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

  • Context: Graduate 
  • Thread starter Thread starter Bobhawke
  • Start date Start date
  • Tags Tags
    Box Operator
Click For Summary
SUMMARY

The discussion centers on the interpretation of the 'box' operator in Wess and Bagger's chiral superfield theory, specifically in chapter 9 regarding the propagators. The 'box' refers to the D'Alembert operator, represented as \Box=\partial^a\partial_a, and is crucial in the equation for the Feynman propagator, \Delta_F(x) = \frac{1}{\Box-m^2}. There is a consensus that the notation may contain a typographical error, suggesting that the expression should either omit the position variable or include the Dirac δ-function for clarity. For a more modern approach, readers are encouraged to consult alternative texts such as those by Gates, Grisaru, Rocek & Seigel, or Buchbinder & Kuzenko.

PREREQUISITES
  • Understanding of supersymmetry concepts
  • Familiarity with Wess and Bagger's "Supersymmetry and Supergravity"
  • Knowledge of differential operators, specifically the D'Alembert operator
  • Basic principles of quantum field theory and propagators
NEXT STEPS
  • Study the derivation of the superfield propagator using superspace variational derivatives
  • Read Gates, Grisaru, Rocek & Seigel's textbook for a modern perspective on superfield theory
  • Explore the implications of the Dirac δ-function in quantum field theory
  • Investigate alternative resources on supersymmetry and supergravity, such as the stringwiki.org page
USEFUL FOR

Students and researchers in theoretical physics, particularly those focusing on supersymmetry, quantum field theory, and chiral superfield theories.

Bobhawke
Messages
142
Reaction score
0
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.
 
Physics news on Phys.org
In and around W&B (9.11) [itex]\Box[/itex] still means the d'Alembertian [itex]\Box=\partial^a\partial_a[/itex], i.e. it's a differential operator.

Below equation (9.11) there's the formula

[itex]\Delta_F(x) = \frac{1}{\Box-m^2}[/itex]

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

[itex]\begin{align}<br /> \Delta_F = \frac{1}{\Box-m^2}\,, <br /> \quad \text{such that:}& \quad<br /> (\Box-m^2)\Delta_F = \mathbf{1}\\<br /> \quad\text{equivalently:}& \quad<br /> (\Box-m^2)\Delta_F(x) = \delta^4(x) <br /> \quad\text{where}\quad <br /> \Delta_F(x) = \Delta_F\cdot\delta^4(x)\,,<br /> \end{align}[/itex]

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

[itex]\Delta_F(x) = \frac{1}{\Box-m^2}\delta^4(x)[/itex]

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

[itex]\tilde{\Delta}_F(k) = \frac{-1}{k^2+m^2}\ .[/itex]

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.
 
Last edited by a moderator:
Thank you very much, that was extremely helpful.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
7K
  • · Replies 0 ·
Replies
0
Views
4K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 309 ·
11
Replies
309
Views
17K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K