epsilon^{012}=+1$) and computes the quantum expectation of, say, two, Wilson loops
\begin{align}
& \left\langle \exp\left(i\oint_{C_1} dx^\mu a_\lambda(x) + i\oint_{C_2} dy^\nu a_\nu(y)\right) \right\rangle_{CS} \nonumber \\[.2cm]
=& \ \exp\left(-\frac{1}{2} \left(\oint_{C_1} dx^\mu \oint_{C_1} dy^\nu + \oint_{C_2} dx^\mu \oint_{C_2} dy^\nu + 2\oint_{C_1} dx^\mu \oint_{C_2} dy^\nu \right) \left\langle a_\mu(x) a_\nu(y) \right\rangle_{CS}\right).
\end{align}
There are three terms in the exponent. The point is one can show the last term would give rise to an exchange phase of $\pi$ while the first two terms would give rise to the Berry phase of a Euclidean $SO(3)$ spinor, hence the name "Bose-Fermi transmutation". My problem is the sign of the phase computed in the literature does not seem right to me. (It is known that Polyakov's original paper has another numerical error besides my problem, so I will point to, say, [GHKL 1989][2], as "literature"). Here I wish someone could help me check it, or clarify if I had any confusion.

I will demonstrate the problem in the last term -- the braiding term -- because its computation involves less steps.

From the above, the last term is equal to (I call it $i\Theta_{br}$ as it will later be equal to the braiding phase)
\begin{align}
i\Theta_{br} =& -\oint_{C_1} dx^\mu \oint_{C_2} dy^\nu \ \langle a_\mu(x) a_\nu(y)\rangle_{CS} \\
=& -\oint_{C_1} dx^\mu \int_\Sigma (d^2 r)^{\nu\rho} \ 2\partial_{[r^\nu|} \langle a_\mu(x) a_{|\rho]}(r) \rangle_{CS}
\end{align}
where in the second equality I used the Stoke's Theorem and $\Sigma$ is a surface bounded by $C_2$. The quantum expectation can be easily computed via
\begin{align}
0 =& \int \mathcal{D}a \ \frac{\delta}{\delta a_\sigma(r)} \left( a_\mu(x) \exp(iS_{CS}+iS_{g.f.})\right) \\
=& \int \mathcal{D}a \ \left(\delta^\sigma_\mu \delta^3(r-x) + a_\mu(x) \frac{i\epsilon^{\sigma\lambda\kappa}}{2\pi} \partial_\lambda a_\kappa(r) + \mbox{(g.f. term)}\right) \exp(iS_{CS}+iS_{g.f.})
\end{align}
("g.f." stands for gauge fixing) which leads to
\begin{align}
2\partial_{[r^\nu|} \langle a_\mu(x) a_{|\rho]}(r) \rangle_{CS} = 2\pi i \epsilon_{\mu\nu\rho} \delta^3(r-x) + \mbox{(g.f. term)}
\end{align}
where my $\epsilon_{012}=+1$ so that $\epsilon_{\sigma\nu\rho} \epsilon^{\sigma\lambda\kappa} = 2\delta^\lambda_{[\nu} \delta^\kappa_{\rho]}$; the gauge fixing term can be omitted since its contribution must vanish when integrated around the Wilson loop which is gauge independent. Therefore
\begin{align}
i\Theta_{br}=-2\pi i \ \oint_{C_1} dx^\mu \int_{\Sigma} (d^2 r)^{\nu\rho} \ \epsilon_{\mu\nu\rho} \ \delta^3(r-x).
\end{align}
The integral just computes the linking number between $C_1$ and $C_2$ (note that everything so far is purely topological and does not involve any geometry). In particular, if $C_2$ forms a circular loop in the $01$-plane, say $\{(\cos(s), \sin(s), 0)| 0\leq s < 2\pi\}$, and $C_1$ threads through it in the $+2$-direction (counter-clockwise braiding), say $\{(0, \cos(s)-1, \sin(s))| 0\leq s < 2\pi\}$, then the integral yields $+1$ and hence the braiding phase is $\Theta_{br}=-2\pi$. (The exchange phase, being half the braiding phase, would be $\Theta_{ex}=-\pi$, hence "Bose-Fermi transmutation".)

I am pretty sure this sign is correct because in the condensed matter "flux attachment" point of view, if a particle has charge $-1$ (or $+1$) under $a$, then $S_{CS}$ attaches $+2\pi$ (or $-2\pi$) $a$-flux to the particle, so another particle going counter-clockwise around it will pick up a phase of $-2\pi$ which equals the $\Theta_{br}$ I computed above.

However, in the literature, the quantum expectation is presented as
\begin{align}
\langle a_\mu(x) a_\nu(y)\rangle_{CS} = \frac{-i}{2} \epsilon_{\mu\nu\rho} \frac{(x-y)^\rho}{|x-y|^3}
\end{align}
(this is not topological but since the literature focused on flat Euclidean space, it is fine). Here my problem comes. From it one finds
\begin{align}
& 2\partial_{[y^{\, \nu}|}\langle a_\mu(x) a_{|\rho]}(y)\rangle_{CS} \\
=& \ \epsilon_{\sigma\nu\rho}\epsilon^{\sigma\lambda\kappa} \partial_{y^{\, \lambda}} \left( \frac{-i}{2} \epsilon_{\mu\kappa\alpha} \frac{(x-y)^\alpha}{|x-y|^3} \right) \\
=& \ \frac{+i}{2}\epsilon_{\sigma\nu\rho} \: \left(\delta^\lambda_\mu \delta^\sigma_\alpha - \delta^\lambda_\alpha \delta^\sigma_\mu\right) \partial_{x^\lambda} \frac{(x-y)^\alpha}{|x-y|^3} \\
=& \ \frac{i}{2} \epsilon_{\mu\nu\rho} (-4\pi) \delta^3(x-y) \ + \ \mbox{($\partial_{x^\mu}$ term)}.
\end{align}
The $\partial_{x^\mu}$ term is a gauge fixing term that vanishes under $\oint dx^\mu$. The first term is OPPOSITE to what I found. Thus, the quantum expectation in the literature leads to a braiding phase of $+2\pi$ when braided once counter-clockwise, which is OPPOSITE to what I computed and argued based on "flux attachment" reasoning.

One might say, these are phases being integer multiples of $2\pi$, so why care about the sign? Because the same sign problem is carried over into the computation of the Berry phase terms too, which depend on the geometry around the loops (due to regularization) and can take arbitrary value. (This Berry phase is the crucial point in the literature.) Related to this, it seems this sign error is also carried over to the more recent literature about the spin connection contribution to flux attachment (and the way they then interpret the fractional quantum Hall "shift"). Moreover, in the anyon context one can also consider $S_{CS}$ whose denominator is not $4\pi$, then the braiding phase can also take arbitrary value.


  [1]: http://www.worldscientific.com/doi/abs/10.1142/S0217732388000398
  [2]: http://www.sciencedirect.com/science/article/pii/037026938991589X


(I also asked this at http://physics.stackexchange.com/questions/280784/sign-error-in-the-bose-fermi-transmutation-literature.)

asked Sep 18, 2016 in Theoretical Physics by Jing-Yuan Chen (20 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...