How do you integrate the Hopf term in 2+1 dimensions?

  • Context: Graduate 
  • Thread starter Thread starter Ian Lovejoy
  • Start date Start date
  • Tags Tags
    Integration Term
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
Ian Lovejoy
Messages
7
Reaction score
0
Hi,

I'm reading Quantum Field Theory in a Nutshell by A. Zee, which is excellent but it is occasionally difficult to fill in the blanks on one's own.

Once such place is the integration of the Hopf term, which results from integrating out the gauge field in a Lagrangian with a Chern-Simons term in 2+1 dimensional spacetime.

The Hopf term is:

[tex]L_{Hopf}=\frac{1}{4\gamma}j_\mu(\frac{\epsilon^{\mu\nu\lambda}\partial_\nu}{\partial^2})j_\lambda[/tex]

From here we are supposed to be able to define a current j representing one particle at rest at the origin while another particle goes halfway around it. Integrating the above with this current we are supposed to obtain [tex]1/4\gamma[/tex]. This defines the fractional statistics of the particles in the system.

Except for a factor of 2 the same treatment appears in section II of this document:

http://arxiv.org/PS_cache/cond-mat/pdf/9501/9501022v2.pdf

It all seems perfectly straightforward but I'm unable to do the integration. For [tex]1/{\partial^2}[/tex] I am using:

[tex]\int{\frac{d^3k}{(2\pi)^3}\frac{-e^{ik(x - y)}}{k^2 + i\epsilon}}[/tex]

No matter what order I do the integration in, I seem to get either an integral that is impossible to do, or an result that is apparently zero. The book and the above reference seem to imply that the result is easily obtained by plugging in the current into the Lagrangian.

Can anyone give me a hint? It would be much appreciated.

Thanks,
Ian
 
Physics news on Phys.org
First point to be noted: You cannot interchange the two currents. If you could, by antisymmetry of the epsilon tensor, the expression is identically zero.

The way the above makes sense is if the derivatives act on the right hand current.

Now, for a static charge the only nonzero component of the current is the zeroth component.
For one that is moving, something else, say the x or the y component is also nonzero.

Moving halfway around: say it moves on two consecutive sides of a square - first along the x and then along the y direction...

does this help?
 
I'm also stuck on this question, simply can't integrate out a closed form. Anyone succeeded doing it?