I am not sure that this notation is specific to supersymmetry theories but I ran into this while studying that.

I see people talking of component fields of a chiral superfield as $\phi$ and $\bar{\psi}$ in the adjoint representation of some gauge group. (One sometimes also seems to use the notation of $\bar{\psi}_+$ to denote which component of the spinor is being picked up) One also denotes the covariant derivative operator as $D$ (which is sometimes written as $D_{++}$ to denote which component of it is being chosen..but I don't understand this notation)

Now one talks of "single trace operators" of the kind $Tr[\phi \bar{\psi}]$, $Tr[\phi \bar{\psi}^2]$, $Tr[\phi^3D \bar{\psi}^4]$ and stuff like that..basically taking arbitrary combinations of powers of the scalar and the fermion and then taking a "gauge trace".

There are lot of things that I don't understand.

  • I am puzzled by the notation of taking powers of a fermionic field. How does one define powers of a spinor? (What is square of a fermion?) Also here in components $\psi$ would have a decomposition like $\psi = \psi_A t_A$ where $\psi_A$ are fermions and $t_A$ runs over the generators of the Lie algebra of a gauge group in the adjoint representation. I guess one is taking a tensor product between the the spinor $\psi_A$ and the matrix $t_A$. But then I don't understand what is a square or any other power of that tensor?

  • I am told that $\bar{\psi}^2 \neq 0$ but $Tr[\bar{\psi}^2] = 0$ I don't understand what is this supposed to mean.

  • One way I can think of the powers could be that $\bar{\psi}$ are after quantization operators on the Hilbert space of the theory and these are powers of that Hilbert space operator. But in this way of thinking I am confused how to intprete the trace over the gauge indices.

  • In this language one wants to think of the supersymmetry operators $Q$ to act on the fields in the following way,


$Q\bar{\psi} = 0$

$Q(DO) = [[\phi,\bar{\psi}],O\} + D(QO)$

where $O$ is some operator and in the first term on the RHS of the last of the above equations the symbol $[,\}$ means that one will take the commutator or the anticommutator depending on whether $O$ is bosonic or fermionic.

The above looks like a very different way of thinking about supersymmetry transformations than the language I am familiar with from the books like that of Weinberg where the action of $Q$ on $\phi$ and $\psi$ is through commutator and anti-commutator respectively or one thinks of infinitesimal supersymmetry transformations as $\delta \phi$ etc.

  • Definitely the last of the above list of commutators is totally unfamiliar to me!

  • What the above notion confuses is how I am supposed to think of the action of $Q$ on composite operators like say $Tr[\phi\bar{\psi}^3]$. Apparently one is supposed to think of this as a fermionic operator since the fermion is being raised to an odd power. (I can't see the full argument here!)

Then the action of $Q$ is supposed to be (dropping the overall trace),

$Q(\phi \bar{\psi}^3) = (Q\phi)\bar{\psi}^3 + \phi (Q \bar{\psi})\bar{\psi}^2 - \phi \bar{\psi}(Q \bar{\psi})\bar{\psi} + \phi \bar{\psi}^2(Q \bar{\psi})$

The signs alternate depending on how many $\bar{\psi}$s has the $Q$ skipped over.

  • I would like to know of explanations for the above way of writing powers of fermionic fields and doing supersymmetry transformations on them. I would be glad if I am directed to some reference which explains the above way of thinking which I haven't come across anywhere else in any book.

  • {Eventually one interested in calculating the "cohomology of $Q$" on the space of all such single trace operators. (for some reason not clear to me people want to drop from this list of operators which are total covariant derivatives) This depends on the theory one is looking it by what the auxiliary field integrates to and appearing on the RHS. I would like to know references along that too and why this is "cohomology" and why is this calculated.}

  • 4
    $\begingroup$ You should try to break your questions down into small, single idea parts. That way they'll be useful to someone thinking about the same thing later on. At the moment, your questions are quite long and are more suited to a forum site (eg physicsforums) than a stackexchange site... $\endgroup$
    – Simon
    Commented Mar 18, 2011 at 5:22

1 Answer 1


Here's my attempt at answering some of your multitude of points...

  • I know you're often looking at $N=2$ susy. $D_{++}$ type covariant derivatives often appear in harmonic superspace, where the ++ indicates the harmonic charge.

  • The spinor components of your gauge field are Lie algebra valued fermionic spin-1/2 fields. That means they take a spacetime point and return an element that's in the Grassmann shell of the direct product of the spin-1/2 module and the Lie algebra. Since it is fermionic, it must be an odd element.

  • If the product is a term in the Lagrangian, then everything must be contracted / traced to yield an invariant object. If the product is not in the Lagrangian, then it depends on context.

  • Products of the Lie algebra elements must be either commutators or traces of quadratic terms. In you're case you're in the adjoint rep so the products are secretly commutators. Cubics etc are combinations of the above. $tr(ABC):=tr(A[B,C])=tr([A,B]C)$ etc...

  • As for "$\bar{\psi}^2 \neq 0$ but $Tr[\bar{\psi}^2] = 0$"... I'm not sure exactly what you mean. Maybe let $\lambda = \lambda^A T_A$ be a anticommuting Lie algebra element. Then the Lie algebra commutator is $[\lambda, \lambda] = \lambda^A \lambda^B [T_A, T_B] \neq 0$ but must have vanishing trace. Or the direct product is $\lambda \lambda = \lambda^A \lambda^B T_AT_B$ which has (assuming an orthogonal basis) the trace ${\rm tr} (\lambda \lambda) = \lambda^A \lambda^B {\rm tr}(T_AT_B) = \lambda^A\lambda^B \delta_{AB}$ vanishes because each component is just an odd Grassmann element and so squares to zero.

    • After that, I think you leave the realm of standard QFT. You seem to want to make the field become a physical state that is a vacuum and so vanishes under supersymmetry?

Maybe you should have a quick skim through the first chapter of Ideas and methods of supersymmetry and supergravity that is called (and has a lot of) "mathematical background". It gives a good description of Grassmann algebras that is often skimmed over in other texts.

  • $\begingroup$ @Simon Can you kindly elaborate or give reference for the following concept you mentioned, "..your gauge field are Lie algebra valued fermionic spin-1/2 fields. That means they take a spacetime point and return an element that's in the Grassmann shell of the direct product of the spin-1/2 module and the Lie algebra.." I am not feeling familiar with this mathematical structure. I guess this explains what is meant by taking square of a fermionic component. $\endgroup$
    – Student
    Commented Mar 21, 2011 at 14:08
  • $\begingroup$ @Simon But I am puzzled if you mean that this $\bar{\psi}_+$ is assigning a particular fermionic component at every point of spacetime. Doesn't that seem to violate lorentz covariance somehow? Kindly explain. $\endgroup$
    – Student
    Commented Mar 21, 2011 at 14:09
  • $\begingroup$ @Anirbit: I'm not sure of one place that says it all in one go. But it's just the combination of what a normal gauge field is with a normal superfield. Read chapters 3.6 and 2.4 of the book I linked to above. BTW, the direct product thing is only local, globally I guess the spinor is a section of the appropriate bundle... but I'm not so comfortable with global aspects. $\endgroup$
    – Simon
    Commented Mar 21, 2011 at 22:42
  • $\begingroup$ @Anirbit "Doesn't that seem to violate lorentz covariance somehow?" No! (and yes). What you are working with here are classical off-shell fields. These fields are not solutions to the equations of motion - instead, they symbolically define the dynamics of the system. Alternatively, you can think of them as the perturbation fields you integrate over inside the path integral. Even their solutions do not have to be Lorentz invariant (look around you, where's the Lorentz invariance in the current solution that you're in?). The only solution we require to Lorentz invariant is the vacuum solution. $\endgroup$
    – Simon
    Commented Mar 23, 2011 at 0:55
  • $\begingroup$ @Simon Thanks for your reply. I will try to locate this book in the library. I am trying to find a way to understand what you said about the meaning of associating every space-time point with a fixed spinor component. Are you saying that this fixing of the $"+"$ component is a statement that one is doing supersymmetry on-shell and hence this is a particular chosen solution? Isn't there any ambiguity about determining what is the $"+"$ component at two different points of the space-time? $\endgroup$
    – Student
    Commented Mar 23, 2011 at 17:55

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