So in the Lagrangian for a SUSY theory we have the F-terms, which I have seen written (e.g., in Stephen Martin's SUSY primer) as

$F^*_i F^i$


$F^i = \frac{\partial W}{\partial \phi^i}$.

I have a few questions about this.

1) What do we do in the case of a gauge multiplet? For instance, in the MSSM we have $Hu = (Hu^+,Hu^0)$. Let's look at the term $y_{jk} Q_j \bar{U_k} Hu$ in the superpotential, where $Q$ is the left-handed SU(2) squark doublet. Should we take

$F^{Hu} = \frac{\partial W}{\partial Hu} = y_{jk} Q_j \bar{U_k}$?

Or should it rather be two terms, like

$F^{Hu^+} = \frac{\partial W}{\partial Hu^+}$

$F^{Hu^0} = \frac{\partial W}{\partial Hu^0}$?

2) If it is $F^{Hu} = \frac{\partial W}{\partial Hu}$, then where Martin has $F^*_i$, should it really be $F^\dagger_i$? Is that implied by the raised and lowered indices?

3) In either case, why is it not a problem that we will get terms like ($y_{33}Q_3\bar{U_3})^*y_{22}Q_2\bar{D_2}$? I'm worried about whether such terms will form gauge invariants and whether they imply vertices that should not be there. I know quarks can mix, but this looks like it describes something like a vertex where a 3rd-gen up-type quark-antiquark pair directly annihilate to a 2nd-gen down-type quark-antiquark pair. Am I misreading this?

Thanks so much for any help! This seems like a straightforward thing to answer but I haven't seen it addressed in the primer, etc.


1 Answer 1


The index $i$ in the potential term $$ V_F = F_i F^*_i$$ is a shortcut for an index that labels all chiral superfields. So if two chiral superfields happen to be organized into an $SU(2)$ doublet, which is the case of $H_u$, we're still having different $i$ for the two components. Effectively, we're summing over the two components of the doublet, too.

Such an $F_iF^*_i$ is clearly $SU(2)$-invariant. The $U$ in $SU(2)$ means unitary and this adjective precisely means that bilinear expressions such as $F_i F^*_i$ are invariant under the transformations. So they are. Here I am assuming that $i$ runs over an orthonormal basis; it should.

The superpotential $W$ itself must be $SU(2)_W$-invariant and gauge-invariant, for that matter. The fields in your Yukawa superpotential that transform nontrivially are $Q$ and $H_u$. Note that in your conventions, neither $Q$ nor $H_u$ is being complex conjugated in the product appearing in $W$. It means that to build an $SU(2)$ invariant, their $SU(2)$ doublet indices have to be contracted via $\epsilon^{ij}$ because there are two equal two-valued indices over there. As you would guess from the conservation of electric charge, the upper (electric charge) component of one doublet is being multiplied by the lower (electric charge) component of the other doublet.

To consolidate the answers,

  1. Yes, $i$ runs over doublets so $F$ is a doublet, too. It has two components and we're summing $F_i F^*_i$ over $i$ which is $SU(2)$-invariant.

  2. Here it is somewhat unclear why you insist on the dagger instead of the asterisk. There may be two basic reasons. First, transpositions and columns-vs-rows. You may discuss whether $F_{H_u}$ which is a doublet should be written as a column or a row. If you want a regular matrix product in $F^*\cdot F$, the first one should be a row and the second one should be a column. However, $F^*_i F_i$ is written in terms of components so we don't need to know whether the components of the doublet should be written in a row or column – it's a pure convention that doesn't affect the formula. So there's no need to indicate the transposition by extra notation. Alternatively, you may prefer $\dagger$ because we're ultimately doing quantum mechanics and we really mean the Hermitian conjugation of all the fields operators – I mean of each component of the doublet separately. Yes, quantum mechanics always requires $\dagger$ where classical theory had a $*$ but it's a convention to use the classical notation with $*$ to distinguish the individual fields, i.e. convention of classical physics, even in quantum mechanics. Especially in the case of SUSY, we're really constructing a classical theory first, use its notation, and then quantize it. Yes, all the starred operators are really Hermitian conjugate ones. Note that $U$ is a singlet which changes nothing.

  3. You may check $SU(2)$ invariance just by carefully tracing which components of the fields, $Q$, $H_u$ as well as $F$, transform as doublets and how bilinear invariants are constructed out of doublets. When you do so right, obeying the rules above, you will see that both $W$ and $V$ are $SU(2)$-invariant. So even your quartic term is manifestly $SU(2)$-invariant because it's again a construction of the algebraic form $T_i T^*_i$ for an $SU(2)$ doublet $T$.

Concerning your unwanted vertices comment, you need to carefully distinguish component fields and the superpartners. The supersymmetric terms in the Lagrangian only produce some interactions for quarks and other for squarks – and different interactions for various combinations. In particular, a cubic superpotential in $d=4$ is renormalizable so you may be sure it only produces renormalizable interactions when rewritten in components. So concerning your quartic term, you will only generate a quartic term in the scalars – squarks – whose dimension is "mass" per factor so the remaining dimensions $m^4$ is still perturbatively renormalizable. There are no $fermion^4$ terms that you get in this way. Instead, even in the component language, you get terms that reproduce $W$ and not $|W|^2$, such as Yukawa terms combining a squark with a quark and higgsino, or a Higgs with two quarks (of course, that's how the masses of quarks are still obtained). Such cubic terms are obtained because there are also quadratic terms in the supersymmetric Lagrangian.

  • 1
    $\begingroup$ I think I didn't get notified that this had been answered or something. Thanks a lot for your detailed response! $\endgroup$
    – gn0m0n
    Sep 24, 2012 at 2:40

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