What is the meaning of 'box' in Wess and Bagger's chiral superfield theory?

[Bobhawke]

Hi,

I apologize if this question has been asked before or if it is dumb, but...

I'm trying to learn some supersymmetry and have been working through Wess and Bagger. However, I have become confused in chapter 9, when they calculate the propagators of a free chiral superfield theory. My confusion is at the set of equations 9.11: They call the position space Feynman propagator the inverse of (box squared + m squared). It might make sense if it was momentum space and box was the momentum, but elsewhere in the book I believe they are using box to mean the D'Alembert operator. This doesn't seem to make sense and all and I can't figure out what's causing the confusion.

Thanks in advance.

 

[Simon_Tyler]

In and around W&B (9.11) $\Box$ still means the d'Alembertian $\Box=\partial^a\partial_a$, i.e. it's a differential operator.

Below equation (9.11) there's the formula

$\Delta_F(x) = \frac{1}{\Box-m^2}$

this probably a typo/oversight of some sort.
Either the $(x)$ is extraneous and it should be interpreted as the operator definition

$\begin{align} \Delta_F = \frac{1}{\Box-m^2}\,, \quad \text{such that:}& \quad (\Box-m^2)\Delta_F = \mathbf{1}\\ \quad\text{equivalently:}& \quad (\Box-m^2)\Delta_F(x) = \delta^4(x) \quad\text{where}\quad \Delta_F(x) = \Delta_F\cdot\delta^4(x)\,, \end{align}$

or (more likely) it was meant to include the Dirac δ-function,

$\Delta_F(x) = \frac{1}{\Box-m^2}\delta^4(x)$

so that using the momentum space representation for the Dirac δ-function you get the normal momentum space propagator

$\tilde{\Delta}_F(k) = \frac{-1}{k^2+m^2}\ .$

By the way:
The approach used in this section is a bit old fashioned / not very common. You can more simply derive the superfield propagator directly using superspace variational derivatives and then, if you want, project out the component propagators. See the standard textbooks by Gates, Grisaru, Rocek & Seigel OR Buchbinder & Kuzenko OR West etc... for more modern / different approaches. More details at http://www.stringwiki.org/w/index.php?title=Supersymmetry_and_Supergravity" is also worth the read.

 

[Bobhawke]

Thank you very much, that was extremely helpful.