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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Is a $SU(2)$ supergauge theory really a $SU(2)$ gauge theory?

+ 3 like - 0 dislike
243 views

Consider $SU(2)$ supergauge theory with $A$, a doublet of two chiral superfields in the fundamental representation.

$$A=
\begin{pmatrix}
\Phi_1\\
\Phi_2
\end{pmatrix}$$
where $\Phi_1$ and $\Phi_2$ are chiral superfields. Since we have said it is in the fundamental representation it transforms

$$A\to{}A'=e^{i\sigma_j\phi_j}A$$

where $\sigma_j$ are Pauli matrices. If $e^{i\sigma_j\phi_j}$ were to be a $SU(2)$ matrix then the $\phi_j$ ought to be real. Nonetheless, this would make $A'$ no longer chiral unless $\phi_j$ are themselves chiral superfields, that is complex functions making $e^{i\sigma_j\phi_j}$ not a $SU(2)$ matrix.

Therefore this is no longer a proper $SU(2)$ theory. What is going on?

asked Nov 27, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ no revision ]
retagged Nov 27, 2015 by dimension10

Why would this transformation make $A'$ not chiral? 

take the super-covariant derivative $\bar{D}_{\dot{\alpha}}$. It needs not vanish in general for such a transformation.

Why can't $\phi$ be both chiral and real?

@JiaYiyang I am a newbie with these things and i would love to hear a better reasoning to the question you pose Jia, but I do know that in 4 space-time dimensions and $\mathcal{N}=1$ SUSY  a (scalar) chiral superfield $\Phi$ accomodates a complex scalar $\phi$, a left chiral Weyl spinor $\psi$ an a auxiliary field $F$ in the following way $\Phi=\phi+\sqrt{2}\theta\psi+\theta^2F+\ldots$  (the dots denote that there are more terms but they are irrelevant for my argument). You see that the fact that $\phi$ being complex makes $\Phi$ complex. Now, I have been wondering if it might be possible to have supermultiplets containnig real scalars, actually I posted a question not so long ago in overflow http://www.physicsoverflow.org/32944/have-supersymmetry-using-real-scalar-instead-complex-scalars and the answer seems to be no.

Yeah I guess reality+chirality=chirality+ antichirality which would make the $\phi$ a constant field, which is too trivial and no longer a local transformation. Let me think about it, and I'm also quite new to Susy.

The peculiarity goes away if we look at the theory in a gauge where only the physical x-space fields remain in the gauge superfield $V$, i.e. Wess-Zumino gauge. This gauge choice completely determines the $\phi$ in your $e^{i\sigma_j\phi_j}$ up to a real part, and this remaining real part is the gauge transformation you can still perform on the matter fields.

1 Answer

+ 4 like - 0 dislike

The kinetic term in the Lagrangian density for a chiral superfield $A$ coupled to a vector superfield $V$ is $A^\dagger e^V A$. The correct gauge transformation of $A$ is given by $A \rightarrow e^{i \phi} A$ where $\phi$ is indeed a chiral superfield to preserve the chirality of $A$, and so with values in the complexified Lie algebra of the gauge group. Under this transformation, $A^\dagger \rightarrow A^\dagger e^{-i \phi^\dagger}$ and so

$A^\dagger e^V A \rightarrow A^\dagger e^{-i \phi^\dagger } e^{V} e^{i \phi} A$.

As the kinetic term has to be gauge invariant, this shows that the correct transformation of the vector superfield $V$ under a gauge transformation is $V \rightarrow V'$ such that $e^{V'} = e^{-i \phi^\dagger } e^{V} e^{i \phi}$ i.e.  $V' = V+i(\phi - \phi^\dagger)+...$ (the ... are here if the gauge group is non-abelian because then one has to be careful about exponentials) and this is indeed the correct gauge transformation which preserves the reality condition satisfied by the gauge superfield, $V=V^\dagger$, which guarantees that the gauge group of the theory is indeed the compact gauge group and not its complexification.

Conclusion: in $\mathcal{N}=1$ $4d$ superfield formalism, gauge transformations are parametrized by chiral superfields, thus preserving chirality of the various charged matter fields, but this chiral superfield enters in the gauge transformation of the gauge field with its conjugate in a way preserving the reality condition of the gauge field.

One standard reference for this kind of questions is the book by  Wess and Bagger, "Supersymmetry and supergravity".

answered Dec 6, 2015 by 40227 (4,660 points) [ no revision ]

But how about the gauge transformation on the matter field? I imagine it should at least reduce to a real-parameter gauge transformation in the x-space, but it's not immediately obvious to me.

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:
$\varnothing\hbar$ysicsOverflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...